A2 Physics Unit 9
Radioactivity
Mr D Powell
Chapter Map
Common Issues
Take care not to mix
units when calculating.
For instance, you may
encounter activities
given in Becquerel's and
count rates stated in
counts per minute; half
lives are often quoted in
reference works using
the most relevant time
unit, be it seconds or
years.
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9.1 Radioactivity
1. How big is the nucleus?
2. How was the nucleus discovered?
3. Why was it not discovered earlier?
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Alpha Decay....
Alpha particles tracks
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Probing the Nucleus – Rutherford Scattering
3
1
By firing alpha particles at a heavy
gold nucleus Rutherford could easily
see that atoms were mostly space
with
a large positive nucleus in the
Rutherford alpha particle scattering experiment
centre.
lead block to
select narrow
beam of alpha
particles
thin gold
foil
microscope to view
zinc sulphide screen,
and count alpha
particles
4
2
nucleus
2
90
1
135
3
50
4
30
vary angle of
scattering
observed
radium
source of
alpha
particles
alpha particle
beam
zinc sulphide
screen, tiny dots of
scattered alpha
light where struck
particles
by alpha particle
5
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5
20
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Scattering Formulae (extension)
There are complex formulae that we can use to work out the distance of
closest approach.
You will
not have to learn these but should appreciate
Rutherford’s
picture of alpha
scattering
some of the maths involved. At the very least the idea of the forces
For involved
calculations
force F =
alpha particle
scattered
2Ze2
40d2
charge +2e

‘aiming error’ b
scattering angle
d
Assumptions:
alpha particle is the He nucleus, charge +2e
gold nucleus has charge + Ze, and is much more
massive than alpha particles
scattering force is inverse square electrical
repulsion
gold nucleus
charge + Ze
equal force F but
nucleus is massive,
so little recoil
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Distance of Closest Approach (extension)
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Distance of Closest Approach (extension)
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Summary Question...
For an alpha particle with an initial KE of 6MeV fired at a gold nucleus find the
distance of closest approach of the alpha and the nucleus...
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9.2 Alpha / Beta / Gamma
1. What are alpha, beta, gamma
2. Why is it dangerous
3. What are the properties of alpha, beta, gamma
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Link them up?
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Guess the field?
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Particle Tracks
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Calibration plateau p.d...
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9.3 More on Alpha / Beta / Gamma
1. What happens to the nucleus in a radioactive change
2. How does the intensity spread out
3. How do we represent the changes in the nucleus in alpha, beta,
gamma.
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How does the intensity spread out
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Lines of stability...
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Emitters....
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Corrected count rates & distances....
We must find the
source centre for
inverse sq law to
work correctly.
d0
d
k
C  C0 
2
d  d 0 
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9.4 The Dangers of Radioactivity
1. Why is ionising radiation
2. What factors determine whether , , are the most dangerous
3. How can exposure to ionising radiation be reduced
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Background Radiation
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Sievert
The sievert (symbol: Sv) is the SI derived unit of dose equivalent. It attempts to reflect the
biological effects of radiation as opposed to the physical aspects, which are characterised by
the absorbed dose, measured in gray. It is named after Rolf Sievert, a Swedish medical
physicist famous for work on radiation dosage measurement and research into the biological
effects of radiation.
1 Sv = 1 J/kg = 1 m2/s2 = 1 m2·s–2
Frequently used SI multiples are the millisievert (1 mSv = 10–3 Sv) and microsievert (1 μSv =
10–6 Sv).
The millisievert is commonly used to measure the effective dose in diagnostic medical
procedures (e.g., X-rays, nuclear medicine, positron emission tomography, and computed
tomography). The natural background effective dose rate varies considerably from place to
place, but typically is around 2.4 mSv/year. For acute (that is, received in a relatively short
time, up to about one hour) full body equivalent dose, 1 Sv causes nausea, 2-5 Sv causes
epilation or hair loss, hemorrhage and will cause death in many cases. More than 3 Sv will
lead to LD 50/30 or death in 50% of cases within 30 days, and over 6 Sv survival is unlikely.
(For more details, see radiation poisoning.)
The collective dose that a population is exposed to is measured in "man-sieverts" (man·Sv).
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9.5 Radioactive Decay
What do we mean by the activity of an isotope Activity A = λN. or
Energy Transfer per second = AE
What is the half-life of an isotope t1/2 = 0.69/λ
Does anything affect radioactive decay
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Rate of Decay is the decay constant...
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Dice Results - 300
Number of sample
350
300
Number of Atoms Left
250
y = 288.79e-0.164x
R² = 0.9922
200
150
100
50
0
0
5
10
15
20
25
Time elaspsed arbitrary
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Gamma 1....
Villard
Paul Villard, a French physicist, is credited with discovering gamma
rays in 1900. Villard recognised them as different from X-rays
(discovered in 1896 by Roentgen) because the gamma rays had a
much greater penetrating depth. It wasn't until 1914 that Rutherford
showed that they were a form of light with a much shorter wavelength
than X-rays.
Roentgen
High-energy photons emitted as one of the three types of radiation
resulting from natural radioactivity are the most energetic form of
electromagnetic radiation, with a very short wavelength (high
frequency). Wavelengths of the longest gamma radiation are less than
10 -10 m, with frequencies greater than 10 18 hertz (cycles per sec).
Gamma Emission occurs primarily after the emission of a decay
particle. Gamma is a form of high energy electromagnetic radiation.
After a particle is ejected from a nucleus the system may have some
slight excess of energy, or exist in a metastable state. This slight excess
of energy is released as gamma. Gamma emission will not change the
isotope or the element. The wavelength of the emitted gamma
radiation will be unique to each isotope. Gamma emission is a
significant health risk.
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Rutherford
Index
Gamma 2
Villard
After a decay reaction, the nucleus is often in an “excited” state.
This means that the decay has resulted in producing a nucleus
which still has excess energy to get rid of. Rather than emitting
another beta or alpha particle, this energy is lost by emitting a
pulse of electromagnetic radiation called a gamma ray.
Roentgen
The gamma ray is identical in nature to light or microwaves, but
of very high energy.
Like all forms of electromagnetic radiation, the gamma ray has no
mass and no charge. Gamma rays interact with material by
colliding with the electrons in the shells of atoms. They lose their
energy slowly in material, being able to travel significant
distances before stopping.
Depending on their initial energy, gamma rays can travel from 1
to hundreds of meters in air and can easily go right through
people.
Rutherford
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Gamma Emission
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Gamma Emission
1.
Gamma is a process
which often
happens in addition
to a beta decay
2.
It can often have
different energies
attached from
differing transitions.
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Alpha Decay
The nucleus is initially in an unstable energy state. An internal
change takes place in the unstable nucleus and an alpha
particle is ejected leaving a decay product. Alpha particles have
a velocity in air of approximately one-twentieth the speed of
light, depending upon the individual particle's energy.
Since the number of protons in the nucleus of an atom
determines the element, the loss of an alpha particle actually
changes the atom to a different element. For example,
polonium-210 is an alpha emitter. During radioactive decay, it
loses two protons, and becomes a lead-206 atom, which is
stable (i.e., nonradioactive).
Examples are;
Americium-241 , Plutonium-236 , Uranium-238
Thorium-232 , Radium-226 , Radon-222 , Polonium-210
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Alpha Decay
The positive charge of alpha particles is useful in some
industrial processes. For example, radium-226 may be
used to treat cancer, by inserting tiny amounts of radium
into the tumorous mass. Polonium-210 serves as a static
eliminator in paper mills and other industries. The alpha
particles, due to their positive charge, attract loose
electrons, thus reducing static charge.
Some smoke detectors take advantage of alpha
emissions from americium-241 to help create an
electrical current.
The alpha particles strike air molecules within a
chamber, knocking electrons loose. The resulting
positively charged ions and negatively charged electrons,
create a current as they flow between positively and
negatively charged plates within the chamber. When
smoke particles enter the device, the charged particles
attract them, breaking the current and setting off the
alarm.
Tumors on liver on the top
left
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Alpha Decay
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Beta Decay
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Protactinium Generator
This experiment uses a 'protactinium generator' to show the exponential decay
of protactinium-234, a grand-daughter of uranium. It has a half-life of just over
a minute, which allows the chance to measure and analyse the decay in short
time.
The bottle contains a complex mixture of substances but when you have shaken
the bottle the and organic layer separates out which contains the protactinium234. This decays with a half-life of 68 seconds and can be monitored.
U 238  Th234  Pa 234  U 234  Th230
U238 T0.5 = 4.47 Billion years
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Protactinium Generator
The Pa234 is a beta emitter 2.3MeV (max) which escapes the tube and leaves
U234 which has a very long alpha half life.
We can monitor the Pa activity in the top layer and any newly spawned Pa from
the bottom layer is prevented from contaminating the top layer.
Hence, we have an initial sample of Pa and can see over a period of time the
change in activity.
70s
years
Pa 234  U 234  Th230
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Graphical Decay
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Real Data
Now try plotting a graph and prove some
formulae
1.
Plot one Loge graph or Ln graph
2.
Plot one normal graph
3.
Show via inspection and
mathematically the formulae below.
Test each one out and see if it is true!
t1/2 = 0.69/λ = ln2/ λ
Time in
seconds
0
20
40
60
80
100
120
140
160
180
200
Activity Recorded
in Bq
210
165
130
110
95
75
65
53
29
28
25
N = N0e-λt
A = A0e-λt
Activity A = λN
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Example Results & Proof of Formulae....
Can you now prove any of these relations by drawing a graph and taking
some readings?
t1/2 = 0.69/λ
N = N0e-λt
A = A0e-λt
Activity A = λN.
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250
Counts in 10 seconds
200
150
y = 213.38e-0.011x
R² = 0.9791
100
50
0
0
50
100
150
200
250
Time in seconds (s)
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Dice Exp...
76 dice were taken and thrown and then the number of dice on a 1 were
counted and removed. The process continued until there were no dice left.
The results were graphed in two different ways by activity and mass
Activity
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Activity
12
10
8
4
14
6
4
3
5
4
1
1
1
1
1
1
1
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Mass
76
64
54
46
42
28
22
18
15
10
6
5
4
3
2
1
1
16
14
12
10
8
6
4
2
0
y = 12.379e-0.181x
R² = 0.8281
0
5
10
15
20
Mass
120
100
y = 105.9e-0.282x
R² = 0.9808
80
60
40
20
0
0
5
10
15
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Index
Time
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Number
300
261
218
194
162
133
110
88
65
54
48
43
37
34
31
25
21
19
17
13
11
300 dice were taken
and thrown and then
the number of dice on
a 1 were counted and
removed. The process
continued until there
were no dice left. The
results were graphed
50
45
40
35
30
25
20
15
10
5
0
y = 43.672e-0.169x
R² = 0.8918
0
5
10
15
20
25
Time elaspsed arbitrary
Number of sample
350
300
Number of Atoms Left
Time Activity
0
39
1
43
2
24
3
32
4
29
5
23
6
22
7
23
8
11
9
6
10
5
11
6
12
3
13
3
14
6
15
4
16
2
17
2
18
4
19
2
20
2
Acticity of Sample
Activity
250
y = 288.79e-0.164x
R² = 0.9922
200
150
100
50
0
0
5
10
15
Time elaspsed arbitrary
20
25
Write a simple definition for..
T0.5 =
=
No =
N=
A=
A0 =
t =
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9.6 Radioactive Decay Theory
Can a source decay completely
What is an exponential decrease
Why is the process random
t1/2 = 0.69/λ
A = λN
N = N0e-λt or A = A0e-λt
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T1/2
1. The number of radioactive nuclei decaying per second (activity) is
proportional to number of nuclei remaining
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Decay Constant
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Decay Constant
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Decay Constant
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Decay Curves...
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Decay Curve
This curve is an exponential decay
It takes the same time for half the
sample to decay i.e. Activity reduces
over time.
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Decay Curves...
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Decay Curves...
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The Decay Equation 1
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Decay Equation 2
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Equivalence of Formulae
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Half - Life
The T0.5 time for an isotope is
defined as the time taken for half
the sample to decay
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Exam Question...
Using the decay equation...
Show that
4 marks...
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Half – Life 2
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Example of using Equation
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T1/2 Examples...
TASK
Can you work out the
decay constant for
each isotope of
carbon?
C9 - 5.457 s-1
C10 – 0.036 s-1
C11 – 5.7 x 10-4 s-1
C12 – 0 s-1
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Example Curve
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9.7 Isotopes in Use
How do we do radioactive dating
What are radioactive tracers
What do we use radioactivity for in hospitals.
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Can you plot this data to find out half life?
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Carbon Dating
14
7
N  n C  p
14
6
C  N  e
14
6
beta decay
14
7
Here is a good example of capture of
a neutron into the nucleus
followed by decay back again via a
beta minus process.
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
Maths for Dating?
Imagine you start with a blob of a man and want to find out the number
of original atoms in the sample which were radioactive.
t1/2 = 0.69/λ
t1/2 = 5700 years so λ = 1.216 x 10-4 years-1 or work out in seconds
N = N0e-λt
Now you know No from the mass spectrometer but you don’t know the
time it has taken from t = 0 i.e. When it died.
You now need to know the activity now (Geiger-Muller tube) and the
activity of a sample of man who is living to compare. If in Bq then λ is in
seconds
Take the ratio of activities & work through to find age, then work back to
find N0 of original sample. (see book page 168)
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Tricks with 2N?
T0.5
N
Series
If you think about the idea of the number of
atoms left after a number n of half lives.
1
1/2
21
2
1/4
22
You can then construct a very useful equation.
3
1/8
23
This enables you to predict the number of
atoms left if you know the original number of
atoms and the T0.5 time.
4
1/16
24
5
1/32
25
Or you can work out the number of T0.5 times it
will take to reduce the number of atoms or
activity to a certain amount.
N  N 0  0.5
n
N
n
 0.5
N0
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9.8 More about Decay Modes
1. What can we tell about radioactive isotopes from an N-Z chart
2. Why don’t naturally occurring isotopes emit
+
radiation?
3. What happens to an unstable nucleus that emits
radiation.
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Technetium Doses...
Also, a commonly-used measure of radioactivity
is the microcurie:
Can you estimate out
the activity per
second?
1 μCi = 3.7×104 disintegrations per second =
2.22×106 disintegrations per minute
How many Bq?
The typical human body contains roughly 0.1 μCi
of naturally occurring potassium-40.
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Stable or Unstable Nuclei
e and unstable nuclei: balance of numbers of protons and neutrons
stable
alpha decay
150
beta decay
electron capture
positron emission
fission
100
N= Z
50
0
0
20
40
60
80
100
proton number Z
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Pauli cliffs
The sides of the valley rise steeply. The Pauli exclusion principle keeps numbers of
protons and neutrons approximately equal. Binding is less strong if either are in excess.
Positron decay
Valley of
Stability
Free particle plains
Z
N
Z–1
N +1
+
Electron decay
–
Z
N
proton number Z
Binding energy taken as zero for
free protons and neutrons.
Z+1
N –1
proton number Z
Bound nuclei have negative
binding energies, down to about
–8 MeV per nucleon
Binding
energy per 0
nucleon/MeV
Fusion hill
–2.5
Larger numbers of
neutrons and protons are
bound more strongly than
smaller numbers, by the –5.0
strong nuclear
interaction.
–7.5
These nuclei go down hill
by nuclear fusion.
0
Coulomb slope
More neutrons than
protons are needed
to overcome
electrical repulsion
of many protons.
25
These nuclei
go down hill
by alpha
decay
50
‘Iron lake’
75
Nuclei near iron in the Periodic
Table are the most strongly bound,
and lie in the bottom of the nuclear
100
proton
valley. Other nuclei tend to ‘run
number Z
downhill’ to become like iron.
0
40
80
neutron number N
120
160
Alpha decay
Z
N

Stable nuclei lie along a narrow band of values of numbers of protons
and neutrons. The more negative the binding energy, the more stable
the nucleus.
Z–2
N –2
proton number Z
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half-life 4.5  109 yr
Decay Chains .....
Decay chain from Th-232
146
142 neutron number N
Th-232

–
Th-228
134

132
Po-216

Pb-212
130
128 Tl-208
–

Bi-212
–
Rn-220

138

Ra-226
Rn-222

Po-218

Th-230
Main source of
naturally
occurring
radium, half-life
1600 yr. Radium
was discovered
by Marie Curie
and her husband
Pierre Curie.
Pb-214
–
Bi-214
–
130
Po-214

128
Pb-210
–

126
140
132
Po-212
U-234

Radon gas, half-life 56 s, 136
from building materials is
an important part of the 134
natural background
radiation
Marie Curie, discoverer with
Pierre Curie of polonium,
named the element after her
native Poland
Pa-234
–
Ra-224

–
142

136
Th-234
144
Ac-228
–
138
U-238


Ra-228
140
neutron number N
126
Pb-208
Bi-210
–
Po-210

proton number Z
81 82 83 84 85 86 87 88 89 90
Tl Pb Bi Po At Rn Fr Ra Ac Th
124
Pb-206 stable
proton number Z
82 83 84 85 86 87 88 89 90 91 92
Pb Bi Po At Rn Fr Ra Ac Th PaIndex
U
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Review...
Why don’t naturally occurring isotopes emit
+
radiation?
1.
they are all manmade by a process of bombarding stable nuclei with
protons.
2.
Hence, protons need to be fast enough to overcome coulomb
repulsion.
3.
If elements gain a proton they actually become more unstable and
not more stable and move away from line. So Copper for example
will never give on + as it would become more unstable.
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9.9 Nuclear Radius
1. Are more massive nuclei wider?
2. How does the radius of a nucleus depend on its mass number A?
3. How dense is the nucleus?
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Small Nucleus....
Nucleons....
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Estimate Nuclear Diameter with “Electron Diffraction”
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Quick Question
hc

E
1.22
sin  
d
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Atomic Scale
Can you estimate the distance of the outer shell electron for a gold atom?
If you are not sure just pick a figure......
3.3 miles
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Atomic Radius....
Various types of scattering experiments
suggest that nuclei are roughly spherical
and appear to have essentially the same
density.
The data are summarised in the
expression called the Fermi model;
r  ro A
1
3
r0 = 1.2 x 10-15m = 1.2fm
r = atomic radius
A = mass number (nucleons)
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Atomic Radius....
Various types of scattering experiments
suggest that nuclei are roughly spherical
and appear to have essentially the same
density.
The data are summarized in the
expression called the Fermi model;
r  ro A
1
3
r0 = 1.2 x 10-15m = 1.2fm
r = atomic radius
A = mass number (nucleons)
Element
A
A1/3
R in fm
Carbon
Oxygen
Silicon
Calcium
Vanadium
Strontium
Indium
12
2.29
2.75
16
2.52
3.02
28
3.04
3.64
40
3.42
4.10
50
3.68
4.42
88
4.45
5.34
115
4.86
5.84
FACT
The diameter of the nucleus is in the range of
1.6fm (1.6 × 10−15 m) (for a proton in light
hydrogen) to about 15 fm (for the heaviest
atoms, such as uranium).
Can you plot a graph and show that this
formulae is correct and the constant is
1.2fm
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Finding R0
Nuclear Radius
8.00
7.00
6.00
R in fm
5.00
y = 1.2x + 7E-14
R² = 1
4.00
3.00
2.00
1.00
0.00
0.00
2.00
4.00
6.00
cuberoot of A ( Nucleon Number)
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Index
Summary & Question
1. Calculate the radius of an oxygen nucleus which has 16 nucleons..
r=r0A 1/3= 1.4 x 10-15 x (16)1/3
r=3.5 x 10-15m (3.5fm)
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Nuclear Density – is massive!
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Reflection high-energy electron diffraction (RHEED)
Reflection high-energy electron diffraction (RHEED) is a technique used to characterise
the surface of crystalline materials.
RHEED systems gather information only from the surface layer of the sample, which
distinguishes RHEED from other materials characterization methods that rely also on
diffraction of high-energy electrons.
Transmission electron microscopy, another common electron diffraction method samples
the bulk of the sample due to the geometry of the system.
Low energy electron diffraction (LEED) is also surface sensitive, but LEED achieves surface
sensitivity through the use of low energy electrons.
http://en.wikipedia.o
rg/wiki/Reflection_hi
gh_energy_electron_
diffraction
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Reflection high-energy electron diffraction (RHEED)
The image shows the setup of the electron gun, sample and detector /CCD
components of a RHEED system. Electrons follow the path indicated by the
arrow and approach the sample at angle θ.
The sample surface diffracts electrons, and some of these diffracted electrons
reach the detector and form the RHEED pattern.
The reflected (specular) beam follows the path from the sample to the detector.
Specular reflection is the mirror-like
reflection of light (or of other kinds of wave)
from a surface, in which light from a single
incoming direction (a ray) is reflected into a
single outgoing direction
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Wave nature of the electron
• Louis de Broglie (1892 - 1987)
• If light can be modeled as a particle or as a wave, can an electron
be modeled as a wave?
• The wavelength of a matter wave (1923) is given by:
• Everyday objects are too massive to give observable
wavelengths; however, electrons are light enough to give
observable wavelengths. Diffraction of electrons was observed
by two groups in 1927, Davisson & Germer and George
Thomson.
• The Bohr model could also be explained using standing waves.
• Whole numbers (1,2,3,etc.) of de Broglie wavelength give the
allowed radii found in the Bohr model.
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Experiments of GP Thompson
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Electron Waves
Two specific examples supporting the wave nature of electrons as suggested in the
DeBroglie hypothesis are the discrete atomic energy levels and the diffraction of electrons
from crystal planes in solid materials. In the Bohr model of atomic energy levels, the
electron waves can be visualized as "wrapping around" the circumference of an electron
orbit in such a way as to experience constructive interference.
Bragg diffraction. Two beams with identical wavelength and phase
approach a crystalline solid and are scattered off two different atoms
within it. The lower beam traverses an extra length of 2dsinθ. Constructive
interference occurs when this length is equal to an integer multiple of the
wavelength of the radiation.
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Microstructure of Nickel Superalloys
• These pictures are taken
using electron diffraction.
They are trying to establish
the structure and if there are
any problems in the
structure of Nickel alloys
used in aeroplane
manufacture. It is important
that any micro fissures are
detected early on.
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Experiments of Davisson and Germer 1927
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Experiments of Davisson and Germer 1927
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De Broglie Summary / Questions
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Exam Questions
Q2 June 07....
The table shows data for some nuclei.
1 eV = 1.6 ×10-19 J & speed of electromagnetic radiation = 3.0 × 108 ms-1
(a) (i) Show that these data support the rule that where R0 is a constant;
R = R0A(1/3)
(ii) The mass of a nucleon is about 1.7 × 10-27 kg. Calculate the density of nuclear matter. (6
marks)
(b) (i) Explain what is meant by the binding energy of a nucleus.
(ii) Show that the total binding energy of a sodium-23 nucleus is about 3 × 10-11 J.
(iii) Calculate the mass-equivalent of this binding energy.
(5 marks)
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Q4 June 05....
Americium-241 ( Am) is a common laboratory source of alpha radiation. It decays
spontaneously to neptunium (Np) with a decay constant of 4.8 × 10-11s-1. A school
laboratory source has an activity due to the presence of americium of 3.7 × 104Bq
when purchased. Avogadro constant = 6.0 × 1023mol-1 one year = 3.2 × 107s
(a) (i) Calculate the half-life, in years, of americium-241. (2 marks)
(ii) Calculate the number of radioactive americium atoms in the laboratory source
when it was purchased.
(2 marks)
(iii) Calculate the activity of the americium in the laboratory source 50 years after
being purchased.
(3 marks)
(iv) Suggest why the actual activity of the sources is likely to be greater than your
answer to part (iii). (1 mark)
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Q4 June 05.... (answers)
(a) (I) half life = 1.44 x 1010 s or half life = 0.69/λ or ln2/λ
Or
450 (449) y (or 460(456)y if they use 3.15 x 107 s = 1y)
II) A = λN = 7.7 x 1014
III)
IV) the decay products/neptunium may also be radioactive
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Q4 June 05....
(b) (i) Use the following data to deduce the energy released in the decay of one
Americium-241 nucleus.
mass of americium-241 nucleus = 4.00171 × 10-25kg
mass of an alpha particle = 0.06644 × 10-25kg
mass of neptunium nucleus = 3.93517 × 10-25kg
speed of electromagnetic radiation in free space = 3.00 × 108ms-1
3 marks
(ii) Explain what is meant by decays spontaneously and how consideration of the
masses of particles involved in a proposed decay helps in deciding whether the
decay is possible.
2 marks
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Q1 June 06....
1 (a) Radioactive lead-214 changes to
lead-206 by a series of decays involving
alpha (α) and negative beta (β.)
emissions. Explain clearly how many
alpha and beta particles are emitted
during this change. (4 marks)
(b) The half-life of lead.214 is 26.8
minutes.
(i) Explain what is meant by half-life.
(ii) Show that the decay constant of lead
214 is approximately 0.026 minute-1.
(iii) Calculate the percentage of the
original number of nuclei of lead-214
left in a sample after a period of 90
minutes. (7 marks)
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Q2 June 07....
(c) Nuclear structure can be explored by bombarding the nuclei with alpha particles.
The de Broglie wavelength of the alpha particle must be similar to the nuclear
diameter. Calculate the energy of an alpha particle that could be used to explore the
structure of manganese-56. (4 marks)
Planck constant = 6.6 × 10-34 J s
mass of an alpha particle = 6.8 × 10-27 kg
wavelength = nuclear diameter of manganese = 9.2 × 10-15 m
momentum = h/λ = 7.17 × 10-20 Js m-1 or (N s)
=h/p or h/mv as p = mv so as Ek = ½mv2 we can sub in;
Ek = p2/2m or can use = h/mv so v = h/m
Using velocity = 1.05-1.06 × 107 ms-1
3.7(4) -3.8(1) × 10-13 J
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Q2 June 07....
(d) (i) State the proton number and nucleon number of the nucleus formed by the
decay of manganese-56.
Proton number......... Nucleon number.........
(ii) The activity of a sample of manganese 56 varies with time according to the
equation
A = A0e-λt
What value should be used for λ in calculations involving manganese-56 when t is in
seconds?
(4 marks)
Z = 26
A = 56
λ = 0.69/half life or 0.69/2.6
7.37 - 7.41 × 10-5 (s-1)
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