Beta Decay Energetics and Conservation Laws

Beta Decay 17
2.1 INTRODUCTION
Beta (β) particles are electrons (e–) or positrons (e+) emitted by the nucleus. In some
cases an orbital electron is captured by the nucleus.
There are three types of β-decay :
1. A neutron gets converted into proton with the emission of an electron (β-) and an anti-
neutrino ( )
–
v
n ⎯→ p + β- + ν
–
2. A proton gets converted into neutron with the emission of a positron (β+) and a
neutrino (υ)
p ⎯→ n + β+ + ν
3. K-capture or electron capture (E.C.)
p + e- ⎯→ n + ν
In all the cases the mass number remains constant but in the first case the atomic
number increases by one unit while in the second and third cases the atomic number
decreases by one unit.
2.2 ENERGETICS OF BETA DECAY
The question of whether a given nuclide will decay by electron emission or positron
emission or K-capture will depend upon the energy available for disintegration.
1. Consider the case of electron (β-) emission for which the reaction is
n ⎯→ p + e– + ν
–
The Q-value for this reaction is given by
Qβ− = (zMA – z+1MA – me
–) c2 ...(2.1)
where M's represents the nuclear rest masses of the parent and daughter respectively and
me
– is the electron mass. The rest mass of the neutrino is zero. The above equation can also
be written in terms of the atomic masses.
zMA = zMA + Zme– ...(2.2)
where M is the atomic rest mass. The binding energy of the electrons being small is
neglected.
zMA = zMA - Zme–
C
HAPTER
2
BETA DECAY

18 Nuclear Physics (T.Y. B.Sc.) (Sem.–VI)
Similarly,
z+1MA = z+1MA - (Z + 1)me– ...(2.3)
Equation (2.1) in terms of atomic masses is given by
Qβ− = (zMA - Zme– - z+1MA + (Z + 1)me– – me–) c2
Qβ− = (zMA - z+1MA) c2 ...(2.4)
Thus the difference between the parent and daughter
atomic masses is the Q-value for β-decay. Hence for β-decay to
be energetically possible
ZMA > Z+1MA ...(2.5)
The decay scheme for β- emitter is shown in Figure 2.1.
2. Consider the case of positron (β+) emission, for which the reaction is
p ⎯→ n + e+ + ν
The Q-value for this reaction in terms of nuclear rest masses is given by
Qβ+ = (zMA – z-1MA – me+) c2 ...(2.6)
where me+ is the rest mass of the positron.
The above equation in terms of atomic masses is
Qβ+ = (zMA – Zme- – z-1MA + (Z – 1) me– – me+) c2
As me- = me+
Qβ+ = (zMA – z-1MA – 2me-) c2 ...(2.7)
Hence for β+ decay to be energetically possible
zMA - z-1MA > 2 me- ...(2.8)
It means that the atomic mass of the parent must be
greater than that of the daughter by at least two electron
masses, or 1.02 MeV. As the Q value for β+ emission is less by a
factor of 2me- the probability for such type of emission is less
than the probability for β- emission.
3. Consider the case of electron capture for which the
reaction is
p + e- ⎯→ n + ν.
The Q value for this reaction in terms of nuclear rest masses is given by
QE.C = [zMA + me- – z-1MA] c2 ...(2.9)
If the atomic masses are considered then
QE.C = [zMA – Zme
– + me
– – z-1MA + (Z – 1) me
-] c2
QE.C = [zMA – z-1MA] c2 ...(2.10)
Hence for electron capture to be energetically possible
zMA > z-1MA ...(2.11)
That is the atomic mass of the parent must be greater
than that of the daughter.
The capture of a K-shell electron depends on the small
probability that the electron be very close to the nucleus. The
result is that as soon as the energy available exceeds 2me
– c2,
positron emission tends to happen more often than orbital
Energy
Q
z+1MAc2
zMAc2
Fig. 2.1 : Decay Scheme of
β– emitter
Q +
z-1MAc2
QE.C.
Energy
zMAc2
2moc2
Fig. 2.2 : Decay scheme of β+
emitter.
Energy
zMAc2
QE.C.
z-1MAc2
Fig. 2.3 : Decay scheme of E.C.

Beta Decay 19
electron capture. Hence K-capture is observed only when the condition (2.11) is satisfied,
but it is not always observed when that condition is fulfilled.
2.3 BETA PARTICLE SPECTRA
A graph of the number of β particles against kinetic energy of the β particles is called
the energy spectrum of β particles. This graph is shown in Figure 2.4.
Fig. 2.4 : Continuous β-particle spectra
As the energy distribution of the emitted particles is continuous (from zero to
maximum) the spectrum is continuous. Every continuous spectrum has a definite
maximum, the height and position of which depend on the nucleus emitting the particles.
There is also a definite upper limit or end point (E0) of energy for the particles emitted by a
nuclide. This end point energy is different for different nuclides. These properties of
continuous spectra are observed for both natural and artificial β-emitters, and when the
emitted particles are electrons or positrons.
In some cases of β-emission, a single continuous spectrum is found which has a single
end point energy. Such a spectrum is called simple. When there are two or more spectra
with different end-point energies and intensities then the spectrum is complex. Many
β-emitters have complex spectra.
Figure 2.5 shows the complex spectra of Cl38. It is composed of three simple spectra
with three different end point energies. The curves are momentum distribution curves.
Fig. 2.5 : The β-particle complex spectra
end point = K.E.max
O E0
Relative
number
of
particles

These spectra can be separated by analysing the curves obtained with a magnetic
spectrometer.
The average energy of the beta particles which form the conti-nuous spectrum can be
calculated from the energy distribution curve. The total energy is found by integrating the
product of the number of particles with a given energy and that energy. The number of
particles is determined by integrating the number of particles with a given energy with
respect to the energy. This number is just the area under the distribution curve. The
average energy is given by
T =
⌡
⌠
0
T0
N (T) T dT
⌡
⌠
0
T0
N (T) dT
...(2.12)
where T0 is the endpoint energy, T is the energy, and N(T) dT is the number of
particles with energies between T and T + dT.
When a beta transformation takes place, the energy of the nucleus decreases by an
amount first equal to the maximum or end point energy of the emitted spectrum.
2.4 DIFFICULTIES IN BETA PARTICLE SPECTRA
In beta decay a parent nucleus with a definite energy state emits an electron and leaves
a product nucleus which is also in a definite energy state. The energy of the emitted
electron, however, is not equal to the difference between the energy of the nucleus before
and after emission. It is found to have any value between zero and the maximum energy of
the continuous spectrum. The average energy of the emitted particle is only about one-
third of the maximum energy. Part of the energy, approximately two-thirds, seems to have
disappeared. Therefore the principle of conservation of energy seems to be violated in the
beta decay process.
A proton, neutron and an electron have intrinsic spins and the value of the spin
angular momentum for each of these particles is
1
2 ". Hence there is a total spin angular
momentum for the nucleus which is given by the vectorial addition of the spin angular
momenta of the protons and neutrons contained in the nucleus. Hence for a nucleus
consisting of even number of nucleons, that is, even A, the total spin angular momentum
should be an even multiple of
1
2 ". If the number of nucleons are odd, that is, odd A, then
the total spin should be an odd multiple of
1
2 ". In beta decay an electron with spin
1
2 " is
emitted by the nucleus. Hence the total spin angular momentum of the even A nucleus
should become an odd multiple of
1
2 " and that of the odd A nucleus should become an
even multiple of
1
2 " . However, in beta decay the total number of nucleons in the nucleus
do not change. That is, the mass number A remains constant. Hence the even multiples
must remain even and the odd multiples must remain odd after the decay.
Thus as explained above there were difficulties about the conservation of energy and
conservation of angular momentum in beta decay.
2.5 PAULI'S NEUTRINO HYPOTHESIS
In order to solve these difficulties in the beta decay process, a third particle was
postulated by Wolfgang Pauli. Enrico Fermi named this particle as the neutrino (υ). It does
not exist inside the nucleus but is created at the time of decay. It has no charge and has

Beta Decay 21
negligible rest mass (almost zero) as compared with the mass of the electron. Its spin
angular momentum is
1
2 ". It carries away the missing energy.
Under this neutrino hypothesis each beta decay process is accompanied by the release
of an amount of energy given by the end point of the beta spectrum. This disintegration
energy is shared among the electron, neutrino and the recoil nucleus. Since the sharing of
energy is in a continuous manner, the energy spectrum of beta particles is a continuous
spectrum. β-transformations are represented by the following processes :
β- emission : n → p + e- + ν
– (anti-neutrino)
β+ emission : p → n + e+ + ν (neutrino)
As two particles (electron and neutrino) are emitted in the beta decay process the total
angular momentum taken out of the parent nucleus is 2 ×
1
2 " or zero. Thus nuclei with
total angular momenta being an even multiple of
1
2 " will have it has an even multiple again
after the decay. Similarly nuclei for odd multiples. Hence the total angular momentum of
the system is conserved.
2.6 PROPERTIES OF A NEUTRINO
To satisfy the principle of conservation of angular momentum, charge and rules of
nuclear statistics, the neutrino is assigned the following properties :
1. It has zero charge.
In beta decay process charge is conserved without the neutrino. Also if it were charged
it would produce ionization which could have been detected. Zero charge implies
negligible magnetic moment.
2. It has almost zero mass.
3. It has spin 1/2 so as to satisfy the law of conservation of angular momentum in beta
decay process.
The neutrino is a fermion.
4. Helicity
The neutrino has an anti-particle called antineutrino (ν
–). The antineutrino has zero
charge, zero mass and spin 1/2. A neutrino differs from an antineutrino by a property that
it exhibits a "handedness" similar to a screw. The spin of the neutrino is always anti-parallel
to its momentum while the spin of the antineutrino is always parallel to its momentum. In
other words, neutrino is a left-handed particle while antineutrino is a right-handed particle.
This is shown in the Fig. 2.6.
Fig. 2.6 : Handedness in neutrino and antineutrino
neutrino
P
S
antineutrino
P
S

To describe the handedness in the neutrino, the idea of helicity is used.
Helicity is defined as H =
–
P ⋅
–
S
|
–
P | |
–
S |
…(2.13)
where P
–
is the momentum and S
–
the spin of the particle.
H = – 1 for a neutrino and H = + 1 for an antineutrino.
The fixed value for helicity imply that both the neutrino and the antineutrino move
with the velocity of light.
2.7 K-CAPTURE PROCESS
In certain elements beta activity is observed in the form a K-capture process. In this
process the nucleus absorbs an electron from the first atomic orbit which is the K-shell. This
electron interacts with a proton and converts it into a neutron.
p + e- ⎯→ n + ν
The effect is similar to the emission of a positron. Therefore the atomic number
decreases by one unit while the mass number remains unchanged. The vacancy created in
the K-shell is filled by an electron which jumps from a higher orbit to the first orbit. In the
process the atom emits a photon of X-Ray. It is observed that the photon corresponds to an
atom of the daughter element. It is due to the emission of such X-Ray photons that this
process could be detected. It is also observed that the intensity of X-Rays decrease
exponentially with time for a given sample. Since this is similar to the results of natural
radioactivity in alpha and beta emission, K-capture or electron capture (E.C.) is also
considered as a process of natural radioactivity.
When an electron is captured an additional angular momentum of
1
2 " is added to the
nucleus. However the total number of nucleons remain constant after the process.
Therefore in order to conserve the total angular momentum a neutrino is emitted by the
nucleus.
2.8 THE DETECTION OF NEUTRINO
Beta decay is a three-body problem. Conservation of momentum therefore demands
that after emission of a β-particle and neutrino, the daughter nucleus recoils in a direction
not exactly opposite to the emitted β particle. If this recoil of the daughter can be detected,
then it would form an indirect evidence of neutrino existence. Figure 2.7 indicates this
recoil of the daughter and the momentum triangle for the β-particle, the daughter nucleus
and the neutrino.
Fig. 2.7 : Recoil of the daughter nucleus after β-emission. The momentum triangle is shown at the
right.
However, the recoil energy is only about a few eVs. This makes experimental
measurements on daughter recoil extremely difficult. Also the neutrino, being chargeless
and massless, interacts extremely weakly with matter. A neutrino with an energy of a few
Daughter
nucleus
p
p
pN

Beta Decay 23
MeV can travel a distance of 100 light years in a chamber containing liquid hydrogen
without any interaction! Inspite of these difficulties, an experiment performed in 1953 gave
conclusive evidence about the neutrino.
COWAN AND REINES EXPERIMENT
C.L. Cowan Jr. and R.F. Reines were the first to observe the direct interaction of free
neutrino in 1953.
The main idea behind this experiment is as follows :
A free neutron decays by the following reaction
0n1 ⎯→ 1p1 + –1e0 + ν
– …(2.14)
This reaction can be inverted by bombarding protons with anti-neutrons. The inverse
reaction is :
ν
– + 1p1 ⎯→ 0n1 + +1e0 …(2.15)
i.e. when protons are bombarded with anti-neutrinos they emit positrons to become
neutrons. Cowan and Reines utilized the large flux of anti neutrinos existing in the
neighbourhood of a nuclear reactor to observe this reaction. Their experimental
arrangement is described below :
Figure 3.8 shows schematically the experimental arrangement.
Fig. 2.8 : Cowan and Reines neutrino detection experiment : Schematic diagram of arrangement.
The protons for the reaction were provided by water in a large plastic tank. About 200
litres of water was stored in this tank. I and II are liquid scintillation detectors in the form of
tanks. Each tank contained about 400 litres of liquid scintillator solution. Cadmium has a
large cross section (probability) of capturing thermal neutrons. Cadmium chloride (CdCl2)
was dissolved in water to provide the cadmium nuclei for neutron detection. Two large
tanks of liquid scintillator were placed on two sides of the water tank. To establish the
reaction given by equation (2.15), the following sequence of events must be identified :
i) An anti-neutrino from the reactor interacts with a proton in the water tank. This
interaction produces a neutron and a positron according to the reaction given by
equation (2.15).

ii) The positron comes to rest quickly and it annihilates with an electron, resulting in two
gamma rays emitted in opposite directions. Each of these gamma rays carries an energy
equal to the electron rest mass 0.511 MeV. These γ-rays can be detected by the liquid
scintillators marked I and II. A large number of photomultiplier tubes are required to
collect the light emitted by the scintillator. Cowan and Reines observed this
annihilation radiation which follows very quickly in an time interval of about 10-9 secs
after the emission of positron.
iii) The neutron as indicated in figure 2.8 diffuses in water (with dissolved CdCl2) and
after many collisions with protons slows down to thermal energies to eventually get
captured by a Cd nucleus. This capture of the neutron takes place much later than the
emission of annihilation γ-rays. The time gap is about 10–5 secs. This capture results in
the emission of γ-rays by the (n, γ) reaction. Usually three gamma rays are emitted and
there is a chance of detecting one γ-ray in the upper detector I and one γ-ray in the
lower detector II. On an average the neutron requires about 6 microseconds to be
captured.
Cowan and Reines succeeded in establishing the sequence of events outlined above
(i), (ii) and (iii). They delayed the annihilation pulse so that it could be displayed at the
beginning of the oscillascope trace. A delay of about 30 µs was required. An electronic
circuit was gated for 20 µs so that it was able to accept a neutron pulse (triggered by
captured γ-rays).
The oscilloscope was photographed each time the sequence of annihilation γ-ray pulse
and neutron pulse was displayed. Figure 2.9 shows this acceptable sequence of events, every
time the oscilloscope was triggered. Cowan and Reines were able to record about
30 events per hour. To check their results they used water without any cadmium in it and
found that the neutron pulses due to capture γ-rays disappeared.
Also when the nuclear reactor was shut down they found that the events mentioned
above could not be observed. The observed sequence of events established reaction given in
equation (2.15). Thus Cowan and Reines experiment proved the existence of neutrino.
Fig. 2.9 : Triggered pulses due to positron annihilation and neutron capture.
2.9 THE VELOCITY AND ENERGY OF BETA-PARTICLES
The velocity, or momentum, of beta particles can be measured by means of the
deflection of the path of the particles in a magnetic field. One kind of instrument which is
often used is the semicircular–focusing, or 180°, magnetic spectrograph which is shown in
Fig. 2.10.

Beta Decay 25
Fig. 2.10 : Schematic diagram of a 180° magnetic spectrograph
As the value of charge to mass ratio is large for electrons small magnetic fields can be
used to deflect beta particles. Because of their large velocities, beta particles have to be
treated relativistically.
The formulas for the velocity and kinetic energy are
v = H r
e
m0
1 – v2/c2 …(2.16)
T = m0 c2
⎣
⎢
⎡
⎦
⎥
⎤
1
1 – v2/c2
– 1 …(2.17)
where m0 is the rest mass of the electron and e is the magnitude of the electronic charge.
It is useful to express the quantities Hr and T as functions of the ratio v/c and equation
(2.16) may be written as
v
c = H r
e
m0c 1 – v2/c2 …(2.18)
Then
Hr (gauss-cm) =
m0c
e
v
c (1 – v2/c2)-1/2 …(2.19)
If the beta particles, after being deflected, are allowed to fall on a photographic plate, a
fogging of the plate is observed which extends for a definite distance and then ceases at a
point characteristic of the beta emitting nuclide. This fogging is continuous, but of non-
uniform intensity along the plate. In addition to this continuous primary spectrum, several
sharp lines are sometimes seen on the plate at positions characteristic of the emitting
substance. These lines form the so-called secondary, or line spectrum, observed for many
but not all beta emitters. A velocity distribution curve or an energy distribution curve can be
obtained using a Geiger counter as the beta particle detector instead of a photographic
plate. The instrument is then called a spectrometer. The counter is placed in a fixed
position i.e. the position of r is fixed, the magnetic field is varied, and the number of beta
particles reaching the counter per unit time is obtained for different values of H. Since
each value of H corresponds to a different value of Hr and hence to a different value of the
kinetic energy, the numbers of particles corresponding to different energy values are
obtained. The continuous spectrum of beta particles energies found for RaE is shown in
Fig. 2.11. In this case no line spectrum is found and the upper limit or maximum energy is
at 1.17 MeV. When a line spectrum is also present, the lines appear as distinct peaks
superimposed upon the continuous distribution curve. This is shown in Fig. 2.12.
Source
Baffles
Chamber wall
Photographic plate
Slit

Fig. 2.11 : The β-particle spectrum of RaE Fig. 2.12 : Beta-particle spectrum of
Au108, showing the line spectrum
2.10 ENERGY LEVELS AND DECAY SCHEMES
Beta-transformations often yield information about the energy levels of the product
nuclei and about decay schemes. These transformations are sometimes accompanied by
γ-rays, and the presence of the γ-radiation means that the product nucleus is formed in an
excited state and passes to its ground state by emitting one or more γ-rays. If no γ-ray is
emitted, the β-transition is directly to the ground state of the product nucleus. In the case of
O15 a positron is emitted, and no γ-ray is observed. The endpoint energy is 1.73 Mev, and
the β-disintegration energy is 1.73 Mev + 1.02 Mev = 2.75 Mev. This follows from the fact,
that two extra electron masses are needed for positron emission to be energetically possible.
It follows that the difference between the ground state energies of O15 and N15 is 2.75 Mev,
which is also the energy equivalent to the difference between the masses of these two
nuclides. The nuclide F20 (Fig. 2.13(a)) emits electrons with an endpoint energy of
5.41 Mev, and also 1.63-Mev γ-rays. The total disintegration energy is 7.04 Mev, and the
product nucleus Ne20 has an excited state 1.63 Mev above the ground state.
The beta-spectra of O15 and F20 are simple and the decay schemes are also simple.
Sometimes, however, two or more groups of β-particles are emitted. The spectrum is the
complex; it can be broken down into two or more simple spectra, and γ-rays are observed.
In the case of O14, in more than 99% of the disintegrations, positrons are emitted with an
end-point energy of 1.84 Mev; 2.30-Mev γ-rays are also observed. The total disintegration
energy is 1.84 Mev + 1.02 MeV + 2.30 Mev = 5.16 Mev, of which 2.86 Mev is the difference in
energy between the ground state of O14 and the excited state of the product nucleus N14.
The N14 nucleus passes to its ground state by emitting a 2.30-Mev γ-ray. In about 0.6% of the
disintegration O14 undergoes a transition directly to the ground state of N14 by emitting
4.1-Mev positrons. The decay scheme is shown in Fig. 2.13(b). The electron-decay of Mg27 is
more complicated : about 70% of the disintegrations correspond to an endpoint energy of
1.78 Mev and about 30% to an endpoint energy of 1.59 Mev; γ-rays are observed with
energies of 0.834 Mev and 1.015 Mev, respectively, and in less than one percent of the
disintegrations, a γ-ray with an energy of 0.18 Mev is observed. Coincidence experiments
show that the 1.78-Mev β-ray and the 0.834-Mev γ-ray belong to the same transition,
(i.e. they are co-incident), and that the 1.59-Mev β-ray and the 1.015-Mev γ-ray belong to the
same transition. A decay scheme consistent with all of these data is shown in Fig. 2.13(c).
The direct transition from the ground state of Mg27 to the ground state of Al27 by electron-
emission is evidently highly forbidden. The electron decay of Cl38 has three groups of
electrons, two γ-rays have been observed, and the decay scheme is shown in Fig. 2.13(d).
9
8
7
6
5
4
3
2
1
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2
l
l
l
l
l l
ll ll
ll
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l
l l
l
l
l
l
Endpoint
Relative
number
of
bets–particles,
N(T)
Kinetic energy of the beta–particles, T(Mev)
0 1000 2000 3000 4000
0
10
20
30
40
Relative
number
of
–particles
Hr (gauss–cm)

Beta Decay 27
A still more complicated scheme is that of 57La140, shown in Fig. 2.13(e); at least four
groups of electrons are involved and at least nine γ-rays have been observed.
The nuclide Cu64 is a particularly interesting case of β-decay because it emits both
electrons and positrons and also undergoes orbital electron capture. In 39% of the
disintegrations, an electron is emitted; the β– - spectrum is simple and the endpoint energy
is 0.57 Mev. The product nucleus Zn64 is formed in its ground state. In 19% of the
disintegrations, a positron is emitted with an endpoint energy of 0.66 Mev; the product
nucleus Ni64 is formed in its ground state. In 42% of the disintegrations, a K-electron is
captured. In nearly all of the captures, the product nucleus Ni64 is formed in its ground
state, but in a small fraction of the K-captures, a γ-ray is observed with an energy of
1.34 Mev. There is, therefore, an excited level of Ni64 1.34 Mev above the ground state. It
has been shown that the γ-ray is observed only in coincidence with the orbital electron
capture, and is not associated with the emission of either the electron or the positron. The
decay scheme of Cu64 is shown in Fig. 2.13(f).
(a) (d)
(b) (e)
(c) (f)
Fig. 2.13 : Decay schemes for β-emitters. (a) F20, (b) O14, (c) Mg27, (d) Cl38, (e) La140,
(f) Cu64.
F20
–
, 5.41 Mev
(1.63 Mev)
1.63
Ne20
0
–
, 2.77 Mev
(16%)
–
, 4.81 Mev
(53%)
–
, 1.11 Mev
(31%)
3.75
2.16
0
A38
Cl38
O14
+
, 1.84 Mev
( > 99%)
2.30
+
, 4.1 Mev
( 0.6%)
N14
0
La140
3.00
–
, 1.67 Mev
( 14%)
–
, 2.15 Mev
( 8%)
–
, 1.15 Mev lower
( 48%)
–
, 1.34 Mev ( 30%)
Ce140
2.52
2.41
2.08
1.60
0
Mg27
1.015
–
, 1.78 Mev
(70%)
Al27
0
–
, 1.59 Mev
(30%)
0.7%
1.834
Cu64
1.34
+
, 0.66 Mev
( 19%);
EC(~ 42%)
0
0
Zn64
Ni64
EC( 0.5%)
–
, 0.57 Mev
( 39%)

SOLVED PROBLEMS
Problem 1 :
Show that the recoil energy of the nucleus undergoing electron capture type of β-decay
is given by
Er =
E2
2Mc2
where E is the total energy release and M is the mass of the recoiling nucleus.
Solution :
In an EC type of β-decay
p + e– ⎯→ n + ν
Since the orbital electron which is captured has negligible momentum, the momentum
of the recoil nucleus will be equal to that of the neutrino
pr = pν =
E
c
Hence the recoil energy Er =
pr
2
2M =
E2
2Mc2
where E is the total energy release and M is the mass of the recoiling nucleus.
Problem 2 :
3Li7 and 4Be7 have atomic masses 7.016005 and 7.016929 a.m.u. Which of them shows
β-activity and of what type? Calculate Q for it.
Solution :
M(Li7) = 7.016005 a.m.u.
M(Be7) = 7.016929 a.m.u.
M(Be7) – M(Li7) = 0.000924 a.m.u.
Consider the reaction
3Li7 ⎯→ 4Be7 + e– + ν
– + Qβ–
For this reaction to take place Qβ– > O.
M(Li7) > M (Be7)
As M(Be7) > M(Li7) this type of decay is not possible.
4Be7 ⎯→ 3Li7 + e+ + ν + Qβ+
For this reaction to take place
M(Be7) – M(Li7) > 2 me–
But M (Be7) – M(Li7) = 0.00924 × 931.5 MeV = 0.861 MeV < 2me– (= 1.022 MeV)
Hence this type of decay is not possible.
4Be7 + e– ⎯→ 3Li7 + ν + QEC
For this reaction to take place, QEC > 0.
M(Be7) > M (Li7)
As this is true, the decay of Be7 to Li7 takes place by electron capture.

Beta Decay 29
Problem 3 :
For a particular neutron decay, the neutrino carries off no energy. If the neutron was at
rest before the decay, calculate the kinetic energies of the proton and the emitted beta
particle.
Given m(p) = 1.007825 a.m.u., m(e) = 0.000548 a.m.u. and m(n) = 1.008665 a.m.u.
Solution :
n → p + e– + ν + Q
Emax = Q = [m(n) – m(p) – m(e–)]
= [1.008665 – 1.007825 – 0.000548]
= 0.000292 × 1.66 × 10-27 × (3 × 108)2
= 4.35196 × 10-14 J
Also Q = Ee– + Ep ({ Eν = 0).
where Ee– and Ep are the kinetic energies of the electron and proton.
From the law of conservation of momentum
pe– = pp = p
∴
p2
2me–
+
p2
2mp
= Q
p2 =
2Q
⎝
⎜
⎛
⎠
⎟
⎞
mp + me–
mp me–
p2 =
2 × 4.35196 × 10-14
1825.8098
1.66 × 10-27
= 7.91351 × 10-44
p = 2.81309 × 10-22 kgm/s
Ep =
p2
2mp
=
7.91347 × 10-44
2 × 1.007825 × 1.66 × 10-27 = 2.365 × 10-17 J
Ee– =
p2
2me–
=
7.91347 × 10-44
2 × 0.000548 × 1.66 × 10-27 = 4.349 × 10-14 J
The kinetic energies of the proton and electron are 2.365 × 10-17 J and 4.349
× 10-14 J respectively.
Problem 4 :
Tritium emits negative β-particles. Represent by an equation the decay process and
calculate the end-point energy of the emitted particles. Given : M( )
3
1 H , M( )
3
2 He and
M( )
0
-1 He as 3.01695 amu, 3.01693 amu and 0.00055 amu respectively.
Solution :
The decay process may be represented as
3
1 H ⎯→
3
2 He +
0
-1 e + v
The emitted β – particle has its energy maximum when it is not shared at all by the
accompanying neutrino. This is the end-point energy e, which is the total energy of
disintegration.
E = ( )
mass of
3
1 H – mass of
3
2 He amu
= (3.01695 – 3.01693) amu

= 2 × 10-5 amu
= 2 × 10-5 × 931.2 MeV = 0.01862 MeV
Problem 5 :
Tritium emits electrons and Mg-23 positrons. Represent the two decay processes by
equations and calculate in each case the end-point energy of the particles emitted. Given
the atomic masses : M(3H) = 3.01695, M( )
0
±1
e = 0.00055; M(3He) = 3.01693,
M(23Na) = 22.99618 and M(23Mg) = 23.0002, all in amu.
Solution :
The first decay process is :
3
1 H →
3
2 He +
0
-1 e + v. When the energy of the neutrino is
zero, the emitted electron has the maximum energy – the total disintegration energy, or the
end-point energy, Ed
Ed = M( )
3
1 H – M( )
3
2 He
= 3.01695 – 3.01693 = 2 × 10-5 amu
= 2 × 10-5 × 931 MeV = 0.0186 MeV
The second decay process is :
22
12 Mg →
23
11 Na +
0
+1 e + v
∴ The end-point energy Ed of positron is,
Ed = M( )
23
12 Mg – M( )
23
11 Na – 2M( )
0
+1 e
= 23.0002 – 22.99618 – 2 × 0.00055
= (23.0002 – 22.99618 – 0.0011) amu
= 0.00292 × 931 MeV = 2.72 MeV
Problem 6 :
Calculate the energy of γ-rays emitted in the β-decay of 28Al. Given Emax = 2.86 MeV,
M(Al28) = 27.98/908 amu, M (Si28) = 27.976929 amu
Solution :
28Al
0
-1e
⎯⎯→ 28Si + Q
(daughter)
Since γ-rays are emitted during the above β decay,
the β-transformation should be leading to an excited
state of a nuclei, 28Si (see figure).
We have 28Al = 27.981908 u.
28Si = 27.976929 u
Q = (27.981908 – 27.976929) u
= 4.979 × 10-3 × 931.5 MeV
= 4.638 MeV
Problem 7 :
The only known nuclei with A = 7 are
7
3 Li, whose atomic mass M3, 7 = 7.01600 u, and
7
4Be, whose atomic mass M4, 7 = 7.01693 u. Which of these nuclei is stable to β-decay? What
process is employed in the β-decay of the unstable nucleus to the stable nucleus?
28
28

Beta Decay 31
Solution :
Since the atomic mass of
7
3 Li is the lowest, it is the nuclecus which is β-stable.
As far as charge conservation is concerned, the β-unstable
7
4 Be could decay into the
stable nucleus either by capturing an atomic electron or by emitting a positron. But as far as
energy conservation is concerned, only electron capture is possible since the difference in
the atomic masses, M4, 7 – M3, 7 = 7.016934 – 7.01600 u = 0.00093 u, is less than two electron
masses, 2me = 0.00110 u. Thus electron capture is the process employed in the β decay of
7
4 Be into
7
3 Li.
Problem 8 :
A free neutron decays into a proton, an electron and an antineutrino. If
M(n) = 1.00898 u, M(p) = 1.00759 and M(e) = 0.00055 u, find the kinetic nenergy shared by
the electron and the antineutrino.
Solution :
The neutron decay may be reprssented as :
n → p = e– +
–
v
Since the total energy is to be conserved, we get
M(n) c2 = M(p)c2 + Ep + M(e)c2 + Ee + M(
–
v) c2 + Ev
Where E’s represent the kinetic energies of the respective particles.
Now, M(
–
v) = 0 and Ep = 0, the proton being at rest. We have,
Ee + Ev = M(n)c2 – M(p)c2 – M(e)c2
= {M(n) – M(p) – M(e)} c2
= (1.00898 – 1.00759 – 0.00055)c2
=
(0.00084 × 1.00 × 10-87 × 9 × 10-16)
1.6 × 10-13 MeV = 0.78 MeV
Problem 9 :
40K decays to 40Ar by electron capture. Assuming that the initial kinetic energy of the
electron and the recoil energy of the nucleus are zero, show that the kinetic energy of the
neutrino is 1.504 MeV.
Given : Mass of 40K = 39.96399 u, Mass of 40Ar = 39.962384u.
Solution :
We have for electron capture
40K + e – 40Ar + v
Q = [M(40K) – M (40Ar)]c2
= (39.963999 u – 39.9623844) 931.5
MeV
u
= 1.504 MeV
∴ The kinetic energy of neutrino = 1.504 MeV
Problem 10 :
11
6 C decays to
11
5 B by positive β emission. What are the maximum and minimum
energies of neutrino? Given : mass of 11C = 11.011433 u, mass of 11B = 11.009305 u.

Solution :
Q = [M(AX) – M(AX’) – 2me]c2
= [11.0114334 – 11.009305 u] 931.5
⎝
⎛
⎠
⎞
MeV
u –2 × 0.511 MeV
= 0.96 MeV
QUESTIONS AND PROBLEMS
1. What is Beta-decay? Explain the salient features of the β-spectrum.
2. Derive the energy conditions under which different types of β-decay can take place.
3. Explain the difficulties encountered in beta decay. Also give arguments which led to
neutrino hypothesis. Why no such particle is emitted in alpha decay?
4. What is meant by electron capture?
5. Describe Cowan and Reines experiment to detect the neutrino.
6. What are the difficulties in experimentally detecting the neutrino?
7. State the properties of the neutrino.
8. Calculate the energy of γ-rays emitted in the β-decay of Al28. Given Emax = 2.86 MeV
and the atomic masses of Al28 and Si28 are 27.981908 a.m.u. and 27.976929 a.m.u.
respectively.
9. The atomic masses of 28Ni64, 29Cu64 and 30Zn64 are respectively 63.927956, 63.929761
and 63.929140 a.m.u. Which of these are β-active and what are the natures of their
β-activity? Calculate Q in each case.
10. A free neutron decays into a proton by the emission of β– particles of maximum kinetic
energy 0.782 MeV. If the rest masses of the electron and neutron are 0.0005486 a.m.u.
and 1.008665 a.m.u. respectively, find the mass of the proton.
11. Calculate the recoil energy of the proton produced in the decay of a free neutron
when the β-particle is emitted with the maximum kinetic energy. The rest masses of
proton, neutron and electron are 1.007825 a.m.u., 1.008665 a.m.u. and 0.000548
a.m.u. respectively.
12. The atomic masses of 18A40, 19K40, 20Ca40 and 21Sc40 are 39.975050, 39.976653,
39.975230 and 39.990250 a.m.u. respectively. Which of these are β-active and what are
the natures of their β-activity?
13. Find the energy of a neutrino in the following K-capture reaction :
55Cs131 + Ð1e0 ⎯→ 54Xe131 + ν
The total energy released in this process is 355 KeV and the binding energy of the
K-electron in Xe131 is 35 KeV. Further the daughter nucleus is formed directly in the
ground state.
14. The nuclide Sm153 emits four groups of β rays, with end point energies of 0.83, 0.72,
0.65 and 0.13 MeV respectively. Gamma rays with energies of 0.1032, 0.0697, 0.172,
0.545 and 0.615 MeV respectively are also emitted. Construct a decay scheme to fit
these data.

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