1
(a)
Which ionizing radiation produces the greatest number of ion pairs per mm in air? Tick (✓)
the correct answer.
α particles
β particles
γ rays
X−rays
(1)
(b)
(i)
Complete the table showing the typical maximum range in air for α and β particles.
Type of radiation
Typical range in air / m
α
β
(2)
(ii)
γ rays have a range of at least 1 km in air.
However, a γ ray detector placed 0.5 m from a γ ray source detects a noticeably
smaller count-rate as it is moved a few centimetres further away from the source.
Explain this observation.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)
(c)
Following an accident, a room is contaminated with dust containing americium which is an
α−emitter.
Explain the most hazardous aspect of the presence of this dust to an unprotected human
entering the room.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 6 marks)
Page 1 of 51
2
(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 m2 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)
(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)
Page 2 of 51
(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)
3
(a)
Bi can decay into
Pb by a β– followed by an α decay, or by an α followed by a β–
decay. One or more of the following elements is involved in these decays:
Write out decay equations showing each stage in both of these decays.
First decay path
Second decay path
(6)
Page 3 of 51
(b)
(i)
Describe how you would perform an experiment that demonstrates that gamma
radiation obeys an inverse square law.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
Explain why gamma radiation obeys an inverse square law but alpha and beta
radiation do not.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(9)
(Total 15 marks)
4
A freshly prepared radioactive source that emits negatively charged beta particles (β–) has an
activity of 120 Bq and a half-life of 12 h.
(a)
(i)
State the effect on the proton number Z and the nucleon number A when a β– particle
is emitted.
...............................................................................................................
...............................................................................................................
(2)
Page 4 of 51
(ii)
Sketch, on the axes below, a graph that shows how the activity varies during the two
days after the source was prepared.
(3)
(b)
(i)
The total energy released in each decay is 5.5 × 10–13 J.
Calculate the initial energy produced each second by the source.
initial energy ..................................... J
(1)
(ii)
Figure 1 shows the energy spectrum for the beta particles emitted in the decay.
It shows that different energy beta particles are possible.
Figure 1
Page 5 of 51
Explain why all the beta particles that are emitted do not have 5.5 × 10–13 J of energy.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(c)
The probability of one of the radioactive atoms decaying each second is
1.6 × 10–5.
How many radioactive atoms are present when the activity is 120 Bq?
number of radioactive atoms ...................................
(1)
(d)
A scientist undertaking an investigation places the freshly prepared source close to
a Geiger-Müller tube as shown in Figure 2 and records a count rate of 50 counts
per second.
Figure 2
State and explain two reasons why the measured count rate is lower than the activity of the
source.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 12 marks)
Page 6 of 51
5
An isotope of technetium is a gamma emitter used by doctors as a tracer in the human body. It is
injected into the patient’s blood stream. Scanners outside the body measure the gamma activity,
enabling the blood flow to be monitored.
(a)
The graph shows the variation of activity with time, t, for a sample of the isotope.
(i)
Use data from the graph to determine the half-life of the technetium isotope.
(3)
(ii)
The decay constant of the technetium isotope is 3.2 × 10−5 s−1. Use data from the
graph and the equation A = λN to calculate the number of nuclei of the radioactive
technetium isotope present at time t = 0.
(2)
(b)
(i)
State why an alpha emitter would not be suitable in this application.
...............................................................................................................
...............................................................................................................
(1)
(ii)
State why the half-life of the technetium isotope makes it suitable for this application.
...............................................................................................................
...............................................................................................................
(1)
Page 7 of 51
(c)
State and explain how the presence of the technetium isotope may do some damage to the
patient.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 9 marks)
6
The actinium series of radioactive decays starts with an isotope of uranium, nucleon (mass)
number 235, proton (atomic) number 92.
Which line in the table shows the nucleon number and proton number of the isotope after the
emission of 5 α particles and 2 β– particles?
Nucleon number
proton number
A
213
82
B
215
80
C
215
84
D
227
87
(Total 1 mark)
7
An alpha particle moves at one-tenth the velocity of a beta particle. They both move through the
same uniform magnetic field at right angles to their motion.
The magnitude of the ratio
is
A
B
C
D
(Total 1 mark)
Page 8 of 51
8
Iodine-123 is a radioisotope used medically as a tracer to monitor thyroid and kidney functions.
The decay of an iodine-123 nucleus produces a gamma ray which, when emitted from inside the
body of a patient, can be detected externally.
(a)
Why are gamma rays the most suitable type of nuclear radiation for this application?
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
In a laboratory experiment on a sample of iodine-123 the following data were collected.
time/h
0
4
8
12
16
20
24
28
32
count-rate
/counts s–1
512
410
338
279
217
191
143
119
91
Why was it unnecessary to correct these values for background radiation?
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 9 of 51
(c)
On the axes provided in the diagram below, complete the graph of count-rate against time.
(2)
(d)
Use your graph to find an accurate value for the half-life of iodine-123.
Show clearly the method you use.
Half-life ..............................
(3)
Page 10 of 51
(e)
Give two reasons why radioisotopes with short half-lives are particularly suitable for use as
a medical tracer.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 11 marks)
9
The radioisotope iodine-131 (
) is used in medicine to treat over-active thyroid glands. It
decays into an isotope of xenon (Xe) by β– emission with a half-life of 8.1 days. The xenon
subsequently emits a γ ray.
(a)
Explain what is meant by:
(i)
isotope;
...............................................................................................................
...............................................................................................................
(2)
(ii)
half-life.
...............................................................................................................
...............................................................................................................
(1)
(b)
Write down the equation which represents the nuclear reaction.
(3)
(c)
Calculate the time (in days) for a sample of iodine to decay to 1% of its initial activity.
(4)
Page 11 of 51
(d)
State and explain which decay product can be detected outside the body during treatment.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 12 marks)
10
The diagram below shows an arrangement used to maintain a constant thickness of sheet paper
or steel as it is being rolled. A radioactive source and detector are used to monitor the thickness.
(a)
Explain briefly how this arrangement works.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
Page 12 of 51
(b)
Alpha, beta or gamma sources could be selected for use in such an arrangement.
State which source should be selected in each case and explain briefly why the others
would not be suitable.
Paper:
Source selected .............................................................................................
Reasons why the others are unsuitable ........................................................
........................................................................................................................
........................................................................................................................
Steel:
Source selected .............................................................................................
Reasons why the others are unsuitable ........................................................
........................................................................................................................
........................................................................................................................
(4)
(c)
Cobalt-60 is commonly used as a source in such applications. This has a half-life of 5.3
years. When fresh the source contains 5.0 × 1020 radioactive atoms.
For it to be useful the source has to have an activity of at least 1.5 × 1012 Bq.
(i)
What is meant by an activity of 1 Bq?
...............................................................................................................
...............................................................................................................
(1)
(ii)
Draw a graph showing the number of radioactive atoms in the source over a period of
3 half-lives. Include suitable scales on the axes.
(2)
Page 13 of 51
(iii)
Determine the decay constant of cobalt-60 in s–1.
(2)
(iv)
After what time will it be necessary to replace the source?
(3)
(Total 15 marks)
11
is an isotope of uranium which decays through a decay chain to the isotope of lead,
which is stable. During the sequence of decays, both alpha particles and beta particles are
emitted.
(a)
(i)
,
represents an alpha particle.
State the values of:
A ....................................................................
Z ....................................................................
(1)
(ii)
represents a beta particle.
State the values of:
A ....................................................................
Z ....................................................................
(1)
(iii)
State and explain how many alpha particles are emitted during the decay chain
between
.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(iv)
State and explain how many beta particles are emitted during the decay chain
between
and
.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
Page 14 of 51
(b)
The decay of uranium can be used to measure the age of rocks. A rock, which contained
N0 atoms of
(i)
when it was formed, will contain NU atoms of the isotope at a later time t.
The relationship between N0, NU and t is:
NU = N0e–λt
State the meaning of the decay constant, λ.
...............................................................................................................
...............................................................................................................
(1)
(ii)
Because the half-life of
is much longer than that of any other member of the
decay chain, it may be assumed that the number of atoms of
present at time t
is equal to the number of atoms
which have decayed. Show that the ratio of the
number of lead atoms (NP) to the number of uranium atoms at time t is given by:
= eλt – 1
(2)
(iii)
The half-life of
is 4.5 × 109 year. Calculate the age of a rock sample which
contains 4.2 parts per million of
and 0.62 parts per million of
.
(4)
(iv)
The above analysis assumes that the rock contained no lead when it was formed.
State and explain the effect on the estimated age if the rock had contained lead when
it was formed.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
Page 15 of 51
(v)
State reasons why the measurement of the amount of uranium in a rock sample
would be inaccurate when performed by a radiation counting technique.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(Total 17 marks)
12
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 × 107 Bq.
(a)
To determine the dose received by a scientist working with the source the number of
gamma ray 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 ray photons per second are incident on each cm2 of the
scientist's body.
(2)
(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.
(3)
(ii)
At what distance from the source could the scientist now work and receive the original
dose of 21 photons per second per cm2 .
(2)
(c)
Explain why it is not necessary to consider the beta particle emission when determining the
dose of radiation the scientist receives.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 9 marks)
Page 16 of 51
13
The graph below shows how the nucleon number A changes with proton number Z for the decay
series that starts with uranium-238. The half-lives of each decay are also shown.
(a)
How many alpha particles and beta particles are emitted when a uranium-238 nucleus
decays to radon-222 (222Rn)?
Number of alpha particles ..............................................................................
Number of beta particles.................................................................................
(2)
Page 17 of 51
(b)
How many neutrons are there in a nucleus of polonium-210 (210Po)?
........................................................................................................................
(1)
(c)
Identify the stable isotope that results from this decay chain.
........................................................................................................................
(1)
(d)
214 g of bismuth-214 (214 Bi) contains 6.0 × 1023 atoms. A sample containing only
bismuth-214 has an initial mass of 0.60 g.
(i)
After what period of time will the mass of bismuth-214 present in the sample be
0.15 g?
(2)
(ii)
Determine the number of bismuth-214 atoms present after this time.
(1)
(iii)
Calculate the activity of the bismuth-214 in the sample after this time.
(4)
(iv)
Explain how the total activity of the sample will be different from the value calculated
in (iii).
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(v)
The bismuth-214 decays into polonium-214. Explain why you would find very little
polonium-214 if you were to analyse the sample.
...............................................................................................................
...............................................................................................................
(2)
(Total 15 marks)
14
(a)
(i)
Explain what is meant by background radiation.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
Page 18 of 51
(ii)
Name a source of background radiation.
...............................................................................................................
(1)
(b)
State the meaning of each of the terms in the equation A = λN
A .....................................................................................................................
........................................................................................................................
λ ......................................................................................................................
........................................................................................................................
N .....................................................................................................................
........................................................................................................................
(3)
(Total 6 marks)
15
A radium-288 nuclide (
) is radioactive and decays by the emission of a β– particle to form
an isotope of actinium (Ac).
(a)
Complete the equation for this decay.
(3)
(b)
β– decay is the result of a neutron within a nucleus decaying into a proton. Describe the
change in the quark sub-structure that occurs during the decay.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
(Total 4 marks)
16
A student has access to a radioactive source that decays by emitting alpha, beta and gamma
radiation. The student wishes to investigate whether the count rate due to the gamma radiation
varies with distance from the source according to an inverse square law and sets up the source
and detector as shown in Figure 1.
Page 19 of 51
(a)
State and explain how the student can ensure that only gamma radiation is detected during
the investigation.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
The corrected count rate due to gamma radiation is 64 counts per second at a distance of
50 mm from the source. Assuming that an inverse square law is obeyed calculate the
expected corrected count rate at a distance of 80 mm from the source.
Count rate at 80 mm ..............................................
(2)
Page 20 of 51
(c)
Using the data from part (b) sketch, on the axes in Figure 2, the graph the student would
expect if an inverse square law were obeyed. The corrected count rate at 50 mm has been
plotted already.
Figure 2
(2)
(Total 6 marks)
17
Thallium (Tl) decays to a stable form of lead (Pb) with the emission of a β– particle. Complete the
equation below for this decay.
(Total 3 marks)
Page 21 of 51
18
(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 the corrected count rate is 0.10 C,
assuming that there is no absorption.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(c)
The activity of a source of b particles falls to 85% of its initial value in 52 s.
Calculate the decay constant of the source.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
Page 22 of 51
(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)
19
(a)
State which type of radiation, α, β or γ,
(i)
produces the greatest number of ion pairs per mm in air,
...............................................................................................................
(ii)
could be used to test for cracks in metal pipes.
...............................................................................................................
...............................................................................................................
(2)
(b)
Specific radioisotope sources are chosen for tracing the passage of particular substances
through the human body.
(i)
Why is a γ emitting source commonly used?
...............................................................................................................
...............................................................................................................
(ii)
State why the source should not have a very short half-life.
...............................................................................................................
...............................................................................................................
Page 23 of 51
(iii)
State why the source should not have a very long half-life.
...............................................................................................................
...............................................................................................................
(3)
(c)
A detector, placed 0.20 m from a sealed γ ray source, receives a mean count rate of 2550
counts per minute. The experimental arrangement is shown in the diagram below. The
mean background radiation is measured as 50 counts per minute.
Calculate the least distance between the source and the detector if the count rate is not to
exceed 6000 counts per minute.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(5)
(Total 10 marks)
20
(a)
(i)
Alpha and beta emissions are known as ionising radiations. State and explain why
such radiations can be described as ionising.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
Page 24 of 51
(ii)
Explain why beta particles have a greater range in air than alpha particles.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(b)
The figure below shows the variation with time of the number of Radon (220Ra) atoms in a
radioactive sample.
(i)
Use the graph to show that the half-life of the decay is approximately 53 s. Show your
reasoning clearly.
(3)
(ii)
The probability of decay (decay constant) for 220Ra is 1.3 × 10–2 s–1. Use data from
the graph to find the activity of the sample at a time t = 72 s.
activity .....................................................
(3)
(Total 10 marks)
Page 25 of 51
21
A radioactive nucleus decays with the emission of an alpha particle and a gamma-ray photon.
(a)
Describe the changes that occur in the proton number and the nucleon number of the
nucleus.
proton number ...............................................................................................
nucleon number .............................................................................................
(2)
(b)
Comment on the relative penetrating powers of the two types of ionizing radiation.
........................................................................................................................
........................................................................................................................
(1)
(c)
Gamma rays from a point source are travelling towards a detector. The distance from the
source to the detector is changed from 1.0 m to 3.0 m.
Calculate
intensity of radiation at 3.0 m
intensity of radiation at 1.0 m
answer .......................................
(2)
(Total 5 marks)
Page 26 of 51
Mark schemes
1
(a)
A α particles
[auto mark question]
1
(b)
(i)
type of
radiation
Typical range
in air / m
α
0.04
β
0.40
Allow students to use their own distance units in the table
α allow 0.03
0.07 m
β allow 0.20
3.0 m.
If a range is given in the table use the larger value.
A specific number is required e.g. not just a few cm.
2
(ii)
reference to the inverse square law of (γ radiation)
or
reference to lowering of the solid angle (subtended by the detector as it moves
away)
or
radiation is spread out (over a larger surface area as the detector is moved
away)
(owtte)
Ignore any references to other types of radiation.
Any contradiction loses the mark. For example, follows inverse
square law so intensity falls exponentially.
1
(c)
dust may be ingested / taken into the body / breathed in
First mark for ingestion not just on the body
causing (molecules in human tissue / cells) to be made cancerous / killed / damaged
by ionisation
Second mark for idea of damage from ionisation
2
[6]
2
(a)
any 2 from:
the sun, cosmic rays, radon (in atmosphere), nuclear fallout (from previous
weapon testing), any radioactive leak (may be given by name of incident) nuclear
waste, carbon-14
1
Page 27 of 51
(b)
(i)
(ratio of area of detector to surface area of sphere)
ratio =
(0.00368)
0.0037
2
(ii)
activity = 0.62/(0.00368 × 1/400) give first mark if either factor is used.
Bq accept s-1 or decay/photons/disintegrations s-1 but not
67000
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]
3
(a)
number correct for alpha (1)
number correct for beta (1)
alpha decay first goes via Tl (1)
numbers correct for Tl (208, 81) (1)
beta decay first goes via Po (1)
numbers correct for Po (212, 84) (1)
6
(b)
(i)
use of GM tube + counter/rate-meter (1)
measurement of count rate (1)
at range of distances + suitable ruler or tape measure (1)
specifies suitable range (1)
determines background & corrects (1)
safety precaution given (1)
graph of count rate or corrected count rate against 1/d2 (1)
max 6
Page 28 of 51
(ii)
gamma not absorbed (1)
spreads uniformly from a point
source/spherically symmetrically (1)
area over which it spreads is proportional
to radius squared (1)
alpha and beta are absorbed in addition to spreading out (1)
max 3
[15]
4
(a)
(i)
Z increases by 1
B1
A remains the same
B1
2
(ii)
Correct curvature starting at 120 Bq
B1
60 (or 0.5 × their start value) at 12 h days later
B1
30 (or half their value at 12 h)
and continuing to fall thereafter approximately
exponentially
B1
3
Page 29 of 51
(b)
(i)
6.6 × 10–11 J (s–1)
(120 × 5.5 × 10–13)
B1
1
(ii)
another particle is emitted in each decay (not gamma
radiation)
or
the nucleus recoils
B1
anti-neutrino emitted (this would get first and second
mark 2 marks)
B1
the other particle/neutrino/antineutrino/nucleus takes
some/varying amounts of the energy
B1
3
(c)
7.5 × 106
B1
1
(d)
Particles are emitted in all directions/particles do not
all go to detector
B1
Detector only detects some of the particles that enter
it/mention of dead time or recovery time
(not detector does not detect all the particles - this adds
nothing)
B1
Some particles are absorbed by the window
B1
max 2
[12]
Page 30 of 51
5
(a)
(i)
correct construction method seen on graph
or quotes appropriate values from the graph
C1
2.0 ×
104
s to 2.4 ×
104
s allow 1 or2 s.f.
A1
repeats and averages
B1
(3)
(ii)
A = 3.7 or 3.8 × 1016 (Bq) at t = 0
C1
1.2 ×
1021
A1
(2)
(b)
(i)
alpha will not penetrate the body or risk to patient from ionisations
B1
(1)
(ii)
long enough half life to make measurements / short enough half life so does not
remain long in body
B1
(1)
(c)
causes ionisations / damage
B1
to cell / cell nucleus / body tissue / DNA
B1
(2)
[9]
6
7
C
[1]
B
[1]
Page 31 of 51
8
(a)
Gamma rays are very penetrating/alpha/beta rays would
not be detected
B1
(outside body)
Gamma rays are less ionising/less hazardous
(to patients)/ alpha/beta are more ionising/more hazardous
B1
2
(b)
Background radiation/count is much smaller/negligible
B1
Random fluctuations in the readings greater than background
B1
2
(c)
Accurate plotting
check all four points (
½ square)
M1
reasonably smooth curve with even point scatter
A1
2
(d)
two or more half-lives averaged
B1
Half-life calculated from best fit line
C1
Half-life = 13
1 hour
A1
3
allow ecf from inaccurate plotting, but straight line = P.E.
Page 32 of 51
(e)
High activity (so only a small sample needed)
B1
Decays quickly
B1
Less risk to patient/other people
B1
(Short half-life ok because) medical test doesn‘t last long
B1
Any two from four
2
[11]
9
(a)
(i)
nuclides or atoms of same element or same proton / atomic number
not just ‘same element’
different mass / nucleon number or numbers of neutrons
B1
B1
(2)
(ii)
time taken for half the sample or half the number of nuclei
(not just ‘atoms’) to decay or activity to fall to half initial value
B1
(1)
(b)
B1
B1
B1
(3)
Page 33 of 51
(c)
A / A0 = 0.01 or A = 1 and A0 = 100
C1
λ = 0.69 / 8.1 or
or 0.086 (0.0856)
attempt to take logs
C1
54 days (53.8)
A1
(for between 6 and 7 half-lives, 50% → 25% → 12.5%)
C1
(4)
(d)
γ radiation detected
M1
β cannot penetrate tissue or γ can penetrate tissue
–
A1
(2)
[12]
10
(a)
thicker material absorbs more particles
B1
count rate (number detected) falls if material is thicker
B1
fall in count rate produces change to adjust process to produce thinner material /
restore to original thickness
allow 1 mark for
‘change in thickness changes count rate and rollers adjust to compensate’
B1
(3)
(b)
use a beta source
M0
alphas would be absorbed by paper
A1
gammas would not be affected
A1
use a gamma source
M0
beta would be absorbed completely
A1
alphas would be absorbed completely
allow beta if candidate includes statement about the steel sheet being thin
A1
(4)
Page 34 of 51
(c)
(i)
1 disintegration / decay / particle emitted per second (per unit time)
not one count per second
B1
(1)
(ii)
correct curvature starting at 5 × 1020;
time scale inserted up to 15 (unit not necessary)
or labelled T1/2, 2T1/2, 3T1/2
M1
sensible scales (not multiples of 3);
correct number of atoms at each half-life;
reasonable curve and unit for time
A1
(2)
(iii)
half-life = 0.69 / decay constant
C1
4.1 – 4.2 × 10–9 (s–1)
A1
(2)
(iv)
A = (–)λN
C1
number of R / A atoms when activity is 1.5 × 1012 Bq = 3.6 × 1020
C1
correct time read from graph
A1
(2.5 y / 920 days / 8.0 × 107s)
or
determines original activity or final number of atoms
2.1 × 1012 Bq or 3.6 × 1020
allow ecf from (iii)
C1
N = N0e–λt or A = A0e–λt
C1
940 d or 2.6 y (answer depends on where rounding off has been done)
A1
(3)
[15]
11
(a)
(i)
A = 4; Z = 2
B1
(1)
(ii)
A = 0; Z = –1
B1
(1)
Page 35 of 51
(iii)
8 alphas
c.a.o.
B1
explanation consistent with their (i) / correct equations including values of A and
Z
B1
(2)
(iv)
6 betas
e.c.f.
B1
explanation consistent with their (i) and (ii) / correct equations including values
of A and Z
B1
(2)
(b)
(i)
probability of a nucleus decaying per unit time or
ratio of activity to number of nuclei (of that type) present or
with terms defined or
with terms defined
B1
(1)
(ii)
NP = N0 – N0e–λt / Nu = (Nu + Np) e–λt / Np = No – Nu
B1
correct manipulation
B1
(2)
(iii)
NP / NU = 0.62 / 4.2
C1
λ = (ln2) / 4.5 ×
or
109
0.69 / 4.5 × 109
C1
correct manipulation of logs
C1
9.0 ×
108
y / 8.96 ×
108
y
A1
look for alternative approach based on No = 4.82 and then
use of N = Noe–λt
(4)
(iv)
age will be overestimated
B1
not all of the Pb is the result of uranium decay or words to that effect
B1
(2)
Page 36 of 51
(v)
change of count rate for U will be small as half life is large
B1
comment about background count (rate) or background radiation
B1
absorption in sample
B1
presence of other radioactive nuclides / daughter nuclei
B1
max 2
[17]
12
(a)
number of gamma ray photons per sec =
(= 6.0 × 106)
B1
correct use of 4πr2; substitution of data
= 21.2
NB they may determine number per m2 and divide by 10 000
B1
(b)
(i)
decay constant = 0.69 / 12 = 0.0575 h–1 or 1.6 × 10–5
(or time =.5 half life)
Cl
dose = 21e–(6×0.0575)
dose = 21 / 20.5
Cl
or new (gamma) activity = 6 × 106 e–(6×0.0575)
or new (total) activity = 3 × 107 e–(6×0.0575)
Cl
15 (gamma rays per cm2 per second) Condone 14.8 – 14.9
(no up)
Al
(ii)
clear attempt to apply inverse square law
Cl
1.3 (1.26) m
Al
Page 37 of 51
(c)
beta particles are more heavily ionising than gamma radiation
or
loses energy rapidly by ionising the air / matter
Bl
beta particle range / penetration (in air) is low
or
beta particle range
is about 30 cm
or
is less than 1.5 m
or
is much lower than gamma radiation
NB: mention of not able to penetrate skin or clothing is talk out
Bl
[9]
13
(a)
number of alpha particles = 4
B1
number of beta particles = 2
B1
(2)
(b)
126
B1
(1)
(c)
Pb-206
B1
(1)
(d)
(i)
number of half lives = 2
or half life = 20 minutes
C1
40 minutes
A1
(2)
(ii)
4.2 × 1020
B1
(1)
Page 38 of 51
(iii)
decay constant = 0.69 / half life (allow e.c.f. from (i))
or
= N0e–λt1 / 2
C1
5.75 × 10–4 s–1 or 5.78 × 10–4 s–1 or 0.0345 min–1
(allow if calculation is done in (ii))
C1
A = λN
C1
2.4 (2.42) × 1017 Bq (or decays per s)
or 1.5 (1.45) × 1019 decays per minute
A1
(4)
(iv)
the (daughter) products are also decaying [or are radioactive]
M1
activity will be greater
A1
(2)
(v)
any 2 of:
polonium-214 has a half life of 1.6 × 10–4 s
B1
decays almost as soon as it is formed or decays very quickly
B1
only some of the bismuth-214 decays via polonium-214
B1
max 2
[15]
14
(a)
(i)
mention of radioactivity / decay / nuclear radiation
B1
ever present / independent of source being in proximity / always there / cannot
be eliminated
B1
(2)
(ii)
radon / rocks / cosmic rays / nuclear fallout / medicine / space / sun
B1
(1)
Page 39 of 51
(b)
A – activity / rate of decay
B1
λ – decay constant / probability of decay
B1
N – number of nuclei (radioactive atoms) present
not number of isotopes / atoms / particles
B1
[6]
15
(a)
Ac
B1
β
B1
suitable anti - neutrino indication (anti - neutrino not required)
B1
(b)
Down quark changes to up quark
B1
[4]
16
(a)
place a sheet of aluminium/metal between source and
detector
Sheet thickness stated 2 to 10 mm thick or several/a
few mm thick plus
M1
only gamma radiation can pass through such a sheet
or alpha and beta will be absorbed/stopped by the sheet
A1
2
(b)
count rate
1/r2 or evidence of C1/C2 = r22/r12
C1
25 counts per second (allow cps or s–1)
(Bq is a up)
A1
2
Page 40 of 51
(c)
their value calculated in (b) plotted correctly and reasonable
attempt to draw correct curvature
C1
Correct point (25 cps) plotted with correct curvature
(mark quality: must not flatten out or rise at end of their
line for large distances)
A1
2
[6]
17
Anti-neutrino indicated appropriately
B1
Beta 0, and –1
B1
Tl 81 Pb 208
B1
[3]
18
(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
Page 41 of 51
(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]
19
(a)
(i)
α (radiation) (1)
(ii)
γ (radiation) (1)
2
(b)
(i)
the radiation needs to pass through the body (to be detected) (1)
(ii)
(otherwise) the activity of the source becomes too weak
(during measurements) (1)
(iii)
the decaying source may remain in the body for a long time
and could cause damage (1)
[or the activity of the source will be low unless a large
quantity is used (T1 / 2 ∝ 1 / λ)]
3
Page 42 of 51
(c)
corrected count rate at 0.2 m (= 2550 - 50) = 2500 (c min-1) (1)
corrected count rate at least distance (= 6000 - 50) = 5950 (c min-1) (1)
use of I =
(or in the form
) (1)
(allow C.E. for using uncorrected count rate)
gives least distance = 0.20 ×
(1)
least distance = 0.13 m (1)
5
[10]
20
(a)
(i)
remove electrons from atoms
B1
by colliding with them/knocking into them
B1
2
(ii)
fewer collisions per metre
B1
because beta particles are smaller/less ionising
(accept smaller charge) not less mass
B1
so lose energy over a larger distance
B1
max 2
(b)
(i)
one correct construction shown on graph (52 s – 53 s)
M1
at least one other determination visible on graph
M1
averaging process shown
A1
3
Page 43 of 51
(ii)
N = 1.08 to 1.1 × 1021 (condone errors in powers of 10)
B1
A = λN
C1
1.4(3) × 1019 Bq (allow decays/second) allow ecf
for minor error in N
A1
or
attempt to find gradient at 72 s
correct extraction from a tangent of graph
1.3 to 1.5 × 1019 Bq (allow decays / second)
3
[10]
21
(a)
(proton) down 2/–2
B1
(nucleon) down 4/–4
B1
2
(b)
gamma greater, must express comparison,
don’t allow simple list
B1
1
Page 44 of 51
(c)
use of inverse-square law
C1
A1
2
[5]
Page 45 of 51
Examiner reports
1
2
4
Most students understood the questions about the range and dangers of ionising radiations but
many failed to gain marks over the details. Part (a) was done well by a majority of students but in
part (b)(i) there was a great deal of uncertainty about the range of alpha and beta particles. It was
common to see alpha particles having ranges over 10 cm. However, almost all students did put
the range of beta particles larger than alpha particles. Part (b)(ii) was done very well with a
majority of students referring to the inverse square relationship between intensity and distance. A
few did contradict themselves by quoting the inverse square relationship but then they talked
about the intensity falling off exponentially. The other successful students discussed the
spreading of the rays. Part (c) was again done well. Most realised that the dust had the potential
to be ingested usually by breathing it in. Some students did struggle with the mark on the
dangers of the ionising radiation. Some gave details of the damage that may be caused but failed
to say that the damage is caused by the ionisation. Others did not explicitly say humans might be
harmed or damaged, they simply said ionisation could occur in the body.
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.
(a)
(b)
(c)
(i)
In this straight-forward question it was surprising how many candidates thought that
the proton number would decrease. Most appreciated that the nucleon number
remained the same.
(ii)
This was usually done well. Common errors were to show the starting activity of 120
Bq at 12 h or to show an activity of 140 Bq at time = 0. Failure to complete the graph
for the 48 h period and very poor line drawing also cost some the third mark.
(i)
That the energy per second would be the activity x the energy per decay was
understood by relatively few candidates. The number of seconds in an hour or a day
appeared in the working of many candidates.
(ii)
Although there were many excellent answers the fact that another particle
(antineutrino) was the reason for the different energies was not as well known as one
would have hoped. Many concentrated only on the recoil of the nucleus. Others wrote
about the energy of emission becoming weaker as the source decayed, the
randomness of radioactive decay or the source becoming charged.
This was usually completed successfully.
Page 46 of 51
(d)
8
There were many good answers. Many candidates appreciated that particles were emitted
in all directions so that only a proportion of those emitted would reach the detector. A
misconception in some answers was that if the source had been shielded then all the
particles would have gone in the same direction (being reflected by the shielding). Other
points about the effect of absorption by the window to the tube and the limitations of the
tube in detecting particles that arrive at the same time, or within a certain time, were also
appreciated by many.
This question gave nearly every candidate the opportunity to gain some marks and a pleasing
number scored very well indeed.
Part (a) returned at least one mark to most candidates, but part (b) was answered poorly. The
examiners were hoping to be told that a typical background count-rate was much smaller than
the figures given in the table and negligible compared to the random fluctuations apparent in
these data.
A surprising number of candidates were unable to plot all four points accurately in part (c), and
some did not even attempt to do so. The better candidates averaged two or more half-life
readings taken from their line of best fit in part (d), although some then lost a mark for an
incorrect unit.
Most candidates provided at least one appropriate reason in answer to part (e).
9
(a)
This was rather disappointingly answered with a number of inconsistencies. Isotope was
rarely used as a comparative term and candidates frequently referred to an isotope as
being ‘an element with the same proton number but different neutron numbers’. Half-life
was often referred to as being ‘the time for the number of atoms / nuclides / molecules to
be reduced to half’. None of these responses was given credit. Candidates should be
encouraged to talk in terms of decay of nuclei or the activity of a sample of material.
(b)
This part was done relatively well, although a minority of candidates was very confused by
the nomenclature used in the equation representing this nuclear reaction.
(c)
This part was well answered with the majority of candidates being able to calculate a value
for the decay constant and the majority of those candidates being able to use their value in
order to calculate a sensible time for the reduction of the activity. Weaker candidates were
awarded a compensation mark for estimating that the time would be between six and
seven half-lives.
(d)
Although the majority of candidates recognised that the γ radiation would be detected
outside the body, few supported their assertion in an unambiguous manner. Weaker
candidates suggested numerous different ‘decay products’ ranging from α and β particles
to iodine and xenon.
Page 47 of 51
10
(a)
There were many well explained answers to this part. However, the question exposed many
misunderstandings. Many candidates wrote that either the source or radioactive particles
passed through the material. Some thought that the material itself was radioactive.
Although many referred to less radiation reaching the detector, it was disappointing how
few referred to the radiation being ‘absorbed’. (Did they think it was reflected?) Many
candidates referred to radiation being detected or not detected as thickness changed rather
than that there being a variation in count rate.
(b)
(i)
To gain credit in either part the correct source first had to be identified. There were
many who stated that alpha sources should be used for paper although the fact that
alpha particles are absorbed by paper and travel only a short distance in air should
be well-known.
(ii)
A beta source was stated by many to be suitable. This was allowed only if they also
stated that the steel would be thin. Many stated or implied in one or other of the two
parts that gamma radiation could pass through anything without any change in
intensity.
(i)
There was a surprising number of incorrect answers to this. Statements such as ’1 Bq
means that one radioactive atom is radiated from the source per second’ or simply
that ‘it is the activity of a source’ were not uncommon. Many associate the value with
the count rate of a detector rather than a property of the source.
(ii)
Poor graph drawing skills cost many candidates a mark here. To gain the first mark
the correct value at t = 0 had to be plotted and indicated at 5 × 1020 and the curvature
had to be correct though not accurate. For the second mark the scale should have
been sensible (e.g. not 5.3, 10.6 etc at the 2 cm grid markings), and the values were
expected to be reasonably accurate at times equal to 1, 2 and 3 half lives.
(iii)
Most candidates did this correctly. Common faults were giving the answer as 0.13
year−1 or as 0.13 s−1.
(iv)
Whilst there were many correct answers, many were confused. A common response
was 1.5 × 10−2 = 5 × 1020 e−λt. These candidates did not appreciate that they needed
to find either the original activity or the final number of radioactive atoms. There were
also many instances where units were mixed. Calculation of the number of atoms
remaining when the activity is 1.2 × 1012 Bq and reading the time from the graph was
an expected, easy route to the answer but this approach was rarely used.
(c)
Page 48 of 51
12
14
15
Only the better candidates scored well on this and many of these obtained full or near full marks.
However, it was disappointing that this was generally a low scoring question.
(a)
The need to identify the approach and then to extract the appropriate data from both the
stem and the question proved to be too demanding for most. Candidates needed to
appreciate that the gamma activity is 1 / 5 of the total activity and that the gamma ray
photons pass through an area of 4πr2. Many tried using the decay formula in this part.
(b)
(i)
Candidates are required to be able to use the decay equations and here the decay
formula could be applied here using the total activity or the gamma activity or the
value 21 given in (a), the last being the easiest approach. Many thought that if the
activity halves in 12 hours it will be reduced by a quarter in 6 hours. Errors made by
those who used the correct approach usually involved an incorrect decay constant or
inability to do the arithmetic.
(ii)
Candidates could gain credit if it was clear that they were applying the inverse square
law but this was far from evident in many scripts.
(c)
The majority of candidates stated that beta particles have a short range but few stated why
this was the case. A significant proportion demonstrated unclear thinking by saying that the
beta particles would not pass through clothes or skin hence talking themselves out of the
mark. A worrying number thought that beta particles were harmless since the skin prevents
them entering the body. Of those who mentioned ionisation a number thought that the beta
particles themselves are ionised.
(a)
(i)
Although many candidates were able to suggest a sensible source for background
radiation few gave a full enough explanation, of what it is, to gain full credit
(ii)
It was unclear in many cases whether or not candidates understood that the source of
background radiation was radioactive decay.
(b)
Most candidates were able to define the symbols in the equation but too often, having
given a correct quantity, the candidate would then give an alternative meaning for the
symbol that was contradictory – inevitably this was penalised. Unsurprisingly, many weak
candidates took λ to mean wavelength.
(a)
The equation was completed well by a large proportion of the candidates but there was a
significant number who could make no valid attempt. The β– was usually correct, but
common errors were to quote
together with an ambiguous symbol for the
antineutrino. Examiners required candidates to make it clear that the third particle was an
anti-neutrino.
16
(b)
This was well answered by the majority who knew that a down quark changes to an up
quark when the neutron decays to a proton.
(a)
Most realised that a metal sheet between the source and detector was needed.
However, many did not give a satisfactory thickness for the metal and others did not say
clearly where it would be placed. Weaker candidates thought that taking a background
count rate and subtracting it was what was expected.
Page 49 of 51
17
19
(b)
Relatively few candidates were able to do this successfully and most showed no
appreciation of the application of the inverse square law to gamma radiation. Many worked
with differences or assumed inverse proportionality. Some of those who obtained the
correct value failed to include the unit.
(c)
It was essential for candidates to show the correct general curvature in this part. For full
credit they needed to have applied the inverse square law successfully in (b) and to have
plotted the correct value on the graph. Many of those who understood the inverse square
law went on to calculate other points correctly when drawing their graph.
There remains a widespread inability to balance this beta-emission equation correctly.
Candidates do not know the correct values of proton and nucleon number for the beta particle,
they could not subsequently balance the equation correctly, and very few indicated the neutrino
fully, often omitting to indicate that it is an anti-neutrino in this case. Few candidates scored three
marks on this comparatively simple question.
The question on nuclear instability performed well on the whole and many candidates gained
high marks. Most of the answers to part (a) had the á radiation correct, but the answer to part (ii)
seemed to be guesswork between β radiation and γ radiation.
In part (b) most candidates scored at least one mark and the majority, two marks, but very few
were awarded all three. The main problem seemed to be part (i) which asked why a ã source
was used. A large number of answers to this were disregarded when candidates referred to the
source and not to the radiation being able to pass through the body. In general the answers to
parts (ii) and (iii) were sensible and overall the majority of candidates seemed to have good
understanding of this part of the course.
The calculation in part (c) provided an accessible five marks, but many candidates forfeited two
marks by not taking into account the mean background radiation. A surprising number corrected
one count only, usually the 2550 count, but failed to apply a correction to the other count rate.
The majority of candidates applied the inverse square law correctly, even if sometimes the
calculation became long winded, but it usually ended with the correct answer. The examiners
were concerned for those candidates who thought that the constant k in the equation was the
Boltzmann constant, and also for those candidates who used the corrected count rate as I and
the uncorrected count rate as Io. Both these fundamental errors indicated complete
misunderstanding of the equation.
20
(a)
(i)
This part was done reasonably well. Some candidates distracted themselves with
ideas about positive and negative ions or with the idea that the alpha particles were
themselves ions.
(ii)
Most of the candidates knew that beta particles were less ionising or that they
suffered fewer collisions. They found it more difficult to express ideas relating to the
number of collisions per millimetre of travel or the consequent notion that alpha
particles lost all of their kinetic energy over a smaller distance than beta particles
Page 50 of 51
(b)
21
(i)
Many candidates could use the graph to find the half life. Some made mistakes over
the interpretation of the scales and others did not show their constructions clearly
enough for credit. Fewer candidates thought it worth finding more than one
determination from the graph and then averaging their answers.
(ii)
This was done quite well but significant figure errors and unit penalties were common
and a surprising number of candidates made powers often mistakes.
(a)
Another simple question for which many gained full marks.
(b)
Candidates were asked for the relative penetrating powers of alpha and gamma radiation.
Far too many produced standard lists of materials that alpha and gamma can penetrate
without giving a summary statement. A significant minority interpreted the question as
requiring a comparison between alpha-particle and beta-particle penetration.
(c)
Roughly 50% of the candidates scored full marks on this question where the normal rules
of significant penalty errors were eased (allowing an answer of -). However, answers of 9
were common even from those acknowledging the correct use of the inverse-square law.
They did not appear to be able to cope with the arithmetic manipulations required.
Page 51 of 51