This document provides information about alpha decay and alpha particles. It discusses: 1) Unstable nuclei attain stability through emission of alpha particles, which are made up of 2 protons and 2 neutrons. 2) Alpha decay involves the emission of an alpha particle from an unstable nucleus, leaving a lighter nucleus. Conservation laws apply. 3) The range of alpha particles is very small, usually only a few centimeters in air or solid materials, due to their high ionization which causes energy loss. Their short range makes them easily stopped.

1. NUCLEAR
PHYSICS
COURSE CODE: USPH603
HSNC UNIVERSITY
DEPARTMENT OF PHYSICS
KC COLLEGE

2. ALPHA DECAY
INTRODUCTON:
Nuclei with certain combination of protons and neutrons have unstable configuration.
[radioactive]
Unstable nuclei attain stability by emission of certain particles.
α- decay: The emission of α-particles by a unstable radioactive nuclei.

3. Alpha particle:
a positively charged particle emitted by a radioactive nucleus.
made up of 2 protons and 2 neutrons.
α-particle is the nucleus of a helium atom [2He4].

4. The process of alpha decay is given as:
ZMA
Z-2MA-4 + 2He4
where ZMA and Z-2MA-4 are the parent and daughter nuclides and
2He4 is the alpha particle.

5. Energetics of Spontaneous Alpha Decay:
The process of alpha decay is given as:
ZXA
Z-2YA-4 + 2He4
where Y is the residual nucleus of mass number A-4 are the parent and atomic number Z-2.
 If the masses of the α-particle [2He4] and the residual nucleus be 𝑀𝛼 and 𝑀1 respectively, and
𝑣𝛼 and 𝑣1 their respective velocities then conservation of momentum requires that:
𝑀𝛼𝑣𝛼 = 𝑀1𝑣1------------------------------------------------------------[1]

6.  If Q is the α-disintegration energy, which is the total energy released in the disintegration
process,
Q =
1
2
𝑀𝛼𝑣α
2
+
1
2
𝑀1𝑣1
2
Using equation [1] and on simplification, we get:
Q = 𝐸𝛼
𝑀1+ 𝑀α
𝑀1

7. Masses of the nuclei in the unit of atomic masses are close to their mass numbers, we can
write 𝑀1= A-4 and 𝑀𝛼 = 4.
Alpha disintegration energy-
Q = 𝐸𝛼
𝐴
𝐴−4
----------------------------------------------------------------[2]
Q can be determined from equation [1] as 𝐸𝛼 can be determined experimentally.
Example: 210Po, the α disintegration energy is 𝐸𝛼 = 5.305 MeV which gives Q = 5.408 MeV.

8. Accurate measurement of α disintegration energy is important from theoretical point of view.
The energy released during the nuclear transformation has its origin in the mass of the
transformation nucleus.
A part of this mass is converted into energy according to Einstein’s mass-energy equivalent
principle……….
The large quantity of energy released in α disintegration energy has also the same origin.

9.  A part of the mass of the disintegrating nucleus is converted into energy as the α disintegration
energy.
Possibility of the α disintegration process:
α disintegration process is possible when the mass M of the disintegrating parent nucleus is greater
then the sum of the masses of the α-particle and the product nucleus.
M > 𝑀𝛼+ 𝑀1
The α disintegration energy is given by:
Q = (M - 𝑀𝛼 - 𝑀1)𝑐2

10. The masses in the above equation are the atomic masses and not the nuclear masses though
the α -emission takes place as a result of the transformation of the nucleus .
This is possible because the electronic masses cancel out in the above equation.

11. Range of α – particles:
 α–particles from natural radio-active elements are easily absorbed in matter.
They can pass through thin paper or a very thin foil of mica or aluminium, but cannot penetrate
a few layers of these.
 α–particles can travel up to a distance of few centimetres from the source in air at S.T.P.
They lose their entire kinetic energy.

12. Range of α–particle:
The monoenergetic α–particle from a given source can travel a definite maximum distance
from the source within a given substance.
The range of α–particle is very small in solid or in a liquid ≈ 10−3
mm for α–energy of few MeV.
Range of α–particle is relatively longer in gases because of low-density of gas.
In case of a gas the range depends on the temperature and the pressure of the gas.

13. With increase of pressure, the range decreases, while it increases with increase of
temperature.
The range of the α–particle depends on their initial velocity or kinetic energy.
Accurate measurements of the ranges of α–particles of different velocities gives the
relationship between the two quantities R = R(v).

14. Experiment to measure the Range of α–particle:
W.H.Bragg [England] – first to determine the range of α–particle.
Measured the ionization produced by α–particle at different distances from the sources along
their path within the medium.
Experimental set-up:
α–particles produced by the source S are collimated by a slit within the plate P.
A and B are two parallel wire-gauzes with a very small gap between them.
The positions of these wire-gauzes can be changed without altering the distance between
them.

15. The collimated beam of α–particles enter the region between A and B.
α–particles ionize the gas through which they pass.
During travel through the gas, an α–particle suffers repeated collisions with the gas atoms and
lose small fraction of energy to these atoms which get ionized.
Large number of ion-pairs is produced between A and B.
Ion pairs produced are attracted to towards these due to potential difference applied between
them which gives rise to ionization current.

16. Ionization current can be measured with the help of electrometer.
The potential difference between A and B is so adjusted that all
ion-pairs produced between them are drawn to them, thereby
producing saturation current.
The saturation current is proportional to the number of ion-pairs
produced between A and B.
Bragg moved the wire-gauzes to different distances from the
source and measured the saturation ion current between them
and plotted it as a function of the mean distance of the gauzes
from the source.

17.  Ionizing power of the α–particles rises with
increasing distance of its travel from the
source.
 The increase is first slow, but is more rapid
afterwards.
 After reaching a maximum the ion current
rapidly begins to go down and falls to zero at
a definite distance from the source known as
the range of α–particles.

18. Fig: Variationof ionizationcurrent withdistance

19.  The steeply falling portion of the ionization current graph bends to the right just before the
current becomes zero.
The bend portion arises due to the straggling of the range.

20. SPECIFIC IONIZATION:
The number ∆𝑛
∆𝑥 of ion-pairs produced per unit length of its travel by an α–particle in a gas at
one atmospheric pressure.
 ∆𝑛 is number of ion-pairs produced in a distance ∆𝑥.
The specific ionization depends on the distance travelled by the α–particle from the source.
Example: For RaC’ α–particles [E = 7.68 MeV], the maximum value of the specific ionization is
about 6000 ion-pairs per millimetre.

21. Specific ionization increases as the distance travelled by the α–particle from the source
increases.
As the α–particle moves farther away from the source, its velocity decreases due to loss of
energy by ionization of the atom of the gas.
The slower α–particle spend longer time near the atoms in the gas.
There is a greater probability of their interaction with the electrons in the atom which is the
cause of ionization of the atoms.

22. For this reason, when α–particles are near the end of their paths the specific ionization is
maximum.

23. Measurement of range of α–particle by measuring the intensity of the beam:
 Another method for the measurement of the range of α–particles is to determine the intensity
of the collimated beam of α–particles at different distances from the source.
This remains unchanged with distance till the end of the path of the α–particles.
M.S.Holloway and M.G.Livingston used this method to determine the α–ranges accurately with
the help of a shallow ionization chamber as the α–detector.

24. A collimated beam of α–particles from a given source falls on the scintillation phosphor (e.g.
ZnS) attached to a photomultiplier.
The photomultiplier tube records the current proportional to the intensity of the α–beam.
The distance between the source and the scintillation detector is gradually increased and graph
of the intensity versus the distance obtained.
The graph is known as integral range curve.

25.  For a monoenergetic beam of α–particles, the intensity
falls abruptly to zero at a definite distance from the source.
The intensity I follows a straight line which has a very steep
and a finite slope.
If all the α–particles having the same initial energy made an
equal number of identical collisions in the absorber the
intensity drops to zero abruptly at a distance d equal to the
range of the α–particles.
 The steep slope should fall off vertically down.

26.  What does the finite slope indicate?????????........... The finite slope shows there is a
statistical fluctuation in the number of collisions suffered by the different α–particles.
The tail of the bent portion is due to straggling of the range.
The linear portion of the graph is extrapolated to zero intensity, we get an extrapolated range.

27. MEAN RANGE [R]: The distance from the source at which the intensity is half the initial
intensity which is a few millimetres shorter than Rex.
The mean range has a value such that 50% of the α–particles in the incident beam have ranges
greater than R while 50% have ranges less than R.
For a given energy of the incident α–particles, the different ranges defined above have slightly
different values.
Example: 210Po α–rays [𝐸𝛼 = 5.3007 MeV], the following values were obtained:
Ionization extrapolated range 𝑅𝑖 = 3.870 cm, extrapolated range 𝑅𝑒𝑥 = 3.897 cm, Mean range:
3.842cm.

28. Range-energy relationship for α–particles:
From the measured values of the ranges and energies of the α–particles, the following
mathematical relationship between the two quantities can be established:
R = 𝑎𝐸3/2
It is an empirical relationship, valid in a limited energy range known as Geiger’s law.
For E in MeV and R in centimetre, the constant a = 0.315.
If v be the velocity in cm/s then v α 𝐸.

29. The relationship for the range in terms of velocity:
R = b𝑣3
where b is another constant: b = 9.416 x 10−28
In the case of a solid, the range Rs[ in centimetre] is related to the range R in air as follows:
𝑅𝑠 =
0.312 𝑅𝐴1/2
𝜌
where 𝜌 is the density of solid of mass number A.

30. Geiger-Nuttall law
Geiger-Nuttall [1911] discovered an empirical relationship between the ranges of the α–
particles and the disintegration constants of the naturally radioactive substances emitting them.
This is known as Geiger-Nuttall law.
The relationship is given by:
log 𝝀 = A + B log R. ----------------------------------------------------[1]
where A and B are constants.
According to this law, the α–particles emitted by the substances having larger disintegration
constants [shorter half-lives] have longer ranges and vice-versa.

31. Equation [1] shows the graph of log 𝜆 and log R is a straight line with a slope B.
For different radioactive series, different straight lines are obtained, which are parallel to one
another, so that B is same for all of them.
Using equation [1] it is possible to determine the disintegration constant and the half life 𝜏 of a
radioactive substance if the range of α–particles emitted by it is known accurately.

32. Figure: Variation of log 𝜆 with log R Figure: Variation of log 𝜏 with 𝐸𝛼

33. The range R α 𝐸3/2, the Geiger-Nuttall can be written as:
log 𝝀 = C + D log E
where C and D are two other constants.

34. The half-life 𝜏 = ln 2 𝜆, the Geiger-Nuttall law can also be expressed by relating the variation
log 𝜏 with log R or log E.
In this case we get straight line graphs, but with negative slopes.

35. IMPORTANT NOTE:
 The ranges of the α–particles from different radioactive isotopes are of the order of few
centimetres in air.
 The half-lives of the radioactive isotope range from less than a millionth of a second to more
than billion[109
] years.

36. Example:
The half life of ThC’ is 3 X 10−7s, while that of 232Th is 1.38 X 1010 years.
The range of the α–particles emitted by them are 8.57 cm and 2.49 cm.
The corresponding energies are 8.78 MeV and 3.97 MeV.
The example shows that for an increase of the α–energy by a factor of 2.24, the half-life
decreases by a factor of 1024
.
Such enormous change of the half-life due to relatively small change of the α–energy can be
explained by the quantum mechanical theory of potential barrier penetration.

37. Alpha Ray Spectra:
 α–particles are emitted with a very high velocity from the radioactive substances.
The velocity of the RaC’ α–particles is 1.92 X 107
m/s which is about 1/16 the velocity of light.
Rutherford and Robinson’s experiment on the determination of q/M of the particles gave their
velocity.
S. Rosenblum designed a magnetic spectrograph to measure the α–particle velocities from
different sources accurately.

38. The instrument is similar to the one used for the measurement of the 𝛽-ray spectra.
In this instrument, a very thin wire on which the radioactive substance is deposited is used as
the source.
The α–particles from the source are collimated by a system of slits.
The collimated beam is subjected to homogenous magnetic field at right angles to the direction
of beam.

39. Figure: Magnetic spectrogarph

40. Under the influence of the magnetic field, a slightly divergent beam of alpha particles describes
a semi-circular paths.
The beam are focussed at one point on the plate for a given energy.
Let B be the magnetic induction field, v be the velocity and R the radius of the semi-circular
path of the α–particles.
The expression we get: Bqv = 𝑀𝑣2 𝑅

41. Rosenblum used magnetic induction fields up to 3.6T [36,000] gauss.
Results:
The velocities of the α–particles emitted by the radioactive substances are of the order of
107
m/s.
For some radio-elements, only one line is obtained in the α–spectrum on the photographic
plate, which shows that they emit α–particles of a single velocity.

42. In some cases a number of parallel lines separated from one another [discrete spectrum] are
obtained, which shows that a number of different mono-energetic groups of α–particles are
emitted by the substances.
Each group has a definite velocity.
The velocities of the different groups are different.
The kinetic energies of the α–particles emitted from naturally radioactive substances are usually
in the range of 4 to 10 MeV.

43. Figure: Semi-circular focussing magnetic spectrograph

44. Short Range Alpha Particles:
α–particles emitted with discrete energies lead to the conclusion that the nucleus has sharply
defined energy levels.
If the difference in the energy of the parent nuclei and the daughter nuclei is sufficiently large
then the daughter nuclei can be formed in the ground state or in one of the excited states.
Short range α–particles of different energies are emitted having different ranges.
The daughter nuclei comes to the ground state by emitting gamma photons of appropriate
energies.

45.  By measurements of 𝛾-energy and α–energy, the energy levels of the nucleus can be found.
Rosenblum showed by using semi-circular focussing magnetic spectrograph that the α–rays
from ThC [𝐵𝑖212] are not monoenergetic.
It consists of several closely spaced mono-energetic groups or α–ray lines

46. Figure: Discrete spectra [Example]

47. The interpretation of complex α–particle spectra in terms of nuclear energy levels is illustrated
by the case of ThC [ Bi212].
It emits six groups of short-range α–particles with energies shown in the figure.

48. Figure: Decay Scheme for the
α–decay of ThC to ThC’

49. Long-Range Alpha Particles:
Long range alpha particles first observed by E. Rutherford and Wood.
Po214 and Po212 emit a few long range particles.
The origin of the long-range α–particles can be interpreted in terms of energy level as shown in
figure.
The nucleus can be in the excited state before the α–disintegration.

50. The excitation can be due to previous disintegration.
Emission of 𝛽 ray leaves the newly formed nucleus in the excited state.
Most probable case: the nucleus goes to the ground state by emitting 𝛾-ray of proper energy.
In some cases, if the life time for α–emitter the is comparable with the life-time for 𝛾-decay,
the newly formed excited nucleus emits α–particles with energy greater than that of the normal
particles.

51. Long-Range
Alpha Particles:

52. The existence of long range α–particles can be explained as being caused by the decay of an
excited nucleus.
The extra energy of the α–particles measures the excitation energy of the initial nucleus.

53. Gamow’s Theory of Alpha Decay:
Results from the experiments on the scattering of the α–particles show that:
 while approaching the nucleus the α–particle is acted by the Coulomb repulsive potential 𝑽𝒄 =
1/r up to the nuclear surface.
Anomalous scattering of the α–particles observed at larger angles in case of some lighter atoms
shows that the nature of the force is different from the Coulomb repulsion for a very close
distance of approach.

54. Present day knowledge of Nuclear structure:
 neutrons and protons forming the nucleus of an atom are very strongly attracted to one
another.
It is a strongly bound structure, known as the nucleus of the atom.
Range of the attractive force: 2 X 𝟏𝟎−𝟏𝟓
m
Nature of the nuclear force holding the protons and the neutrons together: Short range force.

55. In α–disintegration of the heavy nuclei, two protons and two neutrons sometimes form a
cluster known as the α–particle.
Forces acting on the α–particle:
Inside the nucleus, short range nuclear attractive force.
Outside the nucleus , the Coulomb repulsion due to the residual nucleus of positive charge
(Z-2)

56. Figure: Coulomb potential Barrier for α–disintegration.

57. Outside the nucleus:
the Coulomb potential energy 𝑽𝒄 =
2(𝑧−2)𝑒2
4𝜋𝜀0𝑟
is positive.
For r < R, the nuclear radius:
It is an attractive potential,
Exact nature of this potential is not known.

58. Transition from the repulsive Coulomb potential to attractive potential in the nucleus takes
place at the nuclear surface r = R.
The value of the repulsive Coulomb potential is maximum at r = R :
𝑽𝒄 =
2(𝑧−2)𝑒2
4𝜋𝜀0𝑅

59. Example:
222Rn is formed during the α–disintegration of 226Ra (Z=88).
Value of 𝑽𝒔 = 34 MeV.
α–disintegration Q = 4.88 MeV <<< 𝑽𝒔
This is the energy the α–particle has in the disintegrating nucleus.

60. Classical Mechanics explanation:
To escape from the nucleus or to penetrate inside the nucleus: the energy of the α–particle
must be at least equal to 𝑽𝒔
If it is lower, then in some region between the steeply falling curve and the vertical line the
potential energy of the particle will be greater than its total energy. This is the classically
forbidden region.

61. Kinetic energy of the α–particle is positive:
If the α–particle is inside the nucleus
or
At points to the right of line b.
Kinetic energy is negative:
In the region ab where the total energy of the α–particle is less than potential energy.
Known as potential barrier region or classically forbidden region.
α–particle neither can escape nor enter the nucleus.

62. Quantum Mechanical Approach:
Quantum mechanically, such classically forbidden phenomena may occur.
Quantum mechanics - particle is represented by a wave – obeying the Schrodinger equation.
The wave equation for the different regions by substituting the corresponding potentials acting
on the α–particle in these regions.
If these equations are solved with appropriate boundary conditions, then it is found that an α–
particle initially inside the nucleus has a finite probability of coming out of it.

63. Quantum Mechanical Tunnel Effect: the escape of the α–particle from the nucleus as if they
leak out through tunnels in the potential barrier.
 Gives a mathematical relationship between the initial α–energy and the half-life of the
disintegrating nucleus.

64. Figure: