Alpha decay occurs when an unstable nucleus releases an alpha particle to achieve stability. Three key laws are obeyed: conservation of charge, nucleons, and momentum. The alpha particle carries most of the disintegration energy (Q-value) away from the daughter nucleus. Alpha particle velocity and energy can be determined using a magnetic spectrograph which measures particle deflection in a magnetic field. Alpha particles have a well-defined range in a material and become fully absorbed. The Geiger-Nuttall law shows a correlation between an alpha emitter's half-life and the range or energy of its alpha particles.

1. Alpha Decay 1
1.1 INTRODUCTION
Some nuclei have a combination of protons and neutrons which do not lead to a stable
configuration. These nuclei are therefore unstable or radioactive. Unstable nuclei tend to
achieve stability by releasing certain particles. When α-particles are released, the process is
α-decay.
Alpha particle is a positively charged particle emitted by a radioactive nucleus. It is
made up of two protons and two neutrons. Hence α-particle is the nucleus of a helium atom
(2He4).
The process of alpha decay can be given as
ZMA ⎯→ Z-2MA-4 + 2He4 ...(1.1)
where ZMA and
Z-2
MA-4 are the parent and daughter nuclides respectively and 2He4 is the
alpha particle.
1.2 CONSERVATION LAWS OBEYED BY ALPHA-DECAY
Whenever an α-particle is emitted by a radioactive nucleus the following conservation
laws must be obeyed.
i) Conservation of charge and the number of nucleons : In alpha decay, the total
charge and the total number of nucleons must be conserved. If A is the mass number and Z
is the atomic number then
ZMA ⎯→ Z-2MA-4 + 2He4
ii) Conservation of linear momentum : The linear momentum must also be conserved
in the emission of α-particle. If the parent nucleus of mass Mp is at rest, then
initial momentum = 0
If the daughter nucleus of mass Md has a velocity Vd and the a-particle of mass mα has
velocity vα then,
Final momentum = MdVd + mα vα
Hence, MdVd + mα vα = 0
mα vα = – MdVd …(1.2)
The daughter nucleus must have a velocity in a direction opposite to that in which the
α-particle is ejected.
C
HAPTER
1
ALPHA DECAY

2. 2 Nuclear Physics (T.Y. B.Sc.) (Sem.–VI)
iii) Conservation of mass and energy : If E is the energy with which the α-particle is
ejected and Ed is the kinetic energy of the daughter, then according to the principle of
conservation of mass energy :
Mp c2 = Md c2 + mα c2 + Ed + E …(1.3)
1.3 ENERGETICS OF SPONTANEOUS ALPHA DECAY
The total energy change for alpha decay is known as the alpha disintegration energy.
It is assumed that the parent nucleus is at rest. When an alpha particle is emitted, the
product nucleus recoils, carrying with it a certain amount of energy. The principle of
conservation of momentum requires
MV = mα vα ...(1.4)
Where M is the mass of the product nucleus and V is its velocity, mα is the mass of the α-
particle and vα is its velocity.
The α-disintegration energy ⎝
⎛
⎠
⎞
E
ʹ′
α
or Q-value is the sum of the kinetic energies of the
product nucleus and the α-particle.
E
ʹ′
α
=
1
2 MV2 +
1
2 mαv
2
α = Q
E
ʹ′
α
=
1
2 M
⎝
⎜
⎛
⎠
⎟
⎞
mαvα
M
2
+
1
2 mαv
2
α (from 1.4)
E
ʹ′
α
=
1
2 mαv
2
α ⎝
⎜
⎛
⎠
⎟
⎞
mα
M + 1
E
ʹ′
α
= Eα
⎝
⎜
⎛
⎠
⎟
⎞
mα
M + 1
E
ʹ′
α
= Eα
⎝
⎛
⎠
⎞
4
A + 1 ...(1.5)
where Eα =
1
2 mαv
2
α is the kinetic energy of the α-particle, and A is the mass number of the
daughter nucleus.
For large values of A, E
ʹ′
α
≅ Eα and hence the α-particle carries most of the
disintegration energy.
1.4 THE VELOCITY OF ALPHA PARTICLES
The method that gives the most accurate results for the velocity and energy of
α-particles depends on the measurement of the deflection of the paths of the particles in a
magnetic field. When a charged particle moves in a magnetic field, its orbit is a circle whose
radius is determined by the relation
Bqv =
mv2
r …(1.6)
where B is the field strength, q and m are the charge and mass of the particle respectively,
and r is the radius of the orbit. Equation (1.6) may also be written as
v =
q
m Br …(1.7)
The velocity of alpha particle can be determined if the strength of the magnetic field is
known and if the radius of the orbit is measured, since the value of the charge-to-mass ratio
is well known.

3. Alpha Decay 3
Also the kinetic energy E of the alpha particle may be written as E =
1
2 MV2 …(1.8)
A schematic diagram of an apparatus based on this principle is shown in Fig. 1.1.
Fig. 1.1 : Schematic diagram of the deflection chamber of a magnetic spectrograph for
α-particles
The α-particles from a radioactive source emerge in a narrow beam through a defining
slit. A magnetic field of known strength, acting in a direction perpendicular to the plane of
the diagram, bends the α-particles through a angle of 180°. The chamber slits help to
reduce the scattering of α-particles from the top, bottom, and the walls of the chamber.
Particles with the same velocity have semicircular paths of the same radius. They may be
detected with a photographic plate or with counters, and the radius of the path can be
measured. In figure 1.1, the paths of two groups of particles with different velocities are
shown. This instrument is called a magnetic spectrograph and is designed on the principle
of semicircular magnetic focussing.
The velocity v is obtained in cms/sec when B is expressed in gauss, r in cms, and
q
m in
emu/gm. When the relativity correction is taken into account v is replaced by
v = Br
q
m0
1 – v2/c2 …(1.9)
and T = m0c2
⎣
⎢
⎢
⎡
⎦
⎥
⎥
⎤
1
1 –
⎝
⎜
⎛
⎠
⎟
⎞
v2
c2
–1 …(1.10)
where m0 is the rest mass of the α-particle. T is relativistic kinetic energy.
1.5 ABSORPTION OF ALPHA PARTICLES
Range, Ionisation and Stopping Power
The energies of various charged particles, can be determined from the measurements
of their absorption by matter. This is true for α-particles also and some of the features of the
absorption of α-particles are as given below :
1. Alpha particles are easily absorbed; those emitted in radioactive disintegration can
generally be absorbed by a sheet of paper, by an aluminum foil 0.004 cm thick, or by
several centimeters of air.
2. If the particles emitted by a source in air are counted by counting the number of
scintillations on a zinc sulphide screen, it is found that their number stays practically
constant upto a certain distance R from the source, and then drops rapidly to zero.

4. 4 Nuclear Physics (T.Y. B.Sc.) (Sem.–VI)
This distance R is called the range of the particles, and is related to the initial energy of
the particles.
2.5.1 Range of Alpha Particle
Method to determine the range of alpha particles
The measurements of ranges in air can be made with an apparatus shown in Fig. 1.2.
Fig. 1.2 : An apparatus for precise measurements of the range of α-particles
The apparatus consists of a source placed on a moveable block whose distance from
the detector, i.e. a thin screen-walled ionization chamber can be varied. A narrow beam of
alpha particles emitted from the source emerges through a collimating slit, passes through
a known thickness of air, and reaches the detector. The detector is 1 to 2 mm deep.
When the α-particles pass through the ionization chamber, ions are formed in pulses
in the chamber. The voltage pulses induced on the chamber electrode are then amplified
electronically and counted. The counting rate is then determined as a function of distance
between source and detector.
The results of an experiment with α-particles from polonium are shown in Fig. 1.3.
Fig. 1.3 : Range curves for the α-particles from Po212
Diaphragm
Collimated
source on
adjustable
block
Screen-walled
ionization
chamber
3.60 3.70 3.80 3.90 4.00
cm
0
0.2
0.4
0.6
0.8
1.0
Relative
ionization
A
D
C
B
A

5. Alpha Decay 5
Range
The distance through which an α-particle travels in a specified material before
stopping to ionize it is called its range in that material. The range is thus a sharply defined
ionization path-length. The range is usually expressed in cm, in air at 76 cm of mercury
pressure and 15°C.
The range is highest in gases, less in liquids and the least in solids due to more and
more dense packing of the particles.
Plainly, in a gas, the range depends on (i) the initial energy of the α-particle, (ii) the
ionization potential of the gas and (iii) the chances of collision between the α-particles and
the gas particles, that is, on the nature and the temperature and pressure of the gas. With
increase of pressure, the range decreases; it increases if the temperature of the gas is
increased.
The range of α-particles depends on their initial velocity. For monoenergetic α-particle
of velocity V, the range R in standard air is proportional to V3.
∴ R α V3
or R = a V3
where a is a constant.
This relation R = a V3 is known as Geiger Rule or Geiger law and is valid only in a
limited velocity range.
1.5.2 Mean Range and Extrapolated Range
A plot of intensity of a collimated beam of
α-particles vs. different distances from source is as
shown in Fig. 1.4.
For monoenergetic beam of α-particles, the
intensity falls abruptly to zero at a definite
distance d from the source. This d-value gives the
range R of the beam.
The distance from the source at which the
intensity is half the initial intensity is called mean
range R implying 50% of α-particles have ranges
greater than R and 50% have ranges less than R. If
the linear portion ABC of the graph is
extrapolated to zero intensity, we get what is called extrapolated range, Rex which is slightly
more than R.
1.5.3 Specific Ionisation
Due to ionization, there are a large number of ion-pairs generated along the path of
the α-particles in a gas. Their number in unit length of the path at any point is proportional
to the energy lost in the region.
The number of ion-pairs formed per unit path-length at any point in the path of the α-
particle is called specific ionization and is symbolized by I.
Thus the ionization produced by an α-particle at any point in its track is inversely
proportional to its velocity at that point.
Intensity
Distance
Mean
Range
Extrapolated range
A
B
C
D
Fig. 1.4

6. 6 Nuclear Physics (T.Y. B.Sc.) (Sem.–VI)
1.5.4 Straggling
The α-particles of the same initial energy have more or less the same range in matter.
However, a small spread in the values of ranges about a mean value is observed. This
phenomenon is known as the straggling of the
range. A plot of the number of ions per cm along
the path of α-particle vs the range of α-particle is a
curve called Bragg curve as shown in Fig. 1.5. The
Bragg curve indicates that the ionization is fairly
constant over the initial part and rises to the hump
towards the end when the speed of the α-particles
falls below the ionization potential of the gas, the
curve steeply falls down. But the x-axis is not met
abruptly. Near the end of the path it tails off. This
is straggling.
The straggling of range occurs mainly due to
two reasons : (i) there is a statistical fluctuation in
the number of collisions suffered by the different
particles about a mean value in traveling over a
given distance, and (ii) there is also a statistical fluctuation about a mean value in the
energy loss per collision.
1.5.5 Stopping Power
The energy of α-particle progressively decreases as they pass through increasing
thicknesses of matter. The amount of energy loss of an α-particle per unit path length in
the absorber is called the stopping power of the absorber.
The air equivalent of an absorber is the thickness, ta , of standard air that produces the
same energy loss of α-particles as does a given thickness t, of the absorber placed in the
path. The ratio t/ta is called the equivalent stopping power te . The equivalent stopping
power is thus the thickness of the absorber that produces the same energy loss in the
α-particle as does unit thickness of standard air, for since
te = t/ta , we obtain te = t , when ta = 1
The relative stopping power, S, is the ratio 1/te . It is thus the ratio of the energy loss
(of the α-particle) in traversing unit distance in the absorber to that in traversing unit
distance in standard air.
1.6 GEIGER-NUTTAL LAW
In 1911, Geiger and Nuttal made an
experimental study of relation between range
(or energy) of α-particle in air and half-life (or
decay constant) of a number of α-emitters. They
observed that longest lived nuclides (having
smallest value of decay constant) emit the least
energetic α-particle, while the shortest lived
nuclides (having the highest value of decay
constant) emit the most energetic α-particle and
gave following empirical law :
log10 λ = A + B log10 Rα …(1.11)
where λ is the decay constant, Rα the range of α-
particle in air and A and B are constants. If,
therefore, log10 λ are plotted against log10 Rα for
the different α-emitters in three series, three
nearly parallel straight lines are obtained, one for
each series as shown in Fig. 1.6.
range
straggling
ions
per
cm
Fig. 1.5
Po
Th Bi
Po
Rn
Ac
Ra
Th
Pa
Ra
Io
U
Uranium series
Actinium series
Thorium series
(
4
n
+
3
)
(
4
n
+
2
)
(
4
n
)
0.4 0.5 0.6 0.7 0.8
log R (R in cm)
20
15
10
5
0
log
+
20
(
in
s
–1
)
Fig. 1.6 The Geiger-Nuttal Law

7. Alpha Decay 7
In the relation log10 λ = A + B log10 Rα , the constant B has the same value for all the
series while the constant A has different values for each series.
According to Geiger's law Rα ∝ V3 , where V is the velocity of emission of the α-particle.
If E is the energy of the α-particle, then,
E =
1
2 mV2 or E ∝ V2
Rα ∝ E3/2
or log10 Rα ∝
3
2 log10 E
∴ log10 Rα = B' log10 E …(1.12)
Hence in terms of energy Geiger – Nuttal law may be put as
log10 λ = A' + B' log10 E ...(1.13)
Importance
The decay constants of many radioactive substances can be determined using this
empirical law. The ranges of α-particle corresponding to the members of the three series
are found experimentally and from Geiger-Nuttal relation, the decay constants are
calculated.
1.7 ALPHA-RAY SPECTRA
In 1930 it was shown by S. Rosenblum in magnetic deflection experiments that the
normal α-particles emitted by some radioactive nuclides fall into several spaced velocity
groups. Each group has a sharply defined range and energy. The velocities and energies of
the different groups differ so little that the ranges of all the particles lie within the region of
straggling. For this reason the different groups could be separated only in a magnetic
spectrograph of high resolving power. The discrete, closely spaced components of the
α-rays are said to form a spectrum or to show fine structure. The fine structure is because of
the close-spacing of several monoenergetic groups.
Alpha-particle spectra may be divided into three groups :
1. Spectra consisting of a single group or "line". Examples : Rn222, RaA (Po218), RaF
(Po210).
2. Spectra consisting of two or more discrete, closely spaced (in velocity or energy)
components with intensities of the same or of only a slightly different order of
magnitude. Examples : ThC (Bi212), An (Rn219), AcX (Ra223), ThX (Ra224), Pa, Rd
Ac (Th227).
3. Spectra consisting of a main group and groups of much higher energy (long-range)
particles, the latter containing however only a small fraction (10-4 to 10-7) of the
number of particles in the main group. The only examples are : RaC' (Po214) and
ThC' (Po212).
1.8 SHORT-RANGE ALPHA PARTICLES
The fact that α-particles are emitted with discrete energies lead to the conclusion that
the nucleus has sharply defined energy levels. When the energy differences between the
ground state of the parent nucleus and that of the daughter nucleus is sufficiently large
than the daughter nucleus can be formed in the ground state or in one of the excited
states. Hence short range α-particles of different energies and therefore different ranges
are emitted by the parent. The daughter nucleus then comes to the ground state by
emitting gamma photons of appropriate energies. By the measurements of γ energy and α
energy the energy levels of the nucleus can be found.

8. 8 Nuclear Physics (T.Y. B.Sc.) (Sem.–VI)
Rosenblum showed by using semicircular focussing magnetic spectrograph that the α-
rays from ThC (Bi212) are not monoenergetic but consists of several closely spaced mono
energetic groups or α-ray lines.
The interpretation of complex α-particle spectra in terms of nuclear energy levels is
illustrated by the case of ThC (Bi
212
) which emits six groups of short-range α-particles with
energies as shown in figure 1.7.
Fig. 1.7 : Decay scheme for the α-decay of ThC (Bi212) to ThC" (Tl208)
The most energetic particles have a kinetic energy of 6.086 MeV corresponding to an
α-disintegration energy of 6.203 MeV. It is supposed that a ThC (Bi212) nucleus always
releases 6.203 MeV of energy when it decays by α-emission to form a nucleus of ThC"
(Tl208) and that this amount of energy is associated with one of the most energetic
α-particles. The emission of one of these particles is assumed to leave the product nucleus in
its lowest energy state or ground state. All of the disintegration energy has gone into kinetic
energy of the α-particle and the recoil energy of the ThC" (Tl208) nucleus.
Suppose that a ThC (Bi212) nucleus emits an α-particle of the α1 group for which
Eʹ′α1
= 6.163 MeV or 0.040 Mev less than the total available α-disintegration energy. The
ThC" nucleus, which retains the 0.040 MeV of energy, should be left in an excited state. It
might be expected to undergo a transition to the ground state by emitting electromagnetic
radiation. If this is the case, the radiation should appear in the form of γ-rays with an energy
of 0.040 MeV. By extending this analysis the existence of at least five excited states of ThC"
(Tl208) nucleus can be postulated. Experimentally eight γ-rays have been identified.
The energy level diagram for the daughter nucleus is usually shown together with the
decay data for parent nucleus and the result is called the decay-scheme.
(ThC (Bi212) → ThC" (Tl208)).
The diagonal lines represent the different α-particle groups observed in the decay of
ThC (Bi212), the disintegration energies are also shown. The vertical lines represent γ-rays
found experimentally. Each γ-ray corresponds to the energy state. Gamma rays are not
observed for all transitions which seem possible. The absence of these γ-rays may be caused
either their small intensity or by nuclear selection rules which may make the probability of
these transitions very small.
0
(
6
.
2
0
3
M
e
V
)
1
(
6
.
1
6
3
M
e
V
)
2
(
5
.
8
7
4
M
e
V
)
3
(
5
.
7
3
0
M
e
V
)
4
(
5
.
7
1
1
M
e
V
)
5
(
5
.
5
8
4
M
e
V
)
6.203
0.619
0.492
0.473
0.329
0.040
0
(0.040)
(0.288)
(0.328)
(0.472)
(0.452)
(0.144)
(0.164)
(0.432)
(0.144)
ThC" (Tl208
) ThC (Bi212
)

9. Alpha Decay 9
1.9 LONG-RANGE ALPHA PARTICLES
A given nuclide often emits long-range alpha particles with a number of different
energies. RaC' (Po214) and ThC' (Po212) emit a few long range particles having energies
considerably greater than that of main group of particles. The existence of long-range α-
particles was first observed by E. Rutherford and Wood. The origin of the long-range
α-particles emitted by ThC' (Po212) can be interpreted in terms of energy levels as shown in
figure 1.8.
It is possible for a nucleus, before an α-disintegration, to be in an excited state, the
excitation may result from a previous disintegration. The emission of a nuclear β-ray
sometimes leaves the newly formed nucleus in an excited state. In most cases the nucleus
(Po212) goes to the ground state by emitting γ-ray of proper energy. In some cases however
when the life-time for α-decay is comparable with the life-time for γ-decay the newly formed
excited nucleus (Po212) is an α-emitter, and gets rid of its excess energy by emitting
α-particles with greater energy than that of the normal particles. The existence of long-
range α-particles is thus explained as being caused by the decay of an excited nucleus
(Po212). The extra energy of the α-particle measures the excitation energy of the initial
nucleus (Po212).
Fig. 1.8 : Energy level diagram for the ThC (Bi212) → ThC' (Po212) → ThD (Pb208) decays, showing
the origin of long-range α-particles of Po212
1.10 GAMOW'S THEORY OF ALPHA DECAY
Experiments on the Rutherford scattering of alpha particles indicate that Coulomb's
inverse square law of repulsion between the positive charge on the scattering nucleus and
, 0.08 MeV (5%)
, 0.45 MeV (8.5%)
, 0.67 MeV (6%)
, 0.93 MeV (7.5%)
, 1.55 MeV (10%)
, 2.27 MeV (63%)
ThC´ (Po212)
ThC (Bi212)
Energy above ground level
of ThC´ (MeV)
Energy above ground
state of ThD (MeV)
11.20
10.75
10.55
10.29
9.67
8.95
2.19
1.8
1.6
1.34
0.72
0
ThD (Pb208)
(
1
0
.
7
4
6
M
e
V
)
(
1
0
.
6
2
2
M
e
V
)
(
9
.
6
7
5
M
e
V
)
(
8
.
9
4
5
M
e
V
)

10. 10 Nuclear Physics (T.Y. B.Sc.) (Sem.–VI)
that on the scattered alpha particle is maintained down to a distance of the order of the
nuclear radius, beyond which Coulomb's law breaks down.
When an alpha particle approaches a nucleus head-on, its kinetic energy gradually gets
converted to potential energy and it is turned back by the nucleus at the distance of
approach where its kinetic energy is completely converted into potential energy.
Considering the nature of the force experienced by the incident alpha particle a
nucleus can be represented by a potential well. A graph of the potential energy of an alpha
particle as a function of its distance from the centre of the nucleus can have the form as
shown in figure 1.9.
There is a potential barrier for an incident alpha particle upto r = rN. Beyond that there
is a potential well in which the particle can get trapped. Since the nucleons are held
together with strong forces of attraction within the nucleus (where r < rN), the potential
energy of the alpha particle inside the nucleus, whether it exists in the nucleus or is formed
from two protons and two neutrons at the time of its emission, is negative. As there is no
detailed information of the forces which act on the alpha particle when r < rN, the actual
potential energy curve for this region cannot be plotted as can be done for the region r > rN
where the Coulomb forces act.
Fig. 1.9 : Potential energy of an alpha particle as a function of its distance from the centre of the
nucleus.
According to classical mechanics, an incident alpha particle can fall into the potential
well, that is, penetrate the nucleus, only if its energy Eα is greater than the height Vp of the
potential barrier. In the same way an alpha particle can be emitted from within the nucleus
if its energy is greater than the barrier height. Thus, the maximum value of the potential
energy curve, corresponds to the value of the kinetic energy which an alpha particle must
have according to classical mechanics to get into the nucleus from outside, or to escape
from inside the nucleus. However, from the measurement of the ranges of the alpha
particles it has been calculated that particles having kinetic energies very small as compared
to the barrier height can also penetrate the potential barrier.
Example : In case of U238 the height of the potential barrier for alpha particles is
35.6 MeV. Actually the kinetic energy of the emitted alpha particle is found to be 4 MeV,
indicating even when the total energy Eα < Vp then too the alpha particle is able to escape
from the U238 nucleus. For this the alpha particle has to transverse the region rN < r < r1
where its potential energy exceeds its total energy.
Classical physics is unable to explain this phenomenon. However, it can be explained
with the help of quantum mechanics and is called the tunnel effect.
Potential
energy
V
(r
)
– Vo r
Vp
0
E
rN r1
V(r ) = 4 0r
2Ze2

11. Alpha Decay 11
The phenomenon of tunnel effect was explained by George Gamow and independently
by Condon and Gurney. It is known as Gamow's theory of alpha decay.
The probability for tunnel effect or barrier penetration can be obtained by considering
a one dimensional rectangular potential barrier of the form :
V(x) = V0 , 0 < x < L ...(1.14)
= 0 , x < 0 and x > L
Fig. 1.10 : A rectangular potential barrier of height V0 and width L
An alpha particle of total energy E in the region x < 0, is incident upon the barrier in
the direction of increasing x. Let E < V0 , the height of the barrier. The barrier is broken up
into three regions :
Region I : x < 0, V(x) = 0
Region II : 0 < x < L, V(x) = V0
Region III : x > L, V(x) = 0
The Schrodinger equation in region I is
d2 ψI
dx2 +
2m
"2
E ψI
= 0 ...(1.15)
where m is the mass of the alpha particle.
Let k1
2 =
2mE
"2 ...(1.16)
The general solution of equation (1.15) is
ψI (x) = Aei k1x + Be- ik1x ...(1.17)
Aeik1x represents the incident wave and
Be- i k1x represents the reflected wave.
Schrodinger's equation in Region II is
d2 ψII
dx2 +
2m
"2
(E – V0) ψII
= 0 ...(1.18)
Let k2
2 =
2m
"2
(V0 – E)
...(1.19)
The general solution of equation (1.18) is
ψII (x) = Cek2x + De- k2 x ...(1.20)
Cek2x represents the transmitted wave and
De- k2x represents the wave reflected from the other side of the barrier (at x = L).
0 L x
V(x)
V0

12. 12 Nuclear Physics (T.Y. B.Sc.) (Sem.–VI)
In region III, Schrodinger's equation is
d2 ψIII
dx2 +
2m
"2
E ψIII
= 0 ...(1.21)
Since region III has only a transmitted wave, the general solution of equation (1.21) is
ψIII (x) = Fei k1x ...(1.22)
The constants A, B, C, D and F are determined by using the following boundary
conditions :
ψI = ψII and
dψI
dx =
dψII
dx at x = 0 ...(1.23)
ψII = ψIII and
dψII
dx =
dψIII
dx at x = L ...(1.24)
Conditions (1.23) give,
A + B = C + D ...(1.25)
ik1 (A – B) = k2 (C – D) ...(1.26)
Conditions (1.24) give,
Cek2L + De- k2L = Feik1L ...(1.27)
k2 ( )
Cek2L – De- k2L = ik1 Feik1L ...(1.28)
From equations (1.27) and (1.28)
C =
1
2 F
⎝
⎜
⎛
⎠
⎟
⎞
1 +
ik1
k2
e(ik1 - k2)L ...(1.29)
and D =
1
2 F
⎝
⎜
⎛
⎠
⎟
⎞
1 -
ik1
k2
e(ik1 + k2)L ...(1.30)
From equations (1.25) and (1.26)
A =
1
2 C
⎝
⎜
⎛
⎠
⎟
⎞
1 +
k2
i k1
+
1
2 D
⎝
⎜
⎛
⎠
⎟
⎞
1 –
k2
i k1
...(1.31)
Substituting the values of C and D from equations (1.29) and (1.30) in equation (1.31)
gives
A =
1
4 F
⎝
⎜
⎛
⎠
⎟
⎞
1 +
ik1
k2 ⎝
⎜
⎛
⎠
⎟
⎞
1 +
k2
ik1
e(ik1 - k2)L +
1
4 F
⎝
⎜
⎛
⎠
⎟
⎞
1 –
ik1
k2 ⎝
⎜
⎛
⎠
⎟
⎞
1 –
k2
ik1
e(ik1 + k2)L ...(1.32)
The transmission coefficient which specifies the probability that the alpha particle will
be transmitted through the barrier into the region x > L is given by
T =
| F |2
| A |2 ...(1.33)
Since in practice as the barrier is high and wide, k2 L ≥ 1, the first term of equation
(1.32) can be neglected in comparison with the second.
| A |2
| F |2 =
⎝
⎛
⎠
⎞
A
F ⎝
⎛
⎠
⎞
A
F
*
=
1
16 ⎝
⎜
⎛
⎠
⎟
⎞
1 –
ik1
k2 ⎝
⎜
⎛
⎠
⎟
⎞
1 –
k2
ik1 ⎝
⎜
⎛
⎠
⎟
⎞
1 +
ik1
k2 ⎝
⎜
⎛
⎠
⎟
⎞
1 +
k2
ik1
e+ 2k2L
| A |2
| F |2 =
( )
k
2
1 + k
2
2
2
16k
2
1 k
2
2
e
2k2 L

13. Alpha Decay 13
∴ T =
16k
2
1 k
2
2
( )
k
2
1 + k
2
2
2
e– 2k2 L
, k2 L ≥ 1. ...(1.34(a))
Substituting for k1 and k2 from equations (1.16) and (1.19) respectively,
T =
16 E (V0 – E) e–2k2L
V0
2 …(1.34(b))
When 2 k2L ≥ 1, the exponential is extremely small. The factor before the exponential
part is usually of the order of magnitude unity (maximum value is four when k1 . k2). For
order of magnitude calculations, equation (1.34(a)) becomes
T = e– 2k2 L ...(1.35)
This equation gives the probability that an alpha particle with a total energy less than
the barrier height (E < V0) will penetrate the barrier of width L. If the potential is not
constant in the region 0 < x < L, it can be approximated with a series of small steps, each
with a constant potential. Then the total penetration probability is given by
T = e
–2 k2dx
0
L
∫
...(1.36)
For the Coulomb barrier (figure 1.11)
T = exp
⎝
⎜
⎜
⎛
⎠
⎟
⎟
⎞
- 2
2m
"
rN
r1
∫ ( )
V(r) – E 1/2
dr ...(1.37)
where V(r) =
2Ze2
4π ∈0 r
...(1.38)
Fig. 1.11 : Wave mechanical barrier penetration or tunnelling
The probability that the nucleus will emit an alpha particle in unit time is given by the
product of the penetration probability T and the frequency
vα
2rN
with which the alpha
particle bounces back and forth at the potential barrier along a diameter 2rN of the
Incident wave
Transmitted wave
Barrier
width
0 r
V(r)
– V0
rN r1

14. 14 Nuclear Physics (T.Y. B.Sc.) (Sem.–VI)
emitting nucleus with a speed vα. Since the probability of emission of the alpha particle in
unit time is the same as the disintegration constant λ for the process,
λ =
vα
2rN
× T ...(1.39 (a))
log λ = log
vα
2rN
+ log T ...(1.39 (b))
If this relation is compared with the Geiger-Nuttal Law given by equation (1.11),
log λ = A + B log R
It is seen that the range R is related with the velocity vα and hence log T is a constant.
This theory is also able to account why short lived nuclides emit long range alpha
particles. If the nuclide is short lived then equation (1.39 (a)) indicates that the probability
T for barrier penetration for an alpha particle must be high. This is possible if the width of
the barrier is small. Hence it can be inferred that alpha particles which penetrate the
barrier at points of smaller width must have high kinetic energy. Hence their range is long.
SOLVED PROBLEMS
Problem 1 :
Ra226 decays by α-emission to Rn222. The disintegration energy Q = 4.863 MeV.
Assuming the mass of He4 atom to be 4.002603 a.m.u. and M(Ra226) = 226.05432 a.m.u.,
calculate the atomic mass of Rn222.
Solution :
88Ra226 ⎯→ α + 86Rn222 + Q
Q = (M (Ra226) – M(α) – M(Rn222)) × 931.5 MeV
4.863 = (226.05432 – 4.002603 – M(Rn222)) × 931.5 MeV
M(Rn222) = 222.0465 a.m.u.
The atomic mass of Rn222 is 222.0465 a.m.u.
Problem 2 :
The nuclide 84Rn211 emits three groups of alpha particles with kinetic energies of
5.847 MeV, 5.779 MeV and 5.163 MeV respectively. Associated with the α particles are the
γ rays with energies of 0.069 MeV, 0.170 MeV and 0.239 MeV. The mass of 82Pb207 is
207 a.m.u. Construct a decay scheme based on this data.
Solution :
The alpha disintegration energy is given by
E
ʹ′
α
= Eα
⎝
⎜
⎛
⎠
⎟
⎞
mα
M + 1
where Eα is the kinetic energy of the α particle, mα is its mass and M is the mass of the
daughter nuclide.
mα = 4.0000 a.m.u.
M = 207 a.m.u.
When Eα = 5.847 Mev
E
ʹ′
α
= 5.847
⎝
⎛
⎠
⎞
4
207 + 1 = 5.96 MeV = Q1
When Eα = 5.779 MeV
E
ʹ′
α
= 5.779
⎝
⎛
⎠
⎞
4
207 + 1 = 5.891 MeV = Q2

15. Alpha Decay 15
When Eα = 5.613 MeV
E
ʹ′
α
= 5.613
⎝
⎛
⎠
⎞
4
207 + 1 = 5.721 MeV = Q3
The gamma ray energies are :
Q1 - Q2 = (5.96 - 5.891) MeV = 0.069 MeV
Q1 - Q3 = (5.96 - 5.721) MeV = 0.239 MeV
Q2 - Q3 = (5.891 - 5.721) MeV = 0.170 MeV
Fig. 1.12 : Decay Scheme Rn211 → Pb207
Problem 3 :
Estimate the transmission coefficient for a α particle (mass 6.64 × 10-27 kg) having a
kinetic energy of 5 MeV, the height of the barrier being 20 MeV. The width of the barrier is
10 Fermi.
Solution :
The transmission coefficient is given by
T =
16 E (V0 – E) e –2k L
V0
2
Where k =
2m (V0 – E)
"2
E = 5 MeV, V0 = 20 MeV
m = 6.64 × 10-27 kg
L = 10F = 10 × 10-15 m
k =
2 × 6.64 × 10-27 × 15 × 106 × 1.6 × 10-19
(1.054 × 10-34)2 = 1.694 × 1015
e– 2kL = e– 33.86 = 1.971 × 10-15
∴ T =
16 × 5 × 15 × 1.971 × 10-15
(20)2 = 5.913 × 10-15
The alpha particle has the probability of about 5.9 in 1015 collisions against the barrier
wall to penetrate through the barrier.
84Rn211
82Pb207
0.170 MeV
0.069 MeV
0.239 MeV
Q
3
(
5
.
7
2
1
M
e
V
)
Q
2
(
5
.
8
9
1
M
e
V
)
Q
1
(
5
.
9
6
M
e
V
)

16. 16 Nuclear Physics (T.Y. B.Sc.) (Sem.–VI)
QUESTIONS AND PROBLEMS
1. From conservation principles, derive an expression for the alpha disintegration energy.
2. Explain how a magnetic spectrograph can be used to determine the velocity of alpha
particles.
3. What is the Geiger-Nuttal law? Explain its importance in alpha decay.
4. What does the alpha particle spectrum exhibit a fine structure? Explain the origin of
short range and long range alpha particles with appropriate energy level diagrams.
5. Explain Ganow’s theory of alpha decay.
6. Explain the terms : (a) specific ionization, (b) stopping power, (c) Range,
(d) straggling with respect to α-particles.
7. P0
212 α-particles have kinetic energy of 8.776 MeV. Assuming the mass of the α-particle
to be 6.67 × 10–27 kg calculate their velocity.
8. P0
210 α-particles having kinetic energy of 5.3 MeV, are subjected to a magnetic field of
1T. What is the radius of curvature of their tracks? The charge on the α-particle is
3.2 × 10–19 C.
9. Using the data in problems 7 and 8, calculate the alpha disintegration energies for
P0
212 and P0
210.
10. Calculate for an α-particle of energy 5 MeV, the order of magnitude of probability of
leakage through a potential barrier of width 10-12 cm and height 10 MeV.
11. Estimate the transmission coefficient for a α particle of mass 6.64 × 10–27 kg having a
kinetic energy of 5 MeV, the height of the barrier being 15 MeV. The width of the
barrier is 2 × 10-14 m.
12. The nuclide U233 emits six groups of alpha particles, with kinetic energies of 4.816
MeV, 4.773 MeV, 4.717 MeV, 4.655 MeV, 4.582 MeV and 4.489 MeV respectively.
Gamma rays with energies 0.0428 MeV, 0.0561 MeV and 0.099 MeV have also been
reported. Construct a decay scheme based on these data.
13. The nuclide Am241 emits six groups of alpha particles with kinetic energies of 5.534
MeV, 5.500 MeV, 5.477 MeV, 5.435 MeV, 5.378 MeV and 5.311 MeV respectively.
Gamma rays are found, with energies of 0.0264 MeV, 0.0332 MeV, 0.0435 MeV, 0.0555
MeV, 0.0596 MeV, 0.103 MeV and 0.159 MeV. Construct a decay scheme based on
these data.
14. 83Bi212 decays with a half life of 60.5 mins emitting 5 groups of ∝-particles having
energies 6.08 MeV, 6.04 MeV, 5.76 MeV, 5.62 MeV and 5.60 MeV. Show that the
∝-distintegration energies are 6.20 MeV, 6.16 MeV, 5.87 MeV, 5.73 MeV and 5.71 MeV.
Also show that the danghter nucleus is 81TL208. Sketch the energy level scheme for the
daughter nucleus.