Nuclear Physics
UoD Faculty of Science, Department of Physics
2013 - 2014
Dr. Tom Schulman
Lesson 3
Radioactivity
(Lilley p. 14 – 26, Krane p. 160 – 191)
Theory
Radioactivity
Let us review figure 1.2 of lesson 1. According to the nuclear chart of figure 1.2 those nuclei
with a high neutron or proton number are unstable, i.e., radioactive nuclei. If the proton number is
high the repulsive Coulomb force is the reason to the instability. This can be explained by the liquid
drop model of the nucleus. But the reason to the instability of a nucleus with a high neutron number
cannot be understood from the liquid drop model. According to the shell model, however, the Pauli
exclusion principle is valid not only for the electrons of an atom but also for the nucleons of an
atomic nucleus. Therefore, additional neutrons have to settle at higher energy levels. This decreases
the binding energy/nucleon and thus the nucleus with a high neutron number prefers to transfer to a
more stable state through radioactive decay. Radioactive decay can occur through three main
processes:
1. Alpha () decay (figure 3.1) – the nucleus emits an  particle (i.e., a 24 He nucleus), for
238
4
example 92
U 234
90Th 2 He .
2. Beta- (-) decay (figure 3.2) – one neutron in the nucleus converts to a proton by emitting a
14

- particle (i.e., an electron), for example 14
6 C 7 N  e .
3. Beta+ (+) decay (figure 3.2) – one proton in the nucleus converts to a neutron by emitting a
11

+ particle (i.e., a positron), for example 11
6 C 5 B  e .
Figure 3.1
Figure 3.2
- radiation may also be the result of a process called internal conversion. In internal
conversion an excited nucleus emits a gamma ray which interacts with one of its own electrons. The
electron acquires the energy of the gamma ray and escapes from the atom as a - particle.
A proton rich nucleus may capture one of its own electrons. One proton then converts to a
neutron without emitting a positron. Electron capture is followed by the emission of characteristic
X-rays when electrons at higher energy levels fall downwards to fill the empty level of the captured
electron.
Immediately after experiencing any of the radioactive decay processes the nucleus is
generally in an excited state (rotational or vibrational, see Lilley p. 55). The nucleus will then deexcite to its ground state and emit a gamma () ray. As an example of gamma emission figure 3.3
shows the decay scheme of the 20 F (fluorine) isotope. Both an electron and a gamma ray are
emitted in this decay process.
Since in a radioactive process the nucleus transfers from a higher energy state to a lower
energy is released. This energy is acquired and shared by the created particles. For example in 
decay
4
U 228
90Th 2 He
232
92
(3.1)
the released energy
Q  931.48  (M U  M Th  M  ) MeV  931.48  (232.03715  228.02874  4.002603) MeV  5.41 MeV
This energy is shared as kinetic energy between the thorium nucleus and the  particle:
E k , 
M Th
Q  0.983  5.41 MeV  5.32 MeV
M Th  M 
E k ,Th 
M
Q  0.017  5.41 MeV  0.092 MeV
M Th  M 
The created  particles are emitted with this specific kinetic
energy.
One may think that the situation is the same in  decay but
experiments have shown that in  decay the  particles are emitted
with a continuous energy spectra. This problem between the theory
and the experiments was solved by Enrico Fermi who introduced
the idea of a third particle present in  decay – the neutrino.
The  decay formulas above should actually be written as
11
6
C 115B  e   
(3.2)
14
6
C 147N  e   
(3.3)
20
F
-
20
Ne *
20
Ne
5.41 MeV

1.63 MeV
Figure 3.3
Here  is a neutrino and  an antineutrino. In  decay the released energy is divided arbitrarily
between the  particle and the neutrino (the energy acquired by the nucleus is negligible because of
its big mass in comparison to the  particle and the neutrino). The  particle may receive any
energy between 0 and Q. Thus the energy spectrum of the  particles is continuous (figure 3.4).
Figure 3.4
Decay constant and half life
Radioactive decay is a statistical process. If the probability per unit time of the decay of one
nucleus is  the decay rate of a sample with N nuclei is
dN
 N
dt
(3.4)
Thus if at t = 0 the number of parent nuclei is N0 we may write
N
t
dN
t
N N  0  dt  N (t )  N 0 e
0
(3.5)
in other words radioactive decay follows the exponential law. After a time T1/2 called the half life N
= N0/2:
N0
ln 2
 N 0 e T1 / 2  T1 / 2 
2

(3.6)
The unit for radioactive decay is Becquerel: 1 Bq = 1/s. Another unit commonly used is Curie: 1 Ci
= 3.7  1010 Bq.
The mystery of α decay
The energy of emitted alpha particles was a mystery to early investigators because it was
evident that they did not have enough energy, according to classical physics, to escape the nucleus.
Once an approximate size of the nucleus was obtained by Rutherford scattering (lesson 6), one
could calculate the height of the Coulomb barrier at the radius of the nucleus. It was evident that
this energy was several times higher than the observed alpha particle energies. There was also an
incredible range of half lives for the alpha particle which could not be explained by anything in
classical physics. The resolution of this dilemma came with the realization that there was a finite
probability that the alpha particle could penetrate the wall by quantum mechanical tunneling,
Gamow was able to calculate a dependence for the half-life as a function of alpha particle energy
which was in agreement with experimental observations.
Figure 3.5 represents an attempt to model the alpha decay characteristics of polonium-212,
which emits an 8.78 MeV alpha particle with a half-life of 0.3 microseconds. The Coulomb barrier
faced by an alpha particle with this energy is about 26 MeV, so by classical physics it cannot escape
at all. Quantum mechanical tunneling gives a small probability that the alpha can penetrate the
barrier. To evaluate this probability, the alpha particle inside the nucleus is represented by a freeparticle wavefunction subject to the nuclear potential. Inside the barrier, the solution to the
Schrödinger equation becomes a decaying exponential. Calculating the ratio of the wavefunction
outside the barrier and inside and squaring that ratio gives the probability of alpha emission.
Figure 3.5
Examples
3.1 Production of radioactive material
Radioactive nuclei can be produced, e.g., by bombarding some material with protons
accelerated in a particle accelerator. Another method is to place the material in a nuclear reactor
containing free neutrons. The radioactive nuclei are formed by proton or neutron capture.
Let us assume that the number of radioactive nuclei formed per unit time is g. If the decay
constant of the produced radioactive material is  then
dN
dN
 g  N 
 dt
dt
N  g /
If at the beginning the number of the radioactive nuclei is N = 0 then
N
t
dN
g
t
0 N  g /    0 dt  ln( N  g /  )  ln( g /  )  t  N (t )   (1  e )
Thus the maximum (or saturation) number of the radioactive nuclei is g/.
3.2 The life time of the free neutron and proton
In β- decay a proton is converted to a neutron:
n  p  e 
The energy difference is (the mass of the neutrino is zero)
Q  931.48  (mn  m p  me  m ) MeV  931.48  (1.00864  1.00728  0.000549  0) MeV  0.7554 MeV
Since Q is positive energy is released and therefore the free neutron is not stable (the half life of
free neutrons is approximately 13 minutes). But when bound to a nucleus the neutrons do not decay
(why?).
For protons
p  n  e 
Q  931.48  (m p  mn  me  m ) MeV  931.48  (1.00728  1.00864  0.000549  0) MeV  1.778 MeV
which means that a free proton is stable (otherwise the Hydrogen atom could not exist).
3.3 Carbon dating
In the outer regions of the atmosphere high energy cosmic particles interact with the carbon
dioxide molecules and creates radioactive 146C carbon. The ratio of the radioactive CO2 molecules
to the stable ones ( 126CO2 ) is constant and equal to 1.3  10 12 . In living organisms this ratio is also
1.3  10 12 due to the continuous cellular respiration. When an animal or a plant dies the ratio
decreases because of the decay of the 146C isotope.
Let us imagine some archeologists who have found an old garment in a cave of the
Kurdistan mountains. The archeologists would like to know the age of the garment and they turn to
the Physics department of the University of Duhok for help. Are we ready? Of course we are! We
take a small piece of the garment and use the carbon dating method to determine the age of the
garment. Let’s assume that the weight of our piece is 20 g.
First we measure the rate of  particles that are emitted from the sample. Let us say that the
result is 2 counts/s (we measure this many times over a longer period of time and take the average).
The half life of carbon-14 is 5730 years. The decay constant is then

ln 2
 3.836  10 12 s 1
T1 / 2
and the decay rate
N  2 s 1  N  5.214  1011
Here N is the number of the 146C nuclei left in the sample. We then calculate the total number of
carbon nuclei in the sample assuming that the garment is made of cotton. Cotton is 90 % cellulose
with a chemical formula of C6H10O5 and a molecular mass of 162 g/mol. Thus approximately:
N Total  6 
6.022  10 23 mol 1
 20 g  4.461  10 23
162 g / mol
Before the cotton was cut and used the number of carbon-14 nuclei in the sample was then
N 0,14C  1.3  10 12  N Total  5.8  1011
Thus from equation 3.5
N  N 0,14C e t  t 
1

ln
N 0,14C
N
 2.78  1010 s  880 years.
3.4 Modeling the α decay half life of polonium-212 (optional extra example!)
A number of parameters must be calculated to model the barrier penetration which leads to alpha
emission Polonium-212. The nuclear influence is assumed to stop sharply when the emitted alpha
and the reduced nucleus are just touching each other. Using the nuclear radius relationship
(equation 1.1) this distance is
The height of the barrier at the above distance is calculated as the Coulomb potential of point
charges. The charge of the remaining nucleus has been reduced by two by the alpha emission, so
that barrier height is:
The distance at which the Coulomb potential drops to the level of energy of the observed alpha is
So the width of the barrier is
In addition to the tunneling probability calculated below, the alpha emission rate depends upon how
many times an alpha particle with this energy inside the nucleus will hit the walls. The velocity of
the alpha can be calculated from
since an alpha at this energy is nonrelativistic. The frequency of hitting the walls is then
Just for comparison purposes, the expected half-life for a single rectangular barrier equal to the peak
of the barrier in height will be calculated. The tunneling probability for a rectangular barrier of
height 26.2 MeV and width 17.9 fm is
For a given alpha, the probability per second for emission is the product
which may be used in the nuclear decay relationship to obtain the half-life
Obviously this is not a good approximation - it misses by 13 orders of magnitude!!
A better tunneling probability should be obtained by breaking the barrier into five segments and
multiplying the successive tunneling probabilities. The following table was constructed by dividing
the 17.9 fm width into five equal segments with height equal to the midpoint height of the segment.
Segment height
Probability
of tunneling
21.9
1.20 x10^-5
16.4
1.74 x 10^-4
13.2
1.43 x 10^-3
11.0
.98 x 10^-2
9.4
.85 x 10^-1
Product
of
2.47 x 10^-15
probabilities
For a given alpha, the combined tunneling probability per second for emission is the product
which gives half-life
So the model gives a halflife of 0.25 microseconds compared to the experimental halflife of 0.3
microseconds. Not bad! But it must be admitted that this is a fortuitous example. Not all of them
agree this well with just a five segment barrier approximation.
Problems
3.1 Radioactive uranium
U is found in nature, the half life is T1 / 2  4.51  10 9 years. In a sample
238
92
taken from a rock the ratio of the number of parent nuclides 238
92 U to the number of daughter
nuclides is measured to be 0.146. Assuming that there were no daughter nuclides at the time of the
formation of the earth estimate the age of the earth.
3.2 The half life of the iodine-131 isotope is 8 days. In order to obtain an iodide sample with an
activity of 5 mCi how many grams of iodide is required (M = 131 g/mol, NA = 6.022∙1023 mol-1)?
3.2 A radioactive material A with a half life of 5 hours decays to another radioactive material B
with a half life of 1 hour. The material B decays to a stable material C. If at the beginning the
number of the nuclei A is NA,0 find the time dependent functions NA(t), NB(t) and NC(t) and present
them graphically.