Radioactivity and Nuclear Reactions
Radioactivity
Radioactivity was first discovered in 1896 by Henri Becquerel when a photographic plate
wrapped in black paper was exposed when placed in close proximity to a uranium salt. Later,
experiments by Marie and Pierre Curie uncovered other radioactive substances and eventually it
was shown that the radiation from radioactive materials could be classified in three
fundamentally different groups:1
alpha rays or particles (α) — 42 He nuclei
beta rays or particles (β) — electrons (created in and emitted from the nucleus)
gamma rays (γ) — high-energy photons
(From: Six Ideas that Shook Physics, by Thomas Moore)
Afterwards, it was discovered that some radioactive substances emitted positrons, which are
essentially positively charged electrons (antiparticle to the electron). Hence, there are both
negative and positive beta rays: β– – electron; β+ – positron. Emission of β– radiation is far more
common in natural radioactivity.
The phenomenon of radioactivity is due to unstable nuclei. In radioactive processes involving α
or β rays, the radioactive nucleus (parent) emits an α or β particle and is transformed into the
nucleus of a different element (offspring or daughter nucleus). If the offspring is also radioactive,
then it may emit an α particle or a β particle and transform into a different nucleus. This process
of radioactive disintegration or decay will continue until a stable nucleus is obtained. In
radioactive decay processes involving gamma-ray emission, a nucleus that is in an excited state
can lower its energy (and thereby increase its stability) by emitting a gamma photon and in the
process, falling to a lower energy state. Eventually, such transitions will take the nucleus to its
ground state, at which time gamma emissions will cease for that nucleus. Note that in γ decay, no
new nucleus is formed. The energies of γ photons are usually in the range 0.1–10 MeV, which is
significantly higher than the energies of x-ray photons (emitted during atomic- electron
transitions) whose energies are usually in the range 1–100 keV.
1
There are other emissions, e.g., protons, but these are far less common. (See Krane, Mod. Phys.)
1
Radioactive Decay Law
Consistent with the nature of quantum mechanics, one cannot predict precisely when a
radioactive nucleus will decay. One can, however, calculate precisely the probability that any
one nucleus will decay within a given time interval. Consider a sample containing a large
number of identical radioactive nuclei. (For example, an ordinary macroscopic sample with mass
10–100 g will contain ~ 1023 nuclei.) Let the probability per unit time that any one nucleus will
decay be r. This probability depends only on the properties of the nucleus. It does not depend on
the number of nuclei present in the sample.2 Thus, r is a constant for a given radioactive nucleus
for all time. The quantity r is called the decay constant. The decay constant can be precisely
calculated using quantum mechanics. In the following, we assume the decay constant has a
definite (usually different) value for each radioactive nucleus.
Let the number of undecayed nuclei at time t ′ be N ′ . Let ΔN ′ nuclei decay in a short time Δt ′ .
Then from the definition of r,
ΔN ′
r=
Hence,
N′
Δt ′
ΔN ′
Δt ′
ΔN ′
=
Δt ′
N′
.
= rN ′.
Since N ′ is decreasing, ΔN ′ is negative. So we can write
ΔN ′
−
= rN ′,
Δt ′
or,
ΔN ′
= −rN ′,
Δt ′
Taking the limit Δt ′ → 0 , we have
dN ′
= −rN ′.
dt ′
(10.1)
(10.2)
(10.3)
(10.4)
(10.5)
To solve this differential equation, we separate the variables and integrate:
dN ′
∫ N ′ = −r ∫ dt ′.
The limits on the integrals are the initial and final values. At t ′ = 0, N ′ = N 0 . At t ′ = t, N ′ = N.
Hence,
N
t
dN ′
∫ N ′ = −r ∫ dt ′.
N
0
0
This gives ln N − ln N 0 = −rt,
2
This is true for any spontaneous transition process, e.g., atomic transitions involving photon emission, according to
quantum field theory.
2
i.e.,
N = N 0 e− rt .
(Radioactive decay law)
(10.6)
[Show graph of radioactive decay.]
The activity of a sample is the rate at which the radioactive nuclei decay, i.e.,
dN
Activity A =
= rN 0 e− rt .
dt
(10.7)
We can write Eq. (10.7) as
(10.8)
A = A0 e− rt ,
where A0 = rN0 is the initial activity. Note that the activity also follows the exponential decay
law. Activity is measured in becquerels (Bq) or curies (Ci). 1 Bq = 1 decay/s. 1 Ci = 3.7 × 1010
Bq.
Half-Life
The half-life of a radioactive nucleus is the time period taken for half the nuclei originally
present to decay (or for the activity to drop to one-half of its original value).
Thus,
N
− rt
N = 0 = N 0 e 1/2 .
(10.9)
2
Solving for the half-life gives
ln 2 0.693
t1/ 2 =
=
.
(10.10)
r
r
The known half-lives of certain radionuclides can be used to estimate the ages of objects. This is
called radioactive dating.3 Carbon-14 ( 146 C ), which has a half-life of 5730 yrs, is usually used to
estimate the ages of dead organisms (or objects made from such organisms). This is useful for
ages up to about 50,000 years. Uranium-238 and other long-lived radionuclides can be used for
estimating geological ages.
Example: Problem 17.11 from Taylor et al. textbook, 2nd edition
Solution:
N = N 0 e− rt = N 0 e
( )
= N 0 eln 2
− ⎛⎜ tln 2 ⎞⎟ t
⎝ 1/2 ⎠
(−t / t1/2 )
= N 0e
= N0 2
−(ln 2)(t / t1/2 )
−t / t1/2
.
Thus,
N = N0 2
3
−t / t1/2
.
(10.11)
See Physics Today, Oct. 2001, p.32; April 2002, p.13.
3
Example: The half-life of 235U is 7.04 × 108 y. A sample of rock that solidified with the Earth
4.55 × 109 years ago, contains N atoms of 235U. How many 235U atoms did the same rock have at
the time of solidification?
Solution:
N = N0 2
−t / t1/2
.
This gives
( ) = N ⎡⎣2
N0 = N 2
t / t1/2
(4.55×109 y/ 7.04×108 y)
⎤
⎦
= 26.46 N = 88.2N.
Beta Decay
We mentioned previously that radioactivity is due to unstable nuclei. More specifically, a given
nucleus will be unstable if there is a mechanism for it to increase its binding energy and therefore
decrease its rest energy. Radioactive decay represents such a mechanism. In general, the excess
energy in unstable nuclei is due to three conditions:
(1) Neutron number/proton number ratio is not optimal;
(2) Neutron number/proton number ratio is fine, but the nucleus is so large that electrostatic
repulsion between protons makes it energetically favorable for the nucleus to break up;
(3) The nucleus is in an excited state, i.e., in a higher energy state than its ground state.
Condition 1 results in beta decay, condition 2 results in alpha decay, and condition 3 produces
gamma decay. We now describe the beta decay process.
In beta decay, the ratio of neutron number to proton number changes, but the total number of
nucleons remains the same. There are 3 types of beta decay: 4
A
(10.12)
X → ZA+1Y + e− + ν e . (β − decay)
Z
A
Z
X →
−
e +
A
Z
A
Z −1
Y + e+ + ν e . (β + decay)
X →
A
Z −1
Y + ν e . (electron capture)
(10.13)
(10.14)
In β decay, ν e symbolizes an antineutrino. (The subscript "e" indicates that the antineutrino is
−
associated with the electron family of particles.) In β + decay, ν e symbolizes a neutrino. The
neutrino and antineutrino are antiparticles of each other. In electron capture (EC), an inner
atomic electron is captured by the nucleus, which then undergoes beta decay. Examples:
14
22
22
C → 14
N + e− + ν e (β − ); 11
Na → 10
Ne + e+ + ν e (β + ); e− + 57
Co → 57
Fe + ν e (EC).
6
7
27
26
4
Picture, Table 9-3, p. 251, Krane
4
In β– decay, a neutron is converted to a proton and the N/Z ratio is lowered. In β+ decay, a proton
is converted to a neutron and N/Z is increased. In electron capture, a proton is also converted into
a neutron, thereby raising N/Z.
By conservation of energy, the spontaneous beta decay process is only possible if the rest energy
of the initial (parent) nucleus is greater than the sum of the rest energies of the decay products.
Consider β– decay, with parent and offspring in their ground states. Energy conservation requires
that
(10.15)
m A X,nuc c 2 = m A Y,nuc c 2 + me c 2 + mν c 2 + K,
e
Z +1
Z
5
No. of electrons
where K is the kinetic energy of the decay products. In fact, historically, it was the firm belief in
energy conservation that led to the discovery of the neutrino. Neutrinos are neutral and do not
interact electromagnetically. They also do not interact via the strong interaction. Therefore,
neutrinos are extremely difficult to detect – they can pass through light years of solid material
without interacting! The neutrino was not detected in early beta decay experiments, but it was
found that the kinetic energy of the electrons emitted did not have
a fixed value ≈ K as would be expected in a reaction in which
only the offspring nucleus and the much lighter electrons are
produced. In fact, the KE of the electrons had a distribution (see
figure). Also, momentum conservation would be violated if there
were only the offspring nucleus and the electron after the decay.
Pauli solved this problem in 1931 by proposing the existence of the
K
neutrino, which was found twenty-five years later, in 1956. The
energy
distribution of K for the electron is due to the sharing of the total
KE by the electron and the neutrino in different ratios. Modern experiments have shown that the
neutrino has a rest mass, with an upper limit of approximately 0.1 eV/c2. [Sloan Digital Sky
Survey, 2003; Super Kamiokande Collaboration, 1998.] A tritium (3H) beta decay experiment
yielded the best current upper limit of 2.3 eV/c2 (Mainz group, 2001, 2005). These values are
much smaller than the electron's rest mass, which is the next smallest rest mass in the beta-decay
process. For simplicity, we shall therefore take the rest mass of the neutrino to be zero.
With the neutrino having approximately zero rest mass, we can rewrite Eq. (10.15):
m A X,nuc c 2 = m A Y,nuc c 2 + me c 2 + K.
(10.16)
Z +1
Z
If we add the rest energy of Z electrons to both sides (to get atomic masses) and solve for K, we
get
(10.17)
K = (m A X,atom − m A Y,atom )c 2 . (β − decay)
Z
Z +1
The first term in parentheses on the right hand side is the rest mass of the parent atom; the
second term is the rest mass of the offspring atom.
Example: Derive an expression for K in β+ decay.
Solution:
5
K is usually called Q.
5
For β+ decay,
m A X,nuc c 2 = m
Z
A
Z −1Y,nuc
c 2 + me c 2 + K.
Note that the rest mass of the positron is equal to that of the electron. If we add the rest energy of
Z electrons to both sides and solve for K, we get
(10.18)
K = (m A X,atom − m A Y,atom − 2me )c 2 . (β + decay)
Z
For electron capture, one finds that
K = (m A X,atom − m
Z
Z −1
A
Z −1Y,atom
)c 2 .
(electron capture)
(10.19)
[The student should prove this.]
We note that a free neutron should be unstable since mn c 2 > mp c 2 + me c 2 .
In fact, the free neutron is unstable and decays with a half-life of about 15 minutes according to
the reaction: n → p + e− + ν e .
Note that the decay of a free proton to form a neutron, positron and neutrino is energetically
impossible, since mp c 2 < mn c 2 .
Though a free neutron is unstable, a neutron within a nucleus may or may not be stable.
Similarly, though a free proton is stable, a proton in the nucleus can be unstable (resulting in β+
decay).
Note that β decay does not change the value of A. The energetics of β decay can therefore be
understood by considering the rest energies of a group of isobars (same A). In a group of isobars,
the stable nucleus (or nuclei) has the lowest rest energy. If one considers a group of isobars with
neighboring Z values, if N/Z is too large for stability, then β– decay will occur (neutron changes
to a proton). If N/Z is too small for stability, then β+ decay or electron capture will occur (proton
is converted to a neutron). One can actually predict which member of a group of isobars will be
stable by using the expression for the binding energy Eb in the semi-empirical binding energy
formula and minimizing the rest energy as a function of Z. [Rest energy
mA X,atom c 2 = (Nmn + Zm1 H )c 2 − Eb = f(A,Z); df/dZ = 0 gives the value of Z for which the rest
Z
1
energy is minimum, i.e., for a stable nucleus.] (cf. Problem in the homework for previous
chapter.)
Alpha Decay
In alpha decay, a heavy unstable nucleus decays into a less massive offspring nucleus and an
alpha particle:
A
4
(10.20)
X → A−
Y + 42 He.
Z
Z −2
Using the Greek-letter symbol for the alpha particle, Eq. (10.20) can be rewritten as
6
A
Z
An example:
238
92
U→
X→
A− 4
Z −2
Y + α.
(10.21)
Th + 42 He.
234
90
Alpha decay is a spontaneous process and energy is liberated:
K = (m A X,nuc − m A−4 Y,nuc − m 4 He,nuc )c 2 .
(10.22)
If we add the masses of the appropriate number of electrons, we find that
K = (m A X,atom − m A−4 Y,atom − m 4 He,atom )c 2 .
(10.23)
Z −2
Z
2
Z −2
Z
2
If we choose a reference frame in which parent X is at rest before the decay, then it is clear that


pα + pY = 0,
i.e.,
pα = pY .
(10.24)
Kα + K Y = K.
(10.25)
Also,
Alpha decay energies are typically a few MeV, which means that the kinetic energies of the α
particle and the offspring nucleus are much smaller than their rest energies. Thus, the motion can
be treated non-relativistically. Hence,
pα2
pY2
+
= K.
2mα 2mY
Since pα = pY , we have
pα2 ⎛
m ⎞
1 + α ⎟ = K.
⎜
2mα ⎝
mY ⎠
⇒ Kα =
But
mα
mY
≈
pα2
2mα
=
K
.
1 + (mα mY )
4
, where A is the mass number of the parent. Thus, we find
A− 4
⎛ A − 4⎞
Kα ≈ ⎜
K.
⎝ A ⎟⎠
(10.26)
Perhaps the most striking characteristic of alpha decay is that the half-lives of alpha-emitters
vary over 24 orders of magnitude while the α-particle energies vary from about 4 MeV to 9
7
MeV. This rather amazing result is due to the fact that α-emission is due to tunneling or barrier
penetration.
[Show picture of tunneling barrier for alpha particle.]
As you recall, the wave function for a tunneling particle is exponentially decaying:
2mα (U − E )
.
ψ ~ e− ax , where x is the penetration distance and a is given by a =

Thus, the tunneling probability P ∝ e−2ax . Now, the barrier height is just the Coulomb potential
energy when the alpha particle just escapes nucleus. If R is the nuclear radius, then6
2(Z − 2)e2
U max = k
R
We take E = Kα, since the total energy of the α particle is Kα when it is far away from the nucleus
after escape. Note that the energy of the α particle is also Kα inside the nucleus due to energy
conservation. Since the height of the barrier varies with tunneling distance, the tunneling
probability has to be found by integration. However, we can do a rough calculation by using a
barrier of fixed height with half the true maximum barrier height. With this approximation,
P ∼ e−2a( R ′ − R) ,
where
⎡1
⎤
2mα ⎢ U max − Kα ⎥
⎣2
⎦
a=
,

and R′ is the distance from the center of the parent nucleus to where U = Kα. To estimate the
decay constant, we recall that r = probability per unit time. Now when the α particle is trapped in
the nucleus, the period of its wave function, which corresponds to the period of its back and forth
v
motion inside the nucleus, is given by T =
, where v is the speed of the α particle inside the
2R
nucleus. Using typical values of v and R, one finds T ~ 10-22 s.
Thus,
v −2a( R ′ − R)
r ~ P/T =
.
e
2R
Since the parameters a and R′ − R are in the exponent, r is extremely sensitive to changes in
these parameters (note that a depends on Kα ) and this leads to the observed immense range of
values for r and t1/ 2 .
Gamma Decay
[Show Fig. 9.15, Krane]
6
α-particle radius should be added.
8
After α or β decay, the offspring nucleus may be left in an excited state (a nucleon is in an
excited state). The nucleus can reach its ground state by emitting one or more gamma photons.
Gamma energies are usually in the range 100 keV to 10 MeV. Typical half-lives are 10-12 s to
10-9 s. Occasionally, the properties of the two quantum states involved in the transition (e.g.,
large differences in angular momentum) lead to a very long half-life (e.g., Ba-137m that we saw
in the lab). Such states are called isomeric states.
Example 9.10, Krane
12
7
N beta-decays to an excited state of 12
6 C , which subsequently decays to the ground state with
the emission of a 4.43-MeV gamma ray. What is the maximum kinetic energy of the emitted beta
particle?
The decay type is β + decay. To determine the K value for this decay, we first need to find the
mass of the product nucleus 12C in its excited state. In the ground state, 12C has a mass of
4.43 MeV
= 12.004756 u.
12.000000 u, so its mass in the excited state is 12.000000 u +
931.5 MeV/u
Using Eq. (10.18), the K value is therefore
K = (12.018613 u – 12.004756 u – 2 × 0.000549 u) × (931.5 MeV/u) = 11.89 MeV
Notice that we could have just as easily found the K value by first finding the K value for decay
to the ground state, 16.32 MeV, and then subtracting the excitation energy of 4.43 MeV, since
the decay to the excited state has that much less available energy. Neglecting the small correction
for the recoil kinetic energy of the 12C nucleus, the maximum electron kinetic energy is 11.89
MeV.
Natural Radioactivity
All elements except for the least massive (H and He) were produced by nuclear reactions taking
place in the interior of stars. The radioactive ones with relatively short half-lives have long since
decayed to stable elements. However, there are a few elements with half-lives of the same order
of magnitude as the age of the Earth, and so these elements are still present in nature and result in
238
natural radioactivity. The three natural radioactive decay series are for 23290Th, 235
92 U, and 92 U.
These elements decay by α emission producing radioactive offspring, which in turn decay by α
and/or β emission.7 Thus, one has a sequence or series of decays until a stable element is
reached. Note that alpha particles are emitted by the heavy nuclei rather than single protons or
single neutrons because of the relatively high binding energy per nucleon of the 42 He nucleus, or
in other words, its relatively small rest mass. (Recall that for spontaneous decay to be possible,
the total rest mass of offspring plus emitted particles must be less than the rest mass of the
parent.)
Example (From Krane): Compute the total energy released for the 238U → 206Pb chain.
Solution:
This decay chain includes 8 α and 6 β– decays. Thus, for the entire chain:
7
Gamma decay also occurs when nuclei are formed in excited state.
9
∑ K = [m
238
92
U,atom
− m206 Pb,atom − 8m4 H,atom ]c 2 .
82
2
(The masses of the β– particles have been combined with the nuclear masses so that we can use
atomic masses.)
∑ K = [238.050786 u – 205.974455 u – 8 × 4.002603 u] × 931.5 MeV/u = 51.7 MeV
The energy released in natural radioactive decay is partially responsible for the internal high
temperatures of the planets.
10