Atoms and Radiation recap
Make notes on the following
Atomic nomenclature
Rules of nuclear equations
Atomic mass unit
E=mc^2 and mass loss
MeV as a measure of mass
Fission and Fusion
Properties of alpha, beta and gamma including
what it is, how ionising it is, shielding and
penetration, how dangerous it is
Why was polonium-210 a safe, hard to trace and
effective way of killing Alexander Litvinenko
Sources of background radiation.
XKCD comparison of radiation dose
xkcd.com/radiation
Dealing with background radiation in experimental
data – numerical and graphical approaches
Complete and mark the following
Physics factsheet 11
Isaac Physics Skills Exercise J1 & J4
IOP TAP 512-1: Nuclide notation
IOP TAP 512-3: Practice with nuclear equations
IOP TAP 527- 5: Fission – practice questions
IOP TAP 528-2: Fission in a nuclear reactor – how
the mass changes
IOP TAP 528- 4: Fusion questions
AQA exam past paper questions - Mass and energy
Completed?
Completed?
January 2001
Number 11
Radioactivity – an introduction
This Factsheet will explain the nature, properties and effects of
radioactive emissions, the concept of half-life and the hazards and
benefits of radioactivity.
Atoms – a reminder
An atom contains protons, neutrons, and electrons; the properties
of these are shown in the table below.
What is radioactivity?
Particle (symbol)
Mass (atomic mass units)
Charge
Radioactivity involves the spontaneous (in other words, occurring
without outside interference) emission of an alpha or beta particle from
the nucleus of an atom, causing the proton number (see box) of the atom
to change. An individual radioactive decay therefore always involves
one element changing to another element.
proton (p)
neutron (n)
electron (e)
Radioactive decay may also be accompanied by gamma emission. This
is how the nucleus rids itself of excess energy if it is in an excited state
after emitting an alpha or beta particle.
The nucleus of the atom contains the protons and neutrons, which
are known collectively as nucleons. The nucleus is the only part of
the atom involved in radioactivity.
The properties of alpha and beta particles and gamma rays are shown in
Table 1. As the table shows, they differ significantly in their
range – how far they travel, or how easily they are stopped
ionising ability – their ability to “knock” electrons from atoms that
they collide with, turning them into ions.
These are important in considering health risks
Two numbers are used to describe the particles in the nucleus:
A – the nucleon number (or mass number)
Z – the proton number (or atomic number)
So, if an atom has A = 13 and Z = 6, then it would have 6 protons,
and 7 neutrons (since protons + neutrons = A). The proton number
tells you which element the atom is – so for example, any atom
with proton number 6 is a carbon atom.
Table 1. Properties of alpha, beta and gamma radiation.
Type of
Consists of
Range
radiation
alpha
()
helium nuclei – i.e.
2 protons + 2 neutrons
beta
()
fast electrons
gamma
()
very high frequency
(and so high energy)
electromagnetic waves
5 cm in air – can be
stopped by a thick
sheet of paper
Up to a few metres
in air – can be
stopped by a few
millimetres of
aluminium
Singificantly
reduced by several
metres of concrete,
or several cm of
lead
1
1
0
(units of 1.6 10-19C
+1
0
-1
We write an atom of element X with nucleon number A and proton
Ionising
Propertie
s
Highly
ionising
number Z as
A
ZX
- so an atom of oxygen with 8 protons and 8
neutrons in its nucleus is written
Less than
16
8O .
Isotopes of an element have the same proton number, but
different nucleon numbers – so they have the same number of
protons but different numbers of neutrons. Common examples
of isotopes are
14
6C
and
12
6C ,
and
35
17 Cl
and 37
17 Cl .
Where does radioactivity come from?
Weak –
much
less than
Radioactivity comes from both natural and man-made sources. Manmade, “artificial” elements with very large nuclei are always
radioactive, but many naturally occurring elements (or particular
isotopes of elements) are radioactive too.
This naturally occurring radiation means that we are exposed to a low
level of radiation constantly; this is called background radiation.
Tip: In beta emission, the electron comes from the nucleus. It is not one
of the electrons from the atom.
Sources of background radiation include:
What sort of atoms decay?
Atoms that decay radioactively are known as unstable; those that do not
are stable. Atoms may be unstable because:
they are too large (with a larger nucleus than lead)
the balance between protons and neutrons is not right. In small
stable atoms, there are roughly the same number of protons and
neutrons, whilst in large stable atoms, neutrons always outnumber
protons.
This will be covered in more detail in later work on radioactivity.
1
rocks, such as granite
radon gas, which is formed in the ground
cosmic rays, which come from space
artificially produced radioisotopes
Radioactivity – an introduction
Effect of radiaoctive decay on the nucleus – nuclear
equations.
eg: Fluorine is bombarded with particles to produce sodium –22:
22
11 Na
19
4
9F 2
Sodium – 22 then decays by positron emission to form Neon – 22
22
11 Na
Alpha decay removes 2 protons and 2 neutrons from the nucleus
– so Z, the proton number, decreases by 2, and A, the nucleon
number, decreases by 4.
Typical exam question
The following nuclear equation represents the decay of uranium to
238
234
y
thorium:
92 U 90Th + z X
(a) Determine the values of y and z
[2]
(b) Identify the particle X
[1]
(c) Q represents the energy released in this reaction. Which two
forms does this energy take?
[2]
(d) The thorium also decays by - emission forming an isotope of
palladium (Pa). Write the nuclear equation for this decay. [2]
In beta decay, a neutron in the nucleus emits an electron and
changes to a proton. So Z increases by 1 and A stays the same.
Gamma radiation only involves the emission of energy, and so
does not change either A or Z.
Radioactive decay can be represented in nuclear equations:
(a) y = 238 – 234 = 4 z = 92 – 90 = 2 (b) The particle has 4 nucleons and 2 protons
It is an alpha particle or He nucleus. (c) Kinetic energy of the alpha particle Gamma radiation eg: decay of radon 204:
204
86 Rn
200
84 Po
+
The atom produced
after decay –
polonium 200
The atom we start
with – radon 204
4
2
(d)
alpha
particle
0
The particle may also be written as
4
2 He (as it is a
0
1 e (as it is an
electron)
+ - +
(energy ( ))
All the dice are thrown. About one sixth of the dice, or about 1000 dice,
will show a 1 – so these atoms have decayed.
There are 5000 “undecayed” dice left. They are thrown again – again,
about one sixth of them will show a 1, and decay. The decayed dice are
removed, and the process is repeated.
Radioactive decay series
The numbers of dice we’d expect to decay on each throw are shown in
the table and graph below:
In some cases, the atom produced is also radioactive, and so decays in
its turn. This process continues until an atom is produced that will not
decay. If the initial atom has a large nucleus, then the final, nonradioactive atom in the series will be a stable isotope of lead.
Throw
1st
2nd
3rd
4th
5th
6th
234
234
234
230
226
238
90Th 91 Pa 92 U 90Th 88 Ra...
92 U Artificial radioactivity
Radioactive isotopes can be created by bombarding naturally occurring
atoms with particles such as neutrons (used in a nuclear reactor), protons
or particles. Nuclear equations can be written for this type of process
in exactly the same way:
No. of Decays
1000
833
694
579
482
402
Throw
7th
8th
9th
10th
11th
12th
No. of Decays
335
279
233
194
162
135
1200
1000
63
1
64
64
0
29 Cu 0 n 29 Cu 30 Zn 1 “decays”
eg:
234
91 Pa
Imagine you have a large number (say 6000) of normal dice. Each dice
represents a radioactive atom, and we will say a dice has “decayed” (to
give a stable product) if it shows a 1 when it is thrown. Of course, the
dice are not a perfect model – radioactive decay goes on continuously,
not just at set time intervals, and may produce radioactive products.
However, it is close enough to be useful.
helium nucleus).
Tip: Check your finished equation is right by checking that the A
values add up to the same on both sides, and the Z values add up
to the same on both sides.
Always put the A and Z values on for the emitted particles as well
as the atoms. This helps you to make sure you get the equation right.
eg:
Radioactive decay is a random process – it is not possible to predict
when a particular nucleus will decay. We can get a picture of how this
works using another “random” process – throwing dice.
+ 1 The particle may also be written as
234
90Th
Decay law and half life
decay of beryllium 10:
10
10
4 Be 5 B
0
22
10 Ne 1 e
Some artificially produced isotopes decay by emitting a positron –
which is a particle identical to an electron, except that it is positively
charged (in fact, it is the antiparticle of the electron). This is known as
positron emission, or + decay. In positron emission, a proton emits a
positron, thus changing into a neutron, so the proton number decreases
by 1, and the nucleon number remains unchanged.
800
600
400
200
0
0
2
5
10
number of throws
15
Radioactivity – an introduction
Key points to note:
We can predict that the actual results of throwing the dice would
give results very much like these. However, we have no idea which
dice will “decay” on any one throw, or how long a particular dice
will take to “decay”.
The probability of any dice “decaying” is always the same – it is
1
6
. So the number of dice decaying each time will be about
1
6
Tip: If you are told to plot a graph of something against something, the
thing given first (in this case, count rate) goes on the y-axis.
The table below shows the corrected count rates:
Time (s)
0
10
20
Count rate (counts per second) 16.1 13.2 11.0
of
the dice that are left.
The graph produced by the dice has a characteristic shape – it is
called an exponential decay curve. (see Factsheet 10 –
Exponentials and Logarithms – for more on exponentials)
30
9.1
40
7.6
50
6.2
18
16
Radioactivity behaves very much like the dice – there is a constant
probability of a nucleus decaying, so the number of nuclei decaying is
proportional to the total number of them. The number of decays per
second of a sample is called its activity, and is measured in bequerels
(Bq)
Counts per second
14
The number of atoms decaying over a given period is a constant
fraction of (or is proportional to) the number remaining.
This is expressed in an equation as A = N
where A = activity, = decay constant (which is the probability of any
given atom of that element decaying in one second) and N = no. of
undecayed atoms.
12
10
8
6
4
2
0
This equation leads to an exponential decay curve.
0
10
20
30
40
50
Time (seconds)
Tip: You must remember that the atoms decaying do not disappear –
they change into another element, which may or may not be radioactive.
Now we measure two or more half-lives from the graph. For example,
we could choose to find the time taken for the count rate to decline from
16 to 8, and from 14 to 7. We then average the values.
An exponential decay curve has some very important properties; these
will also come up elsewhere in Physics.
From graph:
time taken to decline from16 to 8 is 37 seconds.
time taken to decline from 14 to 7 is 45 – 7.5 = 37.5 seconds.
So our estimate is the average of these values – 37.25=37 sec(2 SF)
(since the original data was given to 2 SF, it would not be appropriate
to use any greater accuracy in the answer).
Activity never decreases to zero
There is a constant half life – for a given radioactive element, the
time taken for the activity to halve will always be the same. (So, for
example, it would take the same time for activity to decrease from
400 Bq to 200 Bq, as for activity to decrease from 200Bq to 100Bq
0.69
, where is the
The half-life in seconds is given by T½ =
Tip: You must always use corrected count rates in any
half-life calculations. If you are told the background count, use it!
decay constant. (This equation will be justified in later studies)
Example 2. A sample of carbon 14 has an activity of 40 Bq
a) After 17 100 years, the activity of this sample will have fallen to
5Bq. Calculate the half life of carbon 14.
Calculations involving half life
b) After how many more years will the activity of this sample have
fallen to approximately 0.078Bq?
Calculations involving half life may require you to:
determine half-life from a graph
determine half life from count rates
determine count-rates or time, given the half life.
The following examples illustrate these.
a) We need to work out how many half-lives are required for the activity
to fall to 5Bq.
1/ 2
/2
/2
20 1
10 1
5
40 T
T
T
Example 1. The table below shows the count rate for a radioactive
isotope. The background count rate is 0.6 counts per second.
Time (s)
0
10
20
30 40 50
Count rate (counts per second) 16.7 13.8 11.6 9.7 8.2 6.8
So 17100 years = 3 half-lives.
So half-life = 17 100 3 = 5700 years.
Plot a graph of corrected count rate against time and use it to
determine the half life of the isotope.
b) 5 2.5 1.25 0.625 0.3125 0.15625 0.078125
So it takes another 6 half-lives = 34 200 years.
3
Radioactivity – an introduction
Typical Exam Question
(a) Sketch a diagram of the apparatus that could be used in
a school laboratory to determine the half-life of a radioactive
sample.
[3]
(b) State the measurements that are required to accurately
determine the half-life.
[2]
(c) Explain how the measurements would be used to determine
the half-life for the sample.
[3]
(a) Diagram should include radioactive source and GM tube
connected to ratemeter
(b) For this sample the count rate should be recorded at intervals of
10s (5-20s) for 300s. (c) Plot a graph of count rate against time
Find, from the graph, the time for the count rate to half
Repeat using different intervals of halving ( 1600 to 800, 1000 to
Average the results.
500 etc.)
Typical Exam Question
The half life of a sample of radioactive material is related to its
decay constant by the equation:
0.69
T1/2 =
(a) Explain the meaning of the symbols:
(i) T1/2
(ii) (b) A sample of 24Na has a half-life of 234 hours.
Calculate the radioactive decay constant for 24Na
[1]
[1]
[2]
(a) (i) T1/2 : Half-life, the average time for the number of undecayed
atoms or nuclei to be reduced to half their initial number. (ii) : Decay constant, the constant of proportionality in the
relationship: activity number of undecayed atoms (b) = ln2 /(234 3600) = 8.2 10 -7s -1
Experiments with radioactivity
Absorption of radiation.
After the background radiation has been measured, the source is
intially placed close to the GM tube and the average reading noted
over a short period. This is repeated, moving the GM tube away from
the source in 5mm steps, until the count rate has returned to the
background level (or the corrected count rate is zero. Materials such
as paper or metal can be placed between the source and the counter to
demonstrate absorption; in this case the GM tube must be less than
5cm from the source.
A Geiger-Müller tube (G-M tube) attached to a ratemeter is used to
detect radioactivity in the school laboratory. It is used to measure the
radioactive count-rate per second (which is proportional to the number
of emissions per second).
Some errors arise in the use of the GM tube because after it has
registered one count, there is a short interval (known as the dead time)
before it can register another, so any decays occurring in this period are
not registered. This problem is most noticeable when the count rate is
high.
Absorption of radiation
The same procedure as for radiation can be followed, except that
the steps by which the GM tube is moved away need to be longer,
since the range of radiation is much greater. Various thicknesses of
aluminium foil can be placed between the source and the counter to
determine the required thickness for total absorption.
Radioactive sources for experiments are always supplied in a holder,
and have low levels of activity. When not in use, they are stored inside a
lead “castle”, in a wooden box; accordingly they will prevent no danger
when stored. The following safety precautions must always be
observed when handling radioactive sources:
Absorption of radiation
Gamma radiation is only absorbed to any extent by lead; the thickness
of lead required can be determined in the same fashion as the
previously described experiments.
always lift with forceps
hold so that the open end is directed away from the body
never bring close to the eyes
Range of radiation in air
Gamma rays are not absorbed to any great extent by air; their
intensity falls off with distance according to an inverse square law –
in other words, the intensity is inversely proportional to the square of
the distance from the source. This gives the equation:
k
I = 2 , where k is a constant.
d
This is demonstrated by placing a GM tube at various distances from
a gamma source and measuring the count rates. The count rates are
corrected for background radiation. A graph is then plotted of
1
. This should give an
corrected count rate (y-axis) against
d2
approximately straight line.
In all experiments, it is necessary to first measure the background
radiation for a short time and obtain an average value. This value is
then subtracted from all future readings to give the count rate due to the
source alone – the corrected count rate.
In any experiment aiming to measure the count rate for just one type of
radiation (in other words, for just , just or just ), it is necessary to
ensure that the other types of radiation do not “interfere”:
To obtain alone, a cover disc of aluminium over the source will
absorb any or produced (NB: is never produced alone, so this
will always be necessary).
An or source can be selected that does not emit radiation, but
Errors – resulting in the graph not being exactly a straight line – are
due to the source itself having a size – in other words, not being a
“point source” to the GM tube’s dead time., and to the random nature
of decay.
it is also necessary to consider any other products in the radioactive
decay series; if an emitter produces a decay product that is also
radioactive, the count-rate of the decay product will also be included
in any measurements. This is addressed by selecting a source for
which all the decay products in the series have much longer halflives (and so a low, roughly constant activity rate) over the time of
the experiment, or much shorter half lives (so almost all of the
atoms decay within a very short time-scale).
Measuring half life
After the background count has been measured, the source is placed
close to the GM tube. Counts are taken at 10 second intervals, and a
graph of corrected count rate (y-axis) against time (x-axis) can be
produced. From the graph, a number of values for the half-life can be
found; an average of these produces a suitable estimate. (In later
work, an alternative, more accurate approach will be used).
Exam Hint: Examination questions about experiments require
essential practical details, such as precautions, how frequently
measurements are taken and sources of error.
4
Radioactivity – an introduction
Exam Workshop
Uses of radioactivity
This is a typical poor student’s answer to an exam question. The
comments explain what is wrong with the answers and how they can be
improved. The examiner’s answer is given below.
238
235
Top number
92 U and 92 U are radioactive isotopes of Uranium.
Carbon dating
Carbon-14 is a naturally occurring beta-emitter with a half-life of 5700
years. It is formed in the atmosphere from nitrogen, due to the action of
cosmic rays, and becomes incorporated in radioactive carbon dioxide.
Bottom number
During photosynthesis, plants and trees take in carbon dioxide from the
atmosphere; this includes carbon-14. The amount of carbon-14 present
as a proportion of the total amount of carbon will be, on average, the
same in a living plant as in the atmosphere as a whole.
(a) State the correct terms, and meanings of, the 'top number' and
'bottom number'.
[4]
top number: number of protons + neutrons
bottom number: number of protons 2/4
However, when the plant dies, it stops interacting with the atmosphere,
so it doesn’t acquire any more carbon-14. The activitity level then
declines exponentially. By measuring the residual activity, it is possible
to estimate how many half-lives there have been since the plant died,
and hence how long ago it lived.
Although the student knows what the numbers mean, s/he has lost
out by not giving the correct terms – not reading carefully, perhaps?
(b) Explain the meaning of the term 'isotope'.
atoms with different numbers of neutrons
[2]
1/2
To obtain both marks, the candidate needs to make it clear that the
isotopes have the same number of protons. This could be written
down directly, or the candidate could make reference to A and Z.
Carbon-14 is particularly suitable for this use due to:
its presence in all living things
the length of its half life – sufficiently short that changes can be
observed over thousands of years, but sufficiently long that there is
still significant residual activity after this period.
(c) U-238 decays via -emission to produce an atom of thorium(Th).
Write a nuclear equation to represent this decay.
[3]
238
234
2/3
U
Th
92
90
This method assumes that the proportion of carbon-14 in the
atmosphere has stayed the same; this depends on whether the amount of
cosmic rays penetrating the atmosphere was the same.
The two marks awarded were for calculation of A and Z for
thorium. The candidate needed to indicate the values of Z and A for
the particle to obtain full marks; clearly s/he knew them, since the
other calculation was correct.
Radioactive tracers
Radioactive tracers are used to follow the path of a compound in a
system such as pipelines or the human body. They rely on the fact that
radioactive isotopes behave identically to non-radioactive ones in
physical and chemical processes. For example, a radioactive tracer can
be used to detect a leak in a pipe, since the count-rate will increase
where the leak occurs as the pipe will block and emissions.
A emitter would not be suitable, since the pipe would not block this.
Isotopes used for this purpose need to have a suitable half-life, so that
the count rate will not become so low as to be almost undetectable
during the course of the investigation.
(d) Uranium-238 has a half-life of 4.5 109 years. Calculate the time
required for the activity of a sample of Uranium-238 to decrease
from 6.04 104 Bq to 1.8875 103 Bq.
6.04 104 1.8875 103 = 32
2 2 2 2 2 = 32, so 5 half-lives
2/3
What the candidate has done is correct, and clearly presented. But
s/he has missed out on the final mark by not reading the question
carefully – it asks for the time in years, not the number of half-lives.
The candidate should be surprised not to use all the information.
Sterilization
Gamma rays can be used to sterilize medical instruments or keep food
fresh for a longer period.
Examiner’s Answers
(a) Top: nucleon / mass number. It’s the no. of nucleons or protons + neutrons in nucleus Bottom: atomic
number / proton number It’s the number of protons in nucleus (b) Atoms with the same number of protons but different numbers of neutrons in their nuclei. 234
4
c) 238
92 U 90 Th 2 (1 each for 234, 90 and 1 for 4 and 2)
Radiotherapy -cancer treatment
Radiotherapy involves using gamma sources to attack cancer cells. It
relies on the cancerous cells being more affected by the radiation than
the normal ones, but obviously the normal cells are affected too, so it
does produce some unpleasant side effects, like those described in the
“Dangers of radioactivity” box. Again, a short half-life is required.
d) 6.04 104 = 1.8875 103 2 2 2 2 2 (method)
so 5 half-lives. So 5 4.5 109 = 2.25 1010 years Dangers of radioactivity
Since radiation is so easily absorbed, it is not dangerous unless the
radioactive source is inside the body. Although radiation is more
penetrating, most of its energy is usually absorbed by clothes, and it is
easy to protect people further by using aluminium shielding. The
greatest danger arises from gamma radiation; although it is not
strongly ionising, it can penetrate deeply into the body.
The hazard represented by a particular radioisotope is therefore
dependent on:
Damage from radiation can include:
radiation burns (like normal burns, but caused by gamma rays)
hair loss
radiation sickness
damage to reproductive organs
delayed effects such as cancer and leukemia
The level of danger depends on the amount of radiation absorbed;
people likely to be exposed to radioactive materials, such as workers
in nuclear power plants, have their radiation dosage carefully
monitored to ensure it does not exceed safe levels.
the nature of its emissions, and of the emissions of its decay
products
its level of activity
its half-life, since long half-life radioisotopes will continue to be
highly active for a long period of time, and hence potentially be a
danger for this time.
This has implications for the disposal of nuclear waste, which includes
long-half-life isotopes. The canisters used to contain nuclear waste
need to be resistant to naturally occurring phenomena like landslides
or earthquakes, since the contents would still represent a danger for
many years to come.
5
Radioactivity – an introduction
Qualitative Questions
1. Give two sources of background radiation
Answers
60
60
0
1. a) 27 Co 28 Ni 1 2. Give two uses for radioactive elements, other than carbon dating.
b)
3. Explain how carbon dating works.
c)
4. Describe the penetrating power of , and radiation.
2. 1.60 10-5 (3SF)
5. Explain what is meant by “half-life”
3. 14 – 15 seconds.
6. Explain what is meant by “activity”, and give its units
4. 108 hours (or 4 days 12 hours)
7. Which form of radiation is potentially the most dangerous to human
beings?
5. 10 minutes
4
210
82 Pb 2 0
01 n 64
30 Zn 1 214
84 Po
63
) 29 Cu
8. Explain why emission occurs
The answers to these questions may be found in the text.
Quantitative Questions
1. Write nuclear equations for the following decays:
a) emission from a
b) emission from a
c)
63
29 Cu
60
27 Co nucleus to produce Nickel (Ni)
214
84 Po nucleus to produce Lead (Pb)
absorbs a neutron, then decays by emission to form zinc (Zn)
2. An isotope has a half-life of 12 hours. Calculate its decay constant.
3. The graph below shows the decline of activity with time for a
radioisotope. Calculate its half-life.
The background count is 2.6.
30
25
counts per second
20
15
10
5
0
0
10
20
30
40
50
60
70
80
90
time/s
Hint: the background count must be allowed for – so, for example, find
the time required for the count to decline from 24 + background to 12 +
background.
Acknowledgements:
This Factsheet was researched and written by Cath Brown
Curriculum Press, Unit 305B The Big Peg, 120 Vyse Street,
Birmingham B18 6NF.
Physics Factsheets may be copied free of charge by teaching staff or
students, provided that their school is a registered subscriber. They may
be networked for use within the school. No part of these Factsheets may
be reproduced, stored in a retrieval system or transmitted in any other
form or by any other means without the prior permission of the
publisher. ISSN 1351-5136
4. A radioisotope has a half-life of 12 hours.
Its initial activity is 1.6 1012 Bq.
Find the time required for its activity to decline to 3.125 109 Bq.
5. The activity of a radioisotope declines from 920Bq to 7.1875Bq in
1 hour 10 minutes. Find its half life, giving your answer in minutes.
6
TAP 512-2: Decay processes
Radioactive decay processes
D decay
Z
N
Z–2
N–2
proton number Z
D
2 fewer protons
2 fewer neutrons
E– decay
E–
Z
N
Z+ 1
N–1
Q
proton number Z
1 more proton
1 less neutron
E+ decay
E+
Z–1
N+1
Q
Z
N
proton number Z
1 less proton
1 more neutron
Jdecay
Z
N
proton number Z
J
same protons
and neutrons
External reference
This activity is taken from Advancing Physics chapter 18, 90O
TAP 512-3: Practice with nuclear equations
1
The isotope
element?
235
U decays into another element, emitting an alpha particle. What is the
This element decays, and the next, and so on until a stable element is reached. The
complete list of particles emitted in this chain is:
235
92 U o [D, E, D, E, D, D, D, D, E, D, E] o X.
What is the stable element X? (You could write down each element in the series, but
there is a quicker way.)
2
The following fission reaction can take place in a nuclear reactor:
235
1
137
95
1
92 U 0 no 55 Cs [ ] Rb > @0 n.
Complete the equation, showing how many neutrons are produced in the reaction. What
is the significance of the number of neutrons produced?
Why are the products of the reaction, caesium-137 and rubidium-95, likely to be
radioactive? What type of decay are these isotopes likely to show?
3
Boron absorbs neutrons with results as follows:
10
5
B 01 n o 73 Li 24D .
Why is boron suitable for use in a control rod?
4
27
Al is irradiated with alpha particles, the products from each
When the isotope 13
aluminium nucleus are a neutron, and a nuclide that emits positrons to give the stable
30
Si
isotope 14 . Write nuclear equations for these two processes.
1
5
Complete the following nuclear equations. In each case describe the decay process:
131
53 I
o
Xe 67
0
31 Ga 1e
11
6C
o
99m
43Tc
6
o
0
1e
Zn
B 01e
o
Tc J .
The Manhattan Project, the development of the atomic bomb, led to the discovery of the
transuranic elements (elements beyond uranium in the periodic table). Plutonium,
element 94, is formed by the bombardment of uranium-238 with neutrons. The nuclear
equations are:
238
1
239
92 U 0 no 92 U
239
239
0
92 Uo 93 Np 1 e
239
239
0
93 Np o 94 Pu 1 e
Complete the following nuclear equation for the formation of americium:
239
94 Pu
2 01 n o
Am 0
1e.
Curium is produced if plutonium-239 is bombarded with alpha particles. If the curium
242
isotope is 96 Cm , complete the equation
239
94 Pu
4
2D
o
If curium is made the target for alpha particle bombardment californium is produced.
Complete the nuclear equation to find the atomic number of californium:
242
96 Cm
4
245
Cf
2 Do
01n.
By firing heavier particles such as carbon or boron ions at the target materials heavier
elements can be synthesised. Complete the nuclear equation (Lw is lawrencium)
252
Cf 95 B o Lw 401n.
One of the transuranic elements is commonly found in the home. Which is this and where
is it used?
2
Practical advice
These questions are intended for homework or practice in class.
Answers and worked solutions
4
1.
The complete decay chain involves the loss of seven alpha particles ( 2
D ) and four beta
0
e
( 1 ).
This represents a loss of 7 u 4, i.e. 28, in mass number and (7 u 2 – 4),
particles
i.e. 10, in atomic number. X is therefore an isotope with mass number (235 – 28), i.e.
208
207, and atomic number (92 – 10), i.e. 82. This is lead, 82 Pb .
2.
235
92 U
1
137
0 no 55 Cs
95
37 Rb
3 01 n.
The reaction produces more neutrons than it absorbs, this will cause a ‘chain reaction’.
To see why the products of the reaction are likely to be radioactive you need to consult
the plot of neutron number against atomic number for the known stable nuclei. For
elements with atomic numbers up to about 30 the number of neutrons in the nucleus is
the same as the number of protons if the nucleus is stable. For higher atomic numbers
the ratio of neutrons to protons gradually increases to 1.5. Look at the position of both
137
95
55 Cs and 37 Rb on the plot and you will see that they both have a considerable excess
of neutrons. They are therefore likely to be radioactive. To become more stable the
nuclides need to decrease the neutron to proton ratio. The emission of a beta particle
does this, increasing the number of protons by one and decreasing the number of
neutrons by one. These isotopes are therefore likely to decay by emitting beta particles.
3.
When boron captures a neutron it is transformed into a stable isotope. If the control rods
are pushed into the reactor more neutrons are absorbed, causing the chain reaction to
slow down. If they are pulled out the chain reaction will proceed more vigorously.
4.
27
13 Al
30
15 P
4
2D
30
14 Si
30
15 P
1
0n
0
1e
5.
131
131
0
53 Io 54 Xe 1e
11
11
6 Co 5 B
beta decay
0
1e
positron emission
67
0
67
electron capture
31 Ga 1 eo 30 Zn
99m
99
nuclear rearrangem ent
43 Tc o 43 Tc
6.
239
1
241
94 Pu 20 no 95 Am 239
4
242
94 Pu 2 Do 96 Cm 242
4
245
96 Cm 2 Do 98 Cf 252
98 Cf
9
257
5 Bo 103 Lw
0
1e
1
0n
1
0n
4 01 n.
Americium is commonly found in homes where it is used in smoke detectors.
External reference
This activity is taken from Advancing Physics chapter 18, 210S
3
TAP 527- 5: Fission – practice questions
What these are for
These questions will give you some simple practice in handling the ideas and calculations that
physicists meet in nuclear fission.
Try these
The process of fission in one type of nuclear reactor proceeds as follows: a nucleus of
235
uranium 92 U captures a single neutron. The resulting nucleus is unstable and splits into two
90
144
or more fragments. These fragments could typically be a pair of nuclei, 36 Kr and 56 Ba for
example. Neutrons are also ejected as a result of the fission. It is these neutrons that go on to
cause subsequent fission events and maintain the chain reaction.
1.
Write down two balanced equations (the first to the unstable uranium; the second to
the final products) that represent this fission process.
2.
Calculate the total mass of the original uranium isotope and the neutron. The table
gives the atomic masses (in atomic mass units) of the particles found in this question.
(1 atomic mass unit (u) { 931 MeV.)
Particle
Mass (u)
1
0n
1.008 665
90
36 Kr
89.919 528
92
36 Kr
91.926 153
96
37 Rb
95.934 284
138
55 Cs
137.911 011
138
56 Ba
137.905 241
144
56 Ba
143.922 941
235
92 U
235.043 923
3.
Calculate the total mass of the four products.
4.
Calculate the change in mass. Does this represent energy gained or lost by the
system?
5.
Convert the mass change into the energy released (in MeV) in the fission event.
6.
These particular barium and krypton isotopes are not the only products possible in
nuclear fission. Repeat the calculation steps 1–5 with the following possible products
caesium-138 and rubidium-96.
Hints
1.
There are two equations, the first for the absorption of the neutron; the second for the
splitting of the unstable nucleus formed in the absorption. Write down all the original
nucleons on the left-hand side of the first equation (do not forget the original neutron).
Put all the products on the right-hand side. Check that all protons, neutrons and
electrons balance. Energy is also an output of the reaction, call it Q.
2.
Add the atomic mass unit values for the uranium and the neutron together.
3.
Add the atomic mass unit values for the barium, krypton and two neutrons together.
5.
Use 'E = 'mc to carry out this conversion. c = 9 x 10
2
2
16
–1
J kg .
Practical advice
This question set provides repetitive practice in handling nuclear mass changes and
conversions between mass and energy. It is suitable for students meeting these ideas for the
first time. There is an energy release / nucleon perspective here – a useful teaching point
when students have completed this task
Answers and worked solutions
1
0n
236
92 U
90
36 Kr
144
56 Ba
2 01 n Q
1.
235
92 U
2.
m = 235.043923 u – 1.008665 u = 236.052 588 u
3.
m = 89.919528 u + 143.922941 u + 1.008665 u + 1.008665 u =235.859 799 u
4.
'm =236.052 588 u – 235.859799 u = 0.192 789 u; energy lost
5.
'E = 0.192 789 u × 931.3 MeV u
6.
m = 137.905241 u + 95.934284 u + 1.008665 u + 1.008665 u =235.856855 u
'm = 236.052 588 u – 235.856855 u = 0.195733 u
'E = 0.195733 u × 931.3 MeV u
–1
–1
=179.49 MeV
= 182.2 MeV
External reference
This activity is taken from Advancing Physics chapter 18, 250S
TAP 528- 4: Fusion questions
Nuclear fusion is the process in which nuclei combine to give heavier elements. In one fusion
reaction, two atoms of deuterium (hydrogen-2) fuse together to give one atom of a helium
isotope (helium-3) together with one other particle.
1.
Write out a balanced equation for this fusion process and say what the fourth particle
is?
2.
Calculate the energy release in this equation. Values you need are in the table.
Particle
Mass (u)
1
0n
1.008 665
2
1H
2.014 102
3
1H
3.016 050
3
2 He
3.016 030
Another possible fusion process is represented by:
2
1H
1
0n
3
1H
(the formation of hydrogen-3, tritium, by a nucleus of deuterium absorbing a neutron). This
equation is certainly balanced. But can it occur in practice?
3.
Calculate the change in mass in this reaction.
4.
Is the reaction possible or not?
Practical advice
These questions revise basic conversions between electron volts and joules and atomic mass
units and kilograms. Students will need to be familiar with gigawatts (GW) and terawatts (TW)
in powers of 10. The questions could be extended either verbally or in writing to ask students
about the volume of deuterium inside the core and about the equivalent volumes of coal or oil
that might be required in a conventional power station. For example, 1 megatonne of coal is
equivalent to
29 x 10
15
J, 1 megatonne of oil is equivalent to 42 x 10
15
J.
Social and human context
The questions provide an opportunity for debate about fusion power generation.
Answers and worked solutions
2
1H
3
2 He
1.
2
1H
1
0n
2.
mass of two hydrogen-2 = 2 × 2.014102 u = 4.028204 u
mass of helium-3 plus neutron = 3.016030 u + 1.008665 u = 4.024695 u
'm = 4.028204 u – 4.024695 u = 0.003509 u
'E = 0.003509 u × 931.3 MeV u
3.
mass of hydrogen-2 plus neutron = 2.014102 u + 1.008665 u = 3.022767
'm =3.022767 – 3.016050 u = 0.006 717 u.
4.
The reaction can occur with a release 'E = 0.006 717 u × 931.3 MeV u
–1
= 3.27 MeV
External reference
This activity is taken from Advancing Physics chapter 18, 250S
–1
= 6.26 MeV
TAP 528-2: Fission in a nuclear reactor – how the mass
changes
Some rather harder questions
These extended questions will test your ability to deal with calculations involving the physics
of nuclear fission.
Use the following conversions and values for some of the questions:
–19
x
1 eV = 1.6 u 10
x
1 atomic mass unit = 1.66 u 10
x
J
–27
8
c = 3 u 10 m s
kg
–1
Particle
Mass (u)
235
92 U
235.043 94
1
1H
1.007 825
3
2 He
3.016 030
1
0n
1.008 665
Try these
Magnox power stations produce about 20 TW h of electrical energy in the UK every year by
fission of uranium. (This energy supplies roughly the electrical needs of Greater London.)
1
The overall efficiency of the process that converts the energy for heating released in
the fission to the final electrical product is 40%. How much energy, in joules, is
produced each second in the company’s reactors?
2
Each fission releases about 200 MeV of energy. How many atoms of 235
92 U need to
fission in each second to produce the heating energy you calculated in question 1?
3
What was the mass of these atoms before they underwent fission?
4
What is the total mass change due to fission in Magnox reactors each second?
In the pressurised water reactor (PWR) the fuel rods do not contain pure 235
92 U . The uranium
comes from mined ore that contains a mixture of
238
92 U
and 235
92 U . The fuel delivered to the
reactor contains 0.7% of 235
92 U . The fuel rod stays in the reactor for about 3 years and is then
removed to allow reprocessing.
This time consider just one reactor with an output of 1 GW.
5
Calculate the number of uranium nuclei disintegrating every second.
6
Calculate the mass of
7
Estimate the mass of
8
Estimate the total mass of both uranium isotopes required in the core for a 3 year
cycle.
9
Is your estimate in question 8 likely to be an upper or a lower limit?
235
92 U
235
92 U
that undergoes fission every second.
required in the core for a 3 year cycle.
Hints
1
Remember the meaning of the term watt-hour. It corresponds to the amount of
energy delivered at a rate of 1 joule per second for 1 hour. Do not forget to include
the efficiency in your calculation.
2
Convert to J from MeV.
3
Use the nucleon number of the uranium and the conversion from atomic mass units
to kilograms.
4
Use 'E = 'mc to calculate this.
7
The information indicates that one-third of the fuel needs to be removed for
reprocessing every year. Your answer to question 6 can be multiplied up to give the
fuel usage in 1 year. This is one-third of the total.
8
The answer to question 7 represents the fuel used, and this is 3% of all the uranium
(both isotopes). Hence the total mass.
2
Practical advice
These questions revise basic conversions between electron volts and joules and atomic mass
units and kilograms. Students will need to be familiar with gigawatts (GW) and terawatts (TW)
in powers of 10. The questions could be extended either verbally or in writing to ask students
about the volume of uranium inside the core and about the equivalent volumes of coal or oil
that might be required in a conventional power station. For example, 1 megatonne of coal is
equivalent to
29 x 10
15
J, 1 megatonne of oil is equivalent to 42 x 10
15
J.
Social and human context
The questions provide an opportunity for debate about fission power generation.
Answers and worked solutions
–1
12
7
19
1.
energy = ((20 u 10 Wh × 3600 J W h ) / (3.16 u 10 )) × (100 / 40) = 5.7 u 10
every second.
2.
number of atoms = (5.7 u 10
atoms.
3.
mass per second = 1.8 u 10
19
26
7.0 × 10
–5
kg s
19
–1
mass change = (5.7 u 10
5.
disintegrations per second =
19
6.
–1
kg s
–27
–1
kg u
=
–1
8 2
6
J s )/ (200 u 10 MeV × 1.6 u 10
26
3.0 × 10
26
J s )/ ((3 u 10 ) kg s ) = about 5 g
mass per second = 7.8 u 10
–5
–1
s × 235.04394 u × 1.66 × 10
–1
JeV ) = 1.8 u 10
–1
4.
((1 u 10
–19
6
J) / (200 u 10 eV × 1.6 u 10
J
–19
–1
19 –1
J eV )) × (100/40) = 7.8 u 10 s
–1
–27
s × 235.04394 u × 1.66 × 10
–1
–5
–1
7.
mass = 3.0 × 10
8.
mass = 2800 kg × (100/0.7) = 400 000 kg
9.
Lower limit.
kg s
7
–1
× 3 years × 3.2 × 10 sy = 2900 kg
External reference
This activity is taken from Advancing Physics chapter 18, 270S
–1
kg u
=
Q1.
(a) With reference to the process of nuclear fusion, explain why energy is released when
two small nuclei join together, and why it is difficult to make two nuclei come together
You may be awarded marks for the quality of written communication in your answer.
......................................................................................................................
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......................................................................................................................
(3)
(b)
A fusion reaction takes place when two deuterium nuclei join, as represented by
mass of 2H nucleus
mass of 3He nucleus
mass of neutron
= 2.01355 u
= 3.01493 u
= 1.00867 u
Calculate
(i)
the mass difference produced when two deuterium nuclei undergo fusion,
.............................................................................................................
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Page 1 of 28
(ii)
the energy released, in J, when this reaction takes place.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(Total 6 marks)
Q2.
You may be awarded marks for the quality of written communication provided in your
answers to part (a)
(a)
In the context of an atomic nucleus,
(i)
state what is meant by binding energy, and explain how it arises,
.............................................................................................................
.............................................................................................................
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.............................................................................................................
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.............................................................................................................
.............................................................................................................
(ii)
state what is meant by mass difference,
.............................................................................................................
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.............................................................................................................
.............................................................................................................
(iii)
state the relationship between binding energy and mass difference.
.............................................................................................................
.............................................................................................................
(4)
Page 2 of 28
(b)
Calculate the average binding energy per nucleon, in MeV nucleon–1, of the zinc nucleus
.
mass of
atom
=
63.92915 u
mass of proton
=
1.00728 u
mass of neutron
=
1.00867 u
mass of electron
=
0.00055 u
......................................................................................................................
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(5)
(c)
Why would you expect the zinc nucleus to be very stable?
......................................................................................................................
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......................................................................................................................
(1)
(Total 10 marks)
Q3.
(a)
(i)
The mass of a nucleus
is M.
If the mass of a proton is m p, and the mass of a neutron is m n, give an expression for
the mass difference ∆m of this nucleus.
∆m = ...................................................................................................
(ii)
Give an expression for the binding energy per nucleon of this nucleus, taking the
speed of light to be c.
.............................................................................................................
.............................................................................................................
(2)
Page 3 of 28
(b)
The figure below shows an enlarged portion of a graph indicating how the binding energy
per nucleon of various nuclides varies with their nucleon number.
(i)
State the value of the nucleon number for the nuclides that are most likely to be
stable. Give your reasoning.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
When fission of uranium 235 takes place, the nucleus splits into two roughly equal
parts and approximately 200 Me V of energy is released. Use information from the
figure above to justify this figure, explaining how you arrive at your answer.
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(5)
(Total 7 marks)
Page 4 of 28
Q4.
In a geothermal power station, water is pumped through pipes into an underground region of
hot rocks. The thermal energy of the rocks heats the water and turns it to steam at high
pressure. The steam then drives a turbine at the surface to produce electricity.
(a)
Water at 21°C is pumped into the hot rocks and steam at 100°C is produced at a rate of
190 kg s–1.
(i)
Show that the energy per second transferred from the hot rocks to the power station
in this process is at least 500 MW.
specific heat capacity of water = 4200 J kg–1 K–1
specific latent heat of steam = 2.3 × 106 J kg–1
.............................................................................................................
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.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
The hot rocks are estimated to have a volume of 4.0 × 106 m 3. Estimate the fall of
temperature of these rocks in one day if thermal energy is removed from them at the
rate calculated in part (i) without any thermal energy gain from deeper underground.
specific heat capacity of the rocks = 850 J kg–1 K–1
density of the rocks
= 3200 kg m–3
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(7)
Page 5 of 28
(b)
Geothermal energy originates as energy released in the radioactive decay of the
uranium isotope
U deep inside the Earth. Each nucleus that decays releases 4.2 MeV.
Calculate the mass of
U that would release energy at a rate of 500 MW.
U
= 4.5 × 109 years
molar mass of
U = 0.238 kg mol–1
half-life of
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(5)
(Total 12 marks)
Q5.
(a) When a nucleus of uranium -235 fissions into barium -141 and krypton -92, the change
in mass is 3.1 × 10–28 kg. Calculate how many nuclei must undergo fission in order to
release 1.0 J of energy by this reaction.
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(2)
Page 6 of 28
(b)
A nuclear power station produces an electrical output power of 600 MW. If the overall
efficiency of the station is 35%, calculate the decrease in the mass of the fuel rods,
because of the release of energy, during one week of continuous operation.
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......................................................................................................................
(4)
(Total 6 marks)
Q6.
(a) Sketch a graph of binding energy per nucleon against nucleon number for the naturally
occurring nuclides on the axes given in the figure below. Add values and a unit to the
binding energy per nucleon axis.
(4)
Page 7 of 28
(b)
Use the graph to explain how energy is released when some nuclides undergo fission and
when other nuclides undergo fusion.
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(3)
(Total 7 marks)
Q7.
The fissile
isotope
of uranium,
been used inAnsome
nuclear
reactors.
It is emits a
normally
produced
by neutron
irradiation
diatio , has
of thorium-232.
irradiated
thorium
nucleus
−
β particle to become an isotope of protactinium.
This isotope of protactinium may undergo β− decay to become
(a)
.
Complete the following equation to show the β− decay of protactinium.
Pa →
+
β + ……..
–
(2)
Page 8 of 28
(b)
Two other nuclei, P and Q, can also decay into
P decays by β+ decay to produce
Q decays by α emission to produce
.
.
.
The figure below shows a grid of neutron number against proton number with the
position of the
isotope shown.
On the grid label the positions of the nuclei P and Q.
Page 9 of 28
(2)
(c)
A typical fission reaction in the reactor is represented by
+
(i)
→
+
+ x neutrons
Calculate the number of neutrons, x .
answer = .............................neutrons
(1)
(ii)
Calculate the energy released, in MeV, in the fission reaction above.
mass of neutron = 1.00867 u
mass of
nucleus = 232.98915 u
mass of
nucleus = 90.90368 u
mass of
nucleus = 138.87810 u
answer = ..................................MeV
(3)
(Total 8 marks)
Q8.
(a)
(i)
Define the atomic mass unit.
...............................................................................................................
...............................................................................................................
(1)
Page 10 of 28
(ii)
State and explain how the mass of a
nucleus is different from the total mass of
its protons and neutrons when separated.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(b)
Explain why nuclei in a star have to be at a high temperature for fusion to take place.
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........................................................................................................................
........................................................................................................................
(3)
(c)
(i)
In massive stars, nuclei of hydrogen
are processed into nuclei of helium
through a series of interactions involving carbon, nitrogen and oxygen called the
CNO cycle.
Complete the nuclear equations below that represent the last two reactions in the
series.
(3)
Page 11 of 28
(ii)
The whole series of reactions is summarised by the following equation.
Calculate the energy, in Me V, that is released.
nuclear mass of
= 4.00150 u
energy ..................................... Me V
(3)
(Total 12 marks)
Q9.
(a)
State what is meant by the binding energy of a nucleus.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
(i)
When a
nucleus absorbs a slow-moving neutron and undergoes fission one
possible pair of fission fragments is technetium
and indium
.
Complete the following equation to represent this fission process.
(1)
Page 12 of 28
(ii)
Calculate the energy released, in MeV, when a single
fission in this way.
binding energy per nucleon of
= 7.59 MeV
binding energy per nucleon of
= 8.36 MeV
binding energy per nucleon of
= 8.51 MeV
nucleus undergoes
energy released ..................................... MeV
(3)
(iii)
Calculate the loss of mass when a
nucleus undergoes fission in this way.
loss of mass ......................................... kg
(2)
Page 13 of 28
(c)
(i)
On the figure below sketch a graph of neutron number, N, against proton number, Z,
for stable nuclei.
proton number, Z
(1)
(ii)
With reference to the figure, explain why fission fragments are unstable and explain
what type of radiation they are likely to emit initially.
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(3)
(Total 12 marks)
Page 14 of 28
M1.
(a) mass difference increases
or B.E. (per nucleon) or stability is greater for nucleus after fusion (1)
(greater) mass difference
or increase in B.E. (per nucleon) implies energy released (1)
both nuclei charged positively or have like charges (1)
electrostatic repulsion (1)
max 3
QWC 2
(b)
(i)
Δm (= 2 × (2.01355) – (3.01493 + 1.00867))
= 3.5 × 10–3 u (1) (5.81 × 10–30 kg)
(ii)
ΔE = 3.5 × 10–3 × 931.3 (MeV) (1)
(= 3.26 MeV)
= 3.26 × 106 × 1.6 × 10–19 = 5.22 × 10–13 (J) (1)
3
[6]
M2.
(a)
(i)
energy
separate nucleons (1)
energy associated with the strong force (1)
(ii)
mass of nucleus < total mass of constituent nucleons (1)
Δm is difference between mass of nucleus and total mass
of nucleons (1)
[Δm = Zmp + (A – Z)m n – m nucleus (1) (1)]
Eb = (Δm)c 2 (1)
[or Eb is energy equivalent of mass defect using E = mc 2]
max 4
QWC 1
(b)
mass of nucleus = 63.92915 – (30 × 0.00055) = 63.91265 (u) (1)
Δm = (30 × 1.00728) + (34 × 1.00867) – 63.91265 (1)
= 0.60053 (u) (1)
Eb = 0.60053 × 931.3 = 559.3 (MeV) (1)
Eb/nucleon =
= 8.74 (MeV/nucleon) (1)
(allow C.E. for Δm and Eb)
5
Page 15 of 28
(c)
nucleus has high value of Eb/nucleon
[or is near maximum of Eb/nucleon vs A curve] (1)
1
[10]
M3.
(a)
(ii)
(i)
∆m = Zmp + (A – Z)m n – M (1)
binding energy per nucleon =
(1)
2
(b)
(i)
A in range 54 → 64 (1)
stability increases as binding energy per nucleon increases (1)
[or binding energy per nucleon is a measure of stability]
[or large binding energy per nucleon shows nucleus is difficult to
break apart]
(ii)
binding energy per nucleon increases from about 7.6 to 8.5 (1)
increase of about 0.9 MeV for 235 nucleons (1)
hence 210 MeV (≈ 200 MeV) in total (1)
5
[7]
M4.
(a)
(i)
heat water to 100 °C, energy (= 190 × 4200 × 79) = 63 (MJ) (1)
vapourise water, energy
(=190 × 2.3 × l06) = 440(MJ) (1)
(437MJ)
energy transferred (per sec) = (437 + 63) MJ (1)
(= 500 MJ)
Page 16 of 28
(ii)
mass of rocks (= 4.0 × 106 × 3200)
= 1.3 × 1010(kg) (1)
(1.28 × 1010)
temperature fall of ΔT in one day, energy removed
(= 1.28 ×1010 × 850 × ΔT) = 1.1 × 1013 ΔT (1)
(1.09 x 1013 AT)
(allow C.E. for value of mass of rocks)
energy transfer in one day (= 500 × 106 × 3600 × 24)
= 4.3 × 1013 (J) (1)
in one day
K (1)
7
(b)
number of nuclei in 1 kg of 238 U =
(1)
activity of lkg of 238U =
(1)
(1)
energy released per sec per kg of 238 U
= 1.2(6) × 107 × 4.2 × 1.6 × 10–13(J) (1)
(8.47 × 10–6(J))
mass of 238Uneeded =
= 5.9(0) × 1013kg (1)
5
[12]
M5.
(a)
for one reaction ΔE (= Δm c 2) = 3.1 × 10–28 × (3.00 x 108)2 (1)
= (2.79 × 10–11J)
number of nuclei required =
= 3.5(8) × 1010 (1)
[or equivalent credit for any other valid method]
2
Page 17 of 28
(b)
output power from reactor =
(1714 MW)
(MW) (1)
energy output from fuel rods in one week
= 1.70 × 109 × 24 × 7 × 3600 (1)
(= 1.03 × 1015 J)
(1)
= 1.14 ×10–2 kg (1)
[or equivalent credit for any other valid method]
4
[6]
M6.
(a)
peak 8.7 (accept 8.0 – 9.2)
in MeV
(or peak 1.4 × 10–12 accept 1.3 – 1.5 × 10–12 in J
at nucleon number 50 – 60
)
accept 50 – 75
sharp rise from origin and moderate fall not below 2/3 of peak height
4
Page 18 of 28
(b)
energy is released/made available when binding energy per
nucleon is increased
in fission a (large) nucleus splits and in fusion (small) nuclei join
the most stable nuclei are at a peak
fusion occurs to the left of peak and fission to the right
max 3
[7]
M7.
(a)
Pa
anti (electron) neutrino
2
(b)
2
(c)
(i)
x=4
1
(ii)
mass defect = [(232.98915 + 1.00867) –
(90.90368 + 138.87810 + 4 × 1.00867)] u
= 0.18136 u
3
energy released (= 0.18136 × 931) = 169 (MeV)
[8]
Page 19 of 28
M8.
(a)
(i)
1/12 the mass of an (atom) of
/ carbon−12 / C12
a reference to a nucleus loses the mark
1
(ii)
separated nucleons have a greater mass
(than when inside a nucleus)
an answer starting with ‘its’ implies the nucleus
because of the (binding) energy added to separate the nucleons or energy is
released when a nucleus is formed (owtte)
marks are independent
direction of energy flow or work done must be explicit
2
(b)
nuclei need to be close together (owtte) for the Strong Nuclear Force to be involved or for
fusion to take place
e.g. first mark – within the range of the SNF
but the electrostatic / electromagnetic force is repulsive (and tries to prevent this)
(if the temperature is high then) the nuclei have (high) kinetic energy / speed (to overcome
the repulsion)
3rd mark is for a simple link between temperature and speed / KE
3
(c)
(i)
15
give the middle mark easily for any e or β with a + in any position
12
3
Page 20 of 28
(ii)
Δmass = 4 × 1.00728 − 4.00150 − (2 × 9.11 × 10−31 / 1.661 × 10−27)
or
Δmass = {4 × 1.00728 − 4.00150 − 2 × 0.00055}(u)
(4×1.00728=4.02912)
1st mark – correct subtractions in any consistent unit. use of mp =
1.67 × 10−27 kg will gain this mark but will not gain the 2nd as it will not
produce an accurate enough result
Δmass = 0.02652(u)
2nd mark - for calculated value
0.02652u
4.405 × 10−29 kg
3.364 × 10−12 J
Δbinding energy (= 0.02652 × 931.5)
{allow 931.3}
Δbinding energy = 24.7 MeV
3rd mark – conversion to Mev
conversion mark stands alone
award 3 marks for answer provided some working shown - no
working gets 2 marks
(2sf expected)
3
[12]
(a) the amount of energy required to separate a nucleus ✓
into its separate neutrons and protons / nucleons ✓
(or energy released on formation of a nucleus ✓
from its separate neutrons and protons / constituents ✓)
M9.
1st mark is for correct energy flow direction
2nd mark is for binding or separating nucleons (nucleus is in the
question but a reference to an atom will lose the mark)
ignore discussion of SNF etc
both marks are independent
2
(b)
(i)
✓
must see subscript and superscripts
1
Page 21 of 28
(ii)
binding energy of U
= 235 × 7.59 ✓ ( = 1784 (MeV))
binding energy of Tc and In
= 112 × 8.36 + 122 × 8.51 ✓
( = 1975 (MeV))
energy released ( = 1975 – 1784) = 191 (MeV) ✓ (allow 190 MeV)
1st mark is for 235 × 7.59 seen anywhere
2nd mark for 112 × 8.36 + 122 × 8.51 or 1975 is only given if there
are no other terms or conversions added to the equation (ignore
which way round the subtraction is positioned)
correct final answer can score 3 marks
3
(iii)
energy released
= 191 × 1.60 × 10−13 ✓
( = 3.06 × 10−11 J)
loss of mass ( = E / c2 )
= 2.91 × 10−11 / (3.00 × 108)2)
= 3.4 × 10−28 (kg) ✓
or
= 191 / 931.5 u ✓ ( = 0.205 u)
= 0.205 × 1.66 × 10−27 (kg)
= 3.4 × 10−28 (kg) ✓
allow CE from (ii)
working must be shown for a CE otherwise full marks can be given
for correct answer only
note for CE
answer = (ii) × 1.78 × 10−30
(2.01 × 10−27 is a common answer)
2
(c)
(i)
line or band from origin, starting at 45° up to Z approximately = 20 reading
Z = 80, N = 110→130 ✓
initial gradient should be about 1 (ie Z = 20 ; N = 15 → 25) and
overall must show some concave curvature. (Ignore slight
waviness in the line)
if band is shown take middle as the line
if line stops at N > 70 extrapolate line to N = 80 for marking
1
Page 22 of 28
(ii)
fission fragments are (likely) to be above / to the left of the line of stability ✓
fission fragments are (likely) to have a larger N / Z ratio than stable nuclei
or
fission fragments are neutron rich owtte ✓
and become neutron or β− emitters ✓
ignore any reference to α emission
a candidate must make a choice for the first two marks
stating that there are more neutrons than protons is not enough for
a mark
1st mark reference to graph
2nd mark – high N / Z ratio or neutron rich
3rd mark beta minus
note not just beta
3
[12]
Page 23 of 28
E1.
Candidates familiar with the principles of nuclear fusion could score all three marks in part
(a) without trouble. The main weaknesses in many scripts were caused by a tendency to write
generally about the mass difference of a nucleus rather than specifically about the increase in
mass difference brought about by the fusion of two light nuclei. Arguments phrased in terms of
the increase in binding energy per nucleon conveyed the most convincing answers. When
addressing the second half of the sentence, a large proportion of the candidates had their
attention distracted by concentrating on the need for a high temperature. They would have been
better advised to focus on the basic physical principle of electrostatic repulsion between two
positively charged nuclei. There was also some confusion with effects attributed to the strong
nuclear force.
In part (b) the calculation of mass difference caused few problems, but the conversion of units in
part (b)(ii) was a bigger hurdle. The main errors were forgetting that 1 MeV is 106 eV (which is
1.6 × 10–13 J), and attempting to convert from eV to J by dividing by e instead of multiplying by e.
E2.
Well-prepared candidates were able to score highly throughout this question. However, the
topic of nuclear binding energy continues to be very poorly understood by many students. The
principal obstacle to progress is a failure to appreciate that binding energy is energy that is
released when a nucleus is formed: it is missing energy rather than energy that a nucleus
possesses. When explaining how it arises, examiners expected candidates to associate binding
energy with the work done on the nucleons by the strong force. A few candidates did this, but far
more strayed into aspects tested later in the question, usually by resorting to mass difference
when answering part (i). Some candidates discussed the whole of parts (i) and (ii) as though
these ideas only have a meaning when dealing with fission (or fusion) reactions.
In part (a) (iii), the equation Eb = (Δm)c 2 was known reasonably well. To gain the available mark,
examiners were looking for some explanation of the terms to accompany any bald statement of
the more basic E = mc 2. A simple statement that 1u is equivalent to 931.3 MeV was not regarded
as worthy of credit.
Full marks were often awarded for the calculation of binding energy per nucleon in part (b). Apart
from arithmetical errors, candidates‘ main problems were failure to account for the electron
masses, and thinking that the binding energy (the penultimate step) is the same as the binding
energy per nucleon (the final step).
Recognition of binding energy per nucleon (rather than just binding energy) as an essential
measure of nuclear stability was the key to success for the single mark in part (c). Some
candidates spotted that 64Zn would be close to 56Fe, and therefore stable, whilst others gave
valuable references to the binding energy per nucleon curve in order to gain the mark.
Page 24 of 28
E3.
The responses to part (a) suggested that students have much less difficulty in calculating a
mass difference when using numbers, than when asked to explain what they are doing using
algebra. Answers to part (a) (ii) were particularly disappointing, demonstrating the ongoing
confusion between binding energy and binding energy per nucleon.
Part (b) was often answered competently, with candidates showing a good understanding of the
role of binding energy per nucleon in determining the stability of a nucleus. Data from the graph
was usually extracted successfully to answer part (b) (ii), but some candidates overlooked the
need to do this and therefore made little progress.
E4.
The calculation, in part (a) (i), of the energy needed to heat the water and to turn it to steam
at 100°C was usually done correctly. Some failed by choosing an incorrect temperature change.
In part (ii), many candidates obtained full marks although some correctly worked out the
temperature change per second, but did not proceed to calculate the temperature change in one
day. A small minority of candidates used an incorrect density formula or an incorrect value of the
specific heat capacity.
In part (b) most candidates knew how to convert MeV into joules correctly and used the given
mass number correctly. They also knew how to calculate the total activity of the rocks from the
energy released per second and the energy released per decay. Many of these candidates
correctly used their activity value and the decay constant, worked out from the half-life, to
calculate the total number of atoms and their mass. Some lost a mark in their calculation of the
decay constant as a result of not converting the half-life into seconds. A significant number of
candidates considered their answer for the total activity to be the total number of atoms and
consequently lost two marks because they failed to use the total activity and the decay constant
to calculate the total number of atoms.
E5.
Both parts of this question could be approached by several routes, and examiners were often
challenged to discover the route by which a candidate had arrived at a correct final answer.
There were frequent, and usually unnecessary, unit conversions into u and MeV and then back
into kg and J. Part (a) was answered well by most candidates.
Part (b) required even greater care with the arithmetic and with powers of ten. The meaning of
the efficiency of the power station was not always appreciated, a common error being to
presume that the reactor output would be 210 MW (which is 35% of 600 MW). Candidates who
made this kind of mistake only suffered a one mark penalty provided the remainder of their
solution hung together in terms of the physics involved.
E6.
Part (a) gave a much greater spread of marks than expected. About one third of candidates
did not attempt to place a unit on the y-scale and less able candidates also could not recall the
correct shape of the graph. At the top end, candidates allowed the graph to fall too steeply as the
nucleon number increased and/or they had the peak in the wrong position. Only the more able
candidates knew the height of the peak.
In part (b) only the more able candidates could use the idea of .binding energy. in a coherent
manner. Less able candidates did not really make any significant points that were worthy of
marks. On a marking point, although the question starts with ‘use the graph..’ it was possible to
score full marks without reference to the graph, as we allowed a reference to high and low
nucleon numbers as being equivalent to being either side of the peak.
Page 25 of 28
E7.
The more able candidates successfully negotiated the majority of this question but the less
able found many pit-falls.
In part (a) most obtained the first mark but then did not obtain the anti-neutrino.
For part (b) some candidates did not identify the position of P. Position Q was easier for
students to identify.
A majority of candidates could balance the number of neutrons in part (c)(i) to obtain the correct
answer x = 4. Those that guessed the answer almost always gave the answer x = 3.
Part (c)(ii) was very discriminating. Less able candidates did not know how to balance the
energies and only scored marks on the conversion from u to MeV. Some did not go directly from
u to MeV and gave many lines of calculation. If correctly performed, they still got the mark for the
conversion, but they had many opportunities to show errors and so tended to be less successful
and missed the mark.
E8.
Very few candidates knew what the atomic mass unit was. A surprising number thought it
was simply another name for nucleon number. The next choice of candidates in giving the
definition of the atomic mass unit was to give an energy or mass equivalence in J, MeV or kg.
In part (a)(ii) a majority knew that the mass of the nucleus was less than its constituents but too
many simply repeated back the question and said they were different. The explanation of why
they are different was very vague by most candidates. The majority thought it was sufficient to
simply state energy and mass are equivalent.
Part (b) was very discriminating. Only the best candidates scored the 3 marks because for most
there was a lack of appreciation that 3 marks usually equates to 3 points needing to be made in
answering the question. Some of the vague statements made included, ‘it takes a great deal of
energy to get fusion to start’ and many also referred to the need to overcome the SNF rather
than an electrostatic force before nuclei could get close enough to fuse.
Completion of the first equation in (c)(i) was done incorrectly by most because they did not pick
up the fact that the emitted particle must be an anti-lepton. The conservation laws covered in
module PHYA1 were often flouted in answering this question. The second equation was done
correctly by most.
The main feature that came out in (c)(ii) was how disorganised a majority of candidates are in
presenting their work. The consequence of this was that it became very difficult to award marks
for incomplete answers when there was just a jumble of figures on the page. The calculation
was difficult for many because they could not decide which units to work in. The more
successful chose to work in atomic mass units. Even here many thought the mass of the
hydrogen nucleus was 1u. In addition candidates answers often lacked precision leading to
rounding errors by using, for example, 1.67 × 10-27 kg rather than 1.673 × 10-27 kg for the mass of
the proton.
Page 26 of 28
E9.
In part (a) was very straightforward for many candidates. Less able candidates did not
elaborate on what the nucleus split up into or they referred to some other splitting such as
fission. Another common error was to talk about the separation of an atom into protons,
neutrons and electrons. Very few candidates got the direction of energy flow wrong.
In part (b)(i) very few candidates referred to particles other than neutrons but a significant
number failed to write down two neutrons in the correct manner.
and
a single were
extremely common.
The follow on part (b)(ii) turned out to be an extremely good discriminator. It highlighted the types
of errors candidates were making. There was a group who did not appreciate the data was given
per nucleon and used the figures without multiplying up by the respective nucleon numbers.
Another significant group added the mass energy of mostly one but sometimes two nucleons
into the proceedings. The last major group got into difficulties because they changed units
unnecessarily or only changed the units of some of the terms. Overall there was a good
percentage of correct answers.
A good proportion of candidates could accomplish the conversion of units required in part (b)(iii)
with relative ease, sometimes from an error carried forward from the previous part. With the
conversion needing two stages using the datasheet there was plenty of opportunity for errors by
dividing instead of multiplying or using an incorrect conversion factor. A separate but common
error was to ignore the answer to part (b)(ii) and simply use the difference in mass of the
nucleons involved, ignoring the binding energy per nucleon completely. Less able candidates did
not really help themselves in this question because very few put words or units to intermediate
stages of the calculation. If they had done so fewer would have lost their way.
A majority of candidates did not score marks in part (c)(i). Many knew the general shape but very
few remembered any details so they had no idea which coordinates to draw their line or band
through. Some gave no real thought to the problem and drew graphs that did not make sense.
For example some graphs went vertical or turned back on themselves.
Part (c)(ii) was very discriminating. Less able candidates simply referred to any radiation that
came to mind and forwarded very little explanation. The bulk of the candidates knew beta minus
radiation was emitted but they were not careful how they expressed their reasons. For example,
stating that there are more neutrons than protons is not sufficient to imply the nuclei are neutron
rich. A majority in this group of candidates made no reference to the graph at all. Many that did
had flawed reasoning. They thought the isotopes were neutron rich because the large number of
free neutrons in the core. Good candidates also got into some difficulty by not reading the
question carefully. Typically these candidates would start their answer with, 'If there are a lot
more neutrons than protons then... but if the neutron to proton ratio is small then…'. These
candidates obviously knew the subject matter but did not score many marks as the question
clearly asks for a choice to be made.
Page 27 of 28
Resource currently unavailable.
Page 28 of 28
The nuclear valley
Make notes on the following
Completed?
Annotated sketch of the nuclear valley (from
advancing physics)
Definition of binding energy
Annotated sketch of binding energy per nucleon
graph
Annotated NZ and AZ graphs
Decays as vectors in NZ and AZ space
Decay paths of proton rich and neutron rich nuclei
Complete and mark the following
Physics factsheet 21
IOP TAP 525-1: Change in energy: Change in mass
IOP TAP 525-3: Fusion in a kettle?
AQA exam past paper questions - Nuclear instability
DFWV
W
K
H
H
)
VLFV
3K\
September 2001
Number 21
Nuclear Transformations & Binding Energy
This Factsheet will explain:
• the concept of binding energy, and how it relates to nuclear
stability;
• how to use mass-energy equivalence in nuclear processes;
• the principles involved in fission and fusion reactions;
Before studying this Factsheet, you should be familiar with basic
atomic structure and equations for nuclear processes including
radioactivity (covered in Factsheet 11 – Radioactivity I).
Units
The standard SI units for mass and energy – the kilogram and joule –
are too large to use conveniently on an atomic scale. Instead, the
unified atomic mass unit (u) and the electronvolt (eV) are used.
1 electronvolt (eV) is the energy transferred to a free electron
when it is accelerated through a potential difference of one volt.
1 unified atomic mass unit (u) =
Binding Energy
The protons and neutrons in the nucleus of an atom are held together by
the strong nuclear force (Factsheet 14 − Particle Physics). So if we
imagine splitting a nucleus up into its separate protons and neutrons, it
would require energy, because we would need to overcome the strong
nuclear force.
Since energy is conserved:
energy put in to
energy of nucleus + separate nucleons
It is necessary to be able to convert these to SI units:
1 eV = charge on electron in coulombs × 1 volt
= 1.602 × 10-19 J
Since the mass in grammes of one carbon-12 atom is its atomic mass
(12) divided by Avagadro’s number (NA = 6.02 × 1023):
total energy of
= separated nucleons
1u =
Since energy is being provided, the separated protons and neutrons must
have more energy in total than the original nucleus. So if the nucleus
was formed from its constituent particles, energy would be released.
As figure 1 shows, the binding energy per nucleon varies considerably
between nuclei.
• Nuclei near the peak of the curve are the most stable.
• The curve peaks at iron – 56.
• The graph of binding energy per nucleon against nucleon number is
similar in form.
Binding energy per nucleon
Instead of looking at the total binding energy of a nucleus, it is often
more useful to consider binding energy per nucleon – in other words,
the total binding energy divided by the total number of nucleons.
For example, for the alpha particle, the total binding energy is 28.4 MeV
Since there are four nucleons (two protons and two neutrons), the
binding energy per nucleon is 28.4 ÷ 4 = 7.1 MeV.
Nuclear processes and binding energy
Processes such as radioactive decay, fission and fusion involve the
nucleons being rearranged into different nuclei. If the new nuclei
produced have a higher binding energy per nucleon, then energy is
given out. All spontaneously occurring processes involve energy
being given out – i.e. the products have greater binding energy per
nucleon than the original nuclei.
The binding energy per nucleon gives an indication of the stability of
the nucleus. A high binding energy per nucleon indicates a high degree
of stability – it would require a lot of energy to take these nuclei apart.
Binding energy per nucleon/MeV
Fig. 1 Binding energy per nucleon plotted against atomic no.
12 / 6.02 × 1023
= 1.66 × 10-24 g = 1.66 ×10-27 kg
12
Exam Hint: You need to know how to work out the relationship
between electronvolts and joules, and between unified atomic
mass units and kilograms, but you do not need to remember the
actual figures.
Binding energy is the energy released when a nucleus is
formed from its constituent particles
10
9
8
7
6
5
4
3
2
1
0
mass of carbon − 12 atom
12
Radioactive decay is a spontaneous process. It always involves a less
stable (i.e. lower binding energy per nucleon) nucleus decaying to form
a more stable nucleus. Energy is therefore always given out in
radioactive decay - in alpha-decay, for example, this energy is largely
in the form of the kinetic energy of the alpha particle.
Iron-56
Nuclear fission involves a heavy nucleus (such as uranium) splitting to
form two smaller nuclei and some neutrons. The nuclei produced will be
nearer the peak of the graph – so energy is released. Fission can only
occur with nuclei to the right of the peak.
0
10
20
30
40
50
Atomic Number (Z)
60
70
80
Nuclear fusion involves small nuclei joining together to form a larger
one – again, some neutrons are usually produced as well. The nucleus
produced will always be nearer the peak of the graph – so again, energy
is released. Fusion can only occur with nuclei to the left of the peak.
90
1
Nuclear Transformations and Binding Energy
3K\VLFV)DFWVKHHW
Example 1: Calculate the energy released (in joules) in the process:
228
224
4
90Th → 88 Ra + 2 He
Mass-energy equivalence
Mass and energy are linked by Einstein’s famous equation:
E = mc2,
E = energy (J), m = mass (kg) and c = speed of light ( ≈3 × 108 ms-1)
Masses: thorium-228= 228.02873u radium - 224 = 224.02020u
helium - 4 = 4.002603u
This equation is most often used in connection with a change in mass. It
tells us, for example that a change in mass of 1kg is equivalent to a
change in energy of 1 × c2 = 9 × 1016 J.
First we must find the mass defect:
Mass defect =( 228.02873 – 224.02020 – 4.002603) u=5.927× 10-3u
Mass defect in kilograms = 5.927 × 10-3 × 1.66 × 10-27
= 9.839 × 10-30 kg
Example: Calculate the energy change, in eV, equivalent to a mass
change of 1u
So energy released = mass defect × c2
= 9.839 × 10-30 × (3 × 108)2 = 8.85 × 10-13J
1u =1.66 × 10-27kg
So energy change in J = 1.66 × 10-27 × c2 = 1.49 × 10-10 J
1 eV = 1.602 × 10-19 J
1.49 × 10 −10
= 9.3× 108eV =930 MeV
So energy change in eV =
1.602 × 10 −19
Typical Exam Question
The nuclear equation shown below represents the decay of radium
to radon with the release of energy.
226
222
4
88 Ra → 86 Rn + 2 He
(a) Show, by appropriate calculations, that radon is more stable
than radium.
[5]
(b) Calculate the mass defect, in kg, for this reaction.
[2]
(c) Calculate the energy released in this reaction.
[2]
(d) Calculate the mass of radium (in kg) required to release 2000 MJ
of energy (ignore energy released by decays of radon)
[3]
Mass defect
This relationship between mass and energy means that since a nucleus
has less energy than its separated nucleons, the mass of the nucleus must
be less than that of its constituent particles.
The mass defect of a nucleus is the difference between the
mass of the nucleus, and the mass of its constituent particles.
In answering this question the following data may be useful
222
Masses: 226Ra = 226.02544u
Rn = 222.01761u
4
He= 4.002603u
1u = 1.661 × 10 -27kg
neutron = 1.008605u
proton = 1.007276u
Avagadro’s number N A = 6.02 ×1023 mol -1
c = 3.0 × 10 8ms –1
The mass defect is related to the binding energy by:
Binding energy (J) = mass defect (kg) × c2
All nuclei have a mass defect, apart from hydrogen – 1, whose nucleus
consists of just a single proton.
The graph of mass defect per nucleon against atomic number is very
similar to the graph of binding energy per nucleon against atomic
number.
(a) To find the binding energy per nucleon of Ra:
Constituent nucleons: 88 protons + 138 neutrons
Mass of constituents = (88 × 1.007276)u + (138 × 1.008605)u
= 227.827778 9
Mass defect (mass of constituents – mass of nucleus):
= 227.827778u − 226.025u = 1.802778u 9
Example. The mass of an alpha particle is 4.00150u. The mass of a
proton is 1.00728u, and the mass of a neutron is 1.00867u
(a) Calculate the mass defect of the alpha particle
(b) Calculate the binding energy of the alpha particle, giving your
answer in MeV.
To find the binding energy per nucleon of Rn:
Mass defect = (86 × 1.007276 + 136 × 1.008605) – 222.01761
= 1.778406u 9
(a) Total mass of constituent particles = 2 × (1.00728 + 1.00867)
= 4.0319u
So mass defect = 4.0319 – 4.00150 = 0.0304u
So forRa, mass defect per nucleon = 1.802778/226 = 7.977 × 10-3u9
for Rn, mass defect per nucleon = 1.778406/222 = 8.011 × 10-3u
(b) 0.0304u = 0.0304 × 1.66 × 10-27kg = 5.0464 × 10-29 kg
Binding energy (J) = 5.0464 × 10-29 × (3 × 108)2 = 4.542 × 10-12J
Binding energy (eV) = 4.542 × 10-12/(1.602 × 10-19) = 2.84 × 107 eV
So binding energy = 28.4 MeV
This tells us that the mass defect per nucleon (or binding energy per
nucleon) is higher for Rn, so it is more stable9
(b) Mass defect =(226.02544− 222.01761–4.002604)u =5.227×10-3u9
Mass defect (kg)=5.227× 10-3 × 1.661 × 10-27=8.68× 10-30kg 9
The mass defect for a reaction may also be considered; this is the
difference between the total mass of the products of the reaction and the
total mass of the reactants. The equation E = mc2 can be used to work
out the energy released , once the mass defect is known.
(c) Energy released = mc2 = 8.68 × 10-30 × (3.0 × 108)29
= 7.8 × 10-13 J9
Since spontaneous nuclear processes always involve energy being given
out:
(d) Number of atoms of Ra required = 2000 MJ/ 7.8 × 10-13 J 9
= 2 × 109 / 7.8 × 10-13 = 2.6 × 1021
Mass of Ra required = 2.6 × 1021 / 6.02 × 1023 × 226 9
= 0.98g = 9.8 × 10-4 kg 9
A spontaneous decay process always results in particles
with a lower total mass.
Calculation of the energy released in nuclear processes
Tip: To convert a number of atoms to a mass:
•
Divide the number of atoms by Avagadro’s number
•
Multiply the answer by the atomic mass
This gives the mass in grammes.
To calculate the energy released in a process, the mass defect in
kilograms should first be calculated, and then the equation E = mc2 used
to find the energy released.
2
Nuclear Transformations and Binding Energy
3K\VLFV)DFWVKHHW
Example 2: Uranium-235 undergoes fission according to the equation
235
1
96
138
1
92 U + 0 n→ 37 Rb + 55 Cs + 2 0 n
(a) Calculate the energy given out by this reaction, giving your answer
in electronvolts
(b) Calculate the energy given out by the fission of 1kg of uranium-235,
giving your answer in MJ.
Masses:
235
92 U
138
55 Cs
8
: 235.044u
96
37 Rb
: 137.920u
1
0n:
: 95.933u
Fission
Fission of uranium – 235 is used to produce power. In this reaction, a
neutron hits the U-235 nucleus, which then splits to produce two smaller
nuclei, more neutrons and energy (mainly in the form of kinetic energy
of the daughter nuclei and neutrons). The smaller nuclei formed may
vary; one example is shown in example 2 on the left and another is
shown below:
235
1
144
90
1
92 U + 0 n → 56 Ba + 36 Kr + 2 0 n
The number of neutrons may also vary; it is generally two or three, with
an average of about 2.6.
1.009u
c = 3.0 × 10 ms-1
(a) Mass defect = (235.044 + 1.009 − 95.933 – 137.920 – 2 × 1.009)u
= 0.182u
Mass defect in kg = 0.182 × 1.66 × 10-27 = 3.0212 × 10-28 kg
Energy given out in joules = 3.0212 × 10-28 × (3 × 108)2
= 2.719 × 10-11J
Energy given out in electronvolts = 2.719 × 10-11/1.602 × 10-19
= 1.7 × 108 eV
The neutrons given off in the fission of the U-235 atom may then go on
to hit other atoms, causing them to undergo fission in turn, which
produces more neutrons, causing more fission etc. This is known as a
chain reaction. To produce power, the chain reaction must be
maintained, but controlled. To maintain the chain reaction, a minimum
of one neutron from each fission reaction should go on to cause another
fission; to control it, there should not be more than one neutron causing
further fission. An uncontrolled chain reaction produces a nuclear bomb.
Tip: When you convert from J to eV, you should always get a
larger number. If you do not, you have probably multiplied by
1.602 × 10-19 instead of dividing.
(b) Need to find number of atoms of uranium-235 in 1kg:
Tip: To convert a mass to a number of atoms:
•
divide the mass in grammes by the atomic mass
•
multiply the answer by Avagadro’s number
To maintain a controlled chain reaction, the following problems must be
overcome:
• If the amount of fission material is too small, too many neutrons will
escape. The minimum acceptable amount of fission material is
called the critical size.
• The neutrons required to cause fission are slow neutrons, but the
neutrons produced by fission are quite fast. So the neutrons
produced have to be slowed down.
• Naturally occurring uranium contains only a small proportion of
U-235; the commonest isotope, U-238, absorbs neutrons without
undergoing fission. So the proportion of U-235 needs to be
increased before uranium can be used as fuel.
Number of atoms = 1000/235 × 6.02 × 1023 = 2.56 × 1024
So total energy released = no. of atoms × energy released per atom
= 2.56 × 1024 × 2.719 × 10-11J
= 6.96 × 1015J
= 6.96 × 109MJ
Typical Exam Question
In a nuclear fusion reaction, a nucleus of deuterium (hydrogen – 2)
coalesces with a tritium nucleus (hydrogen – 3). A helium nucleus is
formed, together with another particle.
Fission is carried out in a thermal reactor (fig 2). The components of
this reactor are used to ensure a controlled chain reaction is maintained.
(a) Identify the other particle formed and write an equation for the
reaction.
[2]
(b) Calculate the energy evolved in the reaction, giving your answer
in MeV
[5]
Fig 2. Thermal Reactor
concrete
Combined mass of deuterium and tritium nuclei = 5.031u
Combined mass of the particles produced in the reaction = 5.011u
1u = 1.661 × 10-27kg
Speed of light c = 3.000 × 108 ms-1
Electronic charge e = 1.602 × 10-19C
(a)
Fission and fusion
Nuclear fission and fusion both release a great deal of energy (see
previous examples); their use as power supplies is therefore of great
interest. Both fission and fusion produce a much greater amount of
energy per kilogram of fuel than do “conventional” energy sources such
as fossil fuel combustion; for example, the fission of 1 kilogram of
uranium-235 releases the same amount of energy as the combustion of
about 25000 kg of coal.
moderator
steam
2
3
4
1 H + 1 H → 2 He + ?
heat
exchanger
Since the Z and A values must balance, the particle produced has
a nucleon number of 1 and proton number of 0 ⇒ it is a neutron.9
control
rods
fuel
rods
So the equation is 12 H + 13H → 24 He+ 01n 9
water
(b) Mass defect = 5.031 – 5.011 = 0.020u9
Mass defect in kg = 0.020 × 1.661 × 10-27 = 3.322 × 10 –29 kg 9
Energy produced = 3.322 × 10-29 × c2 = 2.9898 × 10-12J9
Energy produced in eV = 2.9898 × 10-12/ 1.602 × 10-19 9
= 1.87 × 107eV
= 18.7 MeV9
3
coolant
Nuclear Transformations and Binding Energy
•
•
•
•
•
3K\VLFV)DFWVKHHW
Exam Workshop
This is a typical poor student’s answer to an exam question. The
comments explain what is wrong with the answers and how they can
be improved. The examiner’s answer is given below.
The fuel rods are made of enriched uranium – that is, natural
uranium with extra U-235 added.
The moderator slows down the neutrons produced in fission so that
they can stimulate further fission. It is made of graphite (or
sometimes water). The neutrons are slowed down by their collisions
with carbon nuclei in the moderator; some of their energy is
transferred to the moderator, which gets hot. The slow neutrons are
known as “thermal neutrons”, because they move at speeds
associated with thermal motion; they give the thermal reactor its
name.
The control rods are made of a substance, such as boron, that
absorbs neutrons. They are used to control the chain reaction, to
ensure too many neutrons do not cause fission. They can be raised
or lowered to speed up or slow down the reaction.
The coolant - usually water or pressurized carbon dioxide - is used
to remove energy from the system. The heat generated in the fuel
rods is transferred to the coolant by conduction; this transmits it to
the heat exchanger, where it is used to convert water to high
pressure steam, which is used to drive a turbine to produce
electricity.
The concrete shielding absorbs the nuclear radiation; it is necessary
because of the danger nuclear radiation poses to living creatures.
(a) Explain the purpose of the moderator in a thermal reactor.
To slow down neutrons9 to stop the reaction going too fast8
[2]
1/2
This is a common mistake – the neutrons have to be slowed down
to allow the chain reaction to take place, not to control it.
(b) Suggest a suitable radioactive source for this reactor
Uranium8
[1]
0/1
Not specific enough – only U-235 is suitable, not U-238
(c) Suggest two advantages and two disadvantages of the use of
thermal nuclear reactors rather than fossil fuels to generate
electricity.
[4]
Disadvantages: dangerous8 waste must be stored for a long time9
2/4
Advantages: no pollution8 doesn’t rely on fossil fuels9
“Dangerous” is not specific enough – how and why is it dangerous?
“No pollution” is not specific enough either – the candidate should have
specified which forms of pollution are reduced.
There are environmental hazards associated with nuclear reactors;
they include:
• If radioactive gases or dust escape into the atmosphere, they can be
readily absorbed by humans and animals via food, water or
breathing.
• Used, “spent” fuel rods contain many radioactive decay products;
accordingly they must be stored securely to prevent their being a
hazard. Some of these decay products have very long half-lives, and
hence must be stored for thousands of years.
• If the chain reaction in a fission reactor is not controlled, the reactor
may act as a bomb.
(d) Explain how the energy released by a nuclear fission reaction is
calculated, including a suitable equation. How reliable is this
calculation?
[4]
Using E = mc29
There may be rounding errors, making it inaccurate 8
1/4
The candidate should have realised this was not enough for 4 marks!
Questions on the reliability of a calculation must always be
answered specifically to the context – rounding error can apply to
any calculation in physics.
The benefits of fission reactors include:
• A great deal of energy is produced from a small amount of fuel
• Fission does not produce the gaseous waste products (carbon
dioxide, nitrogen and sulphur oxides) associated with combustion of
fossil fuel.
• Suitable fuel is not in such short supply as fossil fuels.
Examiner’s Answers
(a) Slows neutrons9 so that they are more likely to cause fission.9
(b) Uranium-235 9
(c) Advantages: high energy density, does not produce pollutants
carbon dioxide/nitrogen & sulphur oxides. 99
Disadvantages: Any two of: waste products dangerous, danger of
meltdown, storage of waste products a problem99
(d) There is a mass defect for the reaction 9 which allows the energy
released to be calculated9 by using E = mc29. This calculation
ignores the energy produced by later reactions of fission products
and so may or may not be reliable.9
Fusion
Fusion involves two light nuclei combining to form a heavier one,
together with energy and one or more neutrons. It is the process which
produces energy in the sun. One such reaction involves the fusion of
deuterium (hydrogen-2) and tritium (hydrogen-3) nuclei to form a
helium nucleus, a neutron and energy:
2
3
4
1
1 H + 1H → 2 He + 0 n
Questions
1. Explain what is meant by binding energy and mass defect, and state
an equation connecting them.
Fusion reactors are not yet in existence; it is much more difficult to
achieve and control fusion than fission. This is because it is necessary
to overcome the repulsion between the nuclei. This requires a very high
temperature – 100 million K or above. Normal containers cannot hold
anything as hot as this; containing this material by magnetic fields is
one possibility.
2. Sketch the curve of binding energy per nucleon against proton
number, and explain how it can be used to predict which atoms will
undergo fission and which will undergo fusion.
3. Explain the principles of operation of a thermal reactor.
4. Explain what is meant by nuclear fusion, and give two advantages
of fusion over fission as an energy source.
Fusion has a number of advantages:
• Fusion is a more productive energy source than fission per kilogram
of material.
• The raw materials for fusion can be obtained from sea water.
• The waste products are not radioactive.
• Uncontrolled chain reactions cannot develop.
238
92 U
decays by alpha-emission to give thorium. Calculate the
energy emitted, giving your answer in MeV.
(Masses: U-238, 238.0508u; Th-234, 234.0437u; He-4, 4.0026u)
1u = 1.661× 10-27 kg; c = 3.0 × 108 ms-1
Answers
1 – 4 can be found in the text.
5. Mass defect = (238.0508 – 234.0437 – 4.0026_u = 0.0045u
Mass defect = 0.0045 × 1.66 × 10-27 = 7.47 × 10-30 kg
Energy in joules = 7.47 × 10-30 kg × c2 = 6.723 × 10-13
Energy in eV=6.723×10-13/1.602 × 10-19 = 4.20 × 106eV = 4.20 MeV
5.
This Factsheet was researched and written by Cath Brown.
Curriculum Press, Unit 305B The Big Peg, 120 Vyse Street, Birmingham B18 6NF. Physics Factsheets
may be copied free of charge by teaching staff or students, provided that their school is a registered
subscriber. They may be networked for use within the school. No part of these Factsheets may be
reproduced, stored in a retrieval system or transmitted in any other form or by any other means without the
prior permission of the publisher. ISSN 1351-5136
4
TAP 525-1: Change in energy: Change in mass
Calculating using Erest = mc2
These questions show you how to calculate changes in energy from changes in mass, using
2
Einstein’s relation Erest = mc linking the rest energy of a particle to its mass.
Transmutation of chemical elements
The dream of the ancients was alchemy: turning base metals into gold. Although this is
chemically impossible, at the end of the nineteenth century radioactivity was discovered by
Henri Becquerel. When alpha and beta radiation are emitted atomic nuclei are ‘transmuted’
from one element to another. For example:
238
92 uranium
o 24 alpha 234
90 thorium.
In 1932 using protons (hydrogen nuclei) accelerated through a potential difference of 800 000
V, two English physicists, Cockcroft and Walton, carried out the first artificial transmutation: by
bombarding lithium with the protons they produced two helium nuclei:
1
7
4
1H 3Li o 2
He 24He.
Change in mass
Notice that in both these reactions the mass number and charge (proton number) are
conserved. Energy, however, is only conserved if you take account of changes to the rest
energy – in effect of changes to the masses – of the particles.
In Cockcroft and Walton’s experiment, the masses of the particles are:
x
H: 1.0073 atomic mass units
x
Li: 7.0160 atomic mass units
x
He: 4.0015 atomic mass units.
An atomic mass unit, symbol u, is equal to 1.6605 u 10
1
–27
kg.
Show that the mass decreases in this reaction.
1
7
4
1H 3Li o 2
He 24He.
Calculate 'm in atomic mass units and in kilograms.
Change in energy
2
The energy of the protons was 800 000 electron volts (800 keV). The lithium was in
solid form so the nuclei would only have been vibrating due to thermal energy, less
than an electron volt.
The reaction was captured in this photograph:
Two pairs of alpha particles, emerging in opposite directions, can be seen in the
photograph.
From the range of the tracks through the cloud chamber the energy of the alpha
particles was measured to be 8.5 MeV each.
Show that the total kinetic energy of the particles increases, and calculate 'E in MeV
and in joules.
3
If the increase in kinetic energy comes from the decrease in rest energy you should
2
expect 'E = 'mc . Calculate the ratio of the change in kinetic energy to the change in
–1
mass 'E/'m in J kg .
4
Show that the value of the ratio 'E/'m is approximately consistent with the
2
relationship 'E = 'mc .
16
–1
The large value of c 2 (9 u 10 J kg : use this value from now on in calculations) means that
a small change in mass represents a vast change in rest energy. This relationship between
2
mass and energy is why particle physicists measure masses in MeV / c ; any unit of energy
2
divided by c is a unit of mass.
Creating massive particles
Energy is ‘materialised’ in matter–antimatter production. A photon of electromagnetic radiation
can produce an electron and a positron. In this case, the energy of the photon vanishes and
the rest energy of the particles appears. (This reaction needs to take place near to the
nucleus of a heavy atom to conserve momentum but this is not going to affect your
calculations here.)
In this bubble chamber photograph a photon enters from the bottom. It is uncharged and so
produces no observable track. After some distance the photon disappears and produces the
electron–positron pair. These two charged particles ionise the liquid in the chamber and
bubbles form near the ions and are photographed.
In this case the chamber is filled with liquid hydrogen mixed with liquid neon. It is held under
pressure which is released just as the particles enter the chamber to encourage bubbles to
form and enlarge near the ions.
5
The bubble chamber is in a magnetic field, so charged particles bend due to the force
Bqv on a moving charge. How does the photograph show that the two particles have
opposite charges?
6
The mass of the electron is 5.5 u 10 u. What is the minimum energy photon that will
produce an electron–positron pair? From what part of the electromagnetic spectrum
–34
–1
is this? (Planck constant h = 6.63 u 10 J Hz .)
–4
Nuclear binding energy
If protons and neutrons (together known as nucleons) are bound together in a nucleus, the
bound nucleus must have less energy than the nucleons of which it is made. That is, the rest
energy of the nucleus must be less than the sum of the rest energies of its nucleons. In turn,
this means that the mass of the nucleus must be less than the sum of the masses of its
nucleons.
The simplest compound nucleus is the deuteron, the nucleus of hydrogen-2. It consists of a
proton and a neutron bound together by the strong nuclear force. The masses of these
particles are:
x
proton: 1.0073 u
x
neutron: 1.0087 u
x
deuteron: 2.0136 u.
7
Calculate the difference in mass between a deuteron and one proton and one neutron.
8
Calculate the binding energy of the deuteron in J and in MeV.
9
Calculate the binding energy per nucleon of the deuteron.
10
Express the difference in mass as a percentage of the sum of the masses of the
proton and neutron.
Mass change in nuclear fission
A possible reaction for the nuclear fission of uranium-235 is:
235
1
133
99
92 U 0 no 51 Sb 41Nb
410 n.
The masses of the particles are
x
U-235 = 235.0439 u
x
Sb-133 = 132.9152 u
x
Nb-99 = 98.9116 u
x
neutron (n) = 1.0087 u.
11
Show that the energy change per atom of uranium is about 200 MeV and calculate
'm/m.
Summary
2
Einstein's famous equation Erest = mc reveals a Universe that is not as simple as it seems at
first sight. The mass of a particle is generally a very large part of its total energy. The
existence of rest energy was not suspected until after Einstein had predicted it, because the
change in mass is usually so small, because changes in energy are usually a small fraction of
the rest energy. Only in nuclear reactions where 'm/m ~ 0.1% or more are you able to see
the change in mass, accompanied by what appears to be a huge change in energy.
Hints
1
Compare masses of H plus Li with mass of two He nuclei.
2
Two 8.5 MeV alpha particles come out, but one 800 keV proton goes in.
3
Compare the answers to questions 1 and 2.
4
2
c!
Don’t expect to get exactly the speed of light. Remember to take the square root of
5
What is the difference between forces F and - F?
6
Start with the mass of an electron in atomic mass units. Convert to kilograms. Write
2
down the mass of an electron–positron pair. Use Erest = mc to get the rest energy of
the pair in joules. Then use E = hf.
7
Do this one in the same way as question 1.
8
Erest = mc again. But now use the electron charge to get to electron volts and MeV.
9
How many nucleons in a deuteron?
10
Best to take the difference as a fraction of the mass before.
11
Add up before and after masses in atomic mass units first. Don’t forget there’s one
extra neutron to start with and four extra neutrons afterwards. Then convert mass
changes first to joules and then to MeV.
2
Practical advice
These questions practise the use of the relation between rest energy and mass in various
contexts: nuclear transmutation, creation of particle–antiparticle pairs, nuclear binding and
nuclear fission.
You may need to select only certain groups of questions, depending on what the class has
covered. Alternatively, the whole set could be used for revision.
Some of the questions make extra demands in frequent changes of units, between atomic
mass units, kilograms, joules and electron volts or MeV. You may well need to give extra help
here.
Note the consistent use of the term ‘rest energy’. The rest energy is treated as part of the total
energy. It manifests itself in the mass of a particle. If mass is measured in kilograms and
2
energy in joules, then the conversion is Erest = mc . Remember that the mass, an invariant, is
a physical property of a particle independent of frame of reference.
The questions bring out the fact that the rest energy is a very large fraction of the total energy,
in many cases.
Alternative approaches
You may want to show that energies involved in everyday processes involve negligible
changes in mass. The calculations of percentage change of mass in the questions here
provide a starting point.
Social and human context
It is not possible to ignore the consequences for war and peace of the possibility of tapping
these very large sources of energy.
Answers and worked solutions
1.
Mass of H plus Li 1.0073 u 7.0160 u
Mass of two He
2 u 4.0015 u
8.0233 u
8.0030 u
Difference 'm 8.0030 u 8.0233 u
0.0203 u
So we can find the mass difference in kg:
'm
0.0203 u u1.6605 u10 27 kg
3.3708 u10 29 kg
2.
Increase in energy:
'E
2 u 8.5 MeV 0.8 MeV 16.2 MeV
In joules:
'E
(16.2 u10 6 eV) u (1.6 u10 19 J eV 1 )
2.6 u10 12 J
3.
'E
'm
2.60 u10 12 J
3.37 u10
2
29
kg
2
7.7 u1016 J kg 1 .
16
–1
8
–1
4.
If 'E = 'mc , then c = 7.7 u 10 J kg , so c = 2.8 u 10 m s .
5.
The force on a moving charged particle is Bqv. If the charge q changes sign, the
direction of the force is reversed, so the curvature is opposite.
6.
The mass of an electron or positron is equal to:
(5.5 u 10 4 u) u (1.66 u 10 27 kg )
9.1u 10 31 kg.
2
From Erest = mc , the rest energy of an electron–positron pair is:
E rest
2 u 9.1u10 31 kg u (3 u10 8 m s 1 ) 2
1.6 u10 13 J.
If this energy is supplied by a photon of energy E = hf, then:
f
1.6 u10 13 J
6.63 u10
34
J Hz
2.5 u10 20 Hz.
1
This is the frequency of a gamma ray.
7.
The mass difference is:
2.0136 u (1.0073 u 1.0087 u)
I
0.0024 u.
n kg the mass difference is:
0.0024 u u (1.66 u 10 27 kg )
3.98 u 10 30 kg.
8.
Binding energy 3.98 u 10 30 kg u (3 u 108 m s 1)2
–3.58 u 10 13 J
3.58 u 10 13 J
–
19
1.6 u 10
–2.2 MeV .
9.
J eV
1
–2.2 u 106 eV
The deuteron has two nucleons so the binding energy per nucleon is
–2.2 MeV / 2 = –1.1 MeV.
10.
As a percentage the mass difference is equal to:
0.0024 u
1.0073 u 1.0087 u
1.2 u10 3 u100
0.1% (approximat ely)
11.
Mass after
132.9152 u 98.9116 u (4 u1.0087 u)
Mass difference
236.0526 u 235.8616 u
Change in rest energy
235.8616 u
0.191 u.
0.191 u u (1.66 u10 -27 kg) u (3 u10 8 m s 1 ) 2
1.6 u10
19
J eV
1
The ratio is given by:
'm / m 0.191 u / 236 u
8.1u10 4 ~ 0.1%.
External reference
This activity is taken from Advancing Physics chapter 18, 200S
1.78 u10 8 eV 178 MeV.
TAP 525-3: Fusion in a kettle?
A change of scale
When you are confident with basic calculations of fission and fusion energy changes, you
should work through these questions that try to put the energies of these changes into a more
human scale for you. You will also need to understand the conversion of atomic mass units to
energy and the meaning of the term ‘electron volt’.
Try these
One of the reactions that fuel the stars is the fusion of two protons to give deuterium. In turn
the deuterium goes through a series of reactions, the end product being helium. This is also a
process that releases energy. In this question you are asked to consider the energy that
would be released if all the deuterium in the water contained in an electric kettle were to be
converted by fusion into helium.
The kettle contains 1 litre of water. The data you need are listed below.
1 atomic mass unit (u) = 931 MeV
–19
1 eV = 1.6 u 10
NA = 6.02 × 10
1
2
23
J
–1
mol
Particle
Mass / u
1
1H
1.007 825
2
1H
2.014 102
3
2 He
3.016 030
1
0n
1.008 665
3
2
Two deuterium nuclei 1 H can fuse to give one nucleus of helium 2 He with the
ejection of one other particle. Write down the balanced equation that represents this
reaction.
Calculate the mass change that occurs in this reaction.
3
Convert this energy into joules.
This gives you the energy released when two deuterium nuclei fuse. The next steps take you
through the calculation of the total energy released if all the deuterium in the kettle water were
to fuse to make helium-3. The ratio of deuterium atoms to hydrogen in water is roughly 1 to
7000.
4
What is the mass of 1 mole of water (H = 1 u; O = 16 u roughly)?
5
How many moles of water are contained in the litre?
6
How many molecules of water (H2O) are in the kettle?
7
How many molecules of deuterium oxide (D 2O) are in the kettle?
8
Each heavy water molecule has two atoms of deuterium; what total energy is
released if all the deuterium in the kettle is converted to helium-3?
Now to put this number in a new perspective. It requires 4200 J to increase the temperature
of 1kg of water by 1K.
9
How many litres of water could be heated through 100 K by the fusion energy you
calculated in question 8?
Hints
1
It is important to consider the atomic electrons in this equation. You begin with two,
one for each hydrogen. How many electrons does an un-ionised atom of deuterium
have? So what must one of the emitted particles be? This should lead you to the
other particle.
2
The conversions you need are near the data table in the question.
4
The formula of water shows that there are two hydrogen atoms and one oxygen for
each water molecule.
5
1 litre of water has a mass of 1 kg.
6
1 mole contains 6 x 10 molecules of water.
23
Practical advice
These questions can be modified in many ways, not least by changing the homely example of
a kettle to perhaps a bath full of water or even to Lake Windermere or the local reservoir.
Social and human context
The 6000 litres of heated water may not seem so significant until you realise that this has
3
come from the fusion of deuterium which had an original volume of 0.15 cm .
Answers and worked solutions
1.
2
2
3
1H 1Ho 2 He
01n
2.
'm = (3.016 030 u + 1.008 665 u) – 2 × 2.014 102 u = – 0.0035 u
3.
0.003509 u × 931 × 10 eV u
4.
18 g
5.
1 litre of water has a mass of 1 kg.
–1
6
–19
× 1.6 × 10
–1
number of moles = 1000 g / 18 gmol
23
–1
mol
56 mol × 6.02 × 10
7.
(3.4 ×10 )/7000= 4.9 u 10
8.
energy released = 4.9 ×10
9.
(2.49 ×10 J)/ (4200 J kg K
9
–1
–13
= 5.23 u 10
= 56 mol
25
= 3 u 10
6.
25
J eV
21
21
–13
× (5.23 ×10
–1 –1
9
J) = 2.49 u 10 J
× 100K) = 6000 kg = 6000 litres
External reference
This activity is taken from Advancing Physics chapter 18, 260S
J
Q1.
Figure 1 shows a grid of neutron number against proton number. A nucleus
X is marked.
Figure 1
(a)
Draw arrows on Figure 1, each starting on
following transitions:
X and ending on a daughter nucleus after the
(i)
β– emission
(label this arrow A)
neutron emission (label this arrow B)
electron capture (label this arrow C).
(ii)
Give the equation for electron capture by the nucleus
X.
.............................................................................................................
(4)
Page 1 of 27
(b)
When
Mg decays to
Al by β– decay, the daughter nucleus is produced in one of two
possible excited states. These two states are shown in Figure 2 together with their
corresponding energies.
Figure 2
(i)
Calculate the maximum possible kinetic energy, in J, which an emitted β– particle can
have.
.............................................................................................................
.............................................................................................................
(ii)
The excited aluminium nuclei emit
photon energies in J.
photons. Calculate each of the three possible
.............................................................................................................
.............................................................................................................
.............................................................................................................
(iii)
Calculate the frequency of the most energetic
photon emitted.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
Page 2 of 27
(c)
(i)
State and explain two precautions that should be taken when working with
a sample of
Mg in a school laboratory.
You may be awarded marks for the quality of written communication in your answer.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
Discuss which of the two types of radiation, β– or
would be the more hazardous.
, emitted from a sample of
Mg
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(Total 10 marks)
Page 3 of 27
Q2.
(a)
The lead nuclide
Pb is unstable and decays in three stages through α and β
emissions to a different lead nuclide
Pb. The position of these lead nuclides on a grid of
neutron number, N, against proton number,
b Z, is shown below.
On the grid draw three arrows to represent one possible decay route.
Label each arrow with the decay taking place.
(3)
(b)
The copper nuclide Cu may decay by positron emission or by electron capture to form a
nickel (Ni) nuclide.
Complete the two equations that represent these two possible modes of decay.
positron emission
Cu
electron capture
Cu
(4)
Page 4 of 27
(c)
The nucleus of an atom may be investigated by scattering experiments in which radiation
or particles bombard the nucleus.
Name one type of radiation or particle that may be used in this investigation and describe
the main physical principle of the scattering process.
State the information which can be obtained from the results of this scattering.
You may be awarded marks for the quality of written communication in your answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 10 marks)
Page 5 of 27
Q3.
(a) Sketch, using the axes provided, a graph of neutron number, N, against proton
number, Z, for stable nuclei over the range Z = 0 to Z = 80. Show suitable numerical
values on the N axis.
(2)
(b)
On the graph indicate, for each of the following, a possible position of a nuclide that may
decay by
(i)
α emission, labelling the position with W,
(ii)
β– emission, labelling the position with X,
(iii)
β+ emission, labelling the position with Y.
(3)
Page 6 of 27
(c)
The isotope
decays sequentially by emitting α particles and β– particles,
eventually forming the isotope
. Four α particles are emitted in the sequence.
Calculate the number of β– particles in the sequence.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(d)
A particular nuclide is described as proton-rich. Discuss two ways in which the nuclide
may decay. You may be awarded marks for the quality of written communication in your
answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 10 marks)
Q4.
The age of an ancient boat may be determined by comparing the radioactive decay of
from living wood with that of wood taken from the ancient boat.
A sample of 3.00 × l023 atoms of carbon is removed for investigation from a block of living wood.
In living wood one in 1012 of the carbon atoms is of the radioactive isotope
decay constant of 3.84 × 10–12 s –1.
(a)
, which has a
What is meant by the decay constant?
......................................................................................................................
......................................................................................................................
......................................................................................................................
(1)
Page 7 of 27
(b)
Calculate the half-life of
significant figures.
in years, giving your answer to an appropriate number of
1 year = 3.15 × 107 s
answer = ..................................... years
(3)
(c)
Show that the rate of decay of the
atoms in the living wood sample is 1.15 Bq.
(2)
(d)
A sample of 3.00 × 1023 atoms of carbon is removed from a piece of wood taken
from the ancient boat. The rate of decay due to the
atoms in this sample is 0.65 Bq.
Calculate the age of the ancient boat in years.
answer = ............................ years
(3)
Page 8 of 27
(e)
Give two reasons why it is difficult to obtain a reliable age of the ancient boat from the
carbon dating described.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 11 marks)
Q5.
(a)
State what is meant by the binding energy of a nucleus.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
(i)
When a
nucleus absorbs a slow-moving neutron and undergoes fission one
possible pair of fission fragments is technetium
and indium
.
Complete the following equation to represent this fission process.
(1)
Page 9 of 27
(ii)
Calculate the energy released, in MeV, when a single
fission in this way.
binding energy per nucleon of
= 7.59 MeV
binding energy per nucleon of
= 8.36 MeV
binding energy per nucleon of
= 8.51 MeV
nucleus undergoes
energy released ..................................... MeV
(3)
(iii)
Calculate the loss of mass when a
nucleus undergoes fission in this way.
loss of mass ......................................... kg
(2)
Page 10 of 27
(c)
(i)
On the figure below sketch a graph of neutron number, N, against proton number, Z,
for stable nuclei.
proton number, Z
(1)
(ii)
With reference to the figure, explain why fission fragments are unstable and explain
what type of radiation they are likely to emit initially.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(Total 12 marks)
Page 11 of 27
Q6.
The isotope of uranium,
, decays into a stable isotope of lead,
, by means of a
–
series of α and β decays.
(a)
In this series of decays, α decay occurs 8 times and β– decay occurs n times.
Calculate n.
answer = ...........................................
(1)
(b)
(i)
Explain what is meant by the binding energy of a nucleus.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(ii)
Figure 1 shows the binding energy per nucleon for some stable nuclides.
Figure 1
Use Figure 1 to estimate the binding energy, in MeV, of the
nucleus.
answer = ................................. MeV
(1)
Page 12 of 27
(c)
The half-life of
is 4.5 × 109 years, which is much larger than all the other half-lives of
the decays in the
h series.
A rock sample when formed originally contained 3.0 × 1022 atoms of
and no
atoms.
At any given time most of the atoms are either
atoms in other forms in the decay series.
(i)
or
Sketch on Figure 2 graphs to show how the number of
with a negligible number of
atoms and the number
of
atoms in the rock sample vary over a period of 1.0 × 1010 years from its
formation.
ti
Label your graphs U and Pb.
Figure 2
(2)
(ii)
A certain time, t, after its formation the sample contained twice as many
as
atoms
atoms.
Show that the number of
atoms in the rock sample at time t was 2.0 × 1022.
(1)
Page 13 of 27
(ii)
Calculate t in years.
answer = ................................. years
(3)
(Total 10 marks)
Page 14 of 27
M1.
(a)
(i)
correct arrows: A (1)
B (1)
C (1)
(ii)
e–1 +
+ v e (1)
→
4
(b)
(i)
((4.18 – 1.33) × 10–13) = 2.85 × 10–13 (J) (1)
(ii)
1.33 × 10–13 (J)
0.30 × 10–13 (J)
1.63 × 10
(iii)
–13
for 3 correct values (1)
(J)
(use of ΔE = hf gives) f
= 2.46 × 1020 Hz (1)
(allow C.E. from (b)(ii)if largest value taken)
3
(c)
(i)
((1)for each precaution with reason to max2)
handle with (long) (30 cm) tweezers
because the radiation intensity decreases with distance
store in a lead box (immediately) when not in use to avoid
unnecessary exposure to radiation
[or any sensible precaution with reason]
QWC 2
(ii)
γ rays are more penetrating and are therefore more hazardous
(to the internal organs of the body)
β– particles are more hazardous because they are more ionising (1)
(1) for any argued case for either radiation)
3
[10]
Page 15 of 27
M2.
(a)
(on grid: first arrow to start from
;
arrows must be consecutive;
last arrow must end on
)
arrow showing the change for an α emission (1)
arrow showing the change for a β emission (1)
correct α and two β emissions in any order (1)
3
(b)
(positron emission)
(electron capture)
+ β+ + v e (+Q) (1) (1)
→
+
→
+ v (e) (+Q) (1) (1)
4
(c)
(the following examples may be included)
α particles (1)
coulomb/electrostatic/electromagnetic repulsion
[or K.E. converted to P.E. (as α particle approaches nucleus)] (1)
information:
any of the following: proton number, nuclear charge,
upper limit to nuclear radius
mass of nucleus is most of the mass
of atom (1)
[alternative
(high energy) electron (scattering) (1)
diffraction of de Broglie Waves by nucleus (1)
information:
any of the following: nuclear radius, nuclear density (1)]
3
QWC 2
[10]
M3.
(a) graph passes through N = 10/11 when Z = 10 and N increases as Z
increases (1)
N = 115 → 125 when Z = 80 and graph must bend upwards (1)
2
(b)
(i)
W at Z > 60 just (within one diagonal of a square) below line (1)
(ii)
X just (within one diagonal of a square) above line (1)
(iii)
Y just (within one diagonal of a square) below line (1)
3
(c)
working showing the change due to emission of four α particles (1)
four β– particles (1)
1
Page 16 of 27
(d)
Any two from the following list of processes:
β+
describe the changes to N (up by 1) and Z (down by 1)
[or allow p change to n]
α
move closer to line of stability
[or state the proton to neutron ratio is reduced]
p
only if nuclide is very proton rich
[or electrostatic repulsion has to overcome the strong nuclear force]
[or highly unstable]
[or rare process]
e– capture
describe the changes to N (up by 1) and Z (down by 1)
allow p changes to n
marking: listing two processes (1)
discussing each of the two processes (1) (1)
3
QWC 1
[10]
M4.
(a)
probability of decay per unit time/given time period
or fraction of atoms decaying per second
or the rate of radioactive decay is proportional to the number of (unstable)
nuclei
and nuclear decay constant is the constant of proportionality (1)
1
(b)
use of
=
= ln2/3.84 × 10–12 s (1) (1.805 × 1011 s)
= (1
(1.805 × 1011/3.15 × 107) = 5730 y (1)
answer given to 3 sf (1)
3
Page 17 of 27
(c)
number of nuclei = N = 3.00 × 1023 × 1/1012 (1)
(= 3.00 × 1011 nuclei)
(using
)
rate of decay = 3.84 × 10–12 × 3.00 × 1011 (1)
(= 1.15 Bq)
2
(d)
(N = N0e–λt and activity is proportional to the number of nuclei A  N use of
A = A0e–λt)
0.65 = 1.15 ×
(1)
t = 4720 y (1)
3
(e)
the boat may have been made with the wood some time after the tree was
cut down
the background activity is high compared to the observed count rates
the count rates are low or sample size/mass is small or there is statistical
variation in the recorded results
possible contamination
uncertainty in the ratio of carbon-14 in carbon thousands of years ago
any two (1)(1)
2
[11]
(a) the amount of energy required to separate a nucleus ✓
into its separate neutrons and protons / nucleons ✓
(or energy released on formation of a nucleus ✓
from its separate neutrons and protons / constituents ✓)
M5.
1st mark is for correct energy flow direction
2nd mark is for binding or separating nucleons (nucleus is in the
question but a reference to an atom will lose the mark)
ignore discussion of SNF etc
both marks are independent
2
(b)
(i)
✓
must see subscript and superscripts
1
Page 18 of 27
(ii)
binding energy of U
= 235 × 7.59 ✓ ( = 1784 (MeV))
binding energy of Tc and In
= 112 × 8.36 + 122 × 8.51 ✓
( = 1975 (MeV))
energy released ( = 1975 – 1784) = 191 (MeV) ✓ (allow 190 MeV)
1st mark is for 235 × 7.59 seen anywhere
2nd mark for 112 × 8.36 + 122 × 8.51 or 1975 is only given if there
are no other terms or conversions added to the equation (ignore
which way round the subtraction is positioned)
correct final answer can score 3 marks
3
(iii)
energy released
= 191 × 1.60 × 10−13 ✓
( = 3.06 × 10−11 J)
loss of mass ( = E / c2 )
= 2.91 × 10−11 / (3.00 × 108)2)
= 3.4 × 10−28 (kg) ✓
or
= 191 / 931.5 u ✓ ( = 0.205 u)
= 0.205 × 1.66 × 10−27 (kg)
= 3.4 × 10−28 (kg) ✓
allow CE from (ii)
working must be shown for a CE otherwise full marks can be given
for correct answer only
note for CE
answer = (ii) × 1.78 × 10−30
(2.01 × 10−27 is a common answer)
2
(c)
(i)
line or band from origin, starting at 45° up to Z approximately = 20 reading
Z = 80, N = 110→130 ✓
initial gradient should be about 1 (ie Z = 20 ; N = 15 → 25) and
overall must show some concave curvature. (Ignore slight
waviness in the line)
if band is shown take middle as the line
if line stops at N > 70 extrapolate line to N = 80 for marking
1
Page 19 of 27
(ii)
fission fragments are (likely) to be above / to the left of the line of stability ✓
fission fragments are (likely) to have a larger N / Z ratio than stable nuclei
or
fission fragments are neutron rich owtte ✓
and become neutron or β− emitters ✓
ignore any reference to α emission
a candidate must make a choice for the first two marks
stating that there are more neutrons than protons is not enough for
a mark
1st mark reference to graph
2nd mark – high N / Z ratio or neutron rich
3rd mark beta minus
note not just beta
3
[12]
M6.
(a)
β=6
1
(b)
(i)
the energy required to split up the nucleus
into its individual neutrons and protons/nucleons
(or the energy released to form/hold the nucleus
from its individual neutrons and protons/nucleons
)
2
(ii)
7.88 × 206 = 1620 MeV
(allow 1600-1640 MeV)
1
(c)
(i)
U, a graph starting at 3 × 1022 showing exponential fall passing through
0.75 × 1022 near 9 × 109 years
Pb, inverted graph of the above so that the graphs cross at 1.5 × 1022 near
4.5 × 109 years
2
(ii)
(u represents the number of uranium atoms then)
u = 6 × 1022 – 2u
u = 2 × 1022 atoms
1
Page 20 of 27
(iii)
(use of N = No e-λt)
2 × 1022 = 3 × 1022 × e-λt
t = ln 1.5 / λ
(use of λ = ln 2 / t1/2)
λ = ln 2 / 4.5 × 109 = 1.54 × 10-10
t = 2.6 × 109 years
(or 2.7 × 109 years)
3
[10]
Page 21 of 27
E1.
This question set on the nuclear instability specification yielded reasonable answers but very
few candidates scored full marks. The three transitions required in part (a) were frequently all
correct, although the arrows representing A and C seemed to be randomly chosen in many
scripts. The arrow representing B was correct more often than not. The most common mark for
part (a)(ii) was zero; the majority of candidates did not recognise the need for an electron on the
left hand side of the equation and almost all of those who realised that a neutrino of some sort
was required on the right hand side made it an antineutrino. In most cases the daughter nucleus
was represented by an X and had the wrong subscripts and/or superscripts. Several candidates
attempted the equation in terms of protons, neutrons and positrons. No credit was given for such
an equation.
There were many correct calculations in part (b), although a mark was lost in part (iii) by giving
4.18 × 10–13 (J) as the answer, rather than considering the transition from the 1.63 to the 1.33
energy level. As a consequence of error carried forward, part (iii) usually provided a mark.
It was unfortunate that there occurred a small printing error in the data of figure 2 in part (b)
which involved the factor of 10–13 in the energy levels being obscured. This only occurred in the
option papers for unit 5 and unit 7. The examiners were fully aware of the error and no candidate
lost marks because of it.
Most candidates in part (c) gave two precautions although many stated the same precaution
twice. A sizeable minority only gave the precautions and ignored the ‘explain’ part of the
question. Considering the title and subject coverage of the unit it was surprising to find that many
candidates ignored the radioactive nature of the isotope and laboured under the illusion that
magnesium is wont to burst into a brilliantly white flame if exposed to the
atmosphere/water/dampness, therefore requiring dark glasses, a perspex screen, asbestos
gloves and full breathing equipment. There were some very sensible answers related to the
context of a school laboratory, most of them involving keeping the source at some distance from
the body or handling it with tongs, storing the magnesium in a lead pot when not in use, using the
magnesium only for short periods or using some kind of screening, including a metre or so of
air. Credit was usually given to part (c)(ii) if the penetrating power of the γ radiation or the ability
of the β radiation to cause ionisation was discussed.
E2.
The majority of candidates understood the requirements of part (a) and were able to give
three arrows as a single route, but a significant number did not know the correct direction of both
types of decay. Horizontal and vertical arrows were commonly seen. A common mistake was an
apparent interpretation of the N on the vertical axis to mean nucleon number instead of neutron
number, as the question stated. Several candidates had very little idea how to proceed and drew
random lines which finished well away from the daughter nucleus.
Part (b), where two equations had to be completed was not well done with few candidates
scoring full marks and many scoring zero. The Ni nuclide often appeared as Cu, x or y,
frequently with incorrect values in the subscript and/or superscript and occasionally with different
values in the decay mode. Weaker candidates often put the electron on the wrong side of the
decay equation. Neutrinos appeared in most answers, but most frequently as a neutrino in one
decay mode and an antineutrino in the other.
Page 22 of 27
α particles were the choice of the great majority of candidates for the scattering particles in part
(c) and for many candidates this was their only mark for this section. For most candidates, the
requirement to describe the main physical principle of the scattering process was interpreted as
‘write as much as you can about Rutherford’s scattering experiment’, resulting in answers which
were imprecise and demonstrated little more knowledge or understanding than that covered at
GCSE. The idea that the experiment gave an upper limit for the nuclear radius was missed by
nearly all candidates. The small minority of candidates who described electron diffraction by the
nucleus usually answered the question well, demonstrating appropriate knowledge about the
intensity minimum and the information gained from it. These candidates seemed to have a
deeper understanding of the experiment, perhaps due to the fact that this was a new situation
which had been taught in more detail at this level.
E3.
In scripts where candidates showed no numerical values on the N axis, it was not possible
to judge the value of answers in part (a), so no marks could be awarded. The graph ought to
start in an approximately linear fashion. To satisfy the requirements of the question, it should
continue in an upward curve to meet Z = 80 at an N value around 120. The initial linearity was
generally known, but the features of the upper part of the graph were less familiar to candidates.
Some had no idea about how to respond, their answers bearing more resemblance to NIZ
graphs showing α and β– decay chains.
Attempts at part (b) were usually more successful. Whatever mistakes had been made on the
N/Z curve, examiners simply wanted candidates to show points that were respectively slightly
below (at Z > 60), slightly above, and slightly below whatever line was drawn.
Part (c) attracted many highly successful solutions, with convincing (and often very different)
explanations about why the loss of four a particles had to be accompanied by the emission of
four β– particles in order to form the isotope 206Pb.
Candidates’ responses to part (d) were somewhat variable. It was necessary to identify two
acceptable processes for the first mark; this was then followed by one mark each for the two
discussions. Candidates who referred to β+ emission and to electron capture found it easier to
score the discussion marks than those who chose a emission or proton emission. In the cases
of β+ emission and electron capture, the conversion of a proton into a neutron produces an
immediate and obvious change in the p/n ratio, a emission requires rather more careful
explanation to indicate the effect on this ratio, whilst proton emission is very rare and occurs only
to nuclides with sufficient instability for electrostatic repulsion to overcome the strong force.
Page 23 of 27
E4.
Less than half the candidates could explain the meaning of the decay constant. By contrast
almost all candidates could find the half-life in part (b) and a majority could answer part (c).
Some candidates did not gain credit because they conveniently removed 1012 in their calculation
without showing the division. So lines like, 1.15 × 1012 Bq = 1.15 Bq, were seen. Most candidates
who tackled part (d) using the exponential decay of the activity equation got full marks. Only a
few candidates could not rearrange the equation. By contrast almost all candidates who tried to
use the exponential decay in the number of nuclei got confused. Most had numbers of nuclei on
one side of the equation but activity on the other.
Part (e) did discriminate but only between scoring zero marks or one mark. Very few candidates
attempted two reasons. Most acceptable answers to this question were difficult for the candidate
to express. For example, in question (d) it states that the decay rate due to carbon-14 is 0.65 Bq,
indicating it is a corrected count rate. So an answer to part (e) like, ‘the background can effect
the result’, is not acceptable. This is not the same as saying it is difficult to obtain the results for
the sample activity because the background activity is high in comparison. This example is also
ambiguous in that it suggests the surroundings can influence the rate of decay. Another answer
that was not acceptable was, ‘radioactive decay is random so it’s bound to give false values’. To
gain a mark following this line of thought it was necessary to refer to its effect on the statistics.
The most common answers that candidates found easy to express included the following; the
tree died well before the boat was made; or the boat was repaired later in its life with fresh wood;
or that carbon based microbes died in the wood when the boat was rotting at the end of its
useful life.
Page 24 of 27
E5.
In part (a) was very straightforward for many candidates. Less able candidates did not
elaborate on what the nucleus split up into or they referred to some other splitting such as
fission. Another common error was to talk about the separation of an atom into protons,
neutrons and electrons. Very few candidates got the direction of energy flow wrong.
In part (b)(i) very few candidates referred to particles other than neutrons but a significant
number failed to write down two neutrons in the correct manner.
and
a single were
extremely common.
The follow on part (b)(ii) turned out to be an extremely good discriminator. It highlighted the types
of errors candidates were making. There was a group who did not appreciate the data was given
per nucleon and used the figures without multiplying up by the respective nucleon numbers.
Another significant group added the mass energy of mostly one but sometimes two nucleons
into the proceedings. The last major group got into difficulties because they changed units
unnecessarily or only changed the units of some of the terms. Overall there was a good
percentage of correct answers.
A good proportion of candidates could accomplish the conversion of units required in part (b)(iii)
with relative ease, sometimes from an error carried forward from the previous part. With the
conversion needing two stages using the datasheet there was plenty of opportunity for errors by
dividing instead of multiplying or using an incorrect conversion factor. A separate but common
error was to ignore the answer to part (b)(ii) and simply use the difference in mass of the
nucleons involved, ignoring the binding energy per nucleon completely. Less able candidates did
not really help themselves in this question because very few put words or units to intermediate
stages of the calculation. If they had done so fewer would have lost their way.
A majority of candidates did not score marks in part (c)(i). Many knew the general shape but very
few remembered any details so they had no idea which coordinates to draw their line or band
through. Some gave no real thought to the problem and drew graphs that did not make sense.
For example some graphs went vertical or turned back on themselves.
Part (c)(ii) was very discriminating. Less able candidates simply referred to any radiation that
came to mind and forwarded very little explanation. The bulk of the candidates knew beta minus
radiation was emitted but they were not careful how they expressed their reasons. For example,
stating that there are more neutrons than protons is not sufficient to imply the nuclei are neutron
rich. A majority in this group of candidates made no reference to the graph at all. Many that did
had flawed reasoning. They thought the isotopes were neutron rich because the large number of
free neutrons in the core. Good candidates also got into some difficulty by not reading the
question carefully. Typically these candidates would start their answer with, 'If there are a lot
more neutrons than protons then... but if the neutron to proton ratio is small then…'. These
candidates obviously knew the subject matter but did not score many marks as the question
clearly asks for a choice to be made.
Page 25 of 27
E6.
Even though part (a) needed a little thought almost all students obtained the correct answer.
By contrast part (b)(i) was simply a factual recall question, which was answered poorly by a
significant minority. The main error was for students not to state the energy needs to be given
out or is required, when a nucleus was formed or broken up. It was common to see written, ‘The
energy to keep the nucleus together’. In part (b)(ii) a majority of students simply read the value
from the graph and gave an answer near 7.88 MeV without appreciating the ‘per nucleon’ on the
y-axis of the graph. Part (c)(i) was done well by most students. Some students missed marks
due to a lack of care in choosing specific coordinates for the graphs to pass through. Most
students made a good attempt at part (c)(ii). Part (c)(iii) was more difficult and only the better
student could correctly combine the two equations required to answer the question. A common
mistake made by a few students who looked as if they were going to get the correct answer was
for them to confuse the time units they were using. These students obtained the correct answer
but then multiplied it by 60×60×24×365.
Page 26 of 27
Resource currently unavailable.
Page 27 of 27
Nuclear power stations
Make notes on the following
Nuclear chain reactions in a nuclear bomb
Supercritical, subcritical and critical reaction rates
Simple diagram of a nuclear power station
Role, material choice and construction of the
moderator, control rods, shielding and coolant
Choice of fuel
Calculations of energy density of nuclear fuel as
compared to fossil fuels
Emergency shutdown of nuclear reactors
The nature of radioactive waste
Comparisons of Chernobyl Fukushima Daiichi
incidents
Complete and mark the following
Physics factsheet 58, 113 & 125
Isaac Physics Skills Exercise L2
AQA exam past paper questions - Induced fission
Completed?
Completed?
Physics Factsheet
www.curriculum-press.co.uk
Number 113
Electrical Aspects Of Domestic Wind
Energy Production
CRO trace B shows the output from a simple a.c. generator which
could be connected to a wind turbine. One complete cycle occupies
4 horizontal squares, and the time sensitivity can also be adjusted.
If one horizontal square represents 0.1 seconds, one cycle took 0.4
seconds. The wind turbine was rotating 2.5 times per second,
producing a.c. of frequency 2.5 Hz. The voltage swings between
positive and negative values, driving the electrons first one way
and then the other.
You’ve built your wind turbine in the best possible site, and now
you want it to produce electricity. How does it do this, and what
kind of electrical supply will you get?
Electric current can be either direct or alternating (d.c.
or a.c.). In direct current the electrons flow in one direction
only, while in alternating current they oscillate backwards
and forwards in the cables.
In CRO trace C, the a.c. voltage has been rectified, which means
that the negative half of each cycle has been turned upside down
using diodes. The signal is now d.c., because it is always positive,
but it is far from the smooth signal of a battery. The electrons move
in one direction, but in a jerky manner.
Light-bulbs and heaters don’t mind d.c. or a.c, and motors can be
designed to work with either. But batteries need charging with d.c.
only, and the mains produces 230 V a.c. at a steady frequency of 50
Hz. If you want your wind turbine to charge batteries, you need
d.c., but if you want it to drive mains appliances or connect to the
National Grid, you need a.c. You may need both.
CRO trace D shows the output when the wind was blowing a little
harder. The turbine turned more quickly, producing a higher peak
voltage, but also a higher frequency. A wind turbine connected
directly to a simple generator will produce quite low frequencies
which vary with the speed of the wind.
Telling the difference
Cathode Ray Oscilloscope (C.R.O.) traces of electrical signals are
the only way to understand what you’ve got and modify it to what
you want.
Fig 1. CRO traces
A
Exam Hint: Do not confuse a varying d.c. voltage with a.c.
There is no need for a d.c. voltage to be steady. But it must
always have the current travelling in the same direction. If the
current reverses direction, you have a.c.
B
+
+
−
−
C
Frequency is the number of complete cycles in one
second. Mains frequency is 50Hz, so if you need to connect to
the National Grid, you have to increase the generator speed
with a gear box and also find some way of keeping the frequency
constant.
Worked example:
D
+
+
−
−
CRO trace A shows the output from batteries. The voltage is positive
and the electrons flow steadily in one direction. The CRO sensitivity
can be adjusted to suit the signal being measured. If one vertical
square represents 3 volts, the battery voltage is 6 V. The arrow
indicates the centre line which is at zero volts. Any trace below that
line would show a negative voltage and a current flow in the opposite
direction.
If the wind turbine has a radius of 1.6 m and is rotating at 5 Hz
(5 times per second), what is the speed of the blade tip?
One rotation takes 0.2 seconds, and distance travelled by one
blade tip = 2πr = 10 m, so speed = 10/0.2 or 50 ms-1, a fast speed.
If the turbine itself turned 50 times per second, the speed of the
blade tip would be 500 ms-1. This is faster than the speed of
sound (340 ms-1) so would cause great problems of structural
stress and noise.
The formula for the speed is 2πrf, where f = frequency.
1
113. Electrical Aspects Of Domestic Wind Energy Production
A simple generator
This time the input energy is kinetic, since the wind turns the turbine
and the rectangular coil. A voltage is induced due to the sides of
the coil cutting through the magnetic field lines. Faraday’s Law
says that the induced voltage is proportional to the rate at which
the field-lines are cut, so the faster the turbine turns, the greater will
be the voltage. One rotation of the turbine will give one complete
cycle, and if this takes 1 second the frequency will be 1 Hz.
You will probably have built a d.c. motor at GCSE, similar to the one
in the next diagram.
Fig 2. d.c motor
N
When the coil is horizontal the voltage will be at its peak, as the
sides of the coil are cutting through the field-lines at the maximum
rate. There is no voltage induced when the coil is vertical as no
field-lines are being cut. The commutator and brushes make the
connections reverse every half-turn, and the result is a jerky d.c.
like that shown on CRO trace C on page 1. If you want a.c. you will
need slip rings instead of a commutator, but you still need carbon
brushes which wear out due to sparks and friction.
S
commutator
brushes
Physics Factsheet
So to make your own generator, you can use a motor working in
reverse, turned by the wind. Jerky d.c. can be smoothed using a
capacitor. This charges up as the voltage peak arrives and then
discharges slowly through the load. The effect on the CRO trace is
to make it closer to the output of a battery, though still not very
smooth. Some wind energy enthusiasts have recommended motors
from tape machines, or even built their own motors. Now that you
can buy neodymium magnets, you can get a very strong magnetic
field (about 0.4 Tesla) in a small space.
Force on coil (use
Fleming's left-hand
rule)
The rectangular coil experiences a turning moment or torque due
to the interaction between the electric current and the magnetic
field. The forces are at their maximum when the coil is horizontal,
zero when it is vertical, and the halves of the commutator change
brushes as the coil passes the vertical position, so that the forces
will continue to push the coil in the same direction. The motion is
somewhat jerky. Electrical energy changes to kinetic.
You could also rotate the magnet and have stationary coils, but you
will then generate a.c. and have to use diodes to rectify it if you
want d.c.
Exactly the same apparatus can be used in reverse to make a simple
d.c. generator.
Faraday’s Law says that the induced e.m.f.
Ε = - n (dφ/dt),
Fig 3. Simple d.c generator
where n = number of turns
dφ/dt is the rate of change of flux lines φ (magnetic field lines).
N
The magnetic field strength B is also equal to the number of field
lines per unit area (per m2), so φ = B × A where A is the area being cut
by the coil. You only get an induced voltage with CHANGING
magnetic fields, not static ones. Even though the field is static in
this apparatus, the coil is moving so experiences a changing field.
S
Worked Example 1
Suppose your rectangular coil has 20 turns and is 8 cm long,
with a radius of 2 cm, and rotating at 5 Hz. The magnadur
magnets have a maximum field strength of about 0.1 Tesla,
which means 0.1 flux lines per square metre. What peak voltage
can be achieved?
axel
capacitor
(optional)
The sides of the coil are moving at a speed of :
2πrf or 2πr / 0.2 = 0.6 m s-1.
Induced current
(use fleming's right
hand rule)
When in the horizontal position, each side cuts through an
area A of 2πrf × l = 0.6 × 0.08 = 0.05 m2 per second, which
contains 2π rf × l × B = 0.05 × 0.1 = 0.005 flux lines per
second. (See Fig 4).
output
device (load)
without capacitor
with capacitor
volts
Since there are 20 turns, the induced peak voltage across one
side is 0.005 × 20 = 0.1 V, or 100 mV. Taking both sides together,
their voltages add up, doubling the induced voltage to 200
mV. This is quite good for a tiny school generator, but not
much use for powering anything.
volts
time
time
2
113. Electrical Aspects Of Domestic Wind Energy Production
Physics Factsheet
Worked Example 2
Increase the number of turns to 200 and rotate the coil at 50 Hz.
If the coil is still 8 cm long but has radius 4 cm (i.e. twice as
large), what peak voltage can be achieved with the same magnetic
field?
Worked Example 3
If you have 250 kg of water in your hot water tank, and your wind
turbine produces 500 watts of excess electricity for 2 hours,
what difference will this make to the temperature of your water?
The specific heat capacity of water is 4200 J kg-1 K-1.
The answer will have increased by a factor of 200, (10 × 10 × 2)
so you will get 0.2 × 200 = 40 V, which is quite an improvement.
The full formula for the voltage generated is 4πrf × l × B × n.
Excess electrical energy
= power × time = 500 × 2 × 3600 Joules = 3.6 × 106 J.
Energy taken in by water = mass × s.h.c.x Δϑ where Δϑ = change
in temperature.
So Δϑ = 3.6 × 106 / (250 × 4200) = 3.4oC.
Your hot water tank could easily absorb this much excess energy
but it would not make much difference to the temperature of the
water.
Exam Hint: You can approach these questions by considering
the area of the coil, or by looking at what happens when each
side cuts through the magnetic field. Quite often either
approach is acceptable. But don’t confuse them.
Batteries
Fig 4. Calculating peak voltage
Batteries are a common way of storing electrical energy, but you
will need large, heavy batteries similar to those used in car engines.
They are usually specified by Volts and Ampere-hours (a measure
of how much charge they will hold). Charge Q = current I x time t, so
your battery will supply a large current for a short time or a smaller
current for a longer time.
N
radius r
Worked Example 4
A battery is specified at 12 V, 40 Ampere hours, and it costs £35.
Using Q = I × t, it stores a charge of 40 × 3600 coulombs
= 1.44 × 105 C. The energy stored will be V × Q = 12 × 1.44 × 105 J,
or 1.7 × 106 J. If your wind turbine produces 500 watts of excess
power for 2 hours, as in the previous example, you need to store
3.6 × 106 J, so will need at least 3 of these batteries at a cost of
£105.
Field
strength B
len
gth
l
speed = 2πrf
The situation is not quite as simple as the previous example
suggests, because the batteries have to be stored almost fully
charged, and may produce gases when charging, as well as being
very heavy. You probably need 20 batteries rather than 3, at a cost
of £700.
area A being
swept out per
second = 2πr fl
Inverters
Suppose I generate too much electricity?
Inverters can be purchased for many applications, from about £40
to £1000 depending on the size and quality. If you want to feed
electricity back into the Grid and be paid for doing so, the inverter
may cost as much as the wind turbine itself. An inverter is basically
a switching device that connects the d.c. supply first one way and
then the other at regular intervals. If it is that simple, you will get a
square wave on your CRO, and it can be designed to have the
correct frequency for the Grid but is certainly not the correct shape.
Since the force of the wind is irregular, you may find that you
generate more electrical energy than you need on windy days. You
must then decide what to do with the excess. Some domestic systems
include a bank of batteries which get charged up ready for still
days.
Batteries are quite expensive, and do not last nearly as long as wind
turbines, so you would have to replace them before the wind turbine.
In other systems, you could use the spare electrical energy for
room heaters or water heaters. Or you could feed it back into the
National Grid.
The wave shape has to be carefully modified to be satisfactory. It is
also unlikely that the voltage amplitude will be correct, so you will
also need a transformer to raise the voltage to 230 V. Transformers
only work with changing currents, and not at all with steady d.c., so
you will need both the inverter and the transformer if you want to
run mains appliances such as a washing machine from your storage
batteries.
For the most versatile systems, the turbine generates d.c., since
this can be stored in batteries and overcomes the problem of the
turbine turning at variable speeds and producing variable frequency
a.c. If you then want to drive mains appliances or feed back into the
grid, you have to use the d.c. to produce a.c. at the correct frequency
of 50 Hz. This requires a powerful inverter; quite a lot more
electronics.
3
113. Electrical Aspects Of Domestic Wind Energy Production
Fig 5. Step-up transformer
Exam Hint: We are bombarded these days with the need to use
renewable energy. When asked about wind turbines, etc, make
sure you have ideas on problems with their use as well as
advantages. Both can gain you marks on an exam.
Iron core
Practice Questions
1. (a) If a wind turbine rotates 10 times per second, what is the
frequency of rotation?
(b) What is the time for one rotation?
(c) If the length of the blade is 0.8 m, what is the speed of the
blade tip?
to load
a.c. source
2. To avoid excessive structural stress and noise, a turbine blade
made of a certain material must not have a faster speed at the tip
than 75 m s-1. If the length of the blade is 3.2 m, what is the
limiting rotational frequency of the turbine? (Frequency at which
it must be shut down).
Transformers
If your storage batteries produce a steady d.c. of 12 volts, and the 3. If a wind turbine blade is made of metal, it can cut through the
inverter a.c. output is also 12 V on average, how does a transformer
horizontal component of the Earth’s magnetic field,
increase this to 230V?
(which is 1.7 × 10-5 T) and generate a voltage between its ends. A
certain turbine has a blade length of 1.6 m and is rotating at 5 Hz.
The electric current in the primary coils of the transformer magnetises
Calculate (a) the area swept out per second (b) the voltage
the iron core, as shown in the diagram. This means that the magnetic
generated.
domains in the iron line up in one direction, and iron is so strongly
magnetic that nearly all the field lines remain inside the iron and so 4. On a very windy day, your turbine produces 400 watts of excess
electricity for 8 hours, which is used to heat water. Your hot
pass right round the iron core to the other side, passing through the
water tank contains 50 kg of water. Calculate the rise in temperature
secondary coil. The arrows in the diagram show the directions of
of the water, and comment on the result. (Specific heat capacity
the current and the field lines at one instant in the a.c. cycle. When
of water = 4200 J kg-1 K-1.)
the current changes direction, (as it does 100 times per second at 50
Hz), the domains turn round and line up the other way. So there is a 5. You have one fully charged 12 V, 40 Ampere-hour battery. The
constantly CHANGING magnetic field through the secondary coil,
low-power devices in your home take about 200 watts all the
and this induces another a.c. voltage there also. The size of this
time. (a) What current will the battery supply if it is producing
secondary voltage depends on the numbers of turns on the two
200 watts? (b) How long will it last before it runs down? (c) If
coils. To increase the voltage you will need a step-up transformer. If
you have 20 fully charged batteries, how long will they last?
the input a.c. is a smooth wave, the output will be also, but strange
6. A domestic wind system with all the electronics costs £10 000.
inputs will give strange outputs.
You generate 3500 kWh per year which you would otherwise
have bought at 20p per unit, and sell back to the Grid 1200 kWh
Worked Example 5
per year at 2.5p per unit. (a) How much will you save per year?
To change 12 V to 230 V, you need a turns ratio of 12:230 on the
(b) How much will you recover from the Grid per year? (c) What
two coils. So you could have 120 turns on the primary coil, and
is the payback time of your system?
2300 on the secondary coil. Or 48 on the primary coil, and 920
on the secondary coil. The formula to use is: Vs/Vp = ns/np,
which you will remember from GCSE.
Answers
(a) 10 Hz (b) 0.1 seconds (c) 50 m s-1.
3.7 Hz
(a) 40 m2 s-1. (b) 0.68 mV
Rise in temperature = 55oC. If the water started at 20oC, it is now
at 75oC, which is satisfactory.
But if the wind continued to blow equally hard for another 8
hours, the water would start to boil away. You need a bigger
water tank.
(a) At 12 volts, current supplied = power/volts = 16.7 amps.
(b) 2.4 hours.
(c) 48 hours.
6. (a) £700.
Will you or won’t you go for wind power?
(b) £30.
Clearly the domestic system is a lot more complicated than it appears
(c) 13.7 years.
at first glance. You will need plenty of money, and extra space in
your home for all the extra electronics if you intend to make a
significant difference to your own electricity bill. But if you Acknowledgements:
This Physics Factsheet was researched and written by Hazel Lucas.
understand the principles and want to make a real difference to The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU
global warming, there are careers in engineering and installation of Physics Factsheets may be copied free of charge by teaching staff or students, provided that
their school is a registered subscriber.
systems crying out for you. Renewable energy is the way of the No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted,
future.
in any other form or by any other means, without the prior permission of the publisher.
1.
Notice that this transformer is connected from the inverter to the 2.
output device, not directly from the wind turbine. Since the wind 3.
turbine produces variable voltage amplitude, you would need 4.
variable numbers of turns on the transformer to get exactly 230 V
out, which would be even more complicated. You would also need a
way to control the frequency. This is why most domestic wind
turbines produce d.c. and send the output to batteries first. Large
turbines in wind farms have computers incorporated into the design 5.
to control everything.
ISSN 1351-5136
4
Physics Factsheet
www.curriculum-press.co.uk
Number 125
Energy Efficiency in Power Generation
How can we power our civilisation? Replacing fossil fuel power
stations with renewable sources of energy such as wind and solar
is a possibility, but just how efficient are the different resources?
How efficient are power stations? How efficient are solar cells or
wind turbines? Can we improve the efficiency of power generation?
Example Question 1
Estimate the average energy production for users from a ten turbine
wind farm. Each turbine has a surface area of 100m2 and when
wind speed is sufficient it has an average energy of 1230J/m2/s.
Answer
Wind turbines
The total available wind energy for the turbines is:
10 turbines × 100m2 × 1230J/m2 = 1.23x106J/s.
In one day, this would provide:
1.23 × 106J × 3600 seconds × 24 hours = 1.06 × 1011J or 29520
kWh (1kWh is equivalent to 3.6 × 106J).
Wind turbines extract the kinetic
energy of moving air and the wind is
slower after it has passed through
the blades. The maximum possible
efficiency of a wind turbine is 59.6%.
This is unrelated to the intermittent
availability of the wind. This includes
only the wind in contact within the
blades, not the air passing between
them.
However, if we apply the Betz limit, the maximum possible energy
that could be extracted is 29,520kWh x 0.596 = 17590kWh. In reality,
the overall efficiency for the user is between 10-30% of wind energy,
29520kWh x 0.20 = 5900kWh. In addition, the wind speed is only
suitable 50% of the time, yielding 2700kWh per day for the wind
farm.
Why is there a maximum efficiency? If 100% of the kinetic energy of
the wind were transferred to the blades of the turbine, the wind
would be stopped. No wind would flow through the turbine, passing
around instead! Of course, if no energy is transferred by the turbine,
there’s no point building it. The best compromise between forcing
the air to go around the turbine and stopping the movement of the
air altogether is called the Betz limit, which is a 59.6% transfer of
wind energy to the turbine.
some kinetic energy has been
transferred to the turbine
Solar energy
Solar cells were first created in Bell laboratories during the mid1950s. The design of most solar cells (or photovoltaic cells, PV
cells) has changed little since. This solar cell is a wafer of silicon (a
semiconductor) which is altered in two regions, known as n-type
and p-type. The region where p-type and n-type semiconductors
meet is known as a p-n junction.
moving air
When light of a particular frequency passes through the junction,
it becomes conducting and causes a potential difference across the
wafer. As sunlight has a wide frequency range, a large proportion
of the energy has no effect on the junction and is not absorbed.
The maximum theoretical efficiency for such a cell was calculated in
1961 as being 31%. The record for most efficient PV cell in the lab is
24.7% although most commercially available PV cells rarely do better
than 18%.
The maximum possible efficiency for extracting energy
from the wind to the turbine blades is 59.6%, the Betz limit.
Cross section through solar cell
In reality, wind energy extraction efficiencies of 30-40% are likely
because of the actual construction of the turbine. Once the energy
losses in the generator and power transmission are taken into
account, the overall efficiency of wind turbines from the moving air
to the user is 10-30%! However, of course in this case the “fuel” is
free!
Glass Cover
Coating to reduce reflection
n-type semiconductor
p-type semiconductor
A further disadvantage for wind energy is that the “fuel” is in
intermittent supply. If the wind speed is suitable for energy
production for 50% of the time, then obviously this halves the
production of energy.
backing
The maximum theoretical efficiency of one layer of first
generation photovoltaic cell is 31%.
1
Physics Factsheet
125. Energy Efficiency in Power Generation
Are there any other ways to improve PV cell efficiency? One
possibility is to layer two or three PV cells which are sensitive to
different frequencies of light. This means a greater proportion of
sunlight is absorbed and transformed into electrical energy.
Example: Calculate the maximum theoretical efficiency of an
internal combustion engine. Assume the average gas
temperature is 800oC and the engine is running on a cold day.
The maximum efficiency of a sandwich of PV cells is around 43% for
two and 50% for three layers. However, this is prohibitively expensive
for most uses other than spacecraft solar panels.
Answer: The maximum efficiency according to the Carnot cycle
is given by the temperature difference (800oC-5oC) divided by
the absolute temperature of the hot reservoir
(800oC +273K = 1073K). So the maximum efficiency of this
engine is: 795K / 1073K = 74.1%
Sunlight could be focussed onto solar panels using mirrors or lenses.
Greater intensity light would result in a greater electrical output but
would not make any more efficient use of the incident light.
Stirling cycle
In an internal combustion engine, hot gases are released into the
surroundings. In a Stirling engine, a gas or liquid is permanently
contained. A heat exchanger heats and cools this working fluid,
perhaps driving a piston, doing work. Stirling engines are more
efficient than many open-cycle engines but they are still limited by
the Carnot cycle. They are typically more costly to make but produce
energy more cheaply.
Example question 2
Calculate the area of photovoltaic cells required to provide the
inhabitants of London with electricity. On Earth, the sun typically
provides 1400W/m2. London has a population of approximately
13 million people and the average person consumes around
2800kWh of electricity per year.
Hydrogen fuel cells
Answer
In fuel cells, hydrogen and oxygen undergo a chemical reaction,
creating electricity and leaving water as a by-product. This is quite
different to heat engines, which are limited to a maximum possible
efficiency by the Carnot cycle. The idea of a fuel cell was first
conceived in 1838 but the technology was so complex that the first
practical example was not created until 1959.
The electricity required by the inhabitants of London in one year
is 1.3 × 107 people × 2800kWh = 3.6 × 1010kWh.
Each square metre of PV cell receives 1400W of solar energy
producing 1400W × 0.18 = 252 Watts of electricity with typical
efficiencies.
However, daylight is on average only 12 hours and the solar cell
will receive much less energy when the sun is low in the sky. When
cloudy skies are factored in, 252W/m2 × 0.1 = 25 W/m2 would be
a reasonable average power for a solar cell in the UK.
In one year it would then produce:
25W/m2 × 3600s × 24 hours × 365 hours = 7.9 × 108m2 or 219kWh/m2.
The efficiency of a fuel cell is complex to describe. If you use a fuel
cell to power a car engine, you could measure the kinetic energy of
the car and the energy stored in the fuel gas tanks, and then calculate
the efficiency. This would typically be 40-50%. However, when
you take into account the amount of energy it takes to produce the
hydrogen and oxygen gas, this is closer to 20%.
3.6x1010kWh / 219kWh/m2 = 1.6x108m2 or 160km2. The area of Greater
London is around 1600km2.
Power stations
Before considering the efficiency of a power station, we need to
have a look at each stage of energy production. Usually, a fuel is
burnt, heating water into steam. This steam spins a turbine which
in turn spins either a magnet or a coil inside a generator to produce
electricity. A transformer increases the potential difference of the
electricity and power lines transmit the electricity to the user. Each
stage is obviously less than 100% efficient and the overall efficiency
of a thermal power station is between 30 and 50%.
Carnot’s theorem
In 1824, Nicolas Carnot developed a
theorem which predicted the maximum
efficiency for ANY heat engine. A heat
engine does work on the surroundings as
it goes through a continuous cycle. This
cycle usually consists of a range of
temperature and pressure values. A heat
engine is considered to have two
reservoirs: hot and cold. In an internal
combustion engine, these would be the
hot gases in the piston and the
surrounding air.
Sankey diagram showing energy losses from a coal power station
Chemical
energy from
coal
The Carnot cycle predicted that the
maximum efficiency for any heat engine is
given by the temperature difference
divided by the absolute temperature of the hot reservoir. To make a
heat engine more efficient, the temperature difference must be
increased. A particularly effective way to increase efficiency is to
lower the temperature of the cold reservoir. However, in most heat
engines, this would involve cooling the surroundings!
Electricity
Friction within
turbine and
generator
Hot gases from
chimney
2
Thermal energy to
the surroundings via
cooling towers
Physics Factsheet
125. Energy Efficiency in Power Generation
The transfer of thermal energy from the burning fuel in the furnace to the water in a boiler is around 88%. A further 3% is transferred to heat
as the electricity is distributed through power lines. Transformers are used to step up the electrical voltage to high voltage (400,000V) and
low current, or these losses would be far higher. The power station requires 2% of the total input energy just to operate.
So if the power station is usually only around 40% efficient, where does the rest of the energy go? It is transferred to the surroundings as
heat. After passing through the turbine, the steam is allowed to cool and condense into water. This could occur in a cooling tower or the
water could be pumped back into a nearby river or sea. Gas power stations can achieve 50% efficiency as the hot waste gases are actually
used to spin the turbine.
Combined heat and power
We’ll look at combined heat and power as two parts. Firstly, combined heat and power is one method of improving the efficiency of a
thermal power station. Consider the condenser stage where used steam condenses back into water to return to the boiler for heating. A
large proportion of the original energy is lost to the surroundings. What if this wasted energy could be used? The condenser is a heat
exchanger where the “used” steam transfers thermal energy with cooler water. Rather than just release this warmer water, why not pipe it
to local factories and homes for heating?
Combined heat and power has another guise: there are over 1400 CHP projects around the country. A large building can house its own
power plant. Fuel is burned and a generator produces all of the electricity for the building. However, the waste heat from the power plant
is then used to heat the building. Little energy is lost in the transmission of the electricity to the user due to the proximity of the power
plant. This is particularly useful for large buildings and institutions such as hospitals, universities, schools and large office buildings and
factories.
Questions
1) Use Carnot’s theorem to show why decreasing the temperature of the cold reservoir would be more effective than increasing the hot
reservoir temperature for improving efficiency.
2) Calculate the maximum theoretical efficiency of a power station with a boiler temperature of 603oC. (Take the outside temperature to be
15oC.)
3) Explain why the theoretical maximum efficiency for a wind turbine blade to extract energy from moving air is around 60%.
4) If you assume that all of the heating losses mentioned in a traditional thermal power station are successfully used for local heating,
what is the new efficiency of the power station?
5) Calculate the area of solar panels required to supply your electricity requirements.
Answers
2) 67%
5) About 12.8m2
Acknowledgements:
This Physics Factsheet was researched and written by Jeremy Carter
The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU
Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber.
No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission o
the publisher.
ISSN 1351-5136
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3
( PF ×
3× × Q1.
(a)
(i)
State two physical features or properties required of the shielding to be placed
around the reactor at a nuclear power station.
.............................................................................................................
.............................................................................................................
(ii)
Which material is usually used for this purpose?
.............................................................................................................
(3)
(b)
Describe the effect of the shielding on the rays, neutrons and neutrinos that reach it
from the core of the reactor. Also explain why
h the shielding material becomes radioactive
as the reactor ages. You may be awarded marks for the quality of written communication
provided in your answer.
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(4)
(Total 7 marks)
Page 1 of 25
Q2.
In a nuclear reactor, uranium nuclei undergo induced fission by thermal neutrons. The
reaction is a self-sustaining chain reaction which requires moderation and has to be controlled.
(a)
Explain the meaning of
(i)
induced fission,
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
thermal neutrons,
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(iii)
self-sustaining chain reaction.
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(5)
Page 2 of 25
(b)
You may be awarded marks for the quality of written communication provided in your
answer to parts (b)(i) and (b)(ii).
(i)
Explain what is involved in the process of moderation.
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(ii)
Describe how the rate of fission is controlled in a nuclear reactor.
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(7)
(Total 12 marks)
Q3.
(a)
(i)
Explain why, after a period of use, the fuel rods in a nuclear reactor become
less effective for power production,
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Page 3 of 25
(ii)
more dangerous.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(b)
Describe the stages in the handling and processing of spent fuel rods after they have been
removed from a reactor, indicating how the active wastes are dealt with.
You may be awarded marks for the quality of written communication in your answer.
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(5)
(Total 8 marks)
Page 4 of 25
Q4.
(a) Sketch, using the axes provided, a graph of neutron number, N, against proton
number, Z, for stable nuclei over the range Z = 0 to Z = 80. Show suitable numerical
values on the N axis.
(2)
(b)
On the graph indicate, for each of the following, a possible position of a nuclide that might
decay by
(i)
α emission, labelling the position with W,
(ii)
β– emission, labelling the position with X,
(iii)
β+ emission, labelling the position with Y.
(3)
(c)
Used fuel rods from a nuclear reactor emit β– particles from radioactive isotopes that were
not present before the fuel rod was inserted in the reactor. Explain why β– emitting
isotopes are produced when the fuel roads are in the reactor.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
Page 5 of 25
(d)
A nuclear power station is a reliable source of electricity that does not produce greenhouse
gases but it does produce radioactive waste. Discuss the relative importance of these
features in deciding whether or not new nuclear power stations are needed.
The quality of your written answer will be assessed in this question.
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(6)
(Total 14 marks)
Page 6 of 25
Q5.
(a) In a thermal nuclear reactor, one fission reaction typically releases 2 or 3 neutrons.
Describe and explain how a constant rate of fission is maintained in a reactor by
considering what events or sequence of events may happen to the released neutrons.
The quality of your written communication will be assessed in this question.
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(7)
(b)
Uranium is an α emitter. Explain why spent fuel rods present a greater radiation hazard
than unused uranium fuel rods.
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(3)
(Total 10 marks)
Page 7 of 25
Q6.
The fissile
isotope
of uranium,
been used inAnsome
nuclear
reactors.
It is emits a
normally
produced
by neutron
irradiation
diatio , has
of thorium-232.
irradiated
thorium
nucleus
β− particle to become an isotope off protactinium.
t
This isotope of protactinium may undergo β− decay to become
(a)
.
Complete the following equation to show the β− decay of protactinium.
Pa →
+
β + ……..
–
(2)
(b)
Two other nuclei, P and Q, can also decay into
P decays by β+ decay to produce
Q decays by α emission to produce
.
.
.
The figure below shows a grid of neutron number against proton number with the
position of the
isotope shown.
On the grid label the positions of the nuclei P and Q.
Page 8 of 25
(2)
(c)
A typical fission reaction in the reactor is represented by
+
(i)
→
+
+ x neutrons
Calculate the number of neutrons, x .
answer = .............................neutrons
(1)
(ii)
Calculate the energy released, in MeV, in the fission reaction above.
mass of neutron = 1.00867 u
mass of
nucleus = 232.98915 u
mass of
nucleus = 90.90368 u
mass of
nucleus = 138.87810 u
answer = ..................................MeV
(3)
(Total 8 marks)
Page 9 of 25
Q7.
(a) Describe the changes made inside a nuclear reactor to reduce its power output and
explain the process involved.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
State the main source of the highly radioactive waste from a nuclear reactor.
........................................................................................................................
........................................................................................................................
(1)
(c)
In a nuclear reactor, neutrons are released with high energies. The first few collisions of a
neutron with the moderator transfer sufficient energy to excite nuclei of the moderator.
(i)
Describe and explain the nature of the radiation that may be emitted from an excited
nucleus of the moderator.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(ii)
The subsequent collisions of a neutron with the moderator are elastic.
Describe what happens to the neutrons as a result of these subsequent collisions
with the moderator.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(Total 7 marks)
Q8.
(a)
State what is meant by the binding energy of a nucleus.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 10 of 25
(b)
(i)
When a
nucleus absorbs a slow-moving neutron and undergoes fission one
possible pair of fission fragments is technetium
and indium
.
Complete the following equation to represent this fission process.
(1)
(ii)
Calculate the energy released, in MeV, when a single
fission in this way.
binding energy per nucleon of
= 7.59 MeV
binding energy per nucleon of
= 8.36 MeV
binding energy per nucleon of
= 8.51 MeV
nucleus undergoes
energy released ..................................... MeV
(3)
(iii)
Calculate the loss of mass when a
nucleus undergoes fission in this way.
loss of mass ......................................... kg
(2)
Page 11 of 25
(c)
(i)
On the figure below sketch a graph of neutron number, N, against proton number, Z,
for stable nuclei.
proton number, Z
(1)
(ii)
With reference to the figure, explain why fission fragments are unstable and explain
what type of radiation they are likely to emit initially.
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(3)
(Total 12 marks)
Page 12 of 25
M1.
(a)
(ii)
(i)
thick
high density
material giving minimal fatigue problems after irradiation
any other sensible property e.g. withstands high temperature
any two (1) (1)
(reinforced) concrete (1)
3
(b)
effect of shielding:
rays - intensity (greatly) reduced (1)
neutrons - some absorption (1)
(or speed or energy reduced by collisions) (1)
neutrinos - very little effect (1)
why shielding becomes radioactive:
neutron absorption by nuclei or atoms (1)
makes nuclei (not particles) neutron rich or unstable (1)
become β– emitters and/or emitters
max 4
QWC 2
[7]
M2.
(a)
(i)
splitting of nucleus into two smaller nuclei (1)
brought about by bombardment (1)
(ii)
thermal neutrons have low energies or speeds
(e.g. 0.03 eV) (1)
(iii)
fission reaction gives out neutrons (1)
neutrons (from fission) cause further fissions (1)
self-sustaining when one fission leads to (at least) one
further fission (1)
max 5
(b)
(i)
neutrons from fission are fast (high energy) neutrons
(e.g. 2 MeV) (1)
fission most favourable with low energy neutrons (1)
moderation involves slowing down neutrons (1)
by collision with moderator atoms (1)
large number of collisions required (e.g. 50) (1)
collisions are elastic/k.e. transferred to atoms (1)
suitable moderator material named e.g. graphite, water (1)
moderator must not absorb neutrons (1)
moderator atoms should have (relatively) low mass (1)
QWC 1
Page 13 of 25
(ii)
control involves limiting number of neutrons (1)
excess neutrons absorbed by control rods (1)
suitable control rod material named e.g. boron, cadmium (1)
control rods inserted into reactor to slow reaction rate
(or vice-versa) (1)
max 7
[12]
M3.
(a)
(ii)
(i)
amount of (fissionable) uranium (235) in fuel decreases (1)
fission fragments absorb neutrons (1)
fission fragments are radioactive or unstable (1)
emitting β– and γ radiation (1)
some fission fragments have short half-lives or high activities (1)
Max 3
(b)
moved by remote control (1)
placed in cooling ponds (1)
for several months (1)
[or to allow short T1/2 isotopes to decay]
transport precautions, e.g. impact resistant flasks (1)
separation of uranium from active wastes (1)
high level waste stored (as liquid) (1)
[alternative for last two marks:
rods are buried deep underground
at geologically stable site]
storage precautions, e.g. shielded tanks or monitoring (1)
reference to vitrification (1)
Max 5
[8]
M4.
(a)
graph passes through N = 100 to 130 when Z = 80 (1)
and N = 20 when Z = 20 (1)
2
(b)
(i)
W at Z > 60 just below line (1)
(ii)
X just above line (1)
(iii)
Y just below line (1)
3
Page 14 of 25
(c)
fission nuclei (or fragments) are neutron-rich and therefore
unstable (or radioactive) (1)
neutron-proton ratio is much higher than for a stable nucleus
(of the same charge (or mass)) (1)
β– particle emitted when a neutron changes to a proton
(in a neutron-rich nucleus) (1)
3
(d)
The marking scheme for this part of the question includes an overall
assessment for the Quality of Written Communication (QWC). There
are no discrete marks for the assessment of written communication
but the quality of written communication will be one of the criteria
used to assign the answer to one of three levels.
Level
Descriptor
Mark range
an answer will be expected to meet most of the criteria in the
level descriptor
Good 3
– answer supported by appropriate range of relevant points
– good use of information or ideas about physics, going
beyond those given in the question
– argument well structured with minimal repetition or irrelevant
points
5-6
– accurate and clear expression of ideas with only minor
errors of spelling, punctuation and grammar
Modest 2
– answer partially supported by relevant points
– good use of information or ideas about physics given in the
question but limited beyond this
– the argument shows some attempt at structure
3-4
– the ideas are expressed with reasonable clarity but with a
few errors of spelling, punctuation and grammar
Limited 1
– valid points but not clearly linked to an argument structure
– limited use of information or ideas about physics
– unstructured
1-2
– errors in spelling, punctuation and grammar or lack of
fluency
0
– incorrect, inappropriate or no response
0
Page 15 of 25
examples of the sort of information or ideas that might be used to
support an argument:
•
reduction of greenhouse gas emissions is (thought to be) necessary
to stop global warming (1)
•
long term storage of radioactive waste is essential because the
radiation from it damages (or kills) living cells (1)
•
radioactive isotopes with very long half lives are in the used fuel
rods (1)
•
nuclear power is reliable because it does not use oil or gas from
other countries (1)
•
radioactive waste needs to be stored in secure and safe conditions
for many years (1)
conclusion
either
nuclear power is needed; reduction of greenhouse gases is a greater
problem than the storage of radioactive waste because
1
2
global warming would cause the ice caps to melt/sea levels to
rise (1)
safe storage of radioactive waste can be done (1)
or
nuclear power is not needed; storage of radioactive waste is a
greater problem than reduction of greenhouse gases because
1
2
radioactive waste has to be stored for thousands of years (1)
greenhouse gases can be reduced using renewable energy
sources (1)
[14]
M5.
(a) The candidate’s writing should be legible and the spelling,
punctuation and grammar should be sufficiently accurate for the
meaning to be clear.
The candidate’s answer will be assessed holistically. The answer will be
assigned to one of three levels according to the following criteria.
High Level (Good to excellent): 5 or 6 marks
The information conveyed by the answer is clearly organised, logical and
coherent, using appropriate specialist vocabulary correctly. The form and
style of writing is appropriate to answer the question.
The candidate can explain the role of the moderator and control rods in
maintaining a critical condition inside the reactor. The explanation is given
in a clear sequence of events and the critical condition is defined in terms of
neutrons. To obtain the top mark some other detail must be included. Such
as, one of the alternative scattering or absorbing possibilities or appropriate
reference to critical mass or detailed description of the feedback to adjust
the position of the control rods etc.
Page 16 of 25
Intermediate Level (Modest to adequate): 3 or 4 marks
The information conveyed by the answer may be less well organised and
not fully coherent. There is less use of specialist vocabulary, or specialist
vocabulary may be used incorrectly. The form and style of writing is less
appropriate.
The candidate has a clear idea of two of the following:
the role of the moderator, the role of the control rods or can explain the
critical condition.
Low Level (Poor to limited): 1 or 2 marks
The information conveyed by the answer is poorly organised and may not
be relevant or coherent. There is little correct use of specialist vocabulary.
The form and style of writing may be only partly appropriate.
The candidate explains that a released neutron is absorbed by uranium to
cause a further fission. Alternatively the candidate may explain one of the
following:
the role of the moderator, the role of the control rods or can explain the
critical condition.
The explanation expected could include the following events that
could happen to a released neutron.
a neutron is slowed by the moderator
taking about 50 collisions to reach thermal speeds
then absorbed by uranium-235 to cause a fission event
one neutron released goes on to cause a further fission is the critical
condition
a neutron may leave the reactor core without further interaction
a neutron could be absorbed by uranium-238
a neutron could be absorbed by a control rod
a neutron could be scattered by uranium-238
a neutron could be scattered by uranium-235
max 7
(b)
it is easy to stay out of range or easy to contain an α source or ß/γ have
greater range/are more difficult to screen (1)
most (fission fragments) are (more) radioactive/unstable (1)
and are initially most likely to be beta emitters/(which also) emit γ
radiation/are neutron rich/heavy (1)
ionising radiation damages body tissue/is harmful (1)
max 3
[10]
Page 17 of 25
M6.
(a)
Pa
anti (electron) neutrino
2
(b)
2
(c)
(i)
x=4
1
(ii)
mass defect = [(232.98915 + 1.00867) –
(90.90368 + 138.87810 + 4 × 1.00867)] u
= 0.18136 u
3
energy released (= 0.18136 × 931) = 169 (MeV)
[8]
M7.
(a)
insert control rods (further) into the nuclear core / reactor
a change must be implied for 2 marks
marks by use of (further) or (more)
allow answers that discuss shut down as well as power reduction
which will absorb (more) neutrons (reducing further fission reactions)
If a statement is made that is wrong but not asked for limit the
score to 1 mark (e.g. wrong reference to moderator)
2
(b)
fission fragments / daughter products or spent / used fuel / uranium rods (allow) plutonium
(produced from U-238)
not uranium on its own
1
Page 18 of 25
(c)
(i)
(electromagnetic radiation is emitted)
A reference to α or β loses this first mark
as the energy gaps are large (in a nucleus) as the nucleus de-excites down
discrete energy levels to allow the nucleus to get to the ground level / state
mark for reason
2nd mark must imply energy levels or states
2
(ii)
momentum / kinetic energy is transferred (to the moderator atoms)
or
a neutron slows down / loses kinetic energy (with each collision)
(eventually) reaching speeds associated with thermal random motion or reaches
speeds which can cause fission (owtte)
2
[7]
(a) the amount of energy required to separate a nucleus ✓
into its separate neutrons and protons / nucleons ✓
(or energy released on formation of a nucleus ✓
from its separate neutrons and protons / constituents ✓)
M8.
1st mark is for correct energy flow direction
2nd mark is for binding or separating nucleons (nucleus is in the
question but a reference to an atom will lose the mark)
ignore discussion of SNF etc
both marks are independent
2
(b)
(i)
✓
must see subscript and superscripts
1
(ii)
binding energy of U
= 235 × 7.59 ✓ ( = 1784 (MeV))
binding energy of Tc and In
= 112 × 8.36 + 122 × 8.51 ✓
( = 1975 (MeV))
energy released ( = 1975 – 1784) = 191 (MeV) ✓ (allow 190 MeV)
1st mark is for 235 × 7.59 seen anywhere
2nd mark for 112 × 8.36 + 122 × 8.51 or 1975 is only given if there
are no other terms or conversions added to the equation (ignore
which way round the subtraction is positioned)
correct final answer can score 3 marks
3
Page 19 of 25
(iii)
energy released
= 191 × 1.60 × 10−13 ✓
( = 3.06 × 10−11 J)
loss of mass ( = E / c2 )
= 2.91 × 10−11 / (3.00 × 108)2)
= 3.4 × 10−28 (kg) ✓
or
= 191 / 931.5 u ✓ ( = 0.205 u)
= 0.205 × 1.66 × 10−27 (kg)
= 3.4 × 10−28 (kg) ✓
allow CE from (ii)
working must be shown for a CE otherwise full marks can be given
for correct answer only
note for CE
answer = (ii) × 1.78 × 10−30
(2.01 × 10−27 is a common answer)
2
(c)
(i)
line or band from origin, starting at 45° up to Z approximately = 20 reading
Z = 80, N = 110→130 ✓
initial gradient should be about 1 (ie Z = 20 ; N = 15 → 25) and
overall must show some concave curvature. (Ignore slight
waviness in the line)
if band is shown take middle as the line
if line stops at N > 70 extrapolate line to N = 80 for marking
1
(ii)
fission fragments are (likely) to be above / to the left of the line of stability ✓
fission fragments are (likely) to have a larger N / Z ratio than stable nuclei
or
fission fragments are neutron rich owtte ✓
and become neutron or β− emitters ✓
ignore any reference to α emission
a candidate must make a choice for the first two marks
stating that there are more neutrons than protons is not enough for
a mark
1st mark reference to graph
2nd mark – high N / Z ratio or neutron rich
3rd mark beta minus
note not just beta
3
[12]
Page 20 of 25
E1.
This question was intended to give candidates an opportunity to combine their knowledge of
nuclear properties and reactors, covered extensively in Section 13.5 of the specification, with
their knowledge of particle physics from AS Module 1, to produce well-reasoned answers. Some
were able to do so, but most attempts were too superficial.
Whilst the majority knew that the answer to part (a)(ii) was ‘concrete’, not many could assemble
their thoughts sufficiently to suggest ‘thick’ and ‘dense’ in part (a)(i). Vague statements such as
‘should not let any radiation leak out’ were much more prevalent in candidates’ responses; these
were not considered worthy of credit.
The ability of neutrinos to pass through the shielding was often recognised in part (b), but it is a
misconception that the shielding can prevent all ã radiation and all neutrons from escaping. The
failure to use appropriate technical vocabulary was again a telling weakness in many candidates’
answers. They wrote about γ radiation being slowed down and eventually stopped by the
shielding, when they should have referred to its intensity being reduced. The material was
thought to become radioactive because it captured the radioactive particles, which then
remained radioactive within the shielding. It was satisfying to see that at least some of the
candidates realised that neutron absorption by nuclei in the shielding would make them unstable,
being neutron-rich, causing them to be β– and γ emitters.
E2.
This proved to be a high scoring question for most candidates, who showed better
knowledge of fission processes and nuclear reactors than has sometimes been apparent in
previous years. The main failing in part (a) was a tendency to write about chain reactions
generally, instead of chain reactions involving nuclear fission. This meant that some answers
were not sufficiently specific to the fission process. Fewer candidates than usual stated that
thermal neutrons were ones that had been heated up, and very few indeed thought fission to be
induced by electrons.
In part (b), mixed-up understanding of moderation and control was evident in some scripts.
Occasionally this came out in contradictory statements, such as “moderator rods made out of
carbon absorb the energy and reduce the number of neutrons in the reactor”. When explaining
the kinetic energy lost by the colliding neutrons, candidates need to remember that it is the mass
of the moderating atoms that is crucial rather than their size. The control aspect frequently
tempted candidates to write obsessively about emergency shutdown (and/or meltdown)
procedures, instead of confining their answers to a description of control for a reactor in normal
operation. However, the majority had quite good knowledge of the main principles involved in
both moderation and control. The mark scheme adopted for part (b) gave a good reward to
those who knew the main facts.
Page 21 of 25
E3.
Many answers to part (a), which for full credit required explanations rather than statements,
would have benefited from a more precise use of terminology. In part (a) (i) for example, ‘the fuel
runs out’ is much inferior to ‘the amount of fissionable uranium remaining decreases’. In part (a)
(ii), a mark was awarded for neutron bombardment causing radioactive instability, but the real
reason is that the fission fragments themselves are unstable neutron-rich β and γ emitters.
Furthermore, since some of them have very short half-lives, their activities can be very high.
In part (b), the treatment and handling of spent fuel rods was often understood very well.
However, the need to use remote control, and to take precautions over the transport of used
material, was commonly overlooked. Credit was given equally to procedures involving
reprocessing and to those where the spent rods are simply buried deep underground in stable
rock formations.
E5.
This question was a good discriminator. Most candidates, in part (a), knew how the core of
the reactor functions. Some candidates too readily used the wording of the question as their
answer. Others did not refer to neutrons even though this was asked for in the question. One
example of a phrase given by candidates that did not quite answer the question but sounded
reasonable was, ‘the power levels were kept constant by keeping a constant rate of fission using
control rods’. This offers much of what was in the question itself and it does not refer to
neutrons. The quality of the writing was generally good.
Again question (b) was a good discriminator. The majority of candidates were aware that fission
products are normally unstable because they tend to be neutron rich or that they release beta
and gamma rays. Less able candidates thought used fuel meant that they had undergone alpha
emission.
E6.
The more able candidates successfully negotiated the majority of this question but the less
able found many pit-falls.
In part (a) most obtained the first mark but then did not obtain the anti-neutrino.
For part (b) some candidates did not identify the position of P. Position Q was easier for
students to identify.
A majority of candidates could balance the number of neutrons in part (c)(i) to obtain the correct
answer x = 4. Those that guessed the answer almost always gave the answer x = 3.
Part (c)(ii) was very discriminating. Less able candidates did not know how to balance the
energies and only scored marks on the conversion from u to MeV. Some did not go directly from
u to MeV and gave many lines of calculation. If correctly performed, they still got the mark for the
conversion, but they had many opportunities to show errors and so tended to be less successful
and missed the mark.
Page 22 of 25
E7.
Most candidates were fully aware of the function of the control rods in absorbing excess
neutrons and scored well in part (a). Some candidates said too much by explaining the role of
the control rods to absorb neutrons and the moderator to slow neutrons down but then did not
make it clear which reduced the power. The weaker candidates talked about control rods
controlling the reactions without any further explanation.
Part (b) was answered well by most but it was common to give the answer fuel rods rather than
spent fuel rods.
In (c)(i) the most common answer was ‘gamma rays’ but very few then went on to discuss
energy levels. Some of those that did then spoilt their answer by referring to changing electron
levels.
Part (c)(ii) was a very good question to distinguish between the weak and strong candidates.
The weaker candidates focussed on the wording in question concerning elastic collisions. They
interpreted this to mean the neutrons maintain their kinetic energy or momentum during all
subsequent collisions.
Page 23 of 25
E8.
In part (a) was very straightforward for many candidates. Less able candidates did not
elaborate on what the nucleus split up into or they referred to some other splitting such as
fission. Another common error was to talk about the separation of an atom into protons,
neutrons and electrons. Very few candidates got the direction of energy flow wrong.
In part (b)(i) very few candidates referred to particles other than neutrons but a significant
number failed to write down two neutrons in the correct manner.
and
a single were
extremely common.
The follow on part (b)(ii) turned out to be an extremely good discriminator. It highlighted the types
of errors candidates were making. There was a group who did not appreciate the data was given
per nucleon and used the figures without multiplying up by the respective nucleon numbers.
Another significant group added the mass energy of mostly one but sometimes two nucleons
into the proceedings. The last major group got into difficulties because they changed units
unnecessarily or only changed the units of some of the terms. Overall there was a good
percentage of correct answers.
A good proportion of candidates could accomplish the conversion of units required in part (b)(iii)
with relative ease, sometimes from an error carried forward from the previous part. With the
conversion needing two stages using the datasheet there was plenty of opportunity for errors by
dividing instead of multiplying or using an incorrect conversion factor. A separate but common
error was to ignore the answer to part (b)(ii) and simply use the difference in mass of the
nucleons involved, ignoring the binding energy per nucleon completely. Less able candidates did
not really help themselves in this question because very few put words or units to intermediate
stages of the calculation. If they had done so fewer would have lost their way.
A majority of candidates did not score marks in part (c)(i). Many knew the general shape but very
few remembered any details so they had no idea which coordinates to draw their line or band
through. Some gave no real thought to the problem and drew graphs that did not make sense.
For example some graphs went vertical or turned back on themselves.
Part (c)(ii) was very discriminating. Less able candidates simply referred to any radiation that
came to mind and forwarded very little explanation. The bulk of the candidates knew beta minus
radiation was emitted but they were not careful how they expressed their reasons. For example,
stating that there are more neutrons than protons is not sufficient to imply the nuclei are neutron
rich. A majority in this group of candidates made no reference to the graph at all. Many that did
had flawed reasoning. They thought the isotopes were neutron rich because the large number of
free neutrons in the core. Good candidates also got into some difficulty by not reading the
question carefully. Typically these candidates would start their answer with, 'If there are a lot
more neutrons than protons then... but if the neutron to proton ratio is small then…'. These
candidates obviously knew the subject matter but did not score many marks as the question
clearly asks for a choice to be made.
Page 24 of 25
Resource currently unavailable.
Page 25 of 25
Nuclear medicine
Make notes on the following
Consequences of radiation exposure
The Goiânia accident
Technetium as a medical tracer
Radiotherapy
Comparison of traditional radiotherapy and proton
therapy
Complete and mark the following
Physics factsheet 77 & 82
AQA exam past paper questions - Alpha Beta and
Gamma
Completed?
Completed?
3K\VLFV )DFWVKHHW
www.curriculum-press.co.uk
Number 77
Nuclear Medicine Used For
Treatment
Factsheet ‘Nuclear medicine used for treatment
• biological effects of ionising radiation
• use of nuclear radiation in the treatment of illness
Factsheet, ‘Nuclear medicine used for diagnosis’
• production of radionuclides
• using radionuclides for imaging
• measurement of exposure to radiation
Although X-rays are indistinguishable from γ-rays of the same energy,
Exam Hint : - As always read the question carefully. Very few marks
would be obtained if you wrote about diagnosis when the question asks
for therapy. Sometimes they can be opposite, for example β rays are
useful in killing tumour cells but should be avoided inside the body at
all other times.
γ?
X?
they do not come from the nucleus and so have not been specifically
included. However, all calculations would be identical.
What you should know
properties of:
• α, ✓
• β ✓
• γ ✓ → Factsheet 11 and Qus 1 & 2 on this factsheet
• calculation of half life T½✓
Biological Effects of Radiation
Damage is mainly caused by ionisation.
Ionisation is the removal of electrons from atoms.
α and β particles move at high velocity and can easily knock electrons out
so they are called ionising radiation.
α, β and low energy X and γ-rays are called ‘non-penetrating’ because
they do not pass through tissue.
α
β
Ionising radiation is a beam of high energy particles (waves
are not so highly ionising) that will remove one or more electrons from
an atom in its path.
low energy
X or γ
Each time ionisation occurs the kinetic energy of the particle is reduced, so
the greater the ionisation the shorter the range (distance travelled before
coming to rest).
tissues
They are not used in diagnosis but are useful in therapy.
highly ionising
energy soon lost
Energy
Energy
nothing detected
less ionising energy
travels further
Therapy is treatment to try to make the patient well.
range
danger zone
<1mm
distanced travelled
α radiation is extremely dangerous if it gets inside the body e.g. in the lungs,
β is to be avoided unless it is needed to kill cells.
The purpose of nuclear medicine in therapy is usually to kill cancerous
cells. α and β radiation may be used close to the tumour. They are highly
ionising with a short range so that too many healthy cells are not damaged.
α source
distanced travelled
Worked Example
It takes 5.4 eV to create an ion pair and an α particle creates 5.2×105 ion
pairs per cm of air.
520,000 ion pairs created
5.4 eV
+
needed to make
1 cm
one ion pair
β source
danger zone
The range of α's from a certain source is 2.9 cm, what was their energy
in eV when they started?
Total energy = 2.9 cm × 5.2×105 cm-1 × 5.4 eV
= 8.1 MeV
Alternatively several beams of γ may be arranged to cross where cells need
to be killed.
1
3K\VLFV)DFWVKHHW
77. Nuclear Medicine Used For Treatment
γ and X rays are high frequency, short wavelength (less than 1 nanometre),
electromagnetic radiation. Each photon can remove an electron if it has high
enough energy so they may also be considered as ionising radiation. It is
the energy that is important so that is how they are described e.g a 2.0 MeV
X ray.
These are microscopic events. Larger scale or macroscopic effects are:
Uncontrolled cell division (cancer) may result from either direct or indirect
damage. Organs may fail, e.g. bone marrow, or may be destroyed, e.g. in
skin burns. Genetic abnormalities increase.
E = hf = hc/λ
microscopic
macroscopic
direct damage in cell
indirect damage in cell
cancer
organ failure e.g. bone marrow destruction of
tissue, skin burns, genetic abnormalities
Worked example
Find the wavelength of a 0.35 MeV γ ray
0.35×106 eV = 0.35×106 × 1.6 × 10-19 J = 5.6 × 10-14 J
E=
−34
If the reproductive organs are affected then the effect may be passed on to
the next generation. This is then hereditary damage as distinct from somatic
which just affects the individual.
hc
hc 6.6 ×10 × 3 ×10
∴λ =
=
= 3.5 ×10 −12 m
λ
E
5.6 ×10 −14
8
somatic
hereditary
Ultra-sound, lasers and magnetic resonance imaging (MRI)
do not involve ionising radiations at all and so should be safer.
Some damage may be started by just one event, e.g. cancer or mutation.
This type is called stochiastic (meaning random). There is no safe level of
irradiation.
Computer Tomography (CT) scans use X-rays.
Ionising radiation
α -highly
Non Ionising radiation
Ultra sound
β –slightly less
Laser
high energy γ & X rays - less
Magnetic resonance imaging MRI
Other effects may not be noticeable unless a certain level called a threshold
is reached e.g. skin burns. These are non-stochiastic.
exposure
non stochiastic un-safe
Details of damage
no safe level
Exam Hint:- This section is full of pairs of words,
macroscopic - microscopic
direct - indirect
stochiastic - non-stochiastic
somatic - hereditary
Make sure you know what each means and try to work out which ones
are relevant to the question.
threshold
0
Protection
The effect of ionising radiation is reduced by keeping your distance. If you
double the distance then the effect goes down to one quarter (inverse
square law).
Direct damage:
The radiation may directly ionise an atom in an important molecule within
a cell such as enzymes, DNA or RNA. DNA is particularly important.
The cell will no longer function normally and cell death or mutation may
result.
A
A
Indirect damage
If water molecules in the cell are split into OH− and H+, hydrogen peroxide
(H2 O2 ) may form from two OH−. The hydrogen peroxide may then
damage the DNA in the chromosomes of the nucleus and the cell may die
or be unable to reproduce. This is indirect damage because there is an
extra stage. The cell wall permeability is often affected.
direct damage
radiation
DNA
safe ???
stochiastic
A
A
A
d
2d
Exam Hint: - The inverse square law always applies where there is little
absorption in the medium, e.g. sound, gravity, γ and X rays in air.
indirect damage
radiation
Iα
water
I
d2
1
or 1 = 22
2
I 2 d1
d
makes
Worked Example
A 60Co source provides an exposure rate of 270 μCkg-1s-1 at a position 1.0
m in front of it. Estimate the exposure rate 3.0 m away from the source
ignoring any attenuation in the air.
H 2O 2
chemical damage
DNA
The new position is 3 times farther away
so the new exposure rate will be 1/9 (1/ 32 ) of the value at 1.0 m.
Exposure rate = 270/9 = 30 μCkg-1s-1.
2
77. Nuclear Medicine Used For Treatment
3K\VLFV)DFWVKHHW
Absorption of radiation
Worked Example
80 keV γ rays have μ =690 m-1 in copper.
(a) Calculate the half value thickness
(b) use this answer to find the fraction remaining after travelling
through 3.0 mm.
Shields may be used and the intensity in an absorber follows an exponential
decrease. The same equation is used for X rays or γ rays.
I=
Io =
e=
x =
I = I o e − μx
intensity after travelling distance x in the material.
original intensity (NOT to be confused with the Io in sound
which has value 1 × 10 –12 W m-2 and is the zero level of sound
that can be heard).
linear attenuation coefficient. It is a property of the material but
also depends on the energy of the incident photons.
distance travelled in or the thickness of the material.
(a) x½ =
0.693
0.693
= 690 = 1.0×10-3 m or 1.0 mm
μ
(b) it halves each time it goes through 1.0 mm so 1/8 remains.
Types of therapy
Therapies
Causes
.
radioactive
decay
products
Exam Hint:- the exponent (number up in the air) is always dimensionless
so if x is in cm then μ is the inverse and is cm-1.
Worked Example
A parallel monoenergetic γ-ray beam passes through 3.0cm of a material.
The intensity of the emerging beam is 0.60 of the original. Calculate
a value for μ.
x rays
fast moving
subatomic
n, p
particles
I = 0.60 Io
0.60 Io = Io e-3.0 μ (Io cancels and you don’t need to know it.)
take natural logs both sides, ln (not log, that’s to base 10 not to base e)
ln 0.60 = -3.0 μ
so μ =
ln 0.60 −0.51
=
− 3.0
− 3.0
Unsealed sources
radiopharmaceuticals
(swallowed or injected)
αβγ
Tumour
Sealed sources,
radioactive needles,
patches
Beam therapy
The guiding principle is always that there is no safe dose. All doses should
be kept as low as possible and there should be positive benefit from any
exposure.
= 0.17 cm-1
Exposure is mainly concerned with killing cancerous cells. Luckily dividing
cells are more radiosensitive than non-dividing. So as cancer cells are dividing
rapidly there is a chance of killing them.
For each type of material and energy of the radiation there will be a HalfValue-Thickness x½
A. Unsealed sources
These can only be used if they accumulate in the organ that needs treatment.
A radionuclide is attached to a pharmaceutical (making a radiopharmaceutical)
and the liquid is injected or swallowed.
In the thyroid radioactive Iodine taken orally is used to counteract overactivity
(with a smaller dose than if cancer is present.)
50 % less
ion
radiat
Half Value
The half-value-thickness is the thickness of material required
to reduce the intensity of the radiation to one half the original value.
0.693
(compare with half-life equation Factsheet 11)
x½ = μ
The behaviour of a nuclide introduced into the body
Effective half life of a radioactive substance
As well as the usual reduction in activity with time, physical half-life Tp,
the chemical may be reduced by bodily processes and waste removal or
respiration. This decrease is also exponential so it has a half-life Tb known
as the biological half-life. This varies with the individual and the organ
involved but taking both into account, the effective half-life Te is given by
Intensity I/Wm-2
I0
1
1
1
=
+
Te Tb T p
Worked Example
The physical half life of Iodine 123 is 8.0 days and it is removed from the
body with a half life of 21 days.
I0/2
1
1
1
1 1
=
+
+ = 0.0476 + 0.125 = 0.173(d-1)
=
Te Tb T p 21 8
1
don’t forget this is still
Te
∴ Te = 5.8 days
I0/4
I0/8
0
0
1HVT
2HVT
3HVT
Thickness
4HVT
5HVT
3
77. Nuclear Medicine Used For Treatment
In order to reach the target tumour, healthy tissue must be crossed. To
reduce the damage done on the way in, Multiple beam therapy may be
used where several beams cross at the target
Typical Exam Question
(a) What is meant by the terms (i) radioactive half life (ii) biological
half-life?
(2)
(b) Why does biological half-life depend upon both the organ and
the patient?
(2)
γ ray
(c) Iodine-131 has a half-life of 8 days. Approx. what % will remain
in the body after 1 month. (there is no mention of biological half
life so take the question at face value)
(2)
body
(d) Complete the equation below
(2)
I → 54 Xe + β + γ
131
tumour
(e) Another isotope of iodine I-123 is available which decays by
emission of gamma only. Explain which you would choose for the
treatment of the thyroid by ingestion.
(3)
or the source may rotate so that the target always receives the radiation.
This treatment may be called a gamma knife.
Answers
(a) (i) the average time for half the nuclei to decay or the activity to fall to
½ the initial value. (1)
(NB some exam boards insist on the word average somewhere
because of course they are random processes)
(ii) the average time for half the chemical to be metabolised or excreted
from an organ. (1)
(NB this has nothing to do with radioactivity, it applies to any
chemical)
It is common to split the dose up into fractions and administer them over a
period of some weeks. Healthy cells recover more quickly than cancerous ones.
A cell which was not dividing the first time may be caught the next time, the
odds are higher that a cancerous one will be caught.
Unfortunately the dose that the tumour receives is critical, a small reduction
makes the treatment ineffective and a small increase may do harm.
(b) Different organs will metabolise the chemical at different rates.
Different patients will have different overall metabolic rates which
also depends on what they are doing at the time.
dose too small
cancer cells not all all cancer cells killed and
killed
not too many healthy
cells
(c) There are about 4 half lives (1)
in 1 month so at the end of each half-life there will be 1/2,1/4,1/8,1/16.
1/16 is about 6% (1)(OK to only have 1 sig fig if it says approximately)
(d) a neutron will have changed to a proton and an electron which is ejected.
0
+1
-1
0
n
p
+ e + v
The charge number for I must be 53 (1),
the mass number for Xe is the same at 131 (1)
(e) I would choose I-131. (1) (make sure you say what you mean)
I-123 would be no use, all the radiation would leave the body.
The β is needed (1)
to damage unwanted tissue.(1) (don’t forget to explain WHY)
B. Sealed Sources
dose just right
dose too large
too many healthy
cells killed as well
Qualitative (concept)Test - answers to be found in the factsheet
1. In the equation I = I0e−μx , for the decrease in intensity of radiation
passing through a material, state the name of the constant μ.
2. List 2 factors which affect the value of μ.
3. Why are α and β radiation sometimes used in therapy but never in
diagnosis?
4. What is meant by ionising radiation?
5. Under what condition does the intensity of radiation decrease following
an inverse square rule?
The source of radiation needs to be as close as possible to the tumour to 6. In the equation for decay of radioactive atoms, N = N e−λt , if t is in
o
avoid damage to healthy cells. This may be achieved by surface patches,
seconds then what must be the unit of λ?
inserting needles into the skin, or containers into body cavities. β emitters
are used for short range work. New ideas include inserting tubes into the Quantitative (calculations) test
body and then blowing the radionuclide into them.
1. An α source of activity 30Bq gives an ionisation current of 2.2×10-11A.
When it is not possible to get close, longer range X, γ or electron beams are
(a) How many singly charged ions per sec is this?
used. Cobalt-60 may be used as a source of γ rays of energy of about
1.25 MeV. It is hard to control (the energy cannot be varied, the source
(b) How many ions on average does each a particle cause?
cannot be switched off and the beam is not well defined) and the other
types of radiation are generally preferred.
2. Sample X and Y have the same number of atoms. The half-life of X is 10
No on/off switch
mins and that of Y 20 mins. What is the ratio of activity X to activity Y:
Co-60 No energy adjustment
(a) at the start
No focusing knob
(b) after 20 mins?
On/off switch ✓
X-ray
Energy adjustment ✓
3. With a source of Co60 the γ dose rate is 80 μSvh-1 at 2.0 m away. At
Focusing with difficulty ✓
what distance will it be 20 μSvh-1?
Hospitals are introducing linear accelerators to produce high energy X-ray
beams.
4
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77. Nuclear Medicine Used For Treatment
Quantitative test answers
Exam workshop
This is a typical poor student's answer to an exam question. The comments
explain what is wrong with the answers and how they can be improved.
The examiner's mark scheme is given below.
Q Ne N Q
= ∴ =
t
t
t et
the number of ions per sec is the current divided by the charge on
one ion
2.2 × 10-11
= 1.4 × 108 singly charged ions per sec are produced
N=
1.6 × 10-19
1. (a) I =
(a) Ionising radiation is incident on living matter. Explain with an
example, the terms
(i) direct effect
[2]
1/2
skin burn ✓
8
(b) 30 Bq means 30 particles per sec so 1.4 × 10 = 4.7 × 106
30
ions from each particle.
no explanation, first mark lost
(ii) indirect effect
releases free radicals ✓ damaging cells
2. Activity, A = λN, and λ is inversely proportional to the half-life,
λ = 0.693
T½
[2]
1/2
needs to be specific about the effect of the free radicals – too vague
to get the second mark
the half-life of X is half that of Y so the decay constant of X will be
twice that of Y
λx= 2λy
Ax= λxN =2 λyN = 2Ay so at the start the ratio of the activities is 2 .
After 20 mins. X has N/4 atoms and Y has N/2,
(b) (i) explain why cancerous cells may be killed at a greater rate
than healthy ones.
[2]
healthy cells recover more quickly ✗
0/2
N
2λ
y 4
x =
N =1 so the ratio is now one.
A
λ
y
y 2
although true this does not explain why cancerous cells are more
likely to be killed, no marks.
A
(ii) Explain a method to treat a tumour but keep damage to
healthy cells low
[3]
the beam is focused on the tumour. other areas are shielded.✓
bod
3. If you double the distance away the dose will be one quarter which is
what is required. So 4.0m.
It is not possible to ‘focus’ the beam.
It still passes through healthy tissue which receives the same dose
as the tumour. The material used for shielding should be named.
This might be awarded a benefit of the doubt mark at the examiner’s
discretion.
Examiner’s answers
(a) (i) the radiation ionises the DNA (1)
this leads to mutation/ cancer or the death of the cell (1)
(ii) The interaction of the radiation splits water in the cell into H+ and
OH- / or free radicals, this may form hydrogen peroxide H2O2 (1)
The DNA is damaged or the permeability of the cell membrane
affected.(a mark for mention of cancer or mutation is only allowed
once in the two parts)
(b) (i) radiation is most harmful to cells when dividing (1)
malignant cells divide more often than normal cells (1)
(ii) there are about 4 possible methods here. It is best to select one
and describe it thoroughly. You cannot gain 3 marks for naming
3 different methods and be careful about a mixture of 2.
1st mark
source of radiation
rotated/multiple
beams used
2nd mark
tumour at centre
3rd mark
tumour gets larger
dose than healthy
nearby tissue
lead shielding is placed this stops the radiation the tumour is targeted
around the area
except for a hole over
the target
a small dose called a
fraction is repeated at
interval
healthy cells recover there is more chance of
faster than cancerous catching a cancerous
ones
cell dividing
a radioactive
substance is put near
the tumour
it must be α or β or
have a short half life
Acknowledgements:
This Physics Factsheet was researched and written by Celia Bloor
The Curriculum Press,Bank House, 105 King Street, Wellington, Shropshire, TF1 1NU.
Physics Factsheets may be copied free of charge by teaching staff or students, provided that
their school is a registered subscriber.
No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted,
in any other form or by any other means, without the prior permission of the publisher.
ISSN 1351-5136
because short range or
to quickly stop the
damage
5
3K\VLFV )DFWVKHHW
www.curriculum-press.co.uk
Number 82
The Use of Nuclear Medicine in
Diagnosis
See Factsheet 77 which covers the use of radiation in treatment (therapy)
The nuclide chosen should:
1 emit only γ rays with energy suitable for the camera (energy too low they are absorbed in the body; energy too high - they go through the
detector without registering).
In diagnosis (finding out what is wrong) radioactive material
shows us how the organ is working. Most other images show structure,
not function, and could be used even if the patient was dead.
This factsheet does NOT cover:
(a) X-rays (but don’t forget γ rays behave in the same way)
(b) MRI (magnetic resonance imaging)
(c) Contrast agents to produce better X-ray pictures.
The basic properties of α, β and γ radiation should be known. Half-life and
half-value thickness calculations should be familiar (see Factsheet 11).
have a reasonable half-life so that measurements can be taken but, combined
with elimination from the body, the activity should not remain high for
longer than necessary (see effective half life, Factsheet 77).
3
be easy to obtain.
4
be easy to attach to compounds to make a tracer.
Exam hint Learn this list for some easy marks
Technetium (pronounced Teckneeshum)-99 Tc m
(symbol Tc, the m means metastable - sort of semi stable or in an excited
state) is used for 90% of diagnostic images. It can be included in compounds
which are then called radio-pharmaceuticals and are used as tracers.
Check your specification for exactly what you need to know for your
exam. Read several textbooks which cover the material and learn a few
detailed examples of different techniques.
For information to be able to get out of the body without
taking samples, γ rays are needed.
It has a half-life of 6 hours and emits only γs of 140 keV.
This satisfies all the criteria for a really useful radionuclide.
γ
γ
2
99m
99
43Tc→ 43Tc + γ
half-life 6h
equation 1
(remember γ decay doesn’t alter A or Z of the isotope)
It is produced within the hospital by decay from Molybdenum
body
α
source β
99
m
Mo→9943
Tc + −10β + ν−
half life of the molybdenum is 67h 42
(remember in β decay a neutron changes to a proton)
γ
99
42 Mo .
equation 2
Exam hint Learn these two important equations.
A gamma camera is most commonly used to find out where the ray came
from. The images obtained improved rapidly in the 1990s.
The Molybdenum is produced by a nuclear reactor (either as a fission byproduct or from neutron bombardment). It is sent to the hospital adsorbed
on alumina in a glass column. This forms a generator of 99Tcm sometimes
called a ‘cow’ because it is ‘milked’ each day. (N.B. adsorbed means
sticking to the surface – this is not the same as absorbed)
Exam Hint As always read the question carefully. Very few marks
would be obtained if you wrote about diagnosis when the question asks
for therapy. Sometimes they can be opposite - for example β rays are
useful in killing tumour cells but should be avoided inside the body at
all other times.
saline solution
Choice and production of radio nuclides
99Mo
absorbed on
alumina
Most naturally occurring radio-isotopes have half-lives of many years and
are not useful in medicine. We can produce suitable ones either from the
products left after fission in a nuclear reactor or by placing a target element
in the path of neutrons coming from a nuclear reactor.
evacuated jar
sodium pertechnate solution
This is called neutron irradiation. Absorbing an extra neutron
into the nucleus may make it unstable so that it produces useful radiation.
Exam hint Learn this diagram and practise drawing it.
1
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82. The Use of Nuclear Medicine in Diagnosis
Sodium chloride solution is drawn through the column and, while the
molybdenum remains fixed to the alumina, the technetium is soluble and
what has been produced by decay since the last ‘milking’ is removed for
use with several patients.
Images with the gamma camera
A gamma-emitting radio-isotope enters the body and its position can be
monitored by detecting the emissions from outside.
Changes in γ camera
This is called elution.
output
The ‘generator’ needs to be replaced each week.
electrons
Dynodes
growth and removal of 99m Tc from a generator
PM tube
electrons
Tc
Scintillation
counter
Photo cathode
99m
99 Mo decay
log activity of
1
visible photon
2
Crystal
3
γ rays
Collimator
0
24
48
72
Exam hint Learn this graph and the explanation
γ rays
Time / h
3
Patient
Iodine-123
A large crystal of sodium iodide (the Scintillator) emits a flash
of light (many photons in the visible region of the EM spectrum) when
hit by a γ ray.
In larger hospitals this is produced by a cyclotron. (a particle accelerator)
Antimony-121(symbol Sb) is bombarded with fast moving α particles
accelerated to high speed within the cyclotron.
121
4
123
51 Sb+ 2 He→ 53 I
+ 201n
equation 3
The amplitude of the flash is proportional to the energy of the γ ray. The
flashes are detected by a large number of photo multiplier tubes and the
exact source of each flash worked out. This is then processed to give a
picture of the concentration of isotope at a particular point in the body.
Each point is located in 2 dimensions and has a colour indicating intensity.
Both 99Tcm and 123I are commonly used. (see below in tracers section).
Exam hint Learn this equation
Iodine-123 emits only γs, mainly of energy 159 keV and has a half life of 13
h. It is now replacing Iodine-131 which emits βs as well as γs and has a
much longer half life of 8 days.
Iodine-131
This is produced by irradiation of Tellurium (don’t confuse this symbol Te
with Tc for Technetium!) with neutrons from a nuclear reactor. This
cannot usually be carried out at the hospital and the use of I-131 is being
phased out.
130
1
131
γ
52Te + 0 n→ 52Te +
Gamma camera
electrical output
The Tellurium-131 cannot be separated chemically from the target material
because it is the same element.
Then
131
131
52Te→ 53 I
+ β- + ν
700 electrons
c. 1000 V
equation 4
not to scale
81
increasing
voltage
equation 5
The Iodine can then be separated chemically.
27
9
coated dynode
Typical exam question
When the metastable isotope of technetium-99 is produced by
elution from the Molybdenum isotope, the activities of these
isotopes in the generator before and after elution are as shown in
the graph in fig 3
(a) Explain the shape of the graph (4)
(b) Write the equations for the decay of the two isotopes (2)
99
m
(c) Why is 4 3 Tc the ideal isotope for diagnostic use?
Answers
(a) The amount of Tc rises as the Mo decays but drops to almost zero at
each ‘milking’ and then rises again because the Mo is still decaying.
After about 22hrs a peak is reached and then the Tc is decaying faster
than it is formed. The Mo activity decreases with a gentler curve
because the half life is 67h.
(b) See text eqn 1 & 2
(c) See text under “The nuclide chosen should:”
3
c. 100 V
crystal
scintillates
1 electron
photo cathode
lead collimator
Parts of the gamma camera:
Collimator-this has lead channels which absorb rays not heading straight
along them. A sharper image is obtained as each ray has come from the spot
below the channel. Various shapes are available to cover larger or smaller
body area.
The Crystal - emits many light photons on absorbing a γ photon.It is
about 500 mm in diameter.
2
82. The Use of Nuclear Medicine in Diagnosis
3K\VLFV)DFWVKHHW
Photo multiplier tube –Photo sensitive material on the photo-cathode is
placed close to the crystal. This ejects electrons on receiving the photons
(the photo-electric effect) and the electrons are increased in number by the
multiplier to make a pulse. As each electron enters the tube it is accelerated
to give it more energy. It then hits up to ten coated dynodes and more than
one electron is emitted each time because the electron energy is several times
larger than the work function for the coating. . On reaching the end of the
tube the one electron may now be 106 .
Worked example
10 ml of tracer (1a) is injected into the body and a 10 ml sample
(1b) taken from a different place in the body after 15 mins. when
the tracer should be well dispersed. The activity is 80Bq or 8 Bq
per ml. 10 ml (2a) of the original tracer is diluted to make 1 litre
and the activity of a 10ml sample (2b) also measured after 15
mins. is 200Bq or 20 Bq per ml.
The total activity is the same in both cases because the same size original
sample was used so if the total volume of the fluid in the body is V ml,
then
8V = 20 × 1000 ml
so V = 2500 ml or 2.5 l.
The crystal and the tubes are called a scintillation counter.
Each tube will receive a different proportion of the light from the crystal
and so the origin of the flash can be determined. The coordinates are sent
to the computer which assembles a map of the activity, a dot for each γ
received. Analysis is carried out to remove signals from scattered γs which
would give a false location.
Dilution must be used because the same instrument cannot measure a wide
range of activities. The original undiluted activity would be too high.
Exam Hints and tips
The volume may be given in m3 so a simple number like 5 ml turns into
5×10-6 m3! You should be able to mentally get rid of the complication
and look at ratios.
The problem will probably have a different volume for every sample
but you can’t go wrong if you work out the activity per unit volume
each time. Use the unit which gives sensible sized numbers.
Hint. You should be thinking about this.
If there are 3 secondary electrons emitted each time how many will
there be after 10 dynodes?
Start 1 electron
3 electrons
31
after
1st dynode
nd
2 dynode
9
“
32
3rd dynode
27 “….
33.
5.9 ×104 “
3 10
10th dynode
2. uptake images for diagnosis:
Technetium can be used to study the movement of tracer in many organs,
e.g. liver, lungs, brain and thyroid (originally iodine was used for the latter).
There can be one static picture after some time has elapsed or a series of
images from the start (of the addition of the tracer) giving a dynamic
picture
• The count rate may be compared with that from a model known as a
phantom containing the same amount of radioisotope in a similar
volume. A percentage uptake (amount of tracer in the organ) in the
patient would be calculated.
As it started with one electron this last figure is the ‘gain’ or
multiplying factor of the tube.
A few facts and figures:
The flashes of light are very short and 104 per second can be counted..
Taking the picture could take up to 200s and produce 1 million counts.
The spatial resolution (size of smallest object clearly seen) is
3mm.
Measuring Exposure to radiation –the Units
Exposure - symbol X, unit C kg-1 - the amount of ionisation produced in air
by the radiation.
Tracers
If there are non-radioactive isotopes of the same element in a tracer they
will behave chemically in the same way as the radioactive isotope.
Exposure is the value of the charge of ions (of one sign only)
produced in a unit mass of air by the radiation.
These are called carriers. A carrier-free sample is 100%
radioactive atoms and would have a high activity.
+-
1. dilution method
One use of radioactive substances inserted into the body is to measure the
volume of fluid present, e.g. blood or water. A small quantity of tracer
(Iodine 131 or 123 for blood, tritium 3H for water) is allowed to mix fully
in the body. A sample is withdrawn and the activity compared with a
sample from a prepared mixture of known dilution.
ion
iat
rad
Identical samples
0 1
2
The radiation incident is measured by its power to ionise air. For γ and X
rays this is a good measure of the energy absorbed. This is important. Also
the behaviour of air is close to that of body tissue for these rays. The charge
of one sign (each electron removed leaves a positive ion) per kg of air is
measured.
2a
known
volume
unknown
volume
Q
m
The unit is Coulombs per kilogram, C kg –1 .
(Exposure rate is just the exposure per unit time
in C kg –1 s-1 or C kg –1 h-1 etc)
Exposure = charge / mass of air =
V
1b
++ ions
-
Kg
Comparing samples
1a
+
+
1kg of air
Compare
2b
3
82. The Use of Nuclear Medicine in Diagnosis
Worked example
Equivalent Dose – symbol H (formerly called dose equivalent),
measured in Sv (Sievert)
A radiation worker wears a small ionisation chamber to measure
his exposure over the working week of 40 hrs. It contains
1.0×10-5 m3 of air at STP (Standard Temperature and Pressure).
There is a steady current of 1.2 × 10-14 A all week. Calculate the
exposure.
(density of air at STP = 1.3 kg m-3)
The equivalent dose is the absorbed dose modified by a quality factor Q
which takes account of the type of radiation incident.
The damage is not only dependent on the energy absorbed but on the
density of the ionisation (α is one of the worst, affecting a very small
volume).
Answer
mass , ρ m , m = ρ ×V,
=
v
volume
so mass of air =1.3×1.0×10-5 kg = 1.3×10-5kg
If 1.2×10-14 A flows for 40h,
charge = current ×time, Q=It = 1.2×10-14 A× 40×60×60 s
Q = 1.73nC
Q
1.73 × 10-9 C
∴
=
= 1.33 × 10-4 Ckg-1
1.3 × 10-5 kg
m
this is the exposure X
density =
To take account of this the absorbed dose is multiplied by a quality factor
‘Q’ (not to be confused with charge). For X, b or γ ‘Q’ is about 1, they are
called low LET (linear energy transfer) radiation. For fast neutrons, protons
or α particles (high LET), ‘Q’ is from 5 to 20. A Q of 5 means 5 times the
damage is caused. Dose equivalent is measured in Sieverts (Sv) which is also
J kg-1.
Q values
High LET
Low LET
Absorbed Dose
1
Absorbed Dose – symbol D,
measured in Gy (Gray) or J kg-1
β, X, γ rays
Absorbed dose is the energy absorbed by a certain mass of tissue
when irradiated.
Average Tissue
34 J per Coulomb
exposure
0
kg
1
2
5
10
neutrons
15
20
α particles
H= ‘Q’×D
Bone
150 J per
Coulomb
exposure
Worked example
If the absorbed dose of 0.34 Gy from the previous question was caused
by α particles which have a quality factor ‘Q’ of 20 then the equivalent
dose would be H = 20×0.34 = 6.8 Sv for average tissue and even more
for bone (20×1.5=30 Sv). If the dose was caused by γ the equivalent
dose is the same as the absorbed dose so it would be 0.34 Sv and 1.5 Sv.
0 1
2
kg
Typical Exam question
(a) Define the Gray (1)
(b) A girl of mass 40 kg receives an absorbed dose of 2.6 × 10-7 Gy
from a γ source. Assuming the whole body absorbs the energy
uniformly, estimate the total energy absorbed as a result. (2)
(c) What was the girl’s exposure? The average energy to produce 1
ion pair is 5.44 × 10-18 J. (4)
This applies to all types of radiation and absorber. The unit is called the
Gray (Gy) and 1 Gy is 1 J kg-1. Absorbed dose is usually calculated from
the exposure using a weighting factor f, which is the energy absorbed per
coulomb released. f varies with the photon energy but reaches more than
150 J C-1 for bone irradiated with radiation of energy between 10 keV and
100 keV.
Q
E
E
also f = Q so D = m
D = fX = f m
f has units J C –1 and is the energy required to split ‘1 Coulomb’s
worth’ of ions.
Answer
(a) The Gray is the energy absorbed in Joules per kg of material when
exposed to radiation.
(1) ‘energy absorbed’ is essential
A chest X-ray would give an absorbed dose of about 0.0015 Gy and a
tumour might have to absorb 60 Gy to be destroyed.
(b) If there is 2.6×10-7 Gy or Joules absorbed per kg,
for a 40 kg mass the total energy will be 40×2.6×10-7
(1)
= 1.0 × 10-5 J
(1)
Worked example
(c) NB exposure is total ion charge per kg so you must use the absorbed
dose which is also per kg NOT the total energy.
There will be 2.6 × 10-7 J per kg available to create ion pairs so, per kg,
2.6 × 10-7
energy/ energy to split one pair =
5.4 × 10-18
= 4.8 ×10 10 ion pairs will be created
(1).
Each pair produces a charge e of one sign
(1)
so the total charge produced per kg of one sign is 4.8 ×10 10 × 1.6×10-19 C
(1)
(1)
= 7.6×10-9 C kg-1.
This is the exposure.
Calculate the absorbed dose in tissue and in bone for an exposure
of 0.010 C kg-1.
The number of electrons making 1 C of charge is N = 1/e electrons. The
energy to create one pair of ions in air is about 34 electron Volts (eV) or
34e Joules
So the energy absorbed in an exposure to create 1 C of charge is
N × 34e =1/e × 34e =34 J.
An exposure of 1C kg-1 produces an absorbed dose of 34 Gy
f = 34 J C –1 for air.
The energy absorbed in air by an exposure of 0.010 C kg-1 is
34 × 0.010 = 0.34 J kg-1 or Gy. This is the absorbed dose for average
tissue.
For bone with f = 150 (more energy required to ionise the atoms) the dose
would be 150 × 0.010 = 1.5 J kg-1 or Gy
Five times greater absorbed dose D for the same exposure X.
4
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82. The Use of Nuclear Medicine in Diagnosis
Exam Workshop
This is a typical below average student’s answer to an exam
question. The comments explain what is wrong with the answer
and the examiner’s answer is given below.
(a) Gamma radiation may be detected using a scintillation counter
(comprising the scintillator crystal and a photomultiplier tube).
(i) Draw a diagram of the counter including details of the tube
(4)
(b) the number of photo electrons per second produced at the photo
cathode. (2)
The electrons are multiplied by 1×106 to make the anode current 3
initial n/t ×1×106 = final n/t
initial n/t = 3.2 × 10 –24 / 1×106
= 3.2 × 10 –30
3 ecf
2/2
bright light
(ii) Explain how a radioactive substance can be used to determine
the volume of blood in a person. (4)
A small amount of highly radioactive liquid is injected 3 into a
blood vessel in a person’s arm. The activity is measured just before
the injection is given and then 15 minutes later a same size sample
3 . The activity of the second sample
is taken from the other arm3
is measured and compared with the first. e.g. if it is 1/100 then the
volume it has dispersed into is roughly 100 times larger. 2/4
mirror
mirror
ht
lig
The activities are not measured like this. The student has forgotten
that for all readings to be on the same instrument, you have to dilute
an identical sample with a known volume of water. Two new samples
must be compared after the same time interval because of the decay.
crystal
γ
(iii)What properties should the radioactive substance have to
be used safely for this purpose?
(3)
It should only emit gamma radiation 3 and should have a
short half life.
1/3
The crystal emits a flash of light when a gamma ray hits it. 3
The photomultiplier tube increases the number of photons as they bounce
from mirror to mirror. 2
1/4
The student should give a figure for the half life. They have not made
three points, no mention has been made of attaching the radioactive
element to a suitable chemical compound to make a tracer.
The photocathode has been missed out completely. The photons
emitted from the crystal are changed to electrons which are then
multiplied. There are no mirrors but coated electrodes called dynodes.
The student has only glanced at the diagram. He has not practised it
and tested himself.
Examiner’s answers
(a) (i) see page 2 'gamma camera' .
(b) final current = F × initial current
(c) as student, sometimes a check would be made by doing a count with no
patient present.
(i) (a) n/t = I/e = 20 × 10 –9 / 1.6 ×10-19 = 1.25 × 1011 electrons per
second
(b) 1.25 ×10 11/ 1× 10 6 =1.25 ×10 5 electrons per second
(ii) Inject radioactive tracer
Add 2nd equal amount to known volume of water.
After suitable time take a blood sample.
Compare activity of sample with same volume taken from the
diluted tracer.
(iii) See first page for full list of properties.
(b) What is meant by the multiplication factor of the tube? (1)
final brightness = F × original brightness where F is the multiplication
factor 2
0/1
The student has the right idea but it is the number of electrons which
is increased. There is possibly a discretionary mark here for an error
made previously.
(c) Explain how the effects of background radiation are reduced. (1)
There is lead shielding around the sides of the camera. 3
1/1
(i) If an anode current of 20 nA is produced in a tube with a
multiplication factor of 1.0 × 106 calculate
(a) the number of electrons per second flowing in the circuit
(3)
IA = 20×10-6 A 2
Q = It =ne 3 so n/t = I e
2
n/t = 20×10-6 × 1.6 ×10-19 = 3.2 × 10-24 electrons per sec
Quantitative test
1. If there is a ‘dead’ time between responses to radiation of 0.5 μs what
is the highest count rate that can be accurately counted?
2. A source produces an exposure rate of 400 μC kg-1h-1 at a point 1.0 m
away. At what distance would the exposure rate be 25 μC kg-1 h-1?
Assume no absorption in the intervening medium,
3. A small volume of solution had an activity of 200 Bq when injected into
the bloodstream of a patient. It contained 24Na which has a half-life of
15 h. After 30 h the activity of 1 cm3 of the blood was 8 mBq. Estimate
the volume of blood in the patient.
4. Calculate the absorbed dose for an exposure of 12μCkg-1 if the f value
is 150 JC-1.
2
1/3
First they have put the value for a current of 20 μA instead of nano A
(1×10 –9) Then they have rearranged the equation wrongly, n/t = I/e.
They should have realised you would expect a LARGE number of
electrons.
Acknowledgements:
This Physics Factsheet was researched and written by Celia Bloor
The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU
Physics Factsheets may be copied free of charge by teaching staff or students, provided that
their school is a registered subscriber.
No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted,
in any other form or by any other means, without the prior permission of the publisher.
ISSN 1351-5136
Answers
1. You could just make 2 counts per microsecond so 2×106 counts per sec. is the max.
2. Using the inverse square law, I1/I2 = d22/d12 so 400/25 = d22/12
d22 = 16 so d2 = 4 m. (double the distance away the exposure goes to one quarter).
3. Total remaining activity after 2 half-lives is 50 Bq . 8×10-3 V = 50 V is about 6000 cm 3.
4. D = fX = 150 × 12 ×10-6 Gy = 1.8 × 10 –3 Gy.
5
Q1.
(a) A radioactive source gives an initial count rate of 110 counts per second.
After 10 minutes the count rate is 84 counts per second.
background radiation = 3 counts per second
(i)
Give three origins of the radiation that contributes to this background radiation.
1 ..........................................................................................................
2 ..........................................................................................................
3 ..........................................................................................................
(ii)
Calculate the decay constant of the radioactive source in s–1.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(iii)
Calculate the number of radioactive nuclei in the initial sample assuming that
the detector counts all the radiation emitted from the source.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(7)
Page 1 of 15
(b)
Discuss the dangers of exposing the human body to a source of α radiation. In particular
compare the dangers when the α source is held outside, but in contact with the body, with
those when the source is placed inside the body.
You may be awarded marks for the quality of written communication in your answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 10 marks)
Q2.
(a) Suggest two reasons why an α particle causes more ionisation than a β particle of the
same initial kinetic energy.
You may be awarded marks for the quality of written communication in your answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
Page 2 of 15
(b)
A radioactive source has an activity of 3.2 × 109 Bq and emits α particles, each with kinetic
energy of 5.2 Me V. The source is enclosed in a small aluminium container of mass
2.0 × 10–4 kg which absorbs the radiation completely.
(i)
Calculate the energy, in J, absorbed from the source each second by the aluminium
container.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
Estimate the temperature rise of the aluminium container in 1 minute, assuming no
energy is lost from the aluminium.
specific heat capacity of aluminium = 900 J kg–1 K–1
..............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(5)
(Total 7 marks)
Q3.
(a)
Calculate the radius of the
U nucleus.
r0 = 1.3 × 10–15 m
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(b)
At a distance of 30 mm from a point source of rays the corrected count rate is C.
Calculate the distance from the source at which
h the
t corrected count rate is 0.10 C,
assuming that there is no absorption.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
Page 3 of 15
(c)
The activity of a source of  particles falls to 85% of its initial value in 52 s.
Calculate the decay constant of the source.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(d)
Explain why the isotope of technetium, 99Tc m, is often chosen as a suitable source of
radiation for use in medical diagnosis.
You may be awarded additional marks to those shown in brackets for the quality of written
communication in your answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 10 marks)
Q4.
Potassium-42 decays with a half-life of 12 hours. When potassium-42 decays, it emits β–
particles and gamma rays. One freshly prepared source has an activity of 3.0 × 10 7 Bq.
(a)
To determine the radiation dose absorbed by the scientist working with the source, the
number of gamma rays photons incident on each cm2 of the body has to be known.
One in every five of the decaying nuclei produces a gamma ray photon. A scientist is
initially working 1.50 m from the fresh source with no shielding. Show that at this time
approximately 21 gamma photons per second are incident on each cm2 of the scientist’s
body.
(3)
Page 4 of 15
(b)
The scientist returns 6 hours later and works at the same distance from the source.
(i)
Calculate the new number of gamma ray photons incident per second on each cm2
of the scientist’s body.
number of gamma photons per second per cm2 = ............................
(ii)
Explain why it is not necessary to consider the beta particle emissions when
determining the radiation dose the scientist receives.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(5)
(Total 8 marks)
Q5.
(a) In a radioactivity experiment, background radiation is taken into account when taking
corrected count rate readings in a laboratory. One source of background radiation is the
rocks on which the laboratory is built. Give two other sources of background radiation.
source 1 .........................................................................................................
source 2 .........................................................................................................
(1)
(b)
A γ ray detector with a cross-sectional area of 1.5 × 10–3 m 2 when facing the source is
placed 0.18 m from the source.
A corrected count rate of 0.62 counts s–1 is recorded.
(i)
Assume the source emits γ rays uniformly in all directions.
Show that the ratio
is about 4 × 10–3.
(2)
Page 5 of 15
(ii)
The γ ray detector detects 1 in 400 of the γ photons incident on the facing surface of
the detector.
Calculate the activity of the source. State an appropriate unit.
answer = ................................... unit ...........................................
(3)
(c)
Calculate the corrected count rate when the detector is moved 0.10 m further from the
source.
answer = .......................... counts s–1
(3)
(Total 9 marks)
Q6.
A radioactive source used in a school laboratory is thought to emit α particles and γ radiation.
Describe an experiment that may be used to verify the types of radiation emitted by the source.
The experiment described should allow you to determine how the intensity of radiation varies
with distance in air or with the thickness of suitable absorbers.
Your answer should include:
•
the apparatus you would use and any safety precautions you would take
•
the measurements you would make
•
how the measurements would be used to reach a final decision about the emitted
radiation.
The quality of your written communication will be assessed in your answer.
(Total 6 marks)
Page 6 of 15
M1.
(a)
(i)
origins of background radiation:
cosmic rays
ground, rocks and buildings
air
nuclear weapons testing/nuclear accidents
nuclear power
discharge/waste from nuclear power
medical waste
any three (1) (1)
any two (1)
(ii)
(use of C = C0 e–λt gives) (84 – 3) = (110 – 3) e–λ×600 (1)
(1)
= 4.6(4) × 10–4 (s–1) (1)
(iii)
(use of
= – λN gives)
N=
(1)
= 2.3(1) × 105 (nuclei) (1)
(allow C.E. for value of λ from (ii)
7
(b)
α radiation is highly ionising, hence causes cancer/damage cells/
DNA/kill cells (1)
outside: less damage plus reason
(e.g. absorbed by dead skin some α’s directed away from body) (1)
[or reference to burning]
inside: more damage plus reason
(e.g. all α’s absorbed living tissue will absorb α radiation can reach
vital organs) (1)
3
QWC 1
[10]
M2.
(a) reasons:
α particle has much more mass/momentum than β particle
α particle has twice as much charge as a β particle
α particle travels much slower than a β particle any two (1) (1)
2
QWC 1
(b)
(i)
energy absorbed per sec (= energy released per sec)
= 3.2 × 109 × 5.2 × 106 ×1.6 ×10–19 (1)
= 2.7 ×10–3 (J) (1) (2.66 × 10–3 (J))
Page 7 of 15
(ii)
temperature rise in 1 minute
=
(for numerator) (1) (for denominator) (1)
= 0.90 K (or °C) (1)
(allow C.E. for incorrect value in (i))
5
[7]
M3.
(a)
R (= r0A1/3) = 1.3 × 10–5 × (238)1/3 (1)
= 8.0(6) × 10–15m (1)
2
(b)
(use of inverse square law e.g.
10 =
gives)
(1)
x = 0.095 m (1)
(0.0949 m)
2
(c)
(use of A = Aoexp(–λt gives) 0.85 = 1.0 exp (–λ52) (1)
(1)
= 3.1(3) × 10–3s –1 (1)
3
(d)
it only emits γ rays (1)
relevant properties of γ radiation e.g. may be detected outside
the body/weak ioniser and causes little damage (1)
it has a short enough half-life and will not remain active
in the body after use (1)
it has a long enough half-life to remain active during diagnosis (1)
the substance has a toxicity that can be tolerated by the body (1)
it may be prepared on site (1)
any three (1)(1)(1)
3
[10]
Page 8 of 15
M4.
(a)
number of gamma photons per second =
(= 6.0 × 106) (1)
area of sphere of radius 1.50m (= 4πr2 = 4π × 1.52) = 28.3m2 (1)
number of gamma photons per sec per cm2 =
(1) (= 21(.2))
3
(b)
(i)
decay constant = (
) = 1.60 × 10–5 s –1 (1)
new no. of gamma photons per sec per cm2 = 21(.2)
= 15(.0) (1)
(1)
(or 6 hours is 0.5 half-lives (1) source activity decreases to
2–0.5 of initial activity in this time (1)
new no. of gamma photons per sec per cm2 = 21(.2) × 2–6/12 (1)
= 15(.0) (1))
(ii)
any two of the following points (1)(1)
beta particle range in air is less than 1.5m
beta particle absorbed by air
beta particles lose energy in air more rapidly than gamma photons
beta particles ionise air much more than gamma photons
5
[8]
M5.
(a)
any 2 from:
the sun, cosmic rays, radon (in atmosphere), nuclear fallout (from previous
weapon testing), anyy radioactive leak (may be given by name of incident) nuclear
waste, carbon-14
1
(b)
(i)
(ratio of area of detector to surface area of sphere)
ratio =
0.0037
(0.00368)
2
Page 9 of 15
(ii)
activity = 0.62/(0.00368 × 1/400) give first mark if either factor is used.
67000
Bq accept s-1 or decay/photons/disintegrations s-1 but not
counts s-1
(67400 Bq)
3
(c)
(use of the inverse square law)
or calculating k = 0.020 from I = k/x2
0.26 counts s-1
(allow 0.24-0.26)
3
[9]
M6.
The mark scheme for this part of the question includes an overall assessment for the Quality
of Written Communication (QWC).
QWC
Descriptor
Mark range
High Level (Good to excellent)
The candidate refers to all the necessary apparatus and records the count-rate at various
distances (or thicknesses of absorber). The background is accounted for and a safety
precaution is taken. The presence of an α source is deduced from the rapid fall in the count rate
at 2 – 5 cm in air. The presence of a ɣ source is deduced from the existence of a count-rate
above background beyond 30 -50 cm in air (or a range in any absorber greater than that of beta
particles, e.g. 3 – 6 mm in Al) or from the intensity in air falling as an inverse square of distance
or from an exponential fall with the thickness of a material e.g. lead. The information should be
well organised using appropriate specialist vocabulary. There should only be one or two spelling
or grammatical errors for this mark.
If more than one source is used or a different experiment than the
question set is answered limit the mark to 4
5-6
Page 10 of 15
Intermediate Level (Modest to adequate)
The candidate refers to all the necessary apparatus and records the count-rate at different
distances (or thicknesses of absorber). A safety precaution is stated. The presence of an α
source is deduced from the rapid fall in the count rate at 2 – 5 cm in air and the ɣ source is
deduced from the existence of a count-rate beyond 30 -50 cm in air (or appropriate range in any
absorber, e.g. 3 -6 mm in Al). Some safety aspect is described. One other aspect of the
experiment is given such as the background. The grammar and spelling may have a few
shortcomings but the ideas must be clear.
To get an idea of where to place candidate look for 6 items:
1.Background which must be used in some way either for a
comparison or subtracted appropriately
2.Recording some data with a named instrument
3-4
Low Level (Poor to limited)
The candidate describes recording some results at different distances (or thicknesses of
absorber) and gives some indication of how the presence of α or ɣ may be deduced from their
range. Some attempt is made to cover another aspect of the experiment, which might be safety
or background. There may be many grammatical and spelling errors and the information may be
poorly organised.
3.Safety reference appropriate to a school setting – not lead lined
gown for example
4.Record data with more than one absorber or distances
5.α source determined from results taken
6.ɣ source determined
1-2
Page 11 of 15
The description expected in a competent answer should include a coherent selection of
the following points.
apparatus: source, lead screen, ruler, ɣ ray and α particle detector such as a Geiger Muller tube,
rate-meter or counter and stopwatch, named absorber of varying thicknesses may be used.
safety: examples include, do not have source out of storage longer than necessary, use long
tongs, use a lead screen between source and experimenter.
measurements: with no source present switch on the counter for a fixed period measured by the
stopwatch and record the number of counts or record the rate-meter reading
with the source present measure and record the distance between the source and detector (or
thickness of absorber)
then switch on the counter for a fixed period measured by the stopwatch and record the number
of counts or record the rate-meter reading
repeat the readings for different distances (or thicknesses of absorber).
from results taken
this is a harder mark to achieve
it may involve establishing an inverse square fall in intensity in air
or an exponential fall using thicknesses of lead
if a continuous distribution is not used an absorber or distance in
air that would just eliminate ɣ (30-50cm air / 3-6mm Al) must be
used with and without the source being present or compared to
background
use of measurements:
for each count find the rate by dividing by the time if a rate-meter was not used
subtract the background count-rate from each measured count-rate to obtain the corrected
count-rate
longer recording times may be used at longer distances (or thickness of absorber).
plot a graph of (corrected) count-rate against distance (or thickness of absorber) or refer to
tabulated values
plot a graph of (corrected) count-rate against reciprocal of distance squared or equivalent linear
graph to show inverse square relationship in air
analysis:
the presence of an α source is shown by a rapid fall in the (corrected) count-rate when the
source detector distance is between 2 – 5 cm in air
the presence of a ɣ source is shown if the corrected count-rate is still present when the source
detector distance is greater than 30 cm in air (or at a range beyond that of beta particles in any
other absorber, e.g. 3 mm in Al)
the presence of a ɣ source is best shown by the graph of (corrected) count-rate against
reciprocal of distance squared being a straight line through the origin
6
[6]
Page 12 of 15
E1.
Almost all candidates scored at least one mark for part (a) (i) and many scored both
available marks. Most candidates were familiar with the major sources of background radiation,
although some answers were too vague (e.g. the Earth) or suggested sources which were not
relevant to the situation (e.g. food). In many answers, marks were lost through a failure to
expand on a suggested origin for the background radiation; ‘nuclear’, ‘nuclear power’ and
‘nuclear power stations’ were not sufficiently explicit for the examiner to infer radioactive
discharge from nuclear power stations or remnants of nuclear weapon testing and accidents.
The calculations in parts (a) (ii) and (a) (iii) were generally done very well. The adjustment for the
background count was usually done correctly, although a significant number of candidates did
ignore this effect, or, surprisingly, added the background count. The main error encountered was
attempting to calculate a rate of change from the rate of decay by subtracting the two rates and
dividing by the time. Apart from this most candidates used the correct equation, corrected the
count rates and logged correctly to arrive at the required answer. The calculation in part (iii) was
almost always correct.
In part (b), almost all candidates understood that the α source would present more of a health
hazard inside the body than outside and could explain, in very general terms, why this is so.
Generally, the opportunity for extended writing in this section of the paper does not result in many
very good answers and this was certainly the case this year. Most of the explanations
demonstrated little more knowledge or understanding than an answer to the same question at
GCSE level. Few candidates scored the full 3 marks. Very few candidates made the essential
point that, on the outside, the α radiation would be absorbed by dead skin, which would result in
little more damage than perhaps some burning. Similarly vague statements about the behaviour
of alpha particles within the body gained little credit. Often answers implied that the alpha
radiation would exist for longer inside the body, somehow bouncing about inside causing
damage.
This type of question is a common feature of the nuclear instability section. Students should be
encouraged to think in more depth about the question and write answers based on their A2
knowledge and understanding, being aware of the importance of detail when showing what they
know.
E2.
Many answers in part (a) lacked sufficient detail to gain more than one or two marks.
Although candidates knew that the mass of an α particle was greater than that of a β particle,
few stated that its mass or momentum was much greater. Again, candidates knew that an α
particle had more charge than a β particle but few stated it had twice as much charge. Few
candidates realised that an α particle travels much slower than a β particle with the same kinetic
energy; some even claimed that it travelled faster because it had more momentum. In addition,
candidates often wrote at length about the relative penetrating powers of the two types of
radiation.
In part (b) (i), many candidates gave a correct calculation although some candidates made
arithmetical errors, often in the conversion of MeV to J, or in rounding the answer incorrectly. In
part (ii), the underlying principles behind the calculation were known but some candidates failed
to realise that the increase in temperature in one minute was required.
Page 13 of 15
E5.
A majority of students could not give two clear specific sources of background radiation. The
answers given in response to question part (a) were all too often of a general nature and too
vague to be worthy of a mark. For example, ‘power stations’ or ‘the air’. The answers needed to
be clearer statements like, ‘radioactive material leaked from a power station, or radon gas in the
atmosphere. As only one mark was being awarded only one detailed source gained the mark
provided the second point was in some way appropriate even if poorly stated. Part (b)(i) was a
very good discriminator. More able students realised that a comparison of areas was required to
answer the question. Part (b)(ii) was also a good discriminator. Only the top 20% of students
used the detection efficiency factor as well as the fraction of gamma rays hitting the detector to
obtain the correct answer. Most used only the 1/400 detection efficiency. Students were more
successful in choosing the correct unit. Part (c) was interesting in that students either attempted
the question successfully or they left this section blank.
E6.
On the whole most candidates knew what approach to take and attempted to explain a
suitable experiment. Weak candidates had issues over the language used to answer this
question. Often they would state that the intensity of radiation needs to be calculated rather than
a count rate needs to be recorded. Also they often stated facts instead of describing an
experiment. For example, alpha particles can be stopped by a sheet of paper is a poor substitute
for explaining what data to take and how to interpret the data to arrive at the conclusion that
alpha rays are emitted from the source. Slightly better candidates started to discuss the
background radiation but they did not always carry on to explain how this would be used in the
analysis. Safety in the experiment was usually given but a majority of candidates tended to
overstate the precautions necessary. It was common to see references to remote handling, lead
gowns, and keeping metres away from the source. Only the better candidates could adequately
determine that gamma rays were given out by the source. These either talked about count-rate
falling with the inverse square of distance or they discussed an absorber, which would have
eliminated any beta radiation but still allows some radiation to pass through. The only way to
know radiation passes through is to compare the count-rate with the background radiation. It
was this last point that many candidates missed. Overall candidates seemed to lack planning.
They often missed important considerations and bolted them on at the end. The standard of
English still leaves a lot to be desired. The writing in several cases was virtually illegible and
keywords were often misspelled. Fortunately there were candidates in contrast to this
description who performed the writing task exceedingly well.
Page 14 of 15
Resource currently unavailable.
Page 15 of 15
Probing the nucleus, Nucleosynthesis
Make notes on the following
Completed?
Comparison of Rutherford scattering and electron
diffraction, including experimental set up, results
and advantages and disadvantages
Nucleon number and the radius of the nucleus –
derivation, equation and graph
Nucelosythesis in stars
(https://www.youtube.com/user/CheddarAndJam)
Artificial transmution
Complete and mark the following
Completed?
Physics factsheet 40
Cheddar and Jam worksheets
Isaac Physics Skills Exercise K2 & L7
IOP TAP 521- 6: Rutherford experiment and atomic
structure
IOP TAP 521-7: Rutherford scattering data
IOP TAP 522- 3: Rutherford scattering: directions of
forces
IOP TAP 522- 4: Rutherford scattering: Energy and
closest approach
IOP TAP 522-6: Electrons measure the size of nuclei
AQA exam past paper questions - Evidence for
nucleus
AQA exam past paper questions - Nuclear radius
3K\VLFV )DFWVKHHW
September 2002
Number 40
Probing Matter
This Factsheet will explain:
• some of the techniques which scientists use to investigate matter;
• some of the important discoveries which have been made using the
techniques;
• the importance of this kind of investigation as an illustration of good
science practice.
Geiger and Marsden confirmed their results and realized that there was no
way that the existing model could explain the results obtained.
Before studying this Factsheet you should ensure that you are familiar
with the outline of the work of Rutherford from your GCSE course
The Nuclear Model of the Atom
Rutherford suggested a new model of the atom, which would explain the
experimental results — the “Nuclear” model of the atom.
The alpha scattering experiments
Diffuse negative
cloud
Before the scattering experiments, the accepted model of the atom was of
a homogenous “blob” of positive material with negative “bits” distributed
evenly throughout it, rather like a currant bun. Indeed, it was sometimes
called the “Currant Bun” or “Plum Pudding” model of the atom.
Rutherford realized that the large scattering angles could not be explained as
the sum of a large number of smaller scatterings, because the gold was too
thin. In order to account for the large scattering angles, the atom must
consist of a minute dense centre, in which almost all of the mass and all the
positive charge is concentrated, so that the majority of the particles went
through only slightly deviated by avoiding the central charge/mass
concentration, but some which went closer to the centres would be highly
deviated. He was able to calculate the relative size of the charge/mass
concentration — which he called the “nucleus” — from the proportion of
highly scattered particles. He concluded that the nucleus has a radius of the
order of 10-15 m, compared to the atom with radius of the order of 10-10 m.
Geiger and Marsden carried out an investigation, in which they fired alpha
particles (helium nuclei) at a very thin sheet of gold and measured the
proportion of particles which were scattered through various angles.
Evacuated
vessel
Nucleus containing most of the
mass and all of the positive charge
Ring of
detectors
Exam Hint: You may be asked to draw or complete paths like these.
They are parabolas. The closer the approach to the nucleus, the sharper
the bend.
The neutron remained undetected at this time.
Thin sheet of
gold
The results of this experiment caused the existing model of the atom to be
abandoned in favour of the nuclear model and completely altered our
understanding of matter. It ranks among the most significant investigations
in the history of science.
High energy
beam of alpha
particles
Deep inelastic scattering
•
•
Alpha particles cannot be used to probe deeper into matter. If we wish to
investigate the nature of each of the particles in the nucleus i.e. the proton
and the neutron, we must use high energy electrons. Just as the non-regular
scattering of alpha particles through the atom lead to our understanding of
separate particles making up the atom, so the non-uniform scattering of
electrons from hydrogen nuclei lead to an understanding that each of these
particles is not a “blob”, but also made up of smaller particles.
Rutherford scattering probes the structure of the atom.
It is alpha particles which are used in Rutherford Scattering
The result they expected, based on the model, was that the particles would
go through the gold, but would be deviated through small angles due to
electrostatic repulsions between the positive particles and the positive
charges in the atoms. Most of the particles did behave as expected, but to
their amazement, Geiger and Marsden found that some particles were
deviated by large angles, and some almost came back the way they had
entered.
gold nuclei
Investigations like these led to the discovery of quarks, the particles which
make up the proton and neutron.
This type of scattering is called “inelastic” because interactions between the
particles do not conserve kinetic energy.
Exam Hint: You may need to know more about Fundamental Particles.
On some specifications it is not required for the core syllabus, only as an
optional topic.
paths of alpha particles
1
3K\VLFV)DFWVKHHW
Probing Matter
•
•
•
Exam Workshop
This is a typical weak student’s answer to an exam question. The
comments explain what is wrong with the answers and how they can
be improved. The examiner’s mark scheme is given below.
deep inelastic scattering is used to find out about the make up of
protons and neutrons
high energy electrons are used
kinetic energy is not conserved
(a) The diagram shows a line of gold nuclei, such as might have been
the target for an alpha scattering experiment.
........
........
Typical Exam Question
(a) Complete the following table, which compares two types of
scattering experiment.
Add lines to the diagram to show the paths of the alpha particles.
(3)
Deep inelastic scattering
Incident particles
Target
Electrons
Gold atoms
1/3
(2)
(b) Write a short paragraph describing how the results of the
experiments changed scientific thinking.
Although the candidate has shown deviation, s/he has not shown an
understanding that the closer the line of approach to the nucleus, the
greater the deviation, nor is there any indication of a very large
deviation. At least three paths, one showing a direct approach and
a large deviation should have been shown.
(4)
Incident particles
Alpha scattering
Alpha particles
Deep inelastic scattering
Electrons
Target
Gold atoms
nucleons
(b) Explain why this experiment led to the conclusion that most of
the mass and all of the positive charge of the atom was
concentrated in a small space at the centre.
(3)
Alpha particle scattering:
Before the experiment, scientists believed that positive material was
distributed throughout the atom. The results could only be explained by the
positive charge being concentrated in a tiny volume at the centre – the
nucleus.
The deviations could not have been caused by any other arrangement.
The positively charged alpha particles were repelled by the positive
nuclei.
1/3
The candidate appreciates that the results cannot be explained by
the then accepted model, but does not really say why the nuclear
explanation is the only sensible one.
Deep inelastic scattering:
Before experiments, the proton, neutron and electron were believed to be
fundamental particles. The results showed that these particles were made
up of yet more fundamental particles.
Examiner’s Answers
(a) (as shown on the first page) Three paths, parabolas, showing hardly
any deviation for paths between the nuclei, larger deviation for closer
approach, including one of about 1700.
Questions
1. Describe the alpha scattering experiments.
(b) Positive particles are deviated by the repulsion of the positive nucleus.
Since particles were deviated through different angles, matter must be
distributed unevenly. The large deviations must be caused by regions
of dense mass and charge. Since only relatively few particles were
deviated through large angles the dense regions must be small compared
to the atom as a whole.
2. Why was it important that the vessel was evacuated?
3. (a) The nucleus of carbon-14 has a radius of 2.70 × 10-15 m.
Calculate the volume of this nucleus.
(b) The mass of the nucleus is 2.34 × 10-26 kg.
Calculate the density of the nuclear material of the atom.
(c) Compare this value with that of ordinary materials.
Answers
1. See the text.
2. It is important that the vessel was evacuated so that the alpha particles
were deviated only by the gold atoms, not by extraneous air atoms or
molecules.
Acknowledgements: This Physics Factsheet was researched and written
by Janice Jones.
The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham,
B18 6NF.
3. (a) Volume = 4/3 × 3.142 × (2.70 × 10-15)3 = 8.24 × 10-44m3.
(b) Density = mass/volume = 2.34 ×10-26/(8.24 × 10-44)
= 2.84 ×1017kg/m3.
(c) Although there is a large range of densities for ordinary materials,
this value is many orders of magnitude greater than any of them,
the highest of which is about 104kg/m3.
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permission of the publisher. ISSN 1351-5136
2
When the temperature
is high enough, the
cloud glows dimly,
forming a
________________.
KE
A large, spherical
cloud forms. The ___
decreases, and the
___ and __________
increase.
protostar
electrons
electrostatic repulsion
If the cloud doesn’t have enough
_______, it remains in this way
for the rest of its life, and is
called a __________________.
Each atom is _____________
attracted to each other, and so
is pulled together, creating a
dense region, which attracts
more and more atoms.
nuclear fusion
temperature
hydrogen burning
If the mass is high enough, in the centre of
the cloud, hydrogen atoms are stripped of
their ____________. The nuclei have
enough energy to overcome their
__________________________ and fuse
together.
PE
hydrogen
gravitationally
Brown Dwarf
mass
Missing words:
The fusing of two atoms is called
__________________. Sometimes this
is called ___________________, but
it isn’t burning as we know it.
Gas and dust, mainly made up
of ___________ atoms with
some helium atoms, are
dispersed thinly throughout
space.
Fill in the blanks of each frame using the list of
missing words given. Cut out each frame and arrange
them on your page in order, then stick them down.
The first two frames are already in the right order.
Instructions
_________________________________________
_________________________________________
_________________________________________
_________________________________________
4. Up to which element can nuclear
fusion occur? Elements heavier than
this undergo nuclear fission instead.
What happens in nuclear fission?
2. Write down the 3 nuclear equations described in the previous frame
in the space below.
_________________________
_________________________
_________________________
_________________________
_________________________
1. What is nuclear fusion?
The deuterium
nucleus fuses
with another
proton to form a
helium-3 nucleus.
Two of these
helium-3 nuclei fuse
to form a helium-4
nucleus, and two
more protons.
_________________________________
_________________________________
_________________________________
3. What happens to the mass difference
between the initial 4 protons and the helium
atom created?
2 hydrogen
nuclei/protons join
together to form a
deuterium nucleus, a
positron, and a neutrino.
Answer the questions or follow the instructions in each frame in the spaces provided.
Instructions
__________________.
is called a __________
known as a protostar, but
the star is no longer
Once _________ starts,
of the star’s life, the two are balanced.
against the pull of the ____________, and for most
outward ________________________. This acts
centre, causing the star to expand, and creating an
energy. The atoms start moving away from the star’s
surface gives surrounding atoms more ___________
The energy that travels from the core to the star’s
descriptions in each frame.
on the initial size of the star at the start of its life.
are different ________. What happens next depends
convection zone
photosphere
radiative zone
Red Dwarf
radiation
gravity
convection
fusion
main sequence star
radiation pressure
sizes
Missing Words kinetic
A medium sized main sequence star transports energy via
______________, through the ___________________.
The energy is transported further through the
_____________________ by _______________, to the
surface. When the energy reaches the surface, which is
called the ________________, it is radiated out into
space.
Stars will have collected different amounts of gas, so
by __________ alone.
its core to the surface
transports energy from
main sequence star. It
star is a low-mass, dim
A ________________
Instructions: Fill in the gaps with the words in the bottom right corner (words may be used more than once). Also, illustrate the
1. What happens to the outer
layers and the core?
2. When the temperature gets
high enough, what happens to the
helium in the core?
1. What
happens to the
fusion
process, and
why?
2. What then
happens to the
core?
1. What is the
maximum size of a
low-mass star?
2. What happens to the amount of hydrogen and
helium in the core of a main sequence star towards
the end of its life?
3. What happens to fusion?
1. What happens to
the outer layers of
the star?
2. What happens
to the total mass?
1. What happens to the
balance between the
outward radiation
pressure and the
gravitational inward pull?
2. What happens to the
radius of the star?
1. The star then cools
and shrinks. What
causes the star to stop
shrinking?
2. What is the star now
called?
1. What happens next to the
size of the star?
2. What type of star does it
become?
1. What happens to the
remaining heat of the star?
2. Once the heat has all gone,
what happens to the star?
1. As the star collapses,
what happens to the
pressure and temperature
of the core?
2. The star then expands
again. What causes this
expansion?
and are there to guide you. Cut them out and stick them on your page, and write a description to go with each image. Make sure you
answer every question with each image in your descriptions.
Instructions: In this exercise, you will describe the death of a low-mass star in your own words. The pictures are arranged in order
The diagrams are in the right order, but the descriptions to go with each one aren’t. Cut out each diagram and caption, and match the
correct description to each diagram, and finish off each description.
Instructions
more massive than the Sun. The collapse of
the star would be so great that not even
neutrons can withstand the high pressures.
The core collapses into a singularity and is so
dense that not even light can ___________
_________________________________.
This raises the temperature and
pressure sufficiently such that
______________________________
______________________________
______________________________.
When the core is mostly made up of iron
______________________________
______________________________
one revolution in seconds. They can have
enormous _______________________
______________________________
swelling into a Red Giant, it swells
into a ______________________
Neutron stars spin very rapidly, turning
______________________________
______________________________
when __________________________
We can observe the pulses of radiation
______________________________.
Pulsars are ______________________
neutrons, which are created because
supernova can form _______________
______________________________
______________________________
______________________________
______________________________
because ________________________
of time than a low mass star. This is
main sequence star for a shorter amount
A star with a high mass remains as a
______________________________
______________________________
______________________________
A neutron star is made up entirely of
called a ________________________.
into the Universe in a huge explosion
the star, and the material is flown out
The shockwave sweeps material out from
______________________________
______________________________
______________________________
______________________________
when __________________________
fusion. Elements above iron are created
the Universe are created by just nuclear
All of the naturally occurring elements in
The remaining core left over from the
_________________________________
_________________________________
_________________________________
back outwards because the iron nuclei
For a larger mass star, instead of
______________________________
______________________________
because ________________________
temporarily stops any further collapse
heavier nuclei can begin to fuse. This
growing hotter and denser so that
The core contracts due to gravity,
______________________________
______________________________
______________________________
nuclei ________________________
bounces back outwards because the iron
The collapse of the star recoils and
begins to collapse for the last time.
nuclear fusion finally stops and the star
The collapse of the star recoils and bounces
star if the star was approximately 15 times
is fusing in the core, the core contracts.
nuclei (which cannot fuse together),
A black hole is formed instead of a neutron
When the star runs out of the element it
A large, spherical
cloud forms. The GPE
decreases, and the KE
and temperature
increase.
The fusing of two atoms is called
nuclear fusion. Sometimes this is called
hydrogen burning, but it isn’t burning as
we know it.
When the temperature
is high enough, the
cloud glows dimly,
forming a protostar.
Once fusion starts, the star is
no longer known as a
protostar, but is called a
main sequence star.
Main Sequence Star
A Red Dwarf star is a lowmass, dim main sequence
star. It transports energy
from its core to the surface
by convection alone.
A medium sized main sequence star transports energy via radiation,
through the radiative zone. The energy is transported further
through the convection zone by convection, to the surface. When
the energy reaches the surface, which is called the photosphere, it is
radiated out into space.
Nuclear Fusion
1. Nuclear fusion is a process where lighter nuclei join together to form a heavier nuclei, e.g. 4 hydrogen nuclei become one helium nucleus.
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ଷ
ଷ
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2. ଵଵ ൅ ଵଵ ՜ ଶଵ ൅ ଴ଵ‡ା ൅ ɋ
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3. The mass difference is converted into energy and released.
4. Elements up to Iron can undergo nuclear fusion. In nuclear fission, heavier nuclei split into lighter nuclei.
If the mass is high enough, in the centre of the cloud,
hydrogen atoms are stripped of their electrons. The
nuclei have enough energy to overcome their
electrostatic repulsion and fuse together.
Each atom is gravitationally
attracted to each other, and so
is pulled together, creating a
dense region, which attracts
more and more atoms.
If the cloud doesn’t have enough
mass, it remains in this way for
the rest of its life, and is called a
Brown Dwarf.
Gas and dust, mainly made up
of hydrogen atoms with
some helium atoms, are
dispersed thinly throughout
space.
The Birth of a Star
Answers
The energy that travels from the core to the
star’s surface gives surrounding atoms
more kinetic energy. The atoms start
moving away from the star’s centre,
causing the star to expand, and creating an
outward radiation pressure. This acts
against the pull of the gravity, and for most
of the star’s life, the two are balanced.
Stars will have collected
different amounts of gas, so
are different sizes. What
happens next depends on the
initial size of the star at the
start of its life.
While the outer layers of the Red Giant continue to
expand, the core is still contracting so the
temperature continues to increase. The
temperature gets high enough for helium to start
fusing together, forming a heavier element, carbon.
The star ends up expanding much more than it
did before, and it becomes about a hundred
times bigger than it’s ever been in its life. It has
turned into a Red Giant.
As it can’t produce any more heat, it radiates away
the remaining heat for billions of years. Once the
heat has all gone, it sits as a cold dark mass, called
a Black Dwarf.
The star then begins to cool and shrinks until the
gravitational pull is balanced by the repulsion of the
electrons at the core. It stops shrinking and becomes
a White Dwarf, which is about half as massive as the
Sun, but only slightly bigger than the Earth.
As before, this makes the
temperature and pressure
increase. The temporary heat
creates outward pressure again
and counteracts the inward
force of gravity, pushing the
outer layers of the star
outwards.
Large amounts of matter
are ejected from the outer
layers of the Red Giant,
until only about 20% of the
star’s initial mass remains.
There is no longer anything
generating an outward pressure
to counteract the gravitational
inward pull. This makes the
outer layers of the star begin to
collapse inwards again.
No further fusion
takes place, as there is
not enough mass to
compress the carbon
further to fuse
together. The core
remains stabilised.
If the star is less than 4 times
the mass of the Sun, it is a lowmass, or small star. Towards
the end of the star’s life, it
begins to run out of hydrogen
fuel to fuse together, and the
core is mostly made of helium.
Nuclear fusion in the core
therefore temporarily stops.
Death of a Low Mass Star
When the star runs out of the element it is fusing in
the core, the core contracts. This raises the
temperature and pressure sufficiently such that
heavier elements begin to fuse. This happens for
heavier and heavier elements. When the core is
mostly made up of iron nuclei (which cannot fuse
together), nuclear fusion finally stops and the star
begins to collapse for the last time.
All of the naturally occurring elements in the Universe
are created by just nuclear fusion. Elements above
iron are created when an explosive shockwave is
created. The shockwave travels through the star’s
outer layers, heating the material it encounters to a
high enough temperature that they begin to fuse to
form new elements.
The core contracts due to gravity, growing
hotter and denser so that heavier nuclei can
begin to fuse. This temporarily stops any further
collapse because an outward pressure is
produced when nuclear fusion takes place.
The shockwave sweeps material out
from the star, and the material is flown
out into the Universe in a huge explosion
called a supernova.
For a larger mass star, instead of swelling into a Red
Giant, it swells into a Red Supergiant – just a larger
version of a Red Giant.
The collapse of the star
recoils and bounces back
outwards because the iron
nuclei get crushed
together but the
electrostatic repulsive
force between them
overcome the
gravitational force.
A star with a high mass remains as a main sequence star for a
shorter amount of time than a low mass star.
This is because they are more massive, so the temperatures
and pressures are far greater, and the fuel gets used up much
more quickly, even though there is more of it.
Death of a High Mass Star
Neutron stars spin very
rapidly, turning one
revolution in seconds, and
can have enormous electric
and magnetic fields.
A black hole is formed instead of a neutron star if the star was approximately 15
times more massive than the Sun. The collapse of the star would be so great that not
even neutrons can withstand the high pressures. The core collapses into a
singularity and is so dense that not even light can escape its gravitational pull.
The remaining core left
over from the supernova
can either form a neutron
star, or if it’s massive
enough, can form a black
hole.
A neutron star is made up
entirely of neutrons,
which are created
because of the extremely
high pressure of the
remaining core. Electrons
are forced to combine
with protons, forming the
neutrons.
Pulsars are neutron stars that
pulse with electromagnetic
radiation.We can observe the
pulses of radiation when the
magnetic pole crosses our line of
sight.
TAP 521- 6: Rutherford experiment and atomic structure
1.
Describe briefly the two conflicting theories of the structure of the atom.
2.
Why was the nuclear model of Rutherford accepted as correct?
3.
What would have happened if neutrons had been used in Rutherford’s experiment?
Explain your answer.
4.
What would have happened if aluminium had been used instead of gold in the alpha
scattering experiment? Explain your answer.
5.
What three properties of the nucleus can be deduced from the Rutherford scattering
experiment? Explain your answer.
Practical advice
These questions are to help your students to think about the Rutherford ideas.
Answers and worked solutions
1
The English scientist Thomson suggested that the atom, which is a neutral particle,
was made of positive charge with ‘lumps’ of negative charge inset in it - rather like the
plums in a pudding. For this reason it was known as the Plum Pudding theory of the
atom.
Rutherford explained it this way. He knew that the alpha particles carried a positive
charge so he said that the positive charge of the atom was concentrated in one place
that he called the nucleus, and that the negatively charged particles, the electrons,
were in orbit around the nucleus. Most of the mass was in the nucleus
2
Rutherford’s prediction using the idea of Coulomb law repulsion was verified by
experiment. It also enables experimental values of nuclear charge to be obtained, ie
atomic number.
3
They would not have been repelled so it is unlikely that any would ‘bounce back’.
Some could be absorbed by the nucleus.
4
The charge on the nucleus is much smaller so deflection would be smaller.
See the equation
TAP 521-7: Rutherford scattering data
5
Small, massive and positive.
External reference
This activity is taken from Resourceful Physics
TAP 521-7: Rutherford scattering data
The idea of scattering using Coulomb’s law and a small central positive charge for the atom was
communicated to the Manchester Literary and Philosophical Society in February 1911. His ideas
require that “the scattering due to a single atomic encounter is small” and that “it be supposed
that the diameter of the sphere of positive electricity is minute compared with the diameter and
sphere of influence of the atom”.
The table below shows some of Geiger and Marsden’s results
Counting was carried out for the same time at each angle
deflected angle I number scattered
degrees
15.0
132,000
22.5
27,300
30.0
7,800
37.5
3,300
45.0
1,457
60.0
477
75.0
211
105
70
120
52
135
43
150
33
The actual formula
Number of D particles y falling on unit area deflected by angle Iis given by: -
y
ntb 2 Q cos ec 4 (I / 2)
16r 2
,
where Q is the total number of particles falling on the scattering material, t is the thickness of the
material, n the number of atoms within unit volume of the material, and b given by the formula
below. N is the number of positive charges, e the size of the positive charge, m the mass of an D
particle, u their velocity and E the charge of the D particle.
b
2 NeE
mu 2
1
One of Rutherford’s conclusions was that the number of scintillations per unit area of zinc
sulphide screen is proportional to cos ec 4 (I / 2)
Maths note cosec (I 1/sin (I
What to do
Add extra columns to the table as needed to enable you to draw a graph to test Rutherford’s
conclusion that the number of scintillations per unit area of zinc sulphide screen is proportional to
cos ec 4 (I / 2)
4
As an extension you might like to plot number scattered against 1/I
Write down you conclusions from the graph(s)
2
Practical advice
Some students might like to see Rutherford’s equation and try a test to see how the results come
out. This activity is considered optional. Some websites with papers of the time are given below
for interest
Alternative approaches
A spreadsheet could be used for this activity.
You should find:
x
To a reasonable degree y proportional to cos ec 4 (I / 2)
x
number scattered against 1/I is only proportional at small angles.
4
External references
This activity is based on “The Scattering of DandE Particles By Matter and the Structure of the
Atom. By Professor E RUTHERFORD F.R.S., University of Manchester.“ from which the equation
is quoted and the section in quotation marks at the top of the page.
An abstract of the paper is at: http://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/Rutherford-atom-abstract.html
see also:
Philosophical Magazine, Series 6, Volume 27 March 1914, p. 488 - 498
http://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/Rutherford-1914.html
The paper can also be found in Foundations of Nuclear Physics, Beyer, Robert T (Ed), New York
1949 Dover Publications Inc. pp 111-130. The book also contains papers by Chadwick,
Lawrence, Cockcroft, Gamow and Yukawa amongst others.
Of interest might also be:
On a Diffuse Reflection of the D-Particles, Proc. Roy. Soc. 1909 A vol. 82, p. 495-500 By H.
GEIGER, Ph.D., John Harling Fellow, and E. MARSDEN, Hatfield Scholar, University of
Manchester
http://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/GM-1909.html
and
The Scattering of the D-Particles by Matter by H. GEIGER, Ph.D. Proceedings of the Royal
Society vol. A83, p. 492-504
http://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/Geiger-1910.html
3
TAP 522- 3: Rutherford scattering: directions of forces
Scattering of alpha particles
Rutherford did not have a particle accelerator. Instead he used alpha particles, typically of
energy 5 MeV, from radioactive decay. These questions are about the force of the nucleus on
the alpha particle.
An alpha particle has charge + 2 e, where e is the elementary unit of charge. A nucleus has
charge + Ze, where Z is the number of protons in the nucleus (and the number of electrons in
the atom).
Directions of forces
Path of alpha particle scattered by nucleus
C
B
A
nucleus
The diagram shows an alpha particle approaching a massive nucleus from A. Assume that
the nucleus recoils negligibly as the alpha particle is scattered.
1.
Add to the diagram an arrow showing the direction of the force on the alpha particle
when it is at point A, approaching the nucleus. Label the arrow FA.
2.
Add to the diagram an arrow showing the direction of the force on the alpha particle
when it is at point B, at its closest to the nucleus. Label the arrow FB.
3.
What is the ratio of the magnitudes of the two forces, FA / FB given the distances
shown in the diagram?
4.
Add to the diagram an arrow showing the direction of the force on the nucleus when
the alpha particle is at point B. Label the arrow FN. How does this force compare with
the force FA on the alpha particle at the same instant?
5.
At which point, A, B or C, is the alpha particle travelling slowest?
6.
At which point, A, B or C, is the alpha particle travelling fastest?
7.
The nucleus does in fact recoil a little. Add an arrow labelled ‘recoil’ to show the
direction of recoil you expect as a result of the passage of an alpha particle along the
whole path shown.
Uphill–downhill
The electrical potential gradient around the nucleus can be thought of as like the slope of a
hill. Imagine that you are riding on the alpha particle as it goes by the nucleus. Are you riding
uphill, downhill or momentarily along a contour of the hill:
8.
At A?
9.
At B?
10.
At C?
11.
Is the electric potential at B larger or smaller than the electric potential at A? By what
factor?
Practical advice
These are intended as simple ‘start-up’ questions for the discussion of alpha scattering. They
are mainly qualitative, asking students to think about the direction of forces, and the nature of
changes of kinetic and potential energy. The ‘hill’ metaphor for potential is exploited.
Alternative approaches
Alpha particle orbits generated by a computer simulation could be studied is:
A Useful web site that includes a simulation of alpha particle scattering
http://www-outreach.phy.cam.ac.uk/camphy/nucleus/nucleus_index.htm
The following sections are relevant to this topic
Plum pudding atoms, D scattering, Geiger & Marsden, Shells off tissue paper, Nucleus
Social and human context
The mathematical tools developed by the French mathematicians in the 1700s to deal with
planetary orbits in the solar system were just as useful for alpha particle orbits under a
repulsive force.
Answers and worked solutions
1.
The repulsive force is along the line joining the alpha particle and the nucleus.
Path of alpha particle scattered by nucleus
C
B
FA
A
2.
nucleus
The repulsive force is along the line joining the alpha particle and the nucleus.
Path of alpha particle scattered by nucleus
C
FB
B
FA
nucleus
A
3.
The force at B is four times as large as the force at A, because the distance is halved
2
and the force varies as 1/r .
4.
The force on the nucleus is equal and opposite to the force on the alpha particle. But
because the nucleus is much more massive, it recoils only slightly.
Path of alpha particle scattered by nucleus
C
FB
B
FA
A
nucleus
FN
5.
Particle moves slowest at B, because this is the distance of closest approach, the
particle has been decelerating due to repulsive force. After B it accelerates away.
6.
Here the particle is furthest from the nucleus, the alpha particle has been accelerated
between B and C and is therefore going fastest at C.
7.
The alpha particle path is symmetrical about the line from the nucleus to B. So the net
change of momentum of the nucleus is along this direction.
Path of alpha particle scattered by nucleus
C
FB
B
FA
A
nucleus
recoil
8.
Uphill, because the particle is approaching the nucleus but being pushed away from
it.
9.
Along contour, because the particle is travelling at right angles to the direction of the
force on it.
10.
Downhill, because the particle is travelling away from the nucleus and is being
pushed away from it.
11.
Potential at B larger than potential at A by a factor of 2, because the distance is
halved and the potential varies as 1/r.
External reference
This activity is taken from Advancing Physics chapter 17, 80S
TAP 522- 4: Rutherford scattering: Energy and closest
approach
Scattering of alpha particles
Rutherford did not have a particle accelerator. Instead he used alpha particles, typically of
energy 5 MeV, from radioactive decay. These questions are about how close an alpha
particle can get to different nuclei.
–19
An alpha particle has charge 2e, where e = 1.60 u 10 C. A nucleus has charge Ze, where Z
is the number of protons in the nucleus (and the number of electrons in the atom). The
electrical potential energy of the two charges at a distance r is:
electricalpotentialenergy =
where H0 = 8.85 u 10
–12
2
2e u Ze
4SH0 r
C J
–1
–1
m .
–19
The electrical potential energy in electron volts is obtained by dividing by 1.60 u 10
–1
J eV
Calculating the potential energy
1.
Show that the units of energy from the expression
electricalpotentialenergy =
2e u Ze
4SH0 r
are joules.
2.
Show that the energy in MeV is given by
2Ze
u 106.
4SH 0r
Alpha scattering by gold
This graph shows the energy in MeV of an alpha particle at distances r from a gold nucleus,
Z = 79.
Approach of alpha particle to nucleus
Z = 79 (gold)
25
20
15
10
5
0
0
2
4
6
distance from nucleus /
8
10–14
10
m
–14
3.
Make an arithmetical check to show that at distance r = 1.0 u 10 m the electrical
potential energy is between 20 MeV and 25 MeV, as shown by the graph.
4.
How does the electrical potential energy change if the distance r is doubled?
5.
From the graph, at what distance r will an alpha particle with initial kinetic energy 5
MeV colliding head-on with the nucleus, come to rest momentarily?
6.
From the graph, at what distance r will a 5 MeV alpha particle have lost half its initial
kinetic energy?
7.
From the graph, what energy would an alpha particle need to approach as close as
–14
2.0 u10 m in a head-on collision?
Alpha scattering by tin
The next graph shows, on the same scale as before, the potential energy of an alpha particle
near a nucleus of tin, Z = 50.
Approach of alpha particle to nucleus
Z = 50 (tin)
25
20
15
10
5
0
0
2
4
6
8
10
distance from nucleus / 10–14 m
8.
Why are the values of the potential energy smaller at the same values of r in this
second graph?
9.
At r = 5.0 u 10 m the electrical potential energies of an alpha particle are 4.55 MeV
for gold, Z = 79 and 2.88 MeV for tin, Z = 50. Explain the ratio, 1.58, of the two
energies.
10.
Approximately how close can a 5 MeV alpha particle get to a tin nucleus, in a headon collision?
–14
Alpha scattering by aluminium
The next graph shows the potential energy of an alpha particle close to an aluminium
nucleus,
Z = 13.
Approach of alpha particle to nucleus
Z = 13 (aluminium)
25
20
15
10
5
0
0
2
4
6
distance from nucleus /
11.
8
10–14
10
m
From the graph, how close could a 5 MeV alpha particle get to a nucleus of charge?
Z = 13, in a head-on collision?
–15
The radius of an aluminium nucleus is approximately 3 u 10 m. Does a 5 MeV
alpha particle get close to the nucleus, compared with the dimensions of the nucleus
itself? Could the pattern of scattering be affected?
12.
Heavy ion colliders
Recently, to investigate very high-energy collisions, accelerators have been used to make
head-on collisions between lead nuclei travelling in opposite directions.
13.
How much kinetic energy is needed to get two lead nuclei, Z = 82,
–14
within 1.0 u 10 m of one another? (Assume that electrical forces are the only
forces operating.)
Hints
1.
Treat units like algebraic quantities in the expression for potential energy.
2.
Remember that the charge e coulomb is also the conversion joule per electron volt.
3.
Substitute values in the equation for potential energy.
4.
Remember 1/r.
5.
Read approximately from the graph.
6.
Read approximately from the graph.
7.
Read approximately from the graph.
8.
Look at the equation for electrical potential energy.
9.
Try the ratio 79/50.
10.
Read approximately from the graph.
11.
Read approximately from the graph.
12.
Remember that 10
13.
Substitute in the expression for electrical potential energy. Or start with the answer to
question 3.
–15
–14
is 1/10 of 10
.
Practical advice
The questions focus on the distance of closest approach of an alpha particle to a nucleus.
The approach is through the shape of the 1/r curve of electric potential energy, and the way
the curve varies with radius and charge.
Alternative approaches
Students could explore the electric potential energy close to a nucleus, using a spreadsheet.
Social and human context
With hindsight, Rutherford was very clever to have managed without a particle accelerator.
But how could he have raised the money to build one without knowing what he would find?
Answers and worked solutions
1.
The units are:
CuC
2
C J
2.
1
m 1 u m
= J.
In the expression
electricalpotentialenergy =
2e u Ze
4SH0 r
dividing by e gives
2Ze
4SH 0r
for the energy in eV. Multiply by 10
3.
–6
to get the energy in MeV.
Substituting values gives
EP =
2 u 79 u1.6 u10 19 C
4S u 8.85 u10
12
2
C J
1
1
m u1.0 u10
14
m
u10 6
22.7 MeV.
4.
Halves, because the potential energy is proportional to 1/r.
5.
About 4.6 u 10
6.
Just less than 10.0 u 10
7.
About 12 MeV.
8.
The charge on the nucleus is smaller, so the potential energy is smaller in the same
ratio.
9.
The ratio of the charges, 79 / 50 = 1.58.
10.
About 3 u 10
m.
11.
About 0.75 u 10
–14
–14
m, where the graph reaches 5 MeV.
–14
–14
m.
m.
12.
The alpha particle approaches to about 2.5 times the radius of the nucleus. Attractive
forces between nucleons might begin to be important, and modify the scattering.
13.
Inserting values:
EP =
82 u 82 u1.6 u10 19 C
4S u 8.85 u10
12
2
C J
1
1
m u1.0 u10
14
m
u10 6
External reference
This activity is taken from Advancing Physics chapter 17, 70S
967 MeV.
TAP 522-6: Electrons measure the size of nuclei
Scattering by small particles
Hold a glass plate smeared with a little milk, or dusted with lycopodium powder, in front of a point
source of light and you may see rings of light round the source. The photons are diffracted by
globules of fat in the milk or by the lycopodium spores.
Similarly to diffraction by a small hole of diameter d, there is a first minimum intensity at an angle
T of order of magnitude given by sin T= O / d. (For circular objects or apertures a more exact
expression is sin T = 1.22 O / d.)
Angles and wavelengths
1.
Show that if you see a first dark ring at T = 30q, the circular objects have diameter
approximately twice the wavelength.
2.
Use the expression sin T = 1.22 O / d to find the angle of the first dark ring for particles
four wavelengths in diameter.
Wavelengths for electrons
The de Broglie wavelength O of an electron of momentum p is given by O = h / p, where h is the
–34
–1
Planck constant, 6.6 u 10 J Hz . Since the rest energy of an electron is 0.5 MeV, at energies
of hundreds of MeV, the rest energy can be ignored as part of the total energy E. In this case the
momentum p is given to a good approximation by p = E / c.
3.
Calculate the energy in joules of an electron with energy 100 MeV
(take 1 eV = 1.6 u10
–19
J).
4.
Use the value of the energy from question 3 to calculate the momentum of the electron.
5.
Use the value of the momentum from question 4 to calculate the de Broglie wavelength of
100 MeV electrons.
1
–15
6.
The radius of a single proton or neutron is of the order 1.2 u 10 m. What approximately
is the ratio of the wavelength of the electrons to the diameter of a proton or neutron?
7.
Using the relations p = E / c and O = h / p show that the de Broglie wavelength is
inversely proportional to the energy E.
8.
Use the result of question 7 and the answer to question 5 to show that the de Broglie
–15
wavelength for 400 MeV electrons is about 3.0 u 10 m.
Electron scattering by nuclei
You have seen that electrons of a few hundred MeV have de Broglie wavelengths comparable to
the diameter of a nucleus. Suppose that in an experiment a beam of 400 MeV electrons is
scattered by carbon-12 nuclei. The angle T at which the scattering is first a minimum is 42q, for
which sin T = 0.67.
9.
Calculate the ratio O/ d of the de Broglie wavelength to the diameter of a carbon-12
nucleus.
10.
Use the de Broglie wavelength of 400 MeV electrons from question 8 to show that the
–15
radius of a carbon-12 nucleus is about 2.7 u 10 m.
11.
You might expect the volume occupied by the 12 nucleons of carbon-12 to be 12 times
–15
the volume occupied by one nucleon. The radius of a nucleon is about 1.2 u 10 m.
Show that the ratio of the volumes is about 12 (expect some rounding error in these
figures).
2
12.
–15
A uranium-238 nucleus has a radius of about 7.4 u 10 m. What roughly would be a
good energy of electrons to use to determine its radius by scattering?
Hints
1.
sin 30° = ½.
2.
Substitute in the expression for sin q.
3.
The conversion factor is equal to the magnitude of the charge on the electron.
4.
Use p = E/c.
5.
Use λ = h/p.
6.
Remember that the diameter = 2 x radius.
7.
Substitute p = E/c for p.
8.
Scale down the wavelength in proportion to the increase in energy.
9.
Obtain the ratio λ/d from the value of sin q.
10.
Remember the radius is half the diameter.
11.
The ratio of the volumes is the cube of the ratio of the radii.
12.
Choose a reasonable angle, say 30°.
3
Practical advice
These questions lead students through a numerical example of the measurement of nuclear
dimensions by electron scattering, using the diffraction of the electrons to obtain the scale. Many
students will require help with the powers of ten involved, and the conversion between electron
volts and energy in joules.
The questions are intended primarily as ‘learning questions’ to be gone through slowly. Some
students may profit from tackling them alone, but most will need to be taken through them, and
have the general message pointed out for them.
Social and human context
To measure these tiny distances needed a linear accelerator a few kilometres in length. Large
energies to measure small dimensions are expensive.
Answers and worked solutions
1.
sin 30° = ½ = λ/d approximately.
2.
sin T
1.22
O
d
1.22
4
0.305.
The angle whose sin is 0.305 is 17.8°.
3.
Energy
100 MeV
10 8 eV
10 8 eV u (1.6 u 10 19 J eV 1 )
1.6 u 10 11 J.
4.
p
E
c
1.6 u 10 11 J
3.0 u 10 8 m s 1
0.53 u 10 19 kg m s 1.
4
5.
O
h
p
6.6 u 10 34 J s
0.53 u 10 19 J m 1 s
12 u 10 15 m.
6.
Ratio of wavelength to radius = 12 x 10
–15
m / 1.2 x 10
–15
m = 10;
ratio of wavelength to diameter = 5.
7.
Substitute
p
E
c
in
O
h
p
gives
O
8.
hc
.
E
From
O
hc
E
λ is inversely proportional to E. Since λ for 100 MeV is 12 x 10
–15
3 x 10 m.
–15
9.
Since sin T = 0.67, then 0.67 = 1.22 λ / d, and λ / d = 0.55.
10.
Since λ for 400 MeV is 3 x 10
–15
2.7 x 10 m.
11.
(2.7/1.2) = 11.4. It would be closer to 12 but for rounding errors.
12.
Diameter of the uranium-238 nucleus = 15 x 10
–15
m then λ for 400 MeV is
m, and λ / d = 0.55, then d = 5.5 x 10
–15
m and r =
3
–15
–15
m. Choosing λ
λ = 30 x 10 m, about five times the wavelength of 400 MeV electrons (3 x 10
scale down energy by a factor 5, to say 80 MeV.
5
–15
m). So
External reference
This activity is taken from Advancing Physics chapter 17, 90S
6
Q1.
In an experiment to investigate the structure of the atom, α particles are directed normally at
a thin metal foil which causes them to be scattered.
(a)
(i)
In which direction will the number of α particles per second be a maximum?
.............................................................................................................
(ii)
State what this result suggests about the structure of the atoms in the metal.
.............................................................................................................
.............................................................................................................
(2)
(b)
A small number of α particles are scattered through 180°.
Explain what this suggests about the structure of the atoms in the metal.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(c)
The figure shows the path of an α particle passing near a nucleus.
(i)
Name the force that is responsible for the deflection of the α particle.
.............................................................................................................
(ii)
Draw an arrow on the diagram in the direction of the force on the α particle in the
position where the force is a maximum.
Page 1 of 5
(iii)
The nucleus is replaced with one which has a larger mass number and a smaller
proton number.
Draw on the diagram the path of an α particle that starts with the same velocity and
position as that of the α particle drawn.
(4)
(Total 8 marks)
Page 2 of 5
M1.
(a)
(ii)
(i)
straight on or deflection of zero degrees (1)
the atom consists mainly of open space
[or volume of nucleus is (very much) smaller than volume
of the atom] (1)
2
(b)
most of the mass of an atom is contained in its nucleus
[or the mass of the nucleus is greater than the mass of the α particle] (1)
the nucleus contains a positive charge (1)
the charge is concentrated at the nucleus (1)
max 2
(c)
(i)
electrostatic (force)
[or electromagnetic or coulomb] (1)
(ii)
arrow pointing away from the nucleus
at the closest distance to the nucleus (1)
(iii)
path showing less deflection at all times
4
[8]
Page 3 of 5
E1.
Part (a) was answered well by most candidates but in part (b) only the better ones produced
good answers. These candidates appreciated that in order for backscattering to occur, the á
particle must collide with a particle of greater mass. The fact that repulsion came from having
both a positive nucleus and a positive á particle was widely known.
Part (c) proved to be a good discriminator but it was common to see the force, which was
supposed to act on the á particle, appearing to act on the nucleus.
Page 4 of 5
Resource currently unavailable.
Page 5 of 5
Q1.
The high energy electron diffraction apparatus represented in Figure 1 can be used
to determine nuclear radii. The intensity of the electron beam received by the detector is
measured at various diffraction angles, θ.
Figure 1
(a)
Sketch on the axes below a graph of the results expected from such an electron diffraction
experiment.
(2)
Page 1 of 12
(b)
(i)
Use the data in the table to plot a straight line graph that confirms the relationship
element
radius of
nucleus,
nucleon
number; A
R 10–15m
(ii)
lead
6.66
208
tin
5.49
120
iron
4.35
56
silicon
3.43
28
carbon
2.66
12
Estimate the value of r0 from the graph.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(5)
(c)
Discuss the merits of using high energy electrons to determine nuclear radii rather than
using α particles.
You may be awarded marks for the quality of written communication in your answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 10 marks)
Page 2 of 12
Q2.
The first artificiallyy produced
p
isotope, phosphorus
aluminium isotope,
(a)
, was formed by bombarding an
, with an α particle.
Complete the following nuclear equation by identifying the missing particle.
(1)
(b)
For the reaction to take place the α particle must come within a distance, d, from the
centre of the aluminium nucleus. Calculate d if the nuclear reaction occurs when the α
particle is given an initial kinetic energy of at least 2.18 × 10–12 J.
The electrostatic potential energy between two point charges Q1 and Q2 is equal
to
where r is the separation of the charges and ε0 is the permittivity of free space.
answer = .......................................m
(3)
(Total 4 marks)
Q3.
(a) On the figure below sketch a graph to show how the radius, R, of a nucleus varies with
its nucleon number, A.
(1)
Page 3 of 12
(b)
(i)
The radius of a gold-197 nucleus
is 6.87 × 10–15 m.
Show that the density of this nucleus is about 2.4 × 1017 kg m–3.
(2)
(ii)
Using the data from part (b)(i) calculate the radius of an aluminium-27 nucleus,
.
answer = ...................................... m
(2)
(c)
Nuclear radii have been investigated using α particles in Rutherford scattering experiments
and by using electrons in diffraction experiments.
Make comparisons between these two methods of estimating the radius of a nucleus.
Detail of any apparatus used is not required.
For each method your answer should contain:
•
the principles on which each experiment is based including a reference to an
appropriate equation
•
an explanation of what may limit the accuracy of each method
•
a discussion of the advantages and disadvantages of each method.
Page 4 of 12
The quality of your written communication will be assessed in your answer.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
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........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(6)
(Total 11 marks)
Page 5 of 12
M1.
(a)
graph to show:
electron intensity decreasing with angle of diffraction (1)
to a non-zero first minimum (1)
2
(b)
(i)
last column of table completed correctly (1)
with either
A1/3
5.93
4.93
3.83
3.04
2.29
or
R3/(10 -45m3)
295
165
82.3
40.4
18.8
axes cover more than 50% of graph sheet (1)
all points plotted correctly using labelled axes
(i.e. x -axis A1/3, y -axis R/10–15m or x axis A, y -axis R3/10–45m 3) (1)
(ii)
gradient = r0 (1)
[or gradient = r03]
gives r0 = (1.1 ± 0.1) × 10–15m (1)
5
(c)
electrons are not subject to the strong nuclear force (1)
(so) electron scattering patterns are easier to interpret (1)
electrons give greater resolution
[or electrons are more accurate because they can get closer]
[or α particles cannot get so close to the nucleus because of
electrostatic repulsion] (1)
electrons give less recoil (1)
(high energy) electrons are easier to produce
[or electrons have a lower mass/ larger Q/m, so easier to accelerate] (1)
(in Rutherford scattering) with α particles, the closest distance
of approach, not R is measured (1)
max 3
QWC 1
[10]
Page 6 of 12
M2.
(a)
Al + α →
P+
n
1
(b)
kinetic energy lost by the α particle approaching the
nucleus is equal to the potential energy gain
2.18 × 10–12 =
r = 2.75 × 10–15 (m)
3
[4]
M3.
(a)
graph starting (steeply) near/at the origin and decreasing in gradient
1
(b)
(i)
(use of density = mass/volume)
mark for top line and mark for bottom line
(allow use of 1.66 x 10-27)
Lose mass line mark if reference is made to mass of electrons
= 2.4(2) × 1017 kg m-3
2
Page 7 of 12
(ii)
= 3.54 × 10-15 m
or
m
m
or
volume = mass/density =
= 3.54 × 10-15 m
2
(c)
The candidate’s writing should be legible and the spelling, punctuation and
grammar should be sufficiently accurate for the meaning to be clear.
The candidate’s answer will be assessed holistically. The answer will be assigned to one
of three levels according to the following criteria.
High Level (Good to excellent): 5 or 6 marks
The information conveyed by the answer is clearly organised, logical and coherent, using
appropriate specialist vocabulary correctly. The form and style of writing is appropriate to
answer the question.
The candidate makes 5 to 6 points concerning the principles of the method, the limitations
to the accuracy and the advantages and disadvantages of a particular method
Intermediate Level (Modest to adequate): 3 or 4 marks
The information conveyed by the answer may be less well organised and not fully
coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used
incorrectly. The form and style of writing is less appropriate.
The candidate makes 3 to 4 points concerning the principles of the method, the limitations
to the accuracy and the advantages and disadvantages of a particular method
Low Level (Poor to limited): 1 or 2 marks
The information conveyed by the answer is poorly organised and may not be relevant or
coherent. There is little correct use of specialist vocabulary. The form and style of writing
may be only partly appropriate.
The candidate makes 1 to 2 points concerning the principles of the method, the limitations
to the accuracy and the advantages and disadvantages of a particular method
Page 8 of 12
The explanation expected in a competent answer should include a coherent
selection of the following points concerning the physical principles involved and
their consequences.
principles
• α scattering involves coulomb or electrostatic repulsion
• electron diffraction treats the electron as a wave having a de Broglie wavelength
• some reference to an equation, for example λ = h/mv ; eV = mv2/2 ; Qq/4πεor = Eα ;
sinϴ = 0.61λ/R
• reference to first minimum for electron diffraction
accuracy
• α’s only measure the least distance of approach, not the radius
• α’s have a finite size which must be taken into account
• electrons need to have high speed/kinetic energy
• to have a small wavelength or wavelength comparable to nuclear diameter, the
wavelength determines the resolution
• the wavelength needs to be of the same order as the nuclear diameter for significant
diffraction
• requirement to have a small collision region in order to measure the scattering angle
accurately
• importance in obtaining monoenergetic beams
• cannot detect alpha particles with exactly 180° scattering
• need for a thin sample to prevent multiple scattering
advantages and disadvantages
• α-particle measurements are disturbed by the nuclear recoil
• Mark for α-particle measurements are disturbed by the SNF when coming close to
the nucleus or electrons are not subject to the strong nuclear force.
• A second mark can be given for reference to SNF if they add electrons are leptons or
alpha particles are hadrons.
• α’s are scattered only by the protons and not all the nucleons that make up the
nucleus
• visibility – the first minimum of the electron diffraction is often difficult to determine as
it superposes on other scattering events
6
[11]
Page 9 of 12
E1.
Many candidates had no real idea about the shape of the graph expected in part (a). Straight
lines through the origin - obviously an outright guess - were not uncommon. Amongst the better
efforts which were not fully correct were the single slit diffraction pattern for light, showing
several minima (instead of just one), those minima having zero intensity. A small minority failed
to see that, on Figure 1, θ was shown as the angle between the deviated electron beam and the
incident beam; these candidates drew a symmetrical graph which included the mirror image of
the shape expected.
Answers to part (b), the graphical estimate of ro, were usually the most rewarding part of this
question. Most candidates chose to plot R against A1/3, which gave better-spaced points than the
alternative plot of R3 against A. A common failing amongst those who chose the latter route was
to assume that the gradient would be ro, instead of ro3. Graphical work was generally good, with
well-chosen scales, five nicely plotted points and an acceptable straight line, but the choice of
multiples of 3 for the major scale divisions made plotting a challenging task for some. In an
occasional script, even when an acceptable graph had been drawn, an attempt would be made
to calculate ro from an individual result instead of by use of the gradient. Since the question
required the result to be found from the graph, this was not acceptable. The most serious
misconception in part (b) was, when calculating the data, to assume that A1/3 stands for (A ÷ 3).
An appreciable number of candidates made this error.
The mark scheme adopted for part (c) meant that those candidates who had any familiarity with
high energy electron scattering were easily able to reach the maximum of three marks. Most
answers were, however, very disappointing. The overall impression was that many candidates
were unfamiliar with techniques involved in determining the size of the nucleus, and that some
were relying on the knowledge of á particle scattering gained from GCSE. The main point in the
answer should have been that electrons, unlike á particles, are not subject to the strong nuclear
force. Substantial numbers knew this, but many fewer went on to indicate that it makes electron
scattering patterns easier to interpret. References to electrostatic forces were frequent, but
some candidates stated that electrons would be repelled (or á particles electrically attracted) by
the nucleus. Another common misconception was that the relatively large mass of the á particle
would readily cause it to have a short de Broglie wavelength. Overall, simplistic statements such
as “electrons are very small and can get through tiny spaces in the target” were all too prevalent.
E2.
Part (a) was very straightforward for most candidates but less than half could tackle part (b)
effectively.
Problems were seen at every stage. Some had no idea what was happening at all; some used
the wrong charge on the aluminium nucleus and used 27 × 1.6 × 10.19 C; and some even
changed the equation given in the question to the Coulomb law of force equation by introducing a
squared term for the separation.
Page 10 of 12
E3.
Only the less able students tried to draw graphs of completely the wrong shape by showing
peaks etc. in part (a). A significant minority however failed to get the mark because they drew the
graph with a horizontal asymptote. Part (b)(i) also scored well. Only the bottom 25% had
difficulty over the use of the density equation or the volume of a sphere. Not many students got
caught out by powers of 10 in the calculation but this could have been because of the ‘show that’
nature of the question. Part (b)(i) proved to be much more difficult and only the top third of the
students scored the 2 marks. Some unsuccessful attempts showed the equation for the radius
in terms of the atomic mass number but they did not know where to obtain r o from the
information supplied. Part (c) was a good discriminator and the mean mark was between 3 and
4 out of 6. Two thirds of the students supplied information about alpha particles being scattered
electrostatically. Many hinted at the idea that the least distance of approach is connected to a
measure of the radius of the nucleus. This group of students also referred to electrons behaving
as waves to explain diffraction. The bottom third of students scored poorly because they did not
add much information to what they would have covered at GCSE. It was common to see an
explanation of the scattering distribution of alpha particles and give nothing else. In this way they
almost completely ignored the wording of the question. Students had obviously been taught this
section of the specification in a vast number of different ways. To give students the greatest
benefit, no individual marking point was required for any particular score. Any of the selection of
points listed in the marking scheme were noted and taken into consideration along with the
quality of communication. As a consequence, for example, some students scored full marks
even though they did not refer to any equations. Most students lost marks by not including
enough of the points listed. They did not include many statements that were wrong apart from
one notable exception. A majority of students who gave the equation to find the least distance of
approach for an alpha particle related the initial kinetic energy of the alpha particle with the
Coulomb force expression rather than the potential energy expression.
Page 11 of 12
Resource currently unavailable.
Page 12 of 12
Decay, Dating and long term storage of waste
Make notes on the following
Completed?
Random nature of nuclear decay
Definition of half life
Definition of decay constant
Equation and graph of particle number
Activity and the Becquerel
Equation and graph of activity
Radioactive dating
Using radioactive dating to calculate the age of an
object
Limits of radioactive dating
Benefits of transmutation of radioactive waste
Complete and mark the following
Completed?
Physics factsheet 10, 22 & 36
Isaac Physics Skills Exercise J2 & J3
IOP TAP 514- 2: Half Life
IOP TAP 516-4: Two important dating techniques
IOP TAP 516-2: Radioactive decay with exponentials
IOP TAP 516-1: Decay in theory and practice
AQA exam past paper questions - Radioactive decay
January 2001
Number 10
Exponentials and Logarithms
Fig 2. Graphs of y = 2x, y = 4x and their gradients
Exponentials and logarithms are used in a number of areas of Physics,
including radioactive decay and capacitor charge and discharge.
gradient
of y = 4x
This Factsheet will explain what exponentials and logarithms are, the
rules for their manipulation and how to use these functions on a
calculator. It will also describe the use of logarithmic scales for graphs.
y=2x
Before starting on this Factsheet, you should ensure you are familiar
with the rules for manipulating indices, covered in the Factsheet 8 –
Indices, Standard Form and Orders of Magnitude.
1. The Exponential Function and its graph
Exponentials are related to powers (or indices). To get an idea what is
involved, we will look at the graph of y = 2x.
We can plot some points on this graph by putting in values of x, as in
the table below:
x
y=2x
-3
0.125
-2
0.25
-1
0.5
0
1
1
2
2
4
3
8
You will notice from fig 2 that the gradient of y = 2x is always below the
original curve, and the gradient of y = 4x is always above the original
curve, but that the gradient curves are the same sort of shape as the
originals.
4
16
This suggests that there is perhaps a number between 2 and 4 for which
the gradient curve and the original curve are exactly the same. It turns
out that there is such a number – it is called e and is equal to
2.71828....... Like , e goes on for ever without repeating.
This gives us the graph shown in fig 1.
Fig 1. Graph of y = 2x
15
15
The graph of y = ex is very similar to that of y = 2x or y = 4x (fig 3). ex is
called the exponential function
10
10
Fig 3. Graph and key properties of the exponential function
55
-2
-2
0
0
y = 4x
gradient of y=2x
y = ex
0
2
42
4
We can see some important properties of 2x from this graph:
It is always positive
For negative values of x, it approaches 0
As x gets larger, 2x gets large very rapidly.
(0,1)
You could similarly plot graphs of y = 3x, y = 10x etc. All of these will
have the same 3 properties listed above.
(1, 2.71828....)
Key properties of ex
For each of these graphs, you could find the gradient at any particular
point by drawing a tangent. You could then draw up a table showing the
gradient at a number of different x-values, and plot a graph of this. Fig 2
shows the graphs y = 2x and its gradient, and the graph y = 4x and its
gradient.
1
It is always positive
As x becomes more ve, ex approaches, but doesn’t reach, 0
As x becomes large and +ve, ex gets large very quickly
(more quickly than any power of x)
The gradient of y = ex at any point is equal to the y-value at
that point
Exponentials and Logarithms
In Physics, you will most often meet a negative exponential graph.
These are the same shape as y = e-x (this is not the same as y = -ex). This
looks like the normal exponential graph reflected in the y-axis
2. Logarithms
Logarithms are really another way of writing powers.
For example, we know 102 = 100.
We write this using logarithms as log10 100 = 2, which reads
“log to the base 10 of 100 is 2”
In applications such as radioactive decay or capacitor discharge, the
graph you need will actually be of the form y = Ae-bt, where A and b are
numbers, and t is time – so something like y = 8e-0.6t. Also, you will
usually only need the part of it for which t is positive. Fig 4 gives some
key properties of such graphs.
Similarly, if we write log4 64 = 3, this is another way of writing 43 = 64
So generally:
Fig 4. Negative exponential graphs and their properties
y
loga x = b
x = ab
(0, A)
Manipulating logarithms
There are three laws of logarithms – they are related to the laws of
indices:
y = Ae-bt
t
Key properties of negative exponential graph of the form y = Ae-bt
The “initial” value of y – where the curve crosses the y-axis – is A
As t increases, y decreases rapidly towards 0, but never reaches 0.
Any process modelled by such a graph has a constant “half-life” –
it always takes the same time for the value of y to halve. So, for
example, if y was initially equal to 8, and 5 seconds later was
equal to 4, then in another 5 seconds y will decrease to 2.
ii) e-2x e6x
a
=loga – logb
b
log (if you are dividing the numbers, subtract the logs)
log(an) = nloga
(you can “bring down the power”)
In addition, you need to know the following key facts about logarithms:
Manipulating exponentials
Since exponentials are a special case of powers, you use the same rules
as for manipulating powers. Don’t let the “e” scare you!
Example 1. Simplify: i) (2ex)2
log(ab) = log a + log b
(so if you are multiplying the numbers, you add the logs)
iii) e4x e-3x
You cannot have the log of a negative number or zero
loga a = 1 – so log10 10 = 1. (this is because 10 = 101)
log 1 = 0, for any base of the logarithm (this is because any
number to the power 0 is 1)
You cannot simplify log(x + y) or log(x – y) or anything else
with a + or – sign in the bracket.
i) (2ex)2 = 22 (ex)2 = 4e2x
ii) e-2x e6x = e-2x + 6x = e4x
iii) e4x e-3x = e4x -3x =e7x
As with indices, most problems arise from students trying to invent their
own rules! Stick to the ones above – if they don’t tell you how to
simplify it, then you probably can’t!
Exponentials on your calculator
Your calculator will have a key marked ex (you may need to get at this
by doing 2nd or INV and then lnx). NB This is NOT the key marked
EXP!
Example 2: Express in terms of simpler logarithms
2x y i) log(x2)
ii) log(xy)
iii) log iv) log(3x4)
v) log(Axn)
vi) logaa8
2
It depends on the calculator exactly how this works – to find e , you will
have to do one of:
2 ex
ex 2
Check which it is on your calculator now! You should get 7.389...
i) Since a power is involved, we use the third log law:
log(x2) = 2logx
st
ii)1 law: log(xy) = logx + logy
2x = log(2x) – logy
y iii) 2nd law: log Calculators treat exponentials like indices in terms of order of
operations. So if you do ex 2 3, it will find e2, then multiply it by 3.
= log2 + logx – log y (by 1st law)
iv) 1st law: log(3x4) = log3 + logx4 = log3 + 4logx, by 3rd law
v) 1st law: log(Axn) = logA + logxn = logA + nlogx, by 3rd law
vi) 3rd law: logaa8 = 8logaa = 8 1 = 8.
If you are trying to find the exponential of a negative number, be careful
when you put in the negative sign – you should always get a positive
answer. (Try working out e-2 = 0.135...)
Tip: Many students are tempted to say log(3x4) = 4log(3x). This doesn’t
work because only the x is to the power 4, not the 3.
2
Exponentials and Logarithms
Example 3. Express as a single logarithm
i) log2 + log5 + log 3 ii) log 20 – log 5
Tip. If you are dealing with this sort of equation and you end up trying
to simplify ln(-3ex), or anything else with a number or minus sign in
front of the e – STOP! Rearrange the equation so the esomething is
completely on its own first. You are much less likely to make mistakes
this way!
iii) 2log3 + log8 – log12
i)1st law: log2 + log5 + log3 = log(2 5 3) = log30
ii) 2nd law: log20 – log5 = log(20 5) = log4
iii)3rd law: 2log3 + log8 – log12 = log32 + log8 – log12
= log(32 8) – log12
= log72 – log12 = log(72 12)
= log 6
Finding logs on your calculator
Calculators have buttons for log10 (written log) and natural logarithms
(ln). Do not get the two confused! Calculators vary – with some you
have to press the log or ln button then the number, and with others it’s
the other way round. Check which yours is – try to find ln2 (= 0.693....)
The logarithms that are actually used are
to the base 10 (written log or lg)
to the base e (written ln, which stands for natural logarithm).
Calculators treat logs with indices in the order of operations. If you type
in ln4 2, it will work out ln4, then multiply the answer by 2.
In Physics, you will mainly be concerned with natural logarithms, since
they are helpful when you are dealing with equations that have e in
them.
3. Logarithmic scales
Sometimes graphs are required for data that cover a very wide range of
values – going, for example, from 0.1 to 10000. One example of this is
exponential decay This presents a problem, since if the largest values
are to fit on the graph, it will be very hard to plot the small ones
accurately.
Natural logarithms (lnx)
Fig 5. shows the graph of y = lnx and some of its key properties. You
may notice that the graph looks like the graph of y = ex, but reflected in
the line y = x.
Logarithmic scales are used to overcome this problem. Instead of the
actual value, the logarithm of the actual value is plotted. This can be
done on one or both axes. Logarithms to base 10 are usually used for
this purpose.
Fig 5. Graph and key properties of y = lnx
y
y = lnx
(e, 1)
(1, 0)
This helps because the logarithms will not vary so much in size as the
actual data values – for example, log 0.1 = -1 and log10000 = 4, so it
would be easy to fit both of these on one graph.
x
Example 5. Plot a suitable logarithmic graph of the following data
x
y
1
0.006
2
0.5
3
1
4
3
5
25
6
120
7
900
8
4500
9
10002
10
99870
Only the y-values vary widely, so we only use a logarithmic scale on the
y axis.
First we find the values of logy:
Key properties of lnx
x
y
logy
ln1 = 0
lne = 1
lnex = x for any value of x
lnx is not defined for x0
lnx is negative for x < 1
lnx increases as x gets larger, but only slowly
1
0.006
-2.2
2
0.5
-0.3
3
1
0
4
3
0.48
5
25
1.4
6
120
2.1
7
900
3.0
8
4500
3.7
9
10002
4.0
10
99870
5.0
Then we plot the values of logy against x.
However, on the scale, we mark the original values of y as well as (or
instead of) the values of logy.
This is easy to do when we are using log to base 10, since if logy = 5,
y = 105 etcetera:
Fig 6. Logarithmic scaled graph
We can use natural logarithms to solve equations with an unknown in
the power: (e.g. exponential decay)
y lny
106 6
Example 4. Find the value of t for which 3 = 5e-2t
Step 1. Rearrange the equation so the esomething part is on its own
Dividing by 5, we get 0.6 = e-2t
Step 2. Take the natural logarithm (ln) of both sides of the equation
– do not try to simplify at this stage
ln0.6 = ln(e-2t)
104
4
102
2
100
0
10-2 -2
Step 3. Use laws of logs to simplify
ln0.6 = -2t (by using lnex = x)
10-4 -4
Step 4. Use your calculator to find t
t = ln(0.6) -2 = 0.255 (3SF)
3
x
0
2
4
6
8
10
Exponentials and Logarithms
You may also be given logarithmic graph paper e.g. when plotting
intensity (lux) against LDR resistance ()This can either be logarithmic
on one axis (which we would have used above) or on both axes). Fig 7
shows log-log paper
Questions
1. Sketch the following graphs, showing the coordinates of any points
at which they cross the coordinate axes:
b) y = lnx
c) y = 4e-2x
a) y = ex
Fig 7 Log-log graph paper
2. Simplify the following
b) e7x e1.5x
a) e4x e –6x
1000
c) (e4x)½
d) (3e2x)3
3. a) Given that a = b8, what is logba ?
b) Express the following in terms of simpler logarithms:
100
ii) log iv) log10(100x2)
v) log1
xy z
2
iii) ln(5e2x)
vi) lne4x
c) Express each of the following as a single logarithm
i) log6 – log2 + log5
ii) 6log2 – 2log4 + log ½
iii) –2logx iv) 1 + ln2
10
4. Obtain the solution to each of the following:
b) 0.8 = e-3x
c) 100 = 15e0.5x
a) 2 = 3e4x
1
1
10
100
1000
Answers
1. a), b) see graphs in the text. c) -ve exponential graph, intercept (0, 4)
c) e2x
d) 27e6x
2. a) e-2x b) e5.5x
3. a) 8
b) i) loga + 2logb
ii) logx + logy – 2logz
iii) ln5 + 2x
v) 0
vi) 4x
iv) 2+ 2log10x
c) i) log15
ii) log2
iii) logx-2 iv) ln(2e)
4. a) –0.101 b) 0.0744
c) 3.79 d) 0.304
(all to 3 SF)
5. Either (on log-normal paper):
In the figure, the scales shown go from 1 to 1000. You could change
this, but you must always have the powers of 10 (1000, 100, 10, 1, 0.1
etc) in the same places on the graph paper – so, for example, the scale
on the x-axis could read 0.1, 1, 10, 100 instead.
Example 6. Use log-log graph paper to plot a graph of the following
data
0.09
9.8
0.3
86
3
520
12
1600
48
4600
d) 6e-0.6x – 5 = 0
5. Plot a suitable logarithmic graph for the following data.
x 200 1 500 10 800 99 700 800 000 3 460 000
y 4
7
9
12
16
18
When you use logarithmic graph paper, you plot the actual x and
y-values, but because the divisions in the paper are not of equal size, the
scale is actually logarithmic.
x
y
i) log(ab2)
20
98
9800
15
10
5
We label the x-axis with 0.01, 0.1, 1, 10, 100 and the y-axis with 1, 10,
100 and 1000.
The divisions between 0.01 and 0.1, for example, correspond to
0.02,0.03.... 0.09, so we plot the points accordingly
0
1
100
10000
1000000
100000000
or (on normal graph paper)
Fig 8. Log-log graph
y 20
10000
15
10
1000
5
0
100
00
10
22
10
1044
66
10
1088
x
10
1
0.01
0.1
1
10
Acknowledgements:
This Factsheet was researched and written by Cath Brown
Curriculum Press, Unit 305B The Big Peg, 120 Vyse Street, Birmingham B18
6NF.
Physics Factsheets may be copied free of charge by teaching staff or students,
provided that their school is a registered subscriber. They may be networked for
use within the school. No part of these Factsheets may be reproduced, stored in a
retrieval system or transmitted in any other form or by any other means without
the prior permission of the publisher. ISSN 1351-5136
100
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Number 22
Radioactivity II
Exponential decay curves have some important properties:
• For small values of t the curve falls off quickly but this slows down
as t becomes big; the gradient never = 0 and the curve never cuts the
t axis;
• There is a constant half life – for a given radioactive element, the
time taken for the activity to halve will always be the same. (So, for
example, it would take the same time for activity to decrease from
400 Bq to 200 Bq, as for activity to decrease from 200Bq to 100Bq
This Factsheet will discuss the quantitative treatment of radioactive
decay and explain how to predict which nuclei will decay and how
they will decay.
Before studying this Factsheet, you should be familiar with basic
concepts in radioactivity (covered in Factsheet 11); these include:
• the nature and properties of α, β, and γ radiation
• background radiation and how to correct for it
• half-life
It would also be helpful to be acquainted with exponentials and
logarithms (Factsheet 10).
Equation for exponential decay curves
Equations for exponential curve involve the number “e”. This is a
number between 2 and 3; like π, it cannot be written as an exact decimal
or fraction.
Decay law – a review
From earlier studies, you should recall that:
• radioactive decay is a random process;
•
N = N0 e-λt
A = Ao e-λt
the probability of a nucleus decaying is constant.
These facts lead to:
•
Where N = number of undecayed atoms at time t
No = initial number of undecayed atoms
A = activity at time t
Ao = initial activity
λ = decay constant
The number of nuclei decaying per second (the activity) is
proportional to the total number of nuclei in the sample.
(This is like saying that you’d expect the total number of sixes
obtained by rolling a lot of dice would be proportional to the
number of dice – you’d expect twice as many sixes with 200 dice
as with 100 dice)
Note that both equations have exactly the same form – which is what
would be expected from the fact that A is proportional to N.
This can be expressed as an equation:
A = λN,
Logarithms
To do calculations with exponentials, it is necessary to use natural
logarithms (written as ln on your calculator). ln is the “opposite” of e
– in the same way as ×2 and ÷2 are opposites. We can therefore use
ln to “cancel out” e: - eg ln(e3) = 3. NB: this only works if there is
nothing – like a number or a minus sign – in between the ln and the e.
A = activity
λ = decay constant ( = probability of 1 atom decaying in 1 second)
N = number of undecayed atoms.
Exponential decay
The decay equation tells us that the number of nuclei decaying is
proportional to the total number of nuclei; this may also be written as
This idea is used to solve equations with e in them. The method is:
1. Rearrange the equation to get the part with the esomething on its own
on one side of the equation.
rate of decrease of N = λN
(or for those studying A2 Maths:
(and at any time, A = λN)
dN
= −λN )
dt
2. Work out the other side of the equation, so that it is a single number
3. Take ln of both sides of the equation.
This type of equation (which will also be encountered in other areas of
Physics, notably charge decay for a capacitor) leads to an exponential
decay curve when N (number of undecayed atoms) or A (activity) is
plotted against time:
4. Rearrange to find the unknown.
Worked example: Find x, given that 2 = 5e-0.2x
1. First get the e-0.2x on its own, by dividing by 5:
2÷5 = e-0.2x
N or A
2. Work out the other side
0.4 = e-0.2x
3. Take ln of both sides:
ln 0.4 = lne-0.2x
-0.916 = -0.2x (since ln and e “cancel”)
t
4. x = -0.916/-0.2 = 4.58
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Radioactivity II
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Laws of logarithms
Example 1.
A sample of uranium-238 contains 2.50 × 1023 atoms. Given that the
half-life of uranium-238 is 1.42 × 1019 seconds, find
It can be helpful to be able to simplify logarithms. To do this, use
the following laws:
ln(ab) = lna + lnb
ln(a/b) = lna – lnb
ln(an) = nlna
ln1 = 0
(a) the decay constant, λ;
(b) the initial activity of the sample, in becquerels;
(c) the number of uranium-238 atoms remaining after 3.0 × 1018 s.
Using e and ln on your calculator
Look for ex and lnx on your calculator. They will often be on the
same button – so you’ll need to press SHIFT or 2ND to access one
of them.
(a) λ=
(b) This requires using the equation A = λN, since this is the only way
to relate activity to number of atoms:
On some calculators (including graphicals), you’ll have to press ex
or lnx first, then the number, but on others, you’ll have to put the
number in first, then press ex or lnx. Check which by trying to find:
3
e (you should get 20.0)
A = 4.9 × 10-20 × 2.50×1023 = 1.2 × 104 Bq
ln2 (you should get 0.693…)
(c) Since this asks about a number of atoms, we must use N = N0e-λt
We have No = 2.50 × 1023; t = 3 × 1018
When finding something like e-0.3 × 17, you will need to put brackets
around everything in the power – so you’d press ex then (-0.3 × 17)
(you should get the answer 0.00609…)
So N = 2.50×1023 × e −4.9×10
−20
× 3×10 18
= 2.50 × 1023 × e −0.1464
= 2.2 × 1023 atoms left
Relationship between decay constant and half-life
Using either the equation for activity or for number of undecayed atoms,
it is possible to derive a relationship between half life (T1/2) and the
decay constant (λ).
Example 2
The activity of a radioactive source was measured as 10428 Bq on one
day; at the same time on the following day its activity was 9135 Bq.
If initial activity is A0, then after time T1/2, the activity will be ½A0.
So using the equation for activity, we get:
½A0 = A0 e − λT1/2
(a) Assuming the decay product of the source is stable, determine the
half-life of the source
(b) Explain why it was necessary in part (a) to assume the decay
product was stable
Dividing both sides by A0 gives:
½ = e − λT1/2
(c) Determine the time taken, from the first day, for the activity of the
source to decay to 1000 Bq
Taking logarithms gives:
ln(½) = ln( e − λT1/2 )
(a) Since activity is referred to, we need to use A = Aoe-λt
Ao = 10428; t = 24 × 60 × 60 = 86400s
9135 = 10428e-86400λ
“Cancelling” ln and e gives:
ln (½) = −λT1/2
-0.69 = −λT1/2
0.69 = λT1/2
We need to get the part involving e on its own:
9135/10428 = e-86400λ
0.876 = e-86400λ
Hence:
T1/2 =
0.69
0.69
=
=4.88 ×10-20 (3SF)
Τ 1/2
1.42 × 10 19
Now we must take logs:
ln0.876 = lne-86400λ
-0.132 = -86400λ
λ = 0.132÷ 86400 = 1.53 × 10-6 s-1
0.69
But we want T1/2 =
= 4.5 × 105 s
λ
0.69
0.69
or λ =
λ
Τ 1/2
Exam Hint: You do not need to remember the derivation of this
formula, but you do need to know and be able to use the formula
itself.
(b) If the decay product had not been stable, it would have contributed
to the measured count rate by emitting radiation itself.
Calculations involving half life
(c) Using A = Aoe-λt:
Calculations may require you to:
• determine count-rates or time, given the half life
• determine half life from count rates
• determine numbers of atoms decaying or remaining
• determine half life from a suitable linear graph
1000 = 10428 e −1.53×10
−6
1000/10428 = e −1.53×10
−6
t
−6
t
0.0959 = e −1.53×10 t
ln0.0959 = -1.53 × 10-6 t
t = ln0.0959/(-1.53×10-6) =1.5 × 106 s
The key approach is to concentrate on λ - if you are given the half-life,
find λ first, and if you need to find the half life, first find λ, then use it to
get T1/2.
The following examples illustrate these.
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Radioactivity II
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Finding numbers of atoms from the mass of element
Exam Workshop
You may be told the mass of an element, and need to find the
number of atoms in it. To do this:
•
•
This is a typical poor student’s answer to an exam question. The
comments explain what is wrong with the answers and how they can
be improved. The examiner’s answer is given below.
First divide the mass by the nucleon number of the element
Then multiply the answer by NA − Avagadro’s number – which
is approximately 6.02 × 1023. (You do not have to remember
this number – you will be given it if it is needed)
(a) At the start of an experiment, a sample contains 30.0 μg of
isotope A, which has a half life of 50s, and 60.0μg of isotope B,
which has a half-life of 25s. After what period of time will the
sample contain equal masses of A and B?
(you may assume that the decay products of both A and B are stable) [2]
mass of A: 30
15
1/2
mass of B 60
30
159
Example 3
A sample consists of 0.236g radioactive iron (59Fe). Given that the
half-life of this isotope is 46 days, calculate:
(a) the decay constant, λ;
(b) The initial activity of the sample, in bequerels
The student has clearly understood what is required, but has
neglected to actually write down the time required – s/he should
have read over the question to ensure it was fully answered.
(a) T1/2 = 46 × 24 × 60 × 60 = 3.97 × 106s
0.69
0.69
λ=
=
= 1.7 × 10-7 s-1
Τ 1/2
3.97 × 106
(b) A student says: “If the masses of A and B are equal, their activities
will be equal”.
Explain carefully why this statement is not generally correct. [3]
Because their decay constants are different 9
(b) Number of atoms = 0.236/59 × 6.02 × 1023 = 2.408 × 1021
Initial activity = λN = 1.7 × 10-7× 2.408 × 1021=4.1 ×1014Bq
1/3
Poor exam technique – one statement cannot be worth 3 marks,
and the word “explain” in the question indicates that more is
required.
Determining half-life from a linear graph
It is much easier to determine a quantity from a straight line graph
instead of a curve, since quantities such as the gradient and intercept can
be measured accurately, and a best straight line can be drawn reliably.
(c) Calculate the activity of A two minutes after the start of the
experiment, given that its relative atomic mass is 230
(NA = 6.02 × 1023)
λ = 0.69/50 =0.0149
mass = 30e-120×0.014 9= 5.6μg9
Activity = mass × λ = 0.787 s-18
For A = Aoe-λt (or N = Noe-λt), we need to use logarithms to get the
unknown (λ) and the variable (t) out of the power:
A = Aoe-λt
[6]
3/6
The student has neglected to change mass into number of atoms
– again, poor exam technique – why has NA not been used?
Take logarithms:
lnA = ln(Aoe-λt)
Examiner’s Answers
(a) B declines to15μg in 2 half lives; A declines to 15μg in 1 half life9
So 50 seconds9
(b) Decay rate depends on number of atoms and decay constant9
Number of atoms will be different even if the mass is the same,
because atomic mass will be different.9 Decay constant will be
different as half lives are different.9
(c) λ = 0.69/50 =0.0149
mass = 30e-120×0.014 9= 5.6μg9
Number of atoms = 5.6 × 10-6/ 230 × 6.02 × 10239 = 1.5 × 10169
Activity = λN = 0.014 × 1.5 × 1016=2.1 × 1014 Bq 9
Use the laws of logarithms:
lnA = lnAo + lne-λt
Simplify:
lnA = lnAo − λt
By comparing this to the straight line equation “y = mx + c”, we obtain:
In a graph of lnA (y-axis) against t (x-axis)
The gradient is −λ
The y-intercept is lnA0
Typical Exam Question
The table below shows how the activity of a radioactive sample
reduces with time.
(a) Need to plot lnA against time.
lnA values are:
9.210, 9.202, 9.193, 9.184, 9.174, 9.164, 9.148, 9.143 99
9.22
9.21
9.20
9.19
9.18
9.17
9.16
9.15
9.14
9.13
Time / s
Activity / s-1
0
10000
2
9912
Decay equations
4
9825
6
9738
8
9647
Effect of radiaoctive decay
on the nucleus – nuclear equations.
10
9550
12
9400
14
9350
• Alpha decay removes 2 protons and 2 neutrons from the nucleus
(a) Use–the
plot the
relevant
straight-line
so data
Z, thetoproton
number,
decreases
by 2, graph
and A,that
the will
nucleon
allow
you todecreases
determine
[5]
number,
bythe
4. half-life of the sample.
(b) Use your graph to calculate the decay constant, and hence the
half-life, of the sample
[3]
0
5
axes + labels9 points9 line9
(b) λ = (9.21 – 9.14)/(14 – 0)9 = 5 × 10-39
T1/2 = 0.69/λ = 140s )
3
10
15
Radioactivity II
3K\VLFV)DFWVKHHW
Beta-minus (β−) decay – this involves a neutron in the nucleus emitting
an electron and changing to a proton, so N decreases by 1 and Z
increases by 1. On the N-Z chart, this corresponds to a move downwards
by 1 and to the right by 1.
eg. beryllium-10 decays by β− emission to give boron-10
10
Be
Which nuclei will decay?
Stability of nuclei may be shown on an N-Z chart (fig 1). This shows
the most stable radioactive isotopes as well as the stable nuclei.
Fig 1. N-Z chart
neutron
number (N)
beryllium-10
(Z = 4, N = 6)
10
B
boron-10
(Z = 5, N = 5)
140
Beta-plus (β+) decay – this involves a proton in the nucleus emitting an
positron and changing to a neutron; it only occurs in man-made nuclei.
This results in N increasing by 1 and Z decreasing by 1. On the N-Z
chart, this corresponds to a move upward by 1 and to the left by 1.
eg. sodium-22 decays by β+ emission to give neon-22
130
120
110
100
22
Ne
90
neon-22
(Z = 10, N = 12)
80
22
Na
sodium-22
(Z = 11, N = 11)
70
60
N=Z
Since α decay is the only decay mode to reduce the total number of
nucleons in the nucleus, nuclei that are simply too large to be stable will
always have α decay somewhere in their decay series, although other
decay modes may also be present. Nuclei that are to the left of the line
of stability (i.e. neutron rich) will tend to decay via β− emission, since
this will result in products closer to the line of stability. Proton-rich
small nuclei, lying to the right of the line of stability, can produce
daughter nuclei closer to the line via β+ emission; for large proton-rich
nuclei, α-emission will also bring them closer to the line, since the ratio
of neutrons to protons required for stability is smaller for lower atomic
mass.
50
40
30
20
10
0
0
10
20
30
40
50
60
70
80
90
proton number (Z)
Questions
1. Explain what is meant by the decay constant, and state the relationship
between the decay constant and half life.
Note that small stable nuclei have approximately the same number of
neutrons and protons, but larger nuclei have an increasing proportion of
neutrons. The extra neutrons moderate the effect of the electrostatic
repulsion between the protons
2. Write down the equation for activity at time t.
3. Explain how to find the half-life of a radioisotope from readings of its
activity at 10 second intervals.
4. Radon-224 has a half-life of 55 seconds.
(a) Calculate the activity of a sample of 0.4g radon–224
(b) Find the time required for there to be 0.06g radon-224 remaining
in the sample.
Unstable nuclei may:
• be too large – there are no stable nuclei with Z > 83;
• have too many protons for the number of neutrons (and so be to the
right of the line of stability);
• have too many neutrons for the number of protons (and so be to the
left of the line of stability).
5. A sample of magnesium-28 has an activity of 1.58 × 108 Bq. Ten hours
later, its activity has fallen to 1.13 × 108 Bq. Calculate the half-life of
magnesium-28.
−
6. A radioactive isotope of strontium, 90
38 Sr , decays by β emission to form
an isotope of yttrium (Y).
(a) Write an equation representing this process
Decay modes
Radioactive decay will usually produce a daughter nucleus that is closer
to the line of stability than the parent nucleus.
(b) Would you expect
Radioactive decay may involve:
90
38 Sr
to lie to the left or the right of the line of
stability on an N-Z chart? Explain your reasoning.
Answers
Answers to 1. – 3. can be found in the text
4. (a) λ = 0.69/55 = 0.013 s-1
No of radon atoms = 0.4/224 × 6.02×1023 = 1.1 × 1021
Activity = 0.013 × 1.1 × 1021 = 1.4 × 1019 Bq
(b) 0.06 = 0.4 e-0.013t ⇒ ln0.15 = -0.0126t ⇒ t = 150s
5. Working in hours: 1.13 × 108 = 1.58 × 108 × e−λ10
ln(1.13 × 108/(1.58 × 108))= -λ10 ⇒ -10λ = -0.335 ⇒
λ = 3.35 × 10-2 hours-1 ⇒ T1/2 = 0.69/λ = 20.6 hours
Alpha (α) decay – this removes 2 protons and 2 neutrons from the
nucleus, so both N and Z decrease by 2. On the N-Z chart, this
corresponds to a move of 2 downwards and 2 to the left
eg. radon-204 decays by α emission to give polonium-200
204
Rn
radon-204
(Z = 86, N =118)
6. (a)
200
Po
90
90
0
38 Sr → 39Y + −1e
(b) To the left. Since it decays by β− emission, it will be neutron-rich
compared to stable isotopes.
polonium-200
(Z = 84, N = 116)
Acknowledgements:This Factsheet was researched and written by Cath Brown. Curriculum Press, Unit 305B The Big Peg, 120 Vyse Street, Birmingham B18 6NF.Physics Factsheets may be copied free of charge by
teaching staff or students, provided that their school is a registered subscriber. They may be networked for use within the school. No part o f these Factsheets may be reproduced, stored in a retrieval system or transmitted in
any other form or by any other means without the prior permission of the publisher. ISSN 1351 -5136
4
Physics Factsheet
www.curriculumpress.co.uk
April 2002
Number 36
Comparison of capacitor discharge and
radioactive decay
num ber of undecayed nuclei
number of undecayed nuclei
It is very common that in nature quantities die away with time or decay. It
makes total sense that the more there is of something the faster it will decay.
In radioactivity the more undecayed nuclei there are the greater the probability
that one will decay we write this as A=λ N i.e. the activity is proportional
to the number of nuclei that are present. Of course as more of the nuclei decay
the rate at which they will now decay falls off. This means that the activity
or rate of decay takes the following shape:
One of the characteristics of an exponential decay is that it has a constant
half-life. This means that the quantity that is decaying falls by a constant
ratio in equal time intervals. The half-life time is the time that it takes for
the quantity to decay to half the previously chosen value – it really doesn’t
matter where you start from - the half-life will always be the same.
Worked Example: the activity of a sample of a radioactive material is
500 Bq at a certain time and has fallen to 62.5 Bq after 20 minutes.
Calculate the half-life of the radioactive material.
Answer:
as the number of undecayed nuclei falls the
rate of decay falls too so curve tapers off
in one half-life the activity 250 Bq
in two half-lives the activity 125 Bq
as the num ber of undecayed nuclei falls
the rate of decay falls too so the curve tapers off
in three half-lives the activity 62.5 Bq
three half-lives = 20 minutes
⇒ half-life = 20/3 minutes = 1200/3 seconds = 400 seconds
Exam Hint: don’t forget that in radioactivity experiments the count rates
need to be corrected for background
time
General exponential decay equation
general decay equation
X = X0e−kt
radioactivity
k = λ (decay constant)
count rate
X=R
activity
X=A
capacitor discharge
k = 1/RC (time constant)
number of nuclei
X=N
current
X=I
1
charge
X=Q
voltage
X=V
Physics Factsheet
Comparison of capacitor discharge and radioactive decay.
www.curriculumpress.co.uk
A capacitor stores electric charge. If its plates are connected by a resistor
a current flows and the capacitor discharges.
The atoms in a radioactive substance have unstable nucleii. They do not,
however, all decay together - the decay is a random process with nuclei of
atoms seperately decaying in a random order.
C
R
Initial Charge Q0
The graph of number of active nuclei (N) against time is shown below
The graph of charge (Q) against time is shown below.
charge (Q)
no. of active nuclei remaining
½ No
¼ Qo
½ Qo
One half
life T ½
¼ No
N0
Q0
One half
life T ½
time
Two half
lives
time
Two half
lives
−t/RC
The equation of the graph is N = N0e
The equation of the graph is Q = A0e
The graph of voltage against
Q
time is of the same form, because V =C . The equation of the voltage
−λt
−λt
or A = A0 e
−t/RC
graph is V = V0e
For both graphs λ is known as the decay constant and it can be shown that λ is related to the half life by the equation: λ =
Time
0
T½
2T ½
3T ½
nT½
Charge on capacitor
plates
Q0
½ Q0
¼ Q0
1
8 Q
0
Time
0
T½
2T ½
3T ½
(½)n Q0
nT½
ln2 = 0.693 ≈ 0.7
T½
T½
T½
Number of atoms
still active
N0
½ N0
¼ N0
1
8 N
0
(½)n N0
For different radioactive sources the half lives can vary enormously, from
a fraction of a second to thousands of years. α , β and γ decays all follow
the same equation.
The quantity RC is sometimes referred to as the time constant (τ)
For both cases the size of the rate of decay against time shows a very similar shaped graph to the ones above. The decay rates are given by the following
equations.
-λτ
Decay rate = − 1 Q0e
= − Q
RC
RC
For the capacitor case the rate of decay is known as the discharge current.
Decay rate = −λ N0e
-λτ
= −λΝ
For radioactivity the rate of decay is known as the activity of the source.
Discharge current
Activity of
source
λ No
Q0/RC
time
time
2
Physics Factsheet
Comparison of capacitor discharge and radioactive decay.
www.curriculumpress.co.uk
Worked Examples:
24
A 5 μF capacitor is charged fully and the potential difference between
its plates is 12V. It is then discharged throuygh a 2MΩ resistor. Find each
of the following:
(a)
(b)
(c)
(d)
the time contant;
the decay contant;
the half life for the discharge;
the initial discharge current.
(a)
Time constant
(b)
Decay constant =
1
RC
=
1
10
=
0.1 s
=
1n2
0.1
=
7s
(c)
Half life
(d)
=
=
=
The radioisotope 11Na has a half life of 15 hours. For a freshly prepared
sample of mass 1 gram, find each of the following:
(a)
the number of active atoms initially.
(b)
the number of active atoms after (i) 15h and (ii) 60h.
(c)
the initial activity of the source.
23
(Avagadro's number = 6.02 × 10 atoms per mole.)
(a)
24
1g is 1 mol of 11Na
24
So 1 × 6.02× 1023 = 2.5 × 1022 (= N0)
24
(b)
(i)
After one half-life, N = 1.25 × 1022
(ii)
60h is 4 half-lives
RC
2 × 106 × 5 × 10-6
10 s
22
2.5 × 10
= 1.56 × 1021
4
2
ln2
A0 = λ N0 =
× N0
T 1/2
22
= 0.69 × 2.5 × 10
15 × 60 × 60
N =
(c)
Initial discharge current
= Q 0 = CV0
RC
RC
= V0 (cancelling C)
R
12V
=
2MΩ
= 3.2 × 1017 Bq
Exam Hint: Initial values of quantities are usually denoted by the
subscript "0" (representing at time = 0); hence Q0 , V0 , N0 etc.
= 6 × 10-6 A or 6 μ A
Exam workshop
This is a typical student's answer to an exam question. The comments
explain what is wrong with the answers and how they can be improved.
The examiner's answer and comments are given below.
(c) the initial activity, in Bq, of the sample of radium.
ΔN
Δt =-λN A sample of radium contains 3.5 × 1021 radioactive nuclei. Radium
has a half-life of 1.6 × 103 years. Calculate:
(a) the time taken for the number of nuclei to decay to 1.6 × 1018;
[5]
N = N0 e−λt 1.6 × 1018 = 3.5 × 1021 e-1600t = 1.6 × 103 × 3.5 × 1021 = 5.6 × 1024 Bq -7.69
= 4.8 × 10-3 years ecf
-1600
3/5
Examiner’s comments and answers:
The student’s layout for each of these answers is exemplary and little needs
to be changed in order to make the answers into very good ones.
The student has only made one mistake here; s/he has not converted
the half-life into the decay constant λ ( = 0.69/T1/2). Having made this
mistake, the remainder of the answer is correct and is not further
penalised.
In part (a) calculating the decay constant using λ = 0.69/T1/2 gives a value
of 4.3 × 104 year-1.
Substituting this into the equation N = N0e−λt gives
1.6 × 1018 = 3.5 × 1021 e−43000t
Dividing by 3.5 × 1021 and taking logs to get rid of the “e” gives:
ln(1.65 × 1018/ (3.5 × 1021)) = -4.3 × 10-4 × t
t = 1.8 × 104 year
(b) the number if nuclei that will be present 1000 years from the
present;
[4]
N = N0 e−λt = N0 e−1600000 = 3.5× 1021 × e−1600000 = 3.5× 1021 × 0 = 0 1/3
Two mistakes here - the usual failure to change the half-life into the
decay constant and also the student's unwillingness to work in
seconds. Even with error carried forwards, no credit could be given for
the answer in Bq - since Bq are identical to counts per second. Had
the question not specifically sought an answer in Bq, the student might
well have gained an extra mark for using year-1 in this final answer.
1.6 × 1018
-1600t
(with error carried forward [ecf])
3.5 × 1021 = e
t=
[3]
In part (b) use of the value for ? and the equation N = N0e−λt (again) gives
a value for N of 2.3 × 1021 nuclei.
1/4
The student has made the same mistake in this part and is penalised for
it again - which may seem a bit harsh, but it is another part of the question.
S/he has then gone on to equatie the power of e to zero - which it is not!
It looks like the student's calculator is set to a rounding mode which does
not give answers as powers of ten - since the student's answer is very
small, it has rounded it to zero. Be very careful with the mode to which
you set your calculator ("SC" for scientific is probably the best).
In part (c) the decay constant must be given in s-1 to give an activity in Bq.
λ = 0.69/(1.6 × 103 × 365 × 3600) = 3.3 × 10-10 s-1
ΔN
= -λ N = -3.3 × 10-10 × 3.5 × 1021 = -1.1 × 1012 Bq.
Δt
(the minus sign simply tells us that this is a decay – the gradient of the
graph of N against t is negative).
3
Physics Factsheet
Comparison of capacitor discharge and radioactive decay.
www.curriculumpress.co.uk
Questions
Answers
1 (a) R = R0 e−λt 0.75 = 3.75 e−λt
ln (0.75/3.75) = −λ t = -0.69t/5500
t = 12800 years a)
14
1. (a) Living matter contains the 6C isotope which begins to decay when
the matter dies. The half life of the isotope is estimated to be 5500
years. A sample of wood from an ancient building has an activity.
which registers 0.75 counts per minute, on a ratemeter. If a sample
of newly cut wood of the same mass gives a reading of 3.75 counts
per minute, how old is the specimen?
(b) If a capacitor of capacitance 1F were available, find the resistance
required if the discharge half life is to be the same as that of the
isotope in part (a).
2.
(b)
T1/2 = 5500 (years) × 3600 (seconds) × 24 (hours) × 365 (days)
= 1.73 × 1011 s The longest half life for a capacitor discharge in common use is that
used in a digital clock. If a certain model of clock is disconnected
from its supply it will still show the correct time if reconnection
is made within 20 minutes, because the electronic system can still
operate only half voltage. If a capacitor and a 750k Ω resistor are
connected in parallel with the supply find the capacitance of the
capacitor. Ignore the resistance of the clock.)
C
1
0.69
=
RC
T 1/2
R = (1.73 × 1011)/ (0.69 × 1 (F)) = 2.5 × 1011 Ω 2
V = V0 e−t/RC ½ = e−t/RC
-ln(½ ) = -t/RC
ln2 = t/RC since t = 1200 s
C = 1200/ (0.69 × 750 × 103)
= 2.3 × 10-3 F (= 2.3 mF)
3.
750k Ω
Similarity
Capacitor voltage
Radioactive no. of atoms
(or current, charge) (or activity, count rate)
Exponential decay V = V0e−t/RC
with time, t
CLOCK
Initial value
at t = 0 V0
3.
Describe the similarities between the mathematical explanations
of capacitor discharge and radioactive decay.
Time constant N = N0e−λt
N0
1/RC (R = discharge λ, where
circuit resistance,
λ = decay constant
C = capacitance)
Acknowledgements:
This Physics Factsheet was researched and written by Alfred Ledsham
The Curriculum Press,Unit 305B, The Big Peg,120 Vyse Street, Birmingham, B18 6NF
Physics Factsheets may be copied free of charge by teaching staff or students, provided that
their school is a registered subscriber.
No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted,
in any other form or by any other means, without the prior permission of the publisher.
ISSN 1351-5136
4
TAP 514- 2: Half Life
The half-life of strontium-90 is 27 years. The half-life of sodium-24 is 15 hours.
1.
Sketch two curves on one set of axes to show how the number of atoms of each
change with time.
Two samples are prepared, one containing 10
20
10 atoms of sodium.
20
atoms of strontium and the other containing
2.
Which of the two samples has the highest activity?
3.
A sample of iodine-131, with half-life 8.04 days, has an activity of 7.4 x 10 becquerel.
Calculate the activity of the sample after 4 weeks?
4.
234
(a)
What fraction of a sample remains after 96.4 days?
(b)
What fraction of a sample remains after 241 days?
7
Th has a half life of 24.1 days.
Answers and worked solutions
1.
strontium
1020
sodium
0
15
30
45
60
t/h
2.
Sodium, since a longer half-life means the source is less active.
3.
First find the number of half-lives in 4 weeks:
4 u 7 days
8.04 days
3.48.
The activity will be:
7.4 u 10 7 Bq
2 3.48
6.6 u 10 6 Bq.
4
4
(a)
96.4/24.1=4 so 4 half lives so (1/2) = =1/16 or .0625
(b)
241/24.1=10 so 10 half lives (1/2)
10
= 1/1024 or 9.8 x 10
-4
External references
This activity is taken from Advancing Physics chapter 10, 20S which was an adaptation of
Revised Nuffield Advanced Physics question 15 section F.
TAP 516-1: Decay in theory and practice
Fractions and activity
1.
A
In exponential decay the number of atoms remaining after a given interval of time is
always the same fraction of the number present at the beginning of the time interval.
Therefore, someone might say, however long you wait there will always be some left.
sample of radioactive atoms will last for an infinite time.
Is this reasonable? Explain your answer.
2.
–10
The decay constant for caesium-137 is 7.3 u 10
–1
s .
Calculate the number of atoms in a sample of caesium-137 that has an activity of
5
2.0 u 10 Bq.
Experimental exponentials
No ordinary place on the Earth’s surface is free from ‘background radiation’. In making a
measurement of this background a student set up a GM tube and counter and recorded the
counter reading every 30 s.
3.
Time / s
Count
0
0
30
13
60
31
90
48
120
60
150
75
180
88
210
102
240
119
Using these data, determine the background count per minute. What would you
expect the total count to be after a further 30 seconds?
A radioactive source is set up in front of the same GM tube in the same laboratory. Counter
readings of the activity of the source are taken for 1 minute at time intervals of 1 hour.
Time / h
A / counts min
0
828
1
510
2
320
3
202
4
135
5
95
6
70
7
51
8
41
9
38
10
37
11
31
12
33
13
34
14
29
15
35
–1
4.
What action should be taken to deal with the background count?
5.
Plot a graph to show how the activity of the radioactive substance changes with time
and derive from it three different values of the half-life.
Calculate the mean of these.
6.
Worried by the obvious fluctuations in the final hours of the count, one student
suggests that it might be wise to continue counting for at least as long again.
Is this a good idea? Explain your answer.
7.
There is a slight increase in the count rate towards the end of the experiment.
Is this significant? Explain your answer.
8.
Suggest steps you might take to improve the experiment.
Practical advice
This question compares the theoretical exponential decay with the messy practical situation.
This comparison can be used to bring out some of the characteristics of the models.
Answers and worked solutions
1.
No. The smooth exponential model provides a good description only when there are a
very large number of atoms. However, when the sample is reduced to a few atoms,
the model is a poor fit. How long an atom will last on average is known, but it is not
possible to say how long one particular atom will last.
2.
dN
dt
N
O N
– dN / dt
O
activity
O
2.0 u 10 5 Bq
7.3 u 10 10 s 1
2.7 u 1014 atoms
3.
119 counts
30 counts min 1
4 minutes
After further half minute count 119 30 / 2
background count
134 counts.
–1
4.
Subtract the background count from each of the tabulated counts min .
5.
The half-life is about 80 minutes.
6.
The count rate would not change significantly if the counting were continued for a
longer period. Very little of the substance remains after 8 hours.
7.
No; random fluctuations in the background count rate are probably responsible for
this increase.
8.
The experiment could be repeated and readings taken at intervals closer than 1 hour
during the first few hours.
External reference
This activity is taken from Advancing Physics chapter 10, 30S
TAP 516-2: Radioactive decay with exponentials
1.
The half-life of one radioactive isotope of sodium is 2.6 years. Show that its decay
–9 –1
constant is 8.4 u 10 s .
2.
Calculate the activity of a sample containing one mole of the sodium. (One mole
23
contains 6.02 u 10 atoms.)
A scientist wishes to find the age of a sample of rock. Realising that it contains radioactive
potassium, which decays to give a stable form of argon, the scientist started by making the
following measurements:
decay rate of the potassium in the sample = 0.16 Bq
–6
mass of potassium in the sample = 0.6 u 10
–6
mass of argon in the sample = 4.2 u 10
g
g
3.
The molar mass of the potassium is 40 g. Show that the decay constant O for
–17 –1
9
potassium is 1.8 u 10 s and its half life is 1.2 u 10 years.
4.
Calculate the age of the rock, assuming that originally there was no argon in the
sample and the total mass has not changed. Show the steps in your calculation.
5.
Identify and explain a difficulty involved in measuring the decay rate of 0.16 Bq given
above.
6.
Iodine 124, which is used in medical diagnosis, has a half-life of 4.2 days. Estimate
the fraction remaining after 10 days.
7.
Explain how you would find the half-life of a substance when it is known to be more
than 10 000 years. Assume that a sample of the substance can be isolated.
In an experiment to find the half-life of zinc-63, a sample containing a sample of the
radioactive zinc was placed close to a GM tube and the following readings were recorded.
–1
The background count rate was 30 min .
Time / hours
Counts
/ min
–1
0
259
0.5
158
1.0
101
1.5
76
2.0
56
2.5
49
3.0
37
8.
Plot a graph of count rate against time and use this to find the average time for the
count rate to fall to one-half of its previous value.
9.
Plot a second graph, ln (count rate) against time, and use it to find the half-life.
10.
Discuss which method, 8 or 9, provides a more reliable value.
Practical advice
The questions provide a variety of kinds of practice, at post-16 standard, with radioactive
decay, half life and decay constants.
Answers and worked solutions
1.
O
ln 2
t 12
A
dN
dt
0.693
2.6 y u 3.16 u 10 s y
7
–1
8.4 u 10 9 s –1
2.
– ON
(8.4 u 10 9 s 1 ) u (6.02 u 10 23 )
5.1 u 1015 Bq
3.
O
dN / dt
N
0.16 Bq
[(0.6 u 10
t1 2
6
g)/40 g] u (6.02 u 10 23 )
1.8 u 10 17 s 1
ln 2
O
0.693
1.8 u 10 17 s 1
3.85 u 1016 s
1.2 u 10 9 years
4.
0.6 u 10 6 g
4.8 u 10 6 g
1
8
3 u 1.2 u 109 years
1
so there have been three half-lives.
23
3.6 u 109 years
5.
The background count rate is likely to be higher than 0.16 Bq.
6.
There have been
10 days
4.2 days
2.38 half-lives. The ratio remaining is
1
2
2.38
0.19.
7.
Place a GM tube near the sample and measure the count rate over a long period of
time. Determine a value for dN/dt. Determine N by chemical means, and from that
find O and T1/2.
8.
Subtract the background count rate from all readings. The half-life is about 40
minutes.
first sample half life 200 to 100 in 0.6h
+
second sample half life 160 to 80 in 0.65h
200
third sample half life 120 to 60 in 0.7h
+
average time = 0.65 h | 40 min
0.6 h
100
0.65 h
+
0.7 h
+
+
+
+
0
0
0.5
1.0
1.5
2.0
t/h
2.5
3.0
9.
Corrected
count rate /
–1
min
ln corrected
count rate /
–1
min
229
5.43
128
4.85
71
4.26
46
3.83
26
3.26
19
2.94
7
1.95
5
gradient = –O =
+
t1 =
+
4
+
2
+
3
ln 2 ln 2 × 2.0 h
=
= 0.60 h
O
2.3
= 36 min
+
2.7 – 0.7 = 2.0 h
2
–2.3
2.0 h
+
1
0
0
0.5
1.0
1.5
2.0
t/h
2.5
3.0
10.
The second method is more reliable. By drawing a straight line of best fit in question
9, all the data are being averaged rather than just the three pairs taken from the
curve in question 8.
External reference
This activity is taken from Advancing Physics chapter 10, 90S
TAP 516-4: Two important dating techniques
Radiocarbon dating
This technique was devised by American physicist W. F. Libby in 1949.
14
Carbon-14 ( C) is a naturally occurring radioactive isotope of carbon which is produced when
14
14
neutrons associated with cosmic rays collide with nitrogen ( N) in the atmosphere. The C is
14
12
taken up by living organisms (plants, animals); the ratio of C to C in living tissue remains
14
constant during the life of the organism and depends only on the relative proportions of C
12
12
and C present in the atmosphere. ( C is the `ordinary', stable, abundant isotope of carbon.)
When the organism dies, the ratio of
nitrogen:
14
12
C to C decreases because the
14
14
14
C decays to produce
12
After about 5700 years (the half-life of C), the ratio of C to C falls to half its initial value;
14
12
after a further 5700 years the ratio halves again ± and so on. The C / C ratio therefore
14
12
decreases in a known and predictable way as the sample ages. If the original ratio of C / C
is known for an organic sample, say bone, and the present ratio can be measured, then an
accurate value for the age of the sample can be calculated.
14
12
To measure the ratio of C to C, the sample needs to be ground up, thus destroyed. The
C in a sample is detected and measured via its radioactive emission, and the concentration
of carbon is measured chemically. The method works best on samples containing large
quantities of carbon. Thus wood, charcoal, bone and shells of land and sea animals are good
archaeological samples.
14
Around 100 g of wood or 30 g of charcoal is required to obtain a date. Bone contains a
smaller proportion of carbon so more bone is required than wood (around 1 kg). Around 100 g
of shell is required for dating. The precision of the age measurement decreases with age of
14
14
the sample because the amount of C decreases with time. The half-life of C is 5730 ± 40
years so the method is most reliable for dating samples no more than a few thousand years
old. For a sample about 50 000 years old the uncertainty is about ± 2000 years. The range of
radiocarbon dating is at most 100 000 years.
14
The method rests on the assumption that the sample gains no C after death, and that the
14
12
atmospheric ratio of C / C has remained constant with time. It is therefore necessary to
compare radiocarbon dates against other independent methods of dating such as
dendrochronology.
For an article on the radiocarbon method, see:
Radioactive decay. Dobson, K., Physics Review, Vol. 4, No. 2, pp. 18-21
A good World Wide Web site for details on the method and much besides is:
http://www.c14dating.com/
Dendrochronology
The name of this method comes from the Greek `dendron' (tree) and `chronos' (time).
The cross-section of a tree trunk shows a series of concentric circles known as growth rings.
Each growth ring represents one year of the life of the tree. The thickness of each ring
depends on the climatic conditions in the year that the ring was laid down. Thus the rings
show a distinctive pattern of varying thickness. The same pattern can be seen in timber of
different ages and so a pattern covering a considerable period of time can be constructed.
Once the pattern has been established and tied to known dates, a pattern in a timber of
unknown date can be matched against the established pattern and its date determined. The
first person to suggest dendrochronology as a dating method was Thomas Jefferson, one of
the first American Presidents, who suggested its use in dating Native American burial
mounds.
The method can be used with wood grown since the last Ice Age, about 10 000 years ago.
One of the longest patterns established is that using the Bristlecone Pine which grows in
California. The arid conditions of the area allow samples of the pine to survive in good
condition for thousands of years.
Apart from its use as an absolute method of dating wood, dendrochronology is also used to
calibrate radiocarbon dates. When radiocarbon dates are matched against the tree ring dates,
it
is found that radiocarbon `years' do not equate directly with calendar years because the
14
amount of C in the atmosphere varies slightly. In general, radiocarbon dating gives a young
age for older samples (e.g. a radiocarbon date of 4100 BC may be closer to a real date of
5000 BC).
The method does have its limitations. Climatic conditions vary from place to place and so a
particular pattern can only be used locally. The wood has to be in a good state of
preservation; such wood is difficult to find on archaeological sites. The wood on a site may not
indicate the true age of the site. For example, the wood could have been cut down many
years before its use on a particular site. And not all types of wood can be used since some
types have growth rings of uniform thickness regardless of climatic condition
Practical advice
This activity summarises two important dating techniques used in modern archaeology.
External reference
This activity is taken from Salters Horners Advanced Physics, AS, DIG additional activity 10
Q1.
The radioactive isotope of sodium
has a half life of 2.6 years. A particular sample of
5
this isotope has an initial activity of 5.5 × 10 Bq (disintegrations per second).
(a)
Explain what is meant by the random nature of radioactive decay.
You may be awarded marks for the quality of written communication provided in your
answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(b)
Use the axes to sketch a graph of the activity of the sample of sodium over a period of
6 years.
(2)
(c)
Calculate
(i)
the decay constant, in s–1, of
, 1 year = 3.15 × 107 s
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
Page 1 of 14
(ii)
the number of atoms of
in the sample initially,
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(iii)
the time taken, in s, for the activity of the sample to fall from 1.0 × 105 Bq to
0.75 × 105 Bq.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(6)
(Total 10 marks)
Q2.
(a) A radioactive source gives an initial count rate of 110 counts per second.
After 10 minutes the count rate is 84 counts per second.
background radiation = 3 counts per second
(i)
Give three origins of the radiation that contributes to this background radiation.
1 ..........................................................................................................
2 ..........................................................................................................
3 ..........................................................................................................
(ii)
Calculate the decay constant of the radioactive source in s–1.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
Page 2 of 14
(iii)
Calculate the number of radioactive nuclei in the initial sample assuming that
the detector counts all the radiation emitted from the source.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(7)
(b)
Discuss the dangers of exposing the human body to a source of α radiation. In particular
compare the dangers when the α source is held outside, but in contact with the body, with
those when the source is placed inside the body.
You may be awarded marks for the quality of written communication in your answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 10 marks)
Q3.
The isotope of uranium,
series of α and β– decays.
(a)
, decays into a stable isotope of lead,
, by means of a
In this series of decays, α decay occurs 8 times and β– decay occurs n times.
Calculate n.
answer = ...........................................
(1)
Page 3 of 14
(b)
(i)
Explain what is meant by the binding energy of a nucleus.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(ii)
Figure 1 shows the binding energy per nucleon for some stable nuclides.
Figure 1
Use Figure 1 to estimate the binding energy, in MeV, of the
nucleus.
answer = ................................. MeV
(1)
Page 4 of 14
(c)
The half-life of
is 4.5 × 109 years, which is much larger than all the other half-lives of
the decays in the
h series.
A rock sample when formed originally contained 3.0 × 1022 atoms of
and no
atoms.
At any given time most of the atoms are either
atoms in other forms in the decay series.
(i)
or
Sketch on Figure 2 graphs to show how the number of
with a negligible number of
atoms and the number
of
atoms in the rock sample vary over a period of 1.0 × 1010 years from its
formation.
ti
Label your graphs U and Pb.
Figure 2
(2)
(ii)
A certain time, t, after its formation the sample contained twice as many
as
atoms
atoms.
Show that the number of
atoms in the rock sample at time t was 2.0 × 1022.
(1)
Page 5 of 14
(ii)
Calculate t in years.
answer = ................................. years
(3)
(Total 10 marks)
Q4.
The carbon content of living trees includes a small proportion of carbon-14, which is a
radioactive isotope. After a tree dies, the proportion of carbon-14 in it decreases due to
radioactive decay.
(a)
(i)
The half-life of carbon-14 is 5740 years.
Calculate the radioactive decay constant in yr−1 of carbon-14.
decay constant ..................................... yr−1
(1)
(ii)
A piece of wood taken from an axe handle found on an archaeological site has 0.375
times as many carbon-14 atoms as an equal mass of living wood.
Calculate the age of the axe handle in years.
age ......................................... yr
(3)
Page 6 of 14
(b)
Suggest why the method of carbon dating is likely to be unreliable if a sample is:
(i)
less than 200 years old,
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
more than 60 000 years old.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(Total 6 marks)
Page 7 of 14
M1.
(a) (use of ‘isotope’ instead of ‘nucleus’ not accepted)
there is equal probability of any nucleus decaying,
it cannot be known which particular nucleus will decay next,
it cannot be known at what time a particular nucleus will decay,
the rate of decay is unaffected by the surrounding conditions,
it is only possible to estimate the proportion of nuclei decaying
in the next time interval
any two statements (2)
2
QWC 2
(b)
continuous curve starting at 5.5 × 105 Bq
plus correct 1st half-life (2.6 yrs, 2.75 × 105 Bq (1)
correct 2nd half-life (5.2 years, 1.4 × 105 Bq) (1)
(allow C.E. for incorrect 1st half-life)
2
(c)
(i)
(use of T1/2 =
gives) λ =
(1)
= 8.5 × 10–9 (s–1) (1)
(8.46 × 10–9 (s–1))
(ii)
(use of
= –λN gives) N =
(1)
= 6.5 × 1013 (atoms) (1)
(allow C.E, for value of λ from (i))
(iii)
(use of N = N0e–λt and A
N gives)
=
(1)
= 3.4 × 107 (s) (1)
(allow C.E. for value of λ from (i))
6
[10]
Page 8 of 14
M2.
(a)
(i)
origins of background radiation:
cosmic rays
ground, rocks and buildings
air
nuclear weapons testing/nuclear accidents
nuclear power
discharge/waste from nuclear power
medical waste
any three (1) (1)
any two (1)
(ii)
(use of C = C0 e–λt gives) (84 – 3) = (110 – 3) e–λ×600 (1)
(1)
= 4.6(4) × 10–4 (s–1) (1)
(iii)
(use of
= – λN gives)
N=
(1)
= 2.3(1) × 105 (nuclei) (1)
(allow C.E. for value of λ from (ii)
7
(b)
α radiation is highly ionising, hence causes cancer/damage cells/
DNA/kill cells (1)
outside: less damage plus reason
(e.g. absorbed by dead skin some α’s directed away from body) (1)
[or reference to burning]
inside: more damage plus reason
(e.g. all α’s absorbed living tissue will absorb α radiation can reach
vital organs) (1)
3
QWC 1
[10]
M3.
(a)
β=6
1
Page 9 of 14
(b)
(i)
the energy required to split up the nucleus
into its individual neutrons and protons/nucleons
(or the energy released to form/hold the nucleus
from its individual neutrons and protons/nucleons
)
2
(ii)
7.88 × 206 = 1620 MeV
(allow 1600-1640 MeV)
1
(c)
(i)
U, a graph starting at 3 × 1022 showing exponential fall passing through
0.75 × 1022 near 9 × 109 years
Pb, inverted graph of the above so that the graphs cross at 1.5 × 1022 near
4.5 × 109 years
2
(ii)
(u represents the number of uranium atoms then)
u = 6 × 1022 – 2u
u = 2 × 1022 atoms
1
(iii)
(use of N = No e-λt)
2 × 1022 = 3 × 1022 × e-λt
t = ln 1.5 / λ
(use of λ = ln 2 / t1/2)
λ = ln 2 / 4.5 × 109 = 1.54 × 10-10
t = 2.6 × 109 years
(or 2.7 × 109 years)
3
[10]
M4.
(a)
(i)
λ ( = ln 2 / T1 / 2 = 0.693 / 5740 ) = 1.2 × 10−4 (yr−1) ✓
(1.21 × 10−4 yr−1)
only allow 3.83 × 10−12 s −1 if the unit has been changed
working is not necessary for mark
1
Page 10 of 14
(ii)
(use of Nt = No e– λt and activity is proportional to N
At = Ao e– λt )
0.375 = exp - (1.21 × 10−4 × t) ✓
t=
✓
t = 8100 or 8200(yr) ✓
1st mark substitution, allow EC from (i)
2nd mark rearranging, allow EC from (i)
Allow t / T1 / 2 = 2 n approach
3rd mark no EC (so it is not necessary to evaluate a CE)
so max 2 for a CE
full marks can be given for final answer alone. A minus in the final
answer will lose the last mark
3
(b)
(i)
(it is difficult to measure accurately)
the small drop / change in activity / count-rate
the small change / drop in the ratio of C-14 to C-12 ✓
the activity would be very small / comparable to the background
or the ratio of C-14 to C-12 is too small
or there are too few C-14 atoms
or there is very little decay
or the level of C-14 (in the biosphere) is uncertain (this long ago) ✓
1st mark needs some reference to a change in count-rate or activity
for the mark
be lenient in 2nd mark
in reading a script assume C-14 is the subject. Eg ‘there is little
activity to work with’ scores mark. Also allow any reasonable
suggestion. Eg carbon may have been removed by bonding to
surrounding material
Don't allow, ‘All the carbon has decayed’
2
[6]
Page 11 of 14
E1.
The question on nuclear instability performed well on the whole and many candidates gained
high marks. Answers to part (a) however, were weak, with most candidates achieving only one
of the two available marks. This was disappointing since the examiners had recognised as many
as five possible marking points. The reason for not gaining the two marks was either because
the candidates elaborated unnecessarily on what was essentially one point or because of an
imprecise use of language. Many candidates used ‘randomly’ in the context of their answer,
without really explaining its meaning. Since the same word was part of the subject of the
question, credit was not awarded for such answers. Most candidates gained credit by referring
to the uncertainty in knowing which nucleus would decay or when. The examiners did not accept
statements which referred to an isotope rather than a nucleus.
The graph in part (b) was usually drawn correctly with three points on it: the first at time zero and
the other two at each subsequent half-life.
The calculation in part (c) was well done by most candidates, with many scoring full or almost
full marks. A significant number however, were unable to progress beyond the
exponential decay equation in part (iii) and many had in
E2.
.
Almost all candidates scored at least one mark for part (a) (i) and many scored both
available marks. Most candidates were familiar with the major sources of background radiation,
although some answers were too vague (e.g. the Earth) or suggested sources which were not
relevant to the situation (e.g. food). In many answers, marks were lost through a failure to
expand on a suggested origin for the background radiation; ‘nuclear’, ‘nuclear power’ and
‘nuclear power stations’ were not sufficiently explicit for the examiner to infer radioactive
discharge from nuclear power stations or remnants of nuclear weapon testing and accidents.
The calculations in parts (a) (ii) and (a) (iii) were generally done very well. The adjustment for the
background count was usually done correctly, although a significant number of candidates did
ignore this effect, or, surprisingly, added the background count. The main error encountered was
attempting to calculate a rate of change from the rate of decay by subtracting the two rates and
dividing by the time. Apart from this most candidates used the correct equation, corrected the
count rates and logged correctly to arrive at the required answer. The calculation in part (iii) was
almost always correct.
In part (b), almost all candidates understood that the α source would present more of a health
hazard inside the body than outside and could explain, in very general terms, why this is so.
Generally, the opportunity for extended writing in this section of the paper does not result in many
very good answers and this was certainly the case this year. Most of the explanations
demonstrated little more knowledge or understanding than an answer to the same question at
GCSE level. Few candidates scored the full 3 marks. Very few candidates made the essential
point that, on the outside, the α radiation would be absorbed by dead skin, which would result in
little more damage than perhaps some burning. Similarly vague statements about the behaviour
of alpha particles within the body gained little credit. Often answers implied that the alpha
radiation would exist for longer inside the body, somehow bouncing about inside causing
damage.
This type of question is a common feature of the nuclear instability section. Students should be
encouraged to think in more depth about the question and write answers based on their A2
knowledge and understanding, being aware of the importance of detail when showing what they
know.
Page 12 of 14
E3.
Even though part (a) needed a little thought almost all students obtained the correct answer.
By contrast part (b)(i) was simply a factual recall question, which was answered poorly by a
significant minority. The main error was for students not to state the energy needs to be given
out or is required, when a nucleus was formed or broken up. It was common to see written, ‘The
energy to keep the nucleus together’. In part (b)(ii) a majority of students simply read the value
from the graph and gave an answer near 7.88 MeV without appreciating the ‘per nucleon’ on the
y-axis of the graph. Part (c)(i) was done well by most students. Some students missed marks
due to a lack of care in choosing specific coordinates for the graphs to pass through. Most
students made a good attempt at part (c)(ii). Part (c)(iii) was more difficult and only the better
student could correctly combine the two equations required to answer the question. A common
mistake made by a few students who looked as if they were going to get the correct answer was
for them to confuse the time units they were using. These students obtained the correct answer
but then multiplied it by 60×60×24×365.
E4.
The introductory part (a)(i) was done extremely well with over 90% gaining the mark. A few
got the equation incorrect and fewer still lost their mark by calculating the half-life in s−1 units.
Less able candidates failed at the first step in part (a)(ii) by not using the exponential decay
equation. In other cases it was almost universal that mistakes were made at the substitution
stage. Some common errors were to swap round the abundance of C-12 atoms initially and at a
later time in the equation. Others sometimes added 1 to the proportion 0.375 and used N0 =
1.375 and Nt = .375. Many candidates understood how to process the mathematics and there
were a high percentage of full marks.
Although students appeared to have a good idea of the difficulties in carbon dating young and old
objects in part (b)(i + ii) many could not express their ideas well enough to score marks. It was
common to see, 'In less than 200 years none of the C-14 would have decayed'. Very few
referred to the small difference in activity when comparing a new sample with the 200 year old
sample. Candidates were much more successful in the second part when referring to the much
older sample. Most knew that the activity would be very small and it would be difficult to get
statistically significant data. Several did not gain this last mark because of lack of care in what
they put down on paper. It was common to see variations of, 'There has been so much decay
there is no C-14 atoms left'.
Page 13 of 14
Resource currently unavailable.
Page 14 of 14
Intensity of radiation
Make notes on the following
Derivation of inverse square law
Working safely with radiation – ALARA
Calculating the radiation incident on a fixed area
Complete and mark the following
Physics factsheet 85
AQA exam past paper questions - Safety Aspects
Completed?
Completed?
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Number 85
A2 Questions Radioactivity
Radioactive decay lends itself to a broad range of assessment. In addition to practical investigations, problems can be set to test theory, calculations, graphical
work, and health and environmental issues. Written answers in paragraph form can be demanded (although they still tend to be marked in terms of separate
points made).
Some of the topics that can be assessed include:
1. Types and properties of decay products (including scattering theory and the nuclear atom – Rutherford).
2. The decay process (including sources of radiation [man-made and natural], equations of decay, N-Z graphs, decay chains, etc.).
3. Decay graphs (including ideas of activity, decay constant, half-life, background correction, etc.)
4. Maths based on A=λN, λT½ = 0.693, and equations of the form A=Aoe-λt.
5. The inverse square law for gamma radiation.
6. Environmental issues (safety/waste disposal/half-lives).
7. Medical uses and dangers (this topic has already been covered in some depth in Factsheets 77 and 82).
This Factsheet will deal almost exclusively with answering questions on these topics. There are four exam-style questions to illustrate various points and
demonstrate the methods of solution required. Then there are a number of short questions for you to attempt – the solutions are supplied.
Exam-style question 1
1
(a) Discuss problems involved in safe storage of radioactive waste (3)
(b) Complete this table relating the range, ionising power, and dangers
associated with decay products α, β, and γ.
Decay
Product
Range in
Air
Ionising
Power
Danger to humans
α
Very small
(1)
(2)
β
(1)
Medium
Most dangerous if taken
into body in breathing or
as food / drink.
Less dangerous from
external source.
(1)
γ
Exam Hint: Be clear about the link between range and ionising power
for α, β, and γ decay, and the differences between these decay products.
Exam-style question 2
2. An experiment is set up to confirm that gamma-ray intensity obeys
the inverse square law (I = k / d2).
d
(2)
(1)
radioactive source
detector
Exam-style question 1 solutions
1
(a) Why will this experiment not give the required result for α and β
particles? (1)
(a) half-life very long (1)
possible leakage into environment / water supply (1)
geological instability causing disruption to security of storage (1)
radioactivity may degrade storage vessel (1)
-any three or these or other sensible points.
(b) How can you ensure that alpha or beta particles from the source
cannot reach the detector? (1)
(c) Should the source have a long or a short half-life? Explain your
answer. (2)
Exam Hint: Be prepared to write essay-type answers concerning safety
and environmental issues. Remember, the number and quality of points
you make are important – not the total number of words you write.
1(b)
Decay Range
Product in Air
α
β
γ
Very small
Small (1)
Ionising
Power
Danger to humans
Very large (1)
Most dangerous if taken
into body in breathing or as
food / drink. (1)
Less dangerous from
external source. (1)
Medium
(d) The count rate is taken over a range of values for distance d. Explain
what correction must be made to the recorded count rate in
investigations of this sort, and how the final corrected value is
determined. (2)
(e) These values for corrected count rate (s-1) at various distances are
found. Show graphically that the results obey the inverse square
law. (7)
Most dangerous if taken
into body in breathing or as
food / drink.
Less dangerous from
external source.
Very large (1) Very small (1) Most travel straight
through body (1)
Equally dangerous
externally or internally (1)
d /m
count rate /s
0.20
442
0.40
112
0.60
51
0.80
28
1.00
18
1.20
12
(The third and fourth columns are for your use.)
1
3K\VLFV)DFWVKHHW
85. A2 Questions Radioactivity
Exam-style question 2 solutions
Exam-style question 3
2. (a) some (if not all) particles absorbed by air (1)
(b) metal filter (sheet) in front of source to absorb them (1)
(c) long half-life (1)
to ensure constant (ignoring randomness of decay) activity of source
(1)
(d) correct for background radiation (1)
measure background and subtract from recorded values (1)
(e) count rate = k/d2 , ln(count rate) = ln k – 2 ln d (1)
the logarithmic graph should be a straight line with gradient –2 (1)
This is a small section of the Uranium-238 decay chain:
A(Mass
Po
214
X
212
Pb
210
Y
208
d /m
count rate /s
In (d /m)
In count rate /s
206
0.20
442
-1.61
6.09
204
0.40
112
-0.92
4.72
0.60
51
-0.51
3.93
0.80
28
-0.22
3.33
1.00
18
0.00
2.89
12
0.18
2.48
1.20
Bi
Z
Tl
Pb
Z (Atomic
80
82
84
(a) Identify the decay products in sections X, Y, and Z of the chain. (1)
(b) Could the decay from Po-214 to Pb-206 be accomplished in less than
four steps? (1)
(c) What information does this chart give us about gamma ray emission?
Explain your answer.(2)
(d) Why would the reverse process (decay from Pb to Po) be impossible? (1)
(2 marks for table)
Exam-style question 3 solutions
(a) X = α, Y = β, Z=α (1)
(b) No. (but there could be other pathways possible) (1)
(c) No information about γ rays. (1) They don’t appear in the chart, as
they don’t result in a change in nuclear charge or mass. (1)
(d) Each decay process results in the loss of mass and/or energy from the
nucleus. The reverse process would mean mass and/or energy would
have to be gained in each step. (1)
ln(count rate /s-1)
6.0
5.0
Exam-style question 4
The sketch graph shows the corrected count rate measured for the decay of
a radioactive sample (the sample-detector distance is constant):
4.0
3.7
corrected count rate/s-1
1200
3.0
2.0
500
1.8
1.0
0
-1.8
-1.4
-1.0
-0.6
-0.2
0.2
ln(d/m)
t/s
(a) At time t=0s, what factors will affect the count-rate measured? (3)
(b) If the count-rate is 1200s-1 at t=0s, and is 500s-1 after 10s, find the
decay constant for the radioactive isotope. (2)
(c) What is the half-life of this isotope? (1)
(d) If the measured count-rate is only 1% of the activity of the sample,
find the number of atoms of this radioisotope present after 10s. (2)
(2 marks for graph)
Gradient = -3.7 / 1.8 = -2.1
10
(1)
Exam-style question 4 solutions:
(a) the decay constant (or half-life) of the radioactive isotope
the number of atoms of this radioisotope present
the distance d
the type or size of the detector
the type of decay product (α, β, or γ)
-any 3 of these points, or sensible alternatives
(b) count rate ∝ activity
A = Aoe-λt , 500 / 1200 = e-10λ , λ = 0.088 s-1 . (2)
(c) T½ = 0.693 / λ = 7.9 s (1)
(d) A = 500 × 100 = 5.0 × 104 (1)
A = λN , N = A / λ = 50 000 / 0.088 = 5.7 × 105 atoms. (1)
Alternatively, the marks could be gained by calculating the values for d2,
and plotting count rate against 1 / d2, and obtaining a straight line through
the origin.
2
85. A2 Questions Radioactivity
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Practice Questions
1. What is the SI base unit for the rate of radioactive decay?
2. At a certain stage in a radioactive decay practical, the count rate was found to be 84Bq. If the average background measured was 254 counts in 2
minutes, find the corrected count rate (to the nearest whole number).
3. The following count rates were measured over successive 10s intervals in a beta decay practical. The source-detector distance was fixed, and the
source half-life was 26 years.
162, 188, 163, 152, 183, 171, 159, 173
(a) Find the average count rate in Bq.
(b) Find the maximum percentage error in these results.
(c) What factors were responsible for the variation in readings.
4. Explain in words why the decay constant, λ, is inversely proportional to the half-life, T½.
5. (a) The half-life of a source is 22 years. Find the decay constant, λ.
(b) The decay constant for a radioisotope is 1.4 × 10-4 s-1. Find the half life, T½.
6. Complete the following decay equation:
y-8
y
A
B + ? + ? +γ
x-3
x
Number of Nuclei
7. Here is a sketch graph of simultaneous decay from radioisotopes X and Y:
X
Y
Time
Explain what is happening.
Answers
1. Bq = s-1
2. 254 / (2 × 60) = 2.1,
84 – 2 = 82Bq
3. (a) 169 in 10s = 16.9Bq
(b) (188-169) / 169 = 0.112 = 11.2%.
(c) randomness of decay from source and randomness of background (not just “background”).
4. The decay constant is the probability of decay for each unstable nucleus in the next second. The higher the probability of decay, the shorter time for
half of the nuclei to decay.
5. (a) λ = 0.693 / T½ = 0.693 / (22 × 365 × 24 × 3600) = 9.99 × 10-10s-1.
(b) T½ = 0.693 / λ = 0.693 / 1.4 × 10-4 = 4950s = 1.38 hours.
6. Insert diagram 6.
y
x
A
y-8
B+
x-3
4
0
2 α +
β +γ
2
-1
Acknowledgements:
This Physics Factsheet was researched and written by Paul Freeman
The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU
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ISSN 1351-5136
3
Q1.
(a)
(i)
Explain why, after a period of use, the fuel rods in a nuclear reactor become
less effective for power production,
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(ii)
more dangerous.
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(3)
(b)
Describe the stages in the handling and processing of spent fuel rods after they have been
removed from a reactor, indicating how the active wastes are dealt with.
You may be awarded marks for the quality of written communication in your answer.
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(5)
(Total 8 marks)
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Q2.
(a) Sketch, using the axes provided, a graph of neutron number, N, against proton
number, Z, for stable nuclei over the range Z = 0 to Z = 80. Show suitable numerical
values on the N axis.
(2)
(b)
On the graph indicate, for each of the following, a possible position of a nuclide that might
decay by
(i)
α emission, labelling the position with W,
(ii)
β– emission, labelling the position with X,
(iii)
β+ emission, labelling the position with Y.
(3)
(c)
Used fuel rods from a nuclear reactor emit β– particles from radioactive isotopes that were
not present before the fuel rod was inserted in the reactor. Explain why β– emitting
isotopes are produced when the fuel roads are in the reactor.
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(3)
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(d)
A nuclear power station is a reliable source of electricity that does not produce greenhouse
gases but it does produce radioactive waste. Discuss the relative importance of these
features in deciding whether or not new nuclear power stations are needed.
The quality of your written answer will be assessed in this question.
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(6)
(Total 14 marks)
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Q3.
(a) In a thermal nuclear reactor, one fission reaction typically releases 2 or 3 neutrons.
Describe and explain how a constant rate of fission is maintained in a reactor by
considering what events or sequence of events may happen to the released neutrons.
The quality of your written communication will be assessed in this question.
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(7)
(b)
Uranium is an α emitter. Explain why spent fuel rods present a greater radiation hazard
than unused uranium fuel rods.
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(3)
(Total 10 marks)
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M1.
(a)
(ii)
(i)
amount of (fissionable) uranium (235) in fuel decreases (1)
fission fragments absorb neutrons (1)
fission fragments are radioactive or unstable (1)
emitting β– and γ radiation (1)
some fission fragments have short half-lives or high activities (1)
Max 3
(b)
moved by remote control (1)
placed in cooling ponds (1)
for several months (1)
[or to allow short T1/2 isotopes to decay]
transport precautions, e.g. impact resistant flasks (1)
separation of uranium from active wastes (1)
high level waste stored (as liquid) (1)
[alternative for last two marks:
rods are buried deep underground
at geologically stable site]
storage precautions, e.g. shielded tanks or monitoring (1)
reference to vitrification (1)
Max 5
[8]
M2.
(a)
graph passes through N = 100 to 130 when Z = 80 (1)
and N = 20 when Z = 20 (1)
2
(b)
(i)
W at Z > 60 just below line (1)
(ii)
X just above line (1)
(iii)
Y just below line (1)
3
(c)
fission nuclei (or fragments) are neutron-rich and therefore
unstable (or radioactive) (1)
neutron-proton ratio is much higher than for a stable nucleus
(of the same charge (or mass)) (1)
β– particle emitted when a neutron changes to a proton
(in a neutron-rich nucleus) (1)
3
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(d)
The marking scheme for this part of the question includes an overall
assessment for the Quality of Written Communication (QWC). There
are no discrete marks for the assessment of written communication
but the quality of written communication will be one of the criteria
used to assign the answer to one of three levels.
Level
Descriptor
Mark range
an answer will be expected to meet most of the criteria in the
level descriptor
Good 3
– answer supported by appropriate range of relevant points
– good use of information or ideas about physics, going
beyond those given in the question
– argument well structured with minimal repetition or irrelevant
points
5-6
– accurate and clear expression of ideas with only minor
errors of spelling, punctuation and grammar
Modest 2
– answer partially supported by relevant points
– good use of information or ideas about physics given in the
question but limited beyond this
– the argument shows some attempt at structure
3-4
– the ideas are expressed with reasonable clarity but with a
few errors of spelling, punctuation and grammar
Limited 1
– valid points but not clearly linked to an argument structure
– limited use of information or ideas about physics
– unstructured
1-2
– errors in spelling, punctuation and grammar or lack of
fluency
0
– incorrect, inappropriate or no response
0
examples of the sort of information or ideas that might be used to
support an argument:
•
reduction of greenhouse gas emissions is (thought to be) necessary
to stop global warming (1)
•
long term storage of radioactive waste is essential because the
radiation from it damages (or kills) living cells (1)
•
radioactive isotopes with very long half lives are in the used fuel
rods (1)
•
nuclear power is reliable because it does not use oil or gas from
other countries (1)
•
radioactive waste needs to be stored in secure and safe conditions
for many years (1)
Page 6 of 10
conclusion
either
nuclear power is needed; reduction of greenhouse gases is a greater
problem than the storage of radioactive waste because
1
2
global warming would cause the ice caps to melt/sea levels to
rise (1)
safe storage of radioactive waste can be done (1)
or
nuclear power is not needed; storage of radioactive waste is a
greater problem than reduction of greenhouse gases because
1
2
radioactive waste has to be stored for thousands of years (1)
greenhouse gases can be reduced using renewable energy
sources (1)
[14]
M3.
(a) The candidate’s writing should be legible and the spelling,
punctuation and grammar should be sufficiently accurate for the
meaning to be clear.
The candidate’s answer will be assessed holistically. The answer will be
assigned to one of three levels according to the following criteria.
High Level (Good to excellent): 5 or 6 marks
The information conveyed by the answer is clearly organised, logical and
coherent, using appropriate specialist vocabulary correctly. The form and
style of writing is appropriate to answer the question.
The candidate can explain the role of the moderator and control rods in
maintaining a critical condition inside the reactor. The explanation is given
in a clear sequence of events and the critical condition is defined in terms of
neutrons. To obtain the top mark some other detail must be included. Such
as, one of the alternative scattering or absorbing possibilities or appropriate
reference to critical mass or detailed description of the feedback to adjust
the position of the control rods etc.
Intermediate Level (Modest to adequate): 3 or 4 marks
The information conveyed by the answer may be less well organised and
not fully coherent. There is less use of specialist vocabulary, or specialist
vocabulary may be used incorrectly. The form and style of writing is less
appropriate.
The candidate has a clear idea of two of the following:
the role of the moderator, the role of the control rods or can explain the
critical condition.
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Low Level (Poor to limited): 1 or 2 marks
The information conveyed by the answer is poorly organised and may not
be relevant or coherent. There is little correct use of specialist vocabulary.
The form and style of writing may be only partly appropriate.
The candidate explains that a released neutron is absorbed by uranium to
cause a further fission. Alternatively the candidate may explain one of the
following:
the role of the moderator, the role of the control rods or can explain the
critical condition.
The explanation expected could include the following events that
could happen to a released neutron.
a neutron is slowed by the moderator
taking about 50 collisions to reach thermal speeds
then absorbed by uranium-235 to cause a fission event
one neutron released goes on to cause a further fission is the critical
condition
a neutron may leave the reactor core without further interaction
a neutron could be absorbed by uranium-238
a neutron could be absorbed by a control rod
a neutron could be scattered by uranium-238
a neutron could be scattered by uranium-235
max 7
(b)
it is easy to stay out of range or easy to contain an α source or ß/γ have
greater range/are more difficult to screen (1)
most (fission fragments) are (more) radioactive/unstable (1)
and are initially most likely to be beta emitters/(which also) emit γ
radiation/are neutron rich/heavy (1)
ionising radiation damages body tissue/is harmful (1)
max 3
[10]
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E1.
Many answers to part (a), which for full credit required explanations rather than statements,
would have benefited from a more precise use of terminology. In part (a) (i) for example, ‘the fuel
runs out’ is much inferior to ‘the amount of fissionable uranium remaining decreases’. In part (a)
(ii), a mark was awarded for neutron bombardment causing radioactive instability, but the real
reason is that the fission fragments themselves are unstable neutron-rich β and γ emitters.
Furthermore, since some of them have very short half-lives, their activities can be very high.
In part (b), the treatment and handling of spent fuel rods was often understood very well.
However, the need to use remote control, and to take precautions over the transport of used
material, was commonly overlooked. Credit was given equally to procedures involving
reprocessing and to those where the spent rods are simply buried deep underground in stable
rock formations.
E3.
This question was a good discriminator. Most candidates, in part (a), knew how the core of
the reactor functions. Some candidates too readily used the wording of the question as their
answer. Others did not refer to neutrons even though this was asked for in the question. One
example of a phrase given by candidates that did not quite answer the question but sounded
reasonable was, ‘the power levels were kept constant by keeping a constant rate of fission using
control rods’. This offers much of what was in the question itself and it does not refer to
neutrons. The quality of the writing was generally good.
Again question (b) was a good discriminator. The majority of candidates were aware that fission
products are normally unstable because they tend to be neutron rich or that they release beta
and gamma rays. Less able candidates thought used fuel meant that they had undergone alpha
emission.
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