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UNIT 1 NUCLEAR DECAY
Structure
1.0
Introduction
1.1
Objectives
1.2
Decay (What is decay ?)
1.3
1.2.1
Nature of Beta ray spectrum
1.2.2
Neutrino hypothesis
Fermi theory of Beta decay
1.3.1
Allowed and forbidden transition
1.4
Parity violation in Beta decay
1.5
Concept of helicity
1.6
Multiple transition and selection rules for the decay of the nuclei
1.6.1
Selection rules in Beta decay
1.7
Internal conversion coefficients
1.8
Isomeric nuclei
1.9
Angular correlation of successive decay
1.10
Let Us Sum Up
1.11
Check Your Progress: Possible Answers
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1.0 INTRODUCTION
The phenomenon of radioactive disintegration is the result of instability of nuclei and
their tendency to change into more simple species. In terms of binding energy, we can
say that a nuclide will be energetically stable towards decay by some specified mode
(  emission,  -emission or spontaneous fission) if its binding energy is greater than
the total binding energy of fragments into which it can disintegrate (i.e., if its atomic
mass is smaller than the sum of the masses of the products that are formed as a result
of decay process). Such nuclides can of course be disintegrated if energy is supplied
to them: the energy supplied appears in the form of kinetic energies of the fragments.
It has been observed that nuclei with A > 140 are unstable with respect to  -particle
emission. This is because the emission of  -particle lowers the coulomb energy – the
principle negative energy contribution to the binding energy of heavy nuclei, but does
not change the binding energy appreciably, for  -particle itself is a tightly bound
structure. Proton decay does not occur because, although it lowers the coulomb
energy but simultaneously decreases the nuclear binding energy by an appreciable
Nuclear Decay and Particle Physics
amount except for the proton excess side of  - stability; even in that region 
+
decay leads to proton emission. In this unit, we shall study the  decay process in
detail.
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1.1 OBJECTIVES
The study of the properties of the  particles emitted by the natural and artificial
radioactive nuclides has added greatly to our knowledge of the structure and
properties of atomic nuclei. At present, the main interest is in the information that  particle emission gives about nuclear energy levels and decay schemes, and in the
theory  decay and its relation to other fundamental nuclear problems. Many more
nuclides decay by electron emission, positron emission, and orbital electron capture
than by  particle emission, and the  - decay processes yield information about
hundreds of nuclear species not limited to those with large masses. Consequently,
information about the energy levels and decay schemes of light and intermediate
weight nuclides as well as those in the region of the natural radioactive elements can
be obtained by studying their  radiations.
The main aim of this unit is to study the nuclear decay process. After going through
this unit you should be able to:

Explain beta decay and general features of  -ray spectrum.

Know the Neutrino hypothesis.

Understand the allowed and forbidden transition and parity violation in beta
decay.

Analyze the angular correlation of successive decay.
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1.2 DECAY (What is Decay ?)
The spontaneous decay process in which mass number of nucleus remains unchanged,
but the atomic number changes, is termed as  decay. The change in atomic number
is accomplished by the emission of an electron, emission of a position or by the
capture of an orbital electron. Thus, depending upon the three modes of decay, the  decay are known as  - decay ,  + decay and electron capture (K-capture). The half
2
Nuclear Decay
lives of  - active nuclei range between 0.06 sec to 1018 yrs. The energy of emitted
particle goes up to few Mev.
1.2.1 Nature of beta ray spectrum
When  -ray energies are studied by magnetic spectrographs, it is found that the
energy spectrum is continuous.
Fig 1.  ray spectrum
The continuous spectrum is characterized by the following important properties:
1. Every continuous  -spectrum has a definite maximum, the height and
position of which depend on the nucleus emitting the particle.
2. The particles  + or  - possess some upper limit of energy. This upper limit
or end point energy, as it is called, is different for different nuclides.
3. The emission of  -particles may not be associated with the emission  -rays
e.g. Na24 emits  - rays whereas N13, P32, In114 do not emit  -rays.
In some cases of  -emission a single continuous spectrum is found and therefore a
single end point energy. Such a spectrum is called simple; while some emitters have
got a complex spectra i.e. they emit two or more spectra with different end point
energies.
A question now arises that all nuclear phenomena like the emission of  -particles or
 -rays show the existence of nuclei in different quantum energy states, the emission
of  -rays deny it and instead give a continuous spectrum. There may be several
possibilities for this strange behavior. One possibility may be that the nuclear energy
states involved are not sharp but are so broad as to be essentially a continuum. This is
contradictory to experimental results which shows the nuclear states to be discrete.
Another possibility is that the  -particles do not leave the nucleus during decay with
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Nuclear Decay and Particle Physics
full energy but with only a part of it thus leaving the nucleus in an excited state. Then
the nucleus emits  -rays. But it is found that there are certain nuclides showing
continuous spectra without emitting  -rays. It is also found that the energy of the
emitted electron is not equal to the difference between the energy of the nucleus
before and after emission but it may have any value between zero and maximum
energy of the continuous spectrum. The average energy of the emitted particle is only
1
2
of the maximum energy. The remaining of the energy seems to have disappeared.
2
3
This violates the law of conservation of energy .
The  -particles which are Fermi-particles or Fermions, have spin
1
and when they
2
are emitted from the nucleus, they should change the nuclear spin by
1
. But it was
2
found that nuclei which had an integral spin, either remained with the same spin or an
integral change in spin was involved. Evidently the law of conservation of angular
momentum is also violated .
In the  -decay of certain nuclides it is possible to measure the linear momentum of
both the electron and recoiling nucleus. In these cases it is found that the electron
seldom moves exactly opposite the nucleus as is required for the conservation of
linear momentum. Hence it appears that the  -decay process also violates the
principle of conservation of momentum.
Rutherford made an attempt to explain the paradox. He propounded that the laws of
conservation of energy and momentum do not hold in case of  -disintegration. He
argued this on the basis that since the behavior of light particles in the nucleus could
not be explained on the wave mechanical theory, here was no reason in clinging to the
same laws of conservation of energy and momentum which do not hold in case of  ray continuum.
1.2.2 Neutrino hypothesis
In the continuous beta-ray spectrum, the two alternatives are:
(i)
the conservation laws do not hold,
(ii)
the conservation laws hold good but part of energy and momentum
which are imparted to some form of undetected radiations.
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Nuclear Decay
The first hypothesis was proposed by Bohr and Rutherford but did not find much
favors due to the fact (i) conservation laws have been successful in explaining the
known phenomena (ii) it would make impossible the treatment of nuclear phenomena
in the scheme of quantum mechanires, (iii) Sargent could verify the conservation of
energy law using end point energy.
Hence, only second hypothesis is preferred now-a-days. The hypothesis was put
forward by W.Paull in 1933. According to him in each beta disintegration an
additional particle is emitted, to maintain the conservation laws the particle was
assumed neutral hence named neutrino (little neutral). The neutrino (v) was assumed,
a fermions with intrinsic spin
1
. And it is thought to carry an appropriate amount of
2
energy and momentum in each beta process to conserve these quantities. Further, in
beta decay parent and product nuclei have the same mass number and electron obey
F.D. statistics, so to conserve statistics v was assumed to obey F.D. statistics.
Under, neutrino hypothesis the continuous beta-ray spectrum is explained. It is
maintained that in each beta disintegration the amount of energy released is equal to
end point energy. Thus energy is shared by recoil nucleus, the emitted electron and
the neutrino. The energy carried by neutrino is not fixed and it varies continuously,
leaving thereby a continuously varying energy to the beta particle and hence the
continuous spectrum. At end point, total energy is carried away by the beta particle
and the neutrino carries zero energy.
Now the question is , “whether the two neutral particles emitted in  - and  + decays
are identical ?” To resolve this the analogy is taken from the fact that  - and  +are
antiparticles; therefore, the particles emitted together with  - and  +particles in tow
processes should be antiparticles. This and other evidences support this idea and by
convention the particle emitted in  - decay is taken as antineutrino and in  + decay
the neutrino(v). In this way, Pauli’s hypothesis proved successful in interpreting the
continuous beta spectrum.
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Nuclear Decay and Particle Physics
Check Your Progress 1
Note: a) Write your answers in the proper space.
b) Compare your answers with the ones given at the end of the unit.
1) Discuss the salient features of  ray spectra.
2) State the properties of neutrino.
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1.3 FERMI THEORY OF BETA DECAY
Enrico Fermi propounded a quantitative theory of  -decay. According to this theory.
When a nucleus emits a  particle, its charge changes by one unit while its mass
practically remains unchanged. When the e
jected  -particle is an electron, the number of protons in the nucleus is increased by
one and the number of neutrons is decreased by one. In positron emission, the reverse
process takes place i.e. the number of protons decreases by one and the number of
neutrons increases by one.  -transformations may then be represented by the
following processes:
n  p     v
p  n     v ,
Where v  and v  represent neutrino and antineutrino.
In the above two equations, neutron is not to be considered as composed of a proton,
electron and neutrino but is considered to be transformed into these three particles at
the instant of  emission. Similarly a proton is transformed into a neutron, positron
and neutrino at the time of  + emission. The neutron or proton which is transformed
is not a free particle but is bound to the nucleus by the nuclear forces.
We know from a large number of experimental evidences that electrons do not exist
inside the nucleus, it must be formed at the moment of its emission just as a photon is
formed at the moment of its emission from an atom. The neutrino is also created at the
moment of its emission. These particles are created into states represented by the
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Nuclear Decay
wave functions   and  v . Let us assume that these are functions for plane waves
with moment P and Pv respectively.
   N  .e
ik .r
 v  N v.eik
v .r
Where k 
p
and r represents the space co-ordinate and N is a normalization factor.
h
The probability of emission can be assumed to depend upon the expectation value for
the electron and the neutrino to be at the nucleus i.e. on the factor
2
  (0) . v (0) .
2
It also depends on other factors, whose nature is uncertain.
One such factor is the square modulus of a matrix element M taken between the initial
and final states of the nucleus. This matrix element in its simples from can be taken as
. M  p *Ndr
Assuming that only one nucleon participates.  N represents the initial state of the
nucleon and  P the final state i.e. the proton state. We can also take M to be a vector
having x-component given by
M x  p * xNd ,
Where  x is the x- component of a spin operator. Then
2
M  Mx  My  Mz .
2
2
2
The expression for the probability of emission depends also upon a constant factor g2
which represents the strength of the coupling giving rise to emission and is found to
have value
g = 10-48 to 10-49 g cm.5 sec-2.
Thus the probability of emission per unit time is

Where


2 dn
2
  (o )  y ( o ) M g . ,
h
dE
dn
represents the energy density of the final states and 0 refers to the
dE
location of the nucleus. Let  be the volume of a big box in which we enclose the
system for normalization purposes, then
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Nuclear Decay and Particle Physics

 * .d  1.
1
.

N 

1 ik .r 
e 
1/ 2


There fore

1 ik r
   1/ 2 e  . 


 
Where k  
p
h
, k 
p
h
When the nucleus is at r = 0
 v (0) 
1
   (0).

The number of plane wave states having magnitude of momentum between p and
p+dp with the particle any where in the volume  given by
=
2
p 2 dp
2 2 h3
2

2
p dpv
4 4 h6
 dn 
p dp
2 2 h3
2

pv dpv
 2
2 3
2 h
2
dp .dpv  JdpdE ,
dp dp y ,
where j is the Jacobian.
Using the relation E  cpv  E , J* is found to be equal to
1
2
(mass of neutrino v is
assumed to be zero for this derivation), then
2
2
p p
dn
 2  4 6v dp
dE
4 h c
Substituting the value in equation the probability of emission per unit time
P ( pv , p )dp is
2
2
2  1
  p pv dp
M
g


h 
4 4 h 6c

2
P( Pv P )dp,  
pvc  Ev  E
Using the relation
max
2
 E
To eliminate Pv and replacing P by simply P , we get
P( p)dp 
gM
2
2 h c
3 7 3
E
max


2
 E P 2 dp.
Thus the intensity of emission at p is given by

I ( p)  E
max

2
 E P 2C ( z, p)
Where C(z,p) includes the constant and also the dependence on nuclear charge. This
equation can be written as
8
Nuclear Decay
E
max


 I ( p) 
 E   2

 P C ( z, p) 
Thus there is a linear relationship between P
I ( p)
and E  the  ray energy.
p C ( z. p )
2
A plot of this type is known as Kurie plot.
Fig. 2. Kurie plot
It has been found that the Kurie plots are almost straight lines except at low energies
where they deviate from straight lines. This deviation is accounted for by considering
the distortions at low energies in  - particle wave functions due to the Coulomb
charge of the nucleus .
Allowed and forbidden transition
For beta transition to occur initial and final states of the nucleus are supposed to
satisfy certain definite conditions. The conditions to be satisfied are related with the
conservation of angular momentum and parity and are called selection rules.
Transitions obeying these selection rules are called allowed transitions. On the other
hand the transitions not obeying these selection rules are called forbidden transitions.
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1.4
PARITY VIOLATION IN BETA DECAY
Parity is the fundamental nuclear property, which is more important than spin.
According to quantum mechanics, associated with each particle is a wave function
which depends upon the space coordinates (x, y, z). The parity of the particle decides,
how the wave function behaves when the signs of x, y, z are changed. If the change
of sign of x, y, z does not change the wave function , the particle is said to be in the
even state or is said to have even parity. If on the other hand the wave function also
9
Nuclear Decay and Particle Physics
changes with change in sign, the particle is said to be in odd state and is said to have
negative or odd parity.
Nuclear states are characterized by a definite parity which may be different for
different states of the same nucleus. According to the parity law, two particles which
are the mirror images of each other must obey the same physical rules.
It has been shown that the parity of a nucleus in a given state is related to the value of
the orbital angular momentum L. If L is even the parity is even; if L is odd, the parity
is odd. In general the parity is given by (-1)l . A system of particles will have even
parity when the sum of the numerical values for L for all its particles is even and odd
parity when the sum is odd. Obviously Parity falls in line with energy and angular
momentum of the system. As these quantities are conserved in all physical
phenomena, it was argued that parity is also conserved. The conservation of parity
was also established by use of Schroedinger equation. But, now it has been shown that
in weak interaction of beta decay type, the law of conservation of parity is not obeyed.
The first experiment to demonstrate parity violation was performed in 1957 by Wu,
and others by using Lee-Yang proposal, the sketeh of which is shown in fig.3.
Fig. 3. Experimental demonstration of parity violation
10
Nuclear Decay
They used Co60 grown on the surface of cerium magnesium nitrate. In Co60 under
normal condition, the spins are randomly oriented, because of thermal motions, and
beta particles are emitted in all directions. Co60 was cooled below 0.01 0K and an
external magnetic field of few hundred gauss was applied. As a result, the thermal
motion was reduced to minimum and spins were aligned along the direction of
magnetic field. The magnetic field aligns the atoms of cerium magnesium nitrate. The
alignment produces normal magnetic field which aligns the spins.
The beta particles emitted by polarized nuclei were detected by an anthrecene crystal
mounted 2 cm above the source. The scintillation were transmitted to a photo
multiplier located at the top of the cryostat. To measure the extent of polarisation of
Co60 nuclei. Two NaI  -ray counters were used one in the equatorial plane and other
near the polar position. The observed anisotropyprovided a measure of polarization
and hence of the temperature of the source. After almost 8 minutes the  -anisotropy
disappeared and indicated that aligment had disappeared.
Counting
rate
0
2
4 6
8 10 12
Time (min)
14 16
18
Counting
rate
Time (min)
Fig. 4. Results of parity violation experiment
11
Nuclear Decay and Particle Physics
The counting rate for beta particles were first determined with the field up and then
with the field down. The results indicate
that more electrons are emitted in the direction opposite to that of the magnetic field,
i.e., in the direction opposite to that in which nuclei are aligned. It implies that
electrons are emitted in preferred direction and principle of right-left symmetry is
violated and hence parity is not conserved.
In terms of the pseudoscalar quantity I i . p  , we can say that its average value is
negative. Here, I  is angular momentum of Co60 which does not change sign in parity
operation and p  is linear momentum of beta particle which change sign in parity
operation .
Check Your Progress 2
Note: a) Write your answers in the space given below.
b) Compare your answers with the ones given at the end of the unit.
1) Give Fermi theory of  decay.
2) What is the parity of a P electron. Is it the same as that of an  particle with
angular momentum j = n ?
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1.5 CONCEPT OF HELICITY
As a prelude to the explanation of the non conservation of the parity in above
experiment, we discuss a new physical quantity known as helicity.
Helicity, H represents the state and degree of polarization of particle and photons.
It is defined as
H   1. p1
Where  1. is unit vector in spin direction of particle and p1 is unit vector in the
direction of linear momentum of the particle. Thus to measure the helicity, direction
of motion and direction of spin are measured. The results are:
(i)
particle and anit-particle always have opposite helicity,
12
Nuclear Decay
(ii)
negatrons in beta-decays have H=-v/c, while for positrons H=+v/c,
where v is their velocity .
(iii)
neutrons in beta decay have H=-l and antineutrinos=+1.
It implies that antineutrino has its angular momentum vector aligned with its
momentum vector whereas neutrino has opposite alignment in the process of beta
decay.Now we can explain the violation of conservation of parity in decay of Co 60
which undergoes Gamow Teller transition (5+ state to 4+ state). To conserve angular
momentum, it is necessary that spin vector of both  particle and antineutrino must be
aligned with angular momentum vector of the nucleus. As a result, antineutrino is
emitted in the direction of angular momentum because it behaves as right-handed
screw. Now, since in the decay involved there is anti-correlation between the motions
of the two particles. We conclude that neutrino violates the parity law and electron
compensates for it; thus neutrino is held as a culprit of parity violation.
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1.6 MULTIPLE TRANSITION AND SELECTION RULES
In physics we come across with following four types of interactions1.
The electromagnetic interactions. – These are the well known interactions
which are responsible for phenomena such as Coulomb scattering and gamma
emission. These interactions are treated by Maxwell’s equations and the quantummechanical generalizations of these equations. The charge of the electron is the
universal constant which appears in all electromagnetic interations. Instead of the
elementary charge, we consider the dimension less quantity e 2 / hc 1
1
7 which
2
represents the coupling strength of the electromagnetic interaction. The ordinary
photons are the quanta of the electromagnetic field and they are bosons of zero rest
mass.
2.
The strong nuclear interactions. – These are responsible for the phenomena
such as neutron-proton scattering. The most typical example of a strong interaction is
the Yukawa interaction.
N  N 
In this reaction , a nucleon transforms into a nucleon and a pion. In all interactions of
this type a universal constant which has the same role as electric charge in
electromagnetic field, appears. This constant which is expressed in dimension less
13
Nuclear Decay and Particle Physics
way as f2/hc has a value approximately equal to one. The strong interactions have a
short range of the order of 10-13 cm. According to Yukawa the quanta of the
interaction field, therefore, has a finite rest mass ; the quanta are identified with pimesons which are Bosons of zero spin
3.
The weak interactions – These are responsible for  -decay and allied
phenomena. These interaction are characterized by four participating fermions.
The important examples are –
n  p  e  v
(  - emission )
C 11  B11  e   v 
(Positron emission)
Be 7  e   Li 7  v
(K-capture)
   e  v  v
(  - meson decay).
Weak interactions have a coupling constant G or to be more accurate, like other
coupling constants G2/hc  10-14. In the above four examples we find a neutrino or an
antineutrino participating. But this is not an essential condition for all weak
interactions. The decay of K mesons into pions is also a weak interactions which does
not involve a neutrino or anti-neutrino. It has been proved by Lee and Yang that in
weak interactions, parity is not conserved. This is the major difference between weak
and strong interactions.
4.
The gravitational interactions – These are negligible and beyond scope of
discussion here.
1.4.1 Selection Rules in Beta decay
In beta transition following selection rules has been classified as :
1.
Fermi selection rule
include those transitions in which neither the angular
momentum nor the parity undergoes a change, i.e., I  0 and P  0 Here I is written
for angular momentum and P is for parity.
2. Gamow Teller selection rules include those transition in which there is no change
in parity of the nucleus, but angular momentum is changed, i.e., I  0 or ±1(0-), not
allowed) and P  0 or we can say, that I=0 to I’=0 transition is not allowed because
the spin momentum must be carried away.
Now it is recalled that when electron and neutrino are emitted with zero orbital
angular momentum they have large transition probabilities, for e and  v are large at
14
Nuclear Decay
the nucleus, for s orbits. Thus , l=0 favors allowed transitions while l  0 favours
forbidden transitions. The two possibilities of allowed transitions are mentioned as :
(1)
Beta particle and neutrino are emitted with opposite spins, therefore the total
momentum carried off by the two particles is
S   Sv

l  lv  0,

+
0
0
i.e., there is no change in the spin direction of the nucleon. Thus the angular
momentum of the nucleus remains unchanged. This corresponds to Fermi transition.
(2)
It the two particles are emitted with parallel spins the total angular momentum
carried away by the two particles is
S   Sv +

0
lv  l 1
0
0
i.e., nucleon spin is reversed. The possible changes in the angular momentum of the
nucleus are I  0 , ±1 (with the exception of 0 – 0 transition ).
Fermi transition
I  1
Gamow Teller transition
I  0,1,2 Parity change
parity change
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1.7 INTERNAL CONVERSION COEFFICIENTS
It has been said that the emission of  or  particles leave the nucleus in excited
state. The nucleus restores its normal state by de-excitation through the emission of 
radiation. The other alternative way of de-excitation is internal conversion. The
internal conversion results due to electromagnetic interaction between excited nucleus
and an orbital electron. In internal conversion the energy of interaction ejects one of
the orbital electron. The excited nucleus surrenders its energy to the orbital electron
and hence the kinetic energy of the conversion electron, Te will given by
Te = (Ei-Ef)-Bi
Where Ei and Ef are the energies of initial and final levels and Bi is the binding energy
of electron in its orbit.Since the electrons closest to the nucleus are K-electrons, the
internal conversion of K-shell electrons has the highest probability. If (Ei-Ef) is less
than Bk, the K electron binding energy, then conversion of K electrons becomes
energetically impossible and one observes L electrons conversion and so on.
The vacancy in the shell created due to the conversion electron is filled in by the
electron of higher orbits and subsequent emission of X-radiation and Auger electrons
15
Nuclear Decay and Particle Physics
(The electrons emitted without fluoresence are known as Auger electrons). The
internal conversion electrons produce a series of monoenergetic lines and not a
continuous spectrum as in   emission. The line with lowest energy is TeK=(Ei-Ef)-BK
The next line is
TeL = (Ei-Ef)-BL,
And so on.
Conversion coefficient: The two processes  - emission and internal conversion often
compete (except in 0 – 0 transition) with one another. The relationship between the
two de-excitation processes is expressed by the internal conversion coefficient
 which is equal to ratio of the probability of conversion electron (Po) to the
probability of  emission (Pg), i.e.,  
Po
,
Pg

0
g
Equation also measures the total number of conversion electrons emitted over a given
time divided by the total number of gamma photons emitted in the same transition in
the same time. The conversion co-efficient  is expressed as
  k  L  M  ....,
Where  k , L..... are partial conversion coefficients of
K, L,
electrons ,
respectively. The measurement of conversion coefficient provides information about
(Ei-Ef), for the energy of conversion electrons is also expressed as
T0  E g  ( Ei  E f )  Bi .
Check Your Progress 3
Note: a) Write your answers in the space given below.
b) Compare your answers with the ones given at the end of the unit.
1) What do you mean by Helicity ?
2) Write short notes on internal conversion.
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..........................................................................................................................
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__________________________________________________________________
1.8 ISOMERIC NUCLEI
Theoretical estimates indicate that normally the life-time of  -transition is of the
order of 10-13 second. But there are about 250 known cases in which life-time ranges
16
Nuclear Decay
from 1010 second to several years. In such cases nucleus remain in the excited state for
measurable time and then decay. These retarded transitions are called isomeric
transitions and the states from which they originate are called isomeric state, Because
of the delayed transitions pairs of nuclides exist which are isotopes and isobars, but
nuclear isomers and their existence is known as nuclear isomerism. For example, the
stable isotope of In113 can exist for a finite time state. The meta-stable state is
represented by In113m and it decays with emission of  -ray (half life 105 min). Thus,
In113 and In113m form isomeric pair.
The isomeric states of given isotope differ among other things in energy and angular
momentum. If I is angular momentum in the ground state and Im in the meta-stable
state than I  I m  1 in the unit of h and represents a transition which has a degree of
forbiddenness with respect to  decay.
Fig.5. Nuclear Isomerism of UX2 and UZ
The first example of nuclear isomerism UX2 and UZ is shown in fig.5.
and was
discovered by Hahn. The beta emission of UX1(90Th204) gives an isometric pair UX2 to
UZ by gamma emission (or internal conversion) is highly forbidden, since . Therefore
the usual decay process forUX2 if beta decay giving rise to
82U
234
an  -emitter.
However, a small percentage of UX2 goes UZ by gamma emission with a half life of
13 hours. UZ nucleus transforms to92U234 by beta decay which leaves the nucleus in
excited states. The de-excitation takes place by the emission of  -rays.
The
80m
35Br
has a half-life of 4.4 hours and decays into the ground state by the
emission of the cascade  -photons of energy 0.049 MeV and 0.037 MeV.
17
Nuclear Decay and Particle Physics
Respectively. The ground state has half-life of 18.5 min, which decay by   ,   and
EC processes.The nuclear isomers, one of which has been formed by  - decay are
said to be genetically related. The Br80 isomers and UX2 and UZ isomers are
genetically related. Non genetically related isomers are that of 40In116.
__________________________________________________________________
1.9 ANGULAR CORRELATION OF SUCCESSIVE DECAY
It is known that different multipoles gives rise to different angular distribution of
emitted radiations with respect to different angular momentum direction of emitting
nucleus. Under normal circumstances, the different nuclei are randomly oriented and
therefore, an isotopic distribution of gamma rays results. Thus, emitted radiation
cannot be distinguished in terms of multipoles that give rise to them. However, if the
nuclei are made to orient in one direction so that m1 is known, then multipole from
which the radiations result is proportional to
The angular distribution is proportional to
Slm 
1
l  1  m1 mi  1Ylm1 2
2l  1

 l l  1  m1 ml  1Ylm1  2ml2 Ylm1
2
2

Where Ylm is spherical harmonic. This gives,
3
S10 
sin 2 
2
3
S11 
(1  cos 2  )
16
For the alignment of nuclei following method can be adopted
(1)
(2)
(3)
Let a nucleus decays by a  and  emission and a daughter nucleus is formed in one
of the several possible excited states. The direction of spin axis of the daughter
nucleus in the excited states depends upon the direction of emission of  or 
radiation. If as result of de-excitation, normal state restored then direction of emission
of  rays bears an angular correlation with the direction of emission of  or 
particle. However, the daughter nucleus decays by two or more gamma transitions in
cascade . Then the correlation between two directions is given as;
W  ( )  1  a2 cos 2   a4 cos 4  .....,
18
Nuclear Decay
here only even powers of cos  appear and a2 and a4 ....... are the correlation
coefficient. W ( )d denotes the relative probability that second radiation will be
emitted into a solid angle d , if  is the angle between two directions of emission.
Fig. 6. Angular correlation measurement.
Let  1 and  2 carry the angular momenta I1 and I2 , respectively and I be the spin of
intermediate state in cascade emission of intermediate state in cascade emission of 
rays. In such a case the highest power of cos  is less than or equal to twice the
smallest number of these three number (I1, I2, l). It turns out that angular correlation
cannot occur if I = 0or ½. If something is known about one quantity, we can have
information about other, using selection rules.
If in addition to direction correlation, we know the polarisation of the ray, the parity
change can be inferred and thereby a distinction can be made between the radiations
of same multipole order but different in nature electric or magnetic.
Angular correlation experiments can usually be performed only when transitions with
low multipole orders are involved, for l2 < 2, the life time of the intermediate state is
usually so long that the correlation is destroyed. Thus cos 4  term in W ( )
expression is the highest term that appears in practical cases and only two parameters
need to be determined experimentally.
19
Nuclear Decay and Particle Physics
Check Your Progress 4
Note: a) Write your answers in the space given below.
b) Compare your answers with the ones given at the end of the unit.
1) Write the various selection rules for the  decay.
2) Write short notes on:
(a) Isomeric nuclei
(b)
Angular coorelation
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__________________________________________________________________
1.10 LET US SUM UP
After going through this unit, you have achieved the objectives stated earlier in the
unit. Let us recall what we have discussed so far.

Bucherer in 1909 was the first to prove that  particles are nothing but fast
moving electrons.He determined e/m for  particles and considering the variation
or mass with high velocities with which these particles are emitted, the e/m value
came out to be nearly the same as that of electrons.

Study of  ray spectrum suggest that individual nuclei do not emit the beta
particles of same energy. it implies that conservation of relativistic energy should
be true for individual electrons.

The neutrino hypothesis put forward by W. Pauli in 1933 proved successful in
interpreting the continuous beta spectrum. In view of the idea that electrons are
not contained in nuclei and discovery of neutron in 1932, led Fermi to incorporate
Pauli’s hypothesis to explain beta disintegration. Fermi theory based on the
interaction of nucleon in the nucleus with the electron-neutrino field successfully
explained the beta decay, as the theoretical predictions very well coincide with the
experimentally verified values, a major triumph of Fermi theory.

The law of conservation of parity has been experimentally confirmed with an
accuracy of at least 1:10-8 in strong nuclear interactions in 1957 Lee and Yang
have suggested that in weak interactions like  decay nature does not follow the
law of symmetry and discriminates between left handed and right handed systems,
20
Nuclear Decay
means parity is not conserved and it is experimentally confirmed. To explain the
parity violation a new physical quantity helicity has been studied.

For beta transition to occur are supposed to satisfy some conditions named as
selection rules. Also after  decay the nucleus remains in excited state, the
nucleus restores its normal state through internal conversion. During the
transition, we also have studied isomeric nuclei and angular correlation.

The emission of  particles differs from that of  particles in respect to the
spectrum of the energies of the emitted particles. The most characteristic feature
of the spontaneous  -disintegration of nucleus is the continuous distribution in
energy of the emitted electrons, which is in sharp contrast to the line spectra
observed for  particles. Beta decay is important, not only because of its
relationship to the practical problems of nuclear physics, but also because of the
conceptual problems involved.
__________________________________________________________________
1.11 CHECK YOUR PROGRESS: THE KEY
1. (i) See the section 1.2.1
(ii) See the section 1.2.2
2. (i) See the section 1.2.3
l
(ii) For P electron l = 1 hence parity    1
= -1 (Negative)
4
For  particle j = l = 4 and so parity    1
= + 1(Positive)
3.
(i) See the section 1.3
(ii) See the Section 1.5
4.
(i) See the section 1.4
(ii) See the section 1.6
21
Nuclear Decay and Particle Physics
REFERENCES AND SUGGESTED READINGS
1. “Theory of Beta –Decay” Progress in Nuclear Physics by T. H. R. Skyrme,
Academic Press, New York.
2. Nuclear Physics by Irving Kaplan, Narosa Publishing House.
3. The Atomic Nucleus by R D. Evans, McGraw-Hill Publications.
4. “Beta Ray Spectrometers”, Advances in Electronics by R. W. Hayward,
Academic Press, New York.
5. Nuclear Physics by D. C Tayal.
6. Concept of Modern Physics by Baiser, TMH.
7. Atomic and Nuclear Physics by Brijlal and Subhraininyan.
8. Introduction to atomic and nuclear Physics by Harvey E. White-East west press
New Delhi.
*********
22
___________________________________________________________
UNIT 2
NUCLEAR FORCE AND SCATTERING
Structure
2.0
Introduction
2.1
Objectives
2.2
Basic properties of deuteron
2.2.1 Binding energy, Size and Spin of deutron
2.2.2 magnetic and quadrupole moments
2.3
Existence of excited states of deuteron (Solution of spherically symmetric
square well potential for higher angular momentum states)
2.4
n-p scattering at low energies with specific square well potential
2.5
Results of low energy n-p and p-p scattering
2.6
Spin dependence and scattering length
2.7
Let us sum up
2.8
Check your progress: The key
_____________________________________________________________________
2.0 INTRODUCTION
It is also known that large number of nuclei, available in nature, are stable.
Now the natural question is what bounds the nucleons together in the nucleus to make
it stable? It is further known that most of the nuclei consist of more than one proton –
the positive charged particle; it implies that there exists in the nucleus a forces which
are strong enough to overpower the coulomb repulsion and hold the nucleons
together. The forces which hold the nucleons together are commonly called nuclear
forces and are short range forces as it is evident from the fact that binding energy is
proportional to the number of constituent nucleons.
The nuclear forces cannot be of electromagnetic origin because nuclear forces involve
uncharged particle while electromagnetic forces do not. The purely magnetic forces
cannot account the binding energy of about 8 MeV per nucleon (1.1 MeV per nucleon
in deuteron) in nuclei in general and it is clear that magnetic froes are some hundred
times smaller than nuclear forces. The gravitational forces too cannot explain the
existence of such huge forces because they are about 1036 times smaller. It implies
that nuclear forces cannot be of any of the type discussed so far .
23
If we compare the stable nucleus with the atom then we find that stability of the atom
is governed by predominant central particle but in the case of nucleus there is no
Properties of Nuclei and Scattering
predominant central particle. The forces which hold together the different nucleons
should have mutual forces between the individual nucleons in the ensemble.
It turns out that the nuclear force are strange and of intriguing nature.
Now let us turn to the nuclear interaction which according to Yukawa’s theory, may
be conceived as due to the exchange of a relatively massive particle - the pai meson
or pion with a mass approximately 270 times that of an electron .
Thus nuclear interaction is about 1038 times stronger than the gravitational interaction
and about 1000 times than the electromagnetic interaction and so comes under what
are called ‘strong interaction’. We then infer that none of the only two interactions,
encountered previously, is able to account for the existence of nuclei. The only way
out then is to recognize the existence of another fundamental interaction – the nuclear
interaction (force). Since nuclei are composed of protons and neutrons only which are
packed very densely within the small volume of the nucleus, the heavier nuclei will be
subjected to very strong Columbian repulsive force – the one acting between the
positively charged protons, which tends to tear the nucleus apart. The fact that nuclei
stay as bound systems even in the take of these strong repulsive. Columbian forces, is
a sufficient proof of the great strength of the nuclear forces and that at distances of
the order of nuclear dimensions, it should be attractive in nature .
As a general rule, the wave nature of matter (quantum mechanical principles) is
relevant where the de Broglie wavelength of the particles is of the order of the size of
the system to be studied. So let us compare unclear size with the wavelength of a
nucleon of energy 10 Mev.
h


M N
h
6.6256 1027

2M N E  2 1.67 1024 10 1.6 1019 1/ 2



 9.3 1023 cms.  9.3Fermi 1Frtmi  1013cms.

Which is obviously of the size of the nuclei and hence quantum mechanical
considerations are indeed relevant to the study of nuclei.
Having ascertained that nuclei are quantum mechanical systems composed of
nucleons, it is quite plausible to study the nuclear forces under the simplest possible
24
conditions. The simplest case in which the nuclear force is effective is when there are
only there are two experimentally achievable situations:
1-
When the two nucleons are bound together.
Nuclear Force and Scattering
Of the three possible bound states of a two – nucleon system, di-neutron (nn), diproton (pp) and deuteron (np), nature has provided us with only the deuteron and
the other two are unstable.
2- When the two nucleons are in free state and one is made to impinge on the
other, i.e. The scattering processes.
In practice, it is not possible to make a neutron target and therefore scattering
experiments are limited only to neutron proton (np) scattering and proton-proton (pp)
scattering.
_____________________________________________________________________
2.1 OBJECTIVES
The main aim of this unit is to study the Basic properties of deuteron viz, its binding
energy, its size, spin, magnetic and quadrupole moments etc. After going through this
unit you should be able to:

Understand the various properties of deuteron.

Analyze the existence of excited states of deuteron with the solution of
spherically symmetric square well potential for higher angular momentum
states.

Learn n-p scattering at low energies with specific square well potential.

Comparatively study the results of low energy n-p and p-p scattering.

Know the spin dependence and scattering length.
_____________________________________________________________________
2.2 BASIC PROPERTIES OF DEUTRON
The deuteron consists of a neutron and proton, having charge equal to proton +e,
mass 2.014735 atomic mass units and it obeys the Bose-Einstein statistics. The
experimentally measured properties of deuteron are:
2.2.1 Binding energy, Size and Spin of deutron
Binding energy . (Experimental) = 2.225±.003 Mev.
25
The binding energy of deuteron can be determined from a number of
experiments. The easiest one comprise of allowing slow neutrons to be
captured by protons in a material containing hydrogen i.e. hydrogenous
substances such a paraffin, plastic etc. and measuring the energy of the
Properties of Nuclei and Scattering
emerging  rays. The reaction is called (n-p) capture reaction and may be
written as
0
n1 1 H 1 1 H 2  
Because the neutron carries no charge, the nuclear force binding the deuteron
cannot be electrical. It can also not be to MN = 1.67×10-24 gm. to provide a
2.225 Mev binding energy. Therefore the binding force is of nuclear origin .
Size:
Deuteron radius: The root- mean – square value of deuteron radius is 2.1
Fermi.
Spin:
Spin. 1 (in units of h).
2.2.2 Magnetic and quadrupole moment
Magnetic dipole moment:
The magnetic dipole moment of deuteron is d  0.85735 ± 00003 nuclear magneton.
In a structure made up of particles, one expects the total magnetic moment to be the
vector sum of the magnetic moments due to the orbital motion of the charged
particles. Applying quantum mechanics to describe deuteron we may reasonably
assume the ground state of deuteron to be an S state for which the angular momentum
L = 0. With L=0, the wave function  is spherically symmetrical and for the S state
the angular momentum quantum number l=0, no contribution from orbital motion is
expected to spin. The nucleons are half spin particles and deuteron is known to have a
spin equal to unity which implies that the proton and neutron spins are parallel. In
such a case the proton and neutron magnetic moments should also add up.
Experimental measurements for nucleon magnetic moments give the following
values:
Magnetic moment of proton  p =2.79281 ±0.00004 nm.
Magnetic moment of neutron  n =-1.913148±0.000066 nm.
Sum of the two moment  p   n  = 0.879662±0.00005 nm.
26
Thus deuteron is expected to have a magnetic moment of 0.8797±.00015 nm.
However from experimental measurements deuteron magnetic moment is found to
have a value 0.85733 ± .0002 nm between the expected and the measured values
which is difficult to explain. The simplest interpretation being the at deuteron
Nuclear Force and Scattering
possesses some orbital motion and that our previous assumption of l=0 in the ground
state is not correct. Even this approximate agreement is valid only for the S state .
Others give values very much different from the measured value. Thus magnetic
moment measurements of deuteron establish the following important conclusions.
(1)
In the ground state of deuteron, the proton and neutron spins are
parallel (triple state 3S1.)
(2)
Neutron is a half spin particle.
In the ground state of deuteron, the orbital angular momentum is zero
(l=0, S=1 state ).
Quadrupole moments:
The electric quasrupole moment of deuteron as measured by Rabi et al in a radio
frequency molecular bean method is
Qd = 2.82×10-27 cm.2 or 0.00282 barn,
Which although small but is not zero.
Alternately it can be put as to give
average z2 for proton
1
z2
(1.14)  2
3
r
average r2 for proton
It implies that charge distribution in the ground state is not spherically symmetric
z2 1
because a spherical chare distribution needs a value for Qd=0 or for the 2  . The
r
3
result also indicate that charge distribution is of prolate shape, i.e., elongated along
the z axis. The electric quadruple moment and the magnetic moment discrepancy
cannot be explained by assuming the state to have some other value of l. This suggests
that the wave function contains a mixture of l values. Since total angular momentum
is equal to the vector sum of the orbital and spin angular moments i.e.
L=1+S
27
In a system like deuteron which consists of one proton and one neutron, each having a
spin
1
, the spin quantum number S be given by :
2
S = 1/1+1/1 = 1 or 0.
For L=1 and a maximum value of S=1. from equation l can have only the values 0, 1
and 2 . But the conservation of parity demands that even and odd values of l should
Properties of Nuclei and Scattering
not be simultaneously present in the same wave function and therefore with l=0 , only
l=2 can be present. The wave function then may be written as:
  a0 1s  a2 1d
This means that the system spends a fraction a2 2 of its time in an l=2 state. Therefore
the ground state may be taken to be a mixture of 3S1 and 3D1 states.
The magnetic moment and electric quadrupole moment discrepancies can be fully
accounted for with a2  0.96 and a2  0.04 . This means that deuteron spends 90%
2
2
of the time in an l=0 state and only 4% of the time in an l=2 state. We there fore infer
that the deuteron is not in a purely spherically symmetric state. However at present
we shall assume that the ground state function is spherically symmetric .
Check Your Progress 1
Note: a) Write your answers in the space given below.
b) Compare your answers with the ones given at the end of the unit.
3) Write the basic properties of deuteron.
4) State clearly the definition of nuclear quadrupole moment and discuss the
ground state of the deuteron in the light of the fact that it has small but finite
quadrupole moment.
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___________________________________________________________
2.3 EXISTENCE OF EXCITED STATES OF DEUTRON
28
Extending the calculations of the bound state to cases where the orbital angular
quantum number l is greater than zero leads to a result that deuteron cannot exist in
these states. For the extreme case, binding energy EB  0 , kr0 is still only slightly
greater than  / 2 , since the binding energy EB of the ground state has already been
found negligible compared to the potential well depth V0 . For the first excited state
kr0 would have to be greater than 3 / 2 , since the wave function u(r) would have to
Nuclear Force and Scattering
have a radial node inside the well. But from equation: kr0 must certainly be less than
 for all positive values of binding energy .
We shell here prove that for l  0 no bound state exists. It shall be assumed that the
potential is central and of square well type. The differential equation to be used in
this case l  0 , is which through the substitution u (r )  r (r ) takes the form.
d 2u ( r ) M
 2
dr 2
h

l (l  1)h 2 
E

V
(
r
)


u (r )  0.
Mr 2 

(1)
V(r)
O
Fig. 1. Deuteron wave function of the first excited states
On comparing these equations, we find that it is equivalent to an S-wave radial
equation with potential
Ve ff (r )  V (r ) 
l (l  1)h 2
Mr 2
29
As already discussed the second term on R.H.S. is called the centrifugal potential as
its space derivative gives the classical centrifugal force. This potential is repulsive,
there forces ‘l’ in creases, the binding energy of the lowest bound state decreases.
Returning back to equation and setting l=1 , the next acceptable value of l after 0, we
then get ,
d 2u ( r ) M
 2
dr 2
h
 2h 2 
 2 u (r )  0.
 Mr 
Properties of Nuclei and Scattering
Now E=-EB’ the binding energy of deuteron in the P- state (l=1) and using a square
well potential V (r) = V0’ for r < r0, for the p- state , equation may be written as
and
d 2u ( r ) M
 2
dr 2
h
 '
2h 2 
V

E
'

B
 0
u (r )  0 for
Mr 2 

r  r0
(2)
d 2u ( r ) M
 2
dr 2
h

2h 2 
 EB 
u ( r )  0
Mr 2 

r  r0
(3)
for
Now letting
M

k '   2 V0 ' _ EB '
h

 ME ' 
and  '   2 B .
 h 
The above equation may be written as
and
d 2u ( r )  2 2 
 k '  3 u ( r )  0
dr 2
r 

for
r  r0
d 2u ( r )  2 2 
  '  3 u (r )  0
dr 2
r 

for
r  r0
The least well depth, just repaired to produce this bound state, is the one for which the
binding energy EB’ is just equal to zero, i.e., when  '  0 and k ' 
MV ' / h   k
2
0
0
(say).
If we put k0r = x, the wave equation reduces to
d 2u ( r )
2u (r )
 u (r )  2  0
2
dx
x
for
x  k0 r0
30
and
d 2u (r ) 2u (r )
 2 0
dx 2
x
for
x  k0 r0
The solution of equation with the correct boundary condition becomes
u(r )  A2 x 1 for x  k0 r0 ,
(4)
To solve equation, we make the substitution v = xu(r), so that
d
du (r )
x
u
dx
dx
and
d 2
d 2u ( r )
du (r )
x
2
2
2
dx
dx
dx
Nuclear Force and Scattering
and it can then be re-written as follows
d 2 2 d

   0 for x  k0 r0 .
dx 3 x dx
(5)
Differentiating this equation with respect to x, we get
d 3 2 d 2 2 d d



 0 for x  k0 r0.
dx 3 x dx 2 x 2 dx dx
Dividing this equation by x throughout , we get
1 d 3 2 d 2 2 d 1 d



 0 for x  k0 r0.
x dx 3 x 2 dx 2 x 3 dx x dx
Now since
d 2  1 d  1 d 3 2 d 2 2 d




dx 2  x dx  x dx3 x 2 dx 2 x3 dx'
then the equation may be re-written as
d 2  1 d  1 d
 0 for x  k0 r0.


dx 2  x dx  x dx
Now since u(r)=vx-1, must vanish for x = 0, the solution of above equation is found to
be
1 d
 A1 sin x for x < k0ro.
x dx
Integrating it, we get
  xu(r )  A1 (sin x-x cos x) for x < k0r0,
(6)
To satisfy continuity condition at the boundary (r=r0 or x=k0r0), these solutions
yields,
d
(sin x-x cos x ) = 0 at x = k0r0
dx
31
Or
x sin x = 0 at x = k0r0
Or
k0r0 sin k0r0 = 0 at x = k0r0
The smallest positive root of this equation is k0r0=  . Hence a bound state of the
deuteron for l  can exist only if k0r0<  and this contradicts the previous statement
that k0r0<  . There fore we conclude that no bound states exist for deuteron when l
 , i.e., deuteron does not possess any excited state.
Properties of Nuclei and Scattering
___________________________________________________________
2.4 n-p SCATTERING AT LOW ENERGIES WITH SPECIFIC
SQUARE WELL POTENTIAL
What is Scattering
When an intense and collimated beam of nucleons is born barded on target nuclei the
interactions between incident nucleus and target nuclei takes place. As a result we
may observe the following two possibilities:
(i)
The interactions does not change the incident particles, i.e., incoming
and outgoing particles are the same. The change is in the path of incoming nucleons,
i.e., they are deviated from their original path. This process is known as scattering, In
scattering processes the outgoing particles may have same energy as that of incident
particles or may have the changed energy value. The former is known a elastic
scattering and latter is known as inelastic scattering.
(ii)
The second possibility is that the outgoing particles are different from
the incident particles. Then the interaction process is known as nuclear reaction. In
nuclear reaction we have two alternatives:
(a)
The incident material particles are fully captured by the target and
instead  -radiations (  -photons) are emitted. The situation is termed as radiative
capture (e.g., n-  -reaction).
(b)
In the second alternative the outgoing particles are either charged
particles or some other material particles which are the product of the process itself,
then the process is known as nuclear reaction.
It should be remembered that any of the above alternative may occur, either alone or
with other competing processes. Among the nucleon-nucleon scattering, neutron
32
proton (n-p) scattering is the simplest one, because here the complication due to
coulomb forces are not present.
In (n-p) scattering neutron proton system is analyzed in the state of positive energy,
i.e., in a situation when they are free. In the experiment, a beam of neutrons from an
accelerator
is allowed to impinge on a target containing many essentially free
protons. The simplest substance is hydrogen gas but in some cases other substances
like thin nylon sheet and paraffin are used. Hence, it is natural to think that in target
protons are not free but are bound in molecules. The molecular binding energy is so
small about 1eV, therefore, for the impinging neutrons of energy greater than 1eV,
Nuclear Force and Scattering
protons are treated as free. The presence of electrons also do not affect the process
because they are too light to cause any appreciable trouble to incoming neutrons.
When neutrons impinge on protons, some of them are captured to form deuteron and
balance of energy is radiated in the form of  rays; but the great majority of neutrons
undergo elastic scattering. In the process, the interactions between two nucleons is of
such a order that the neutrons changed their velocities in magnitude as well as in
direction.
The proton - proton (p-p) scattering is due to the presence of coulomb repulsion
between two protons. The presence of coulomb repulsion increases the change of
direction the account of which is made in estimating the nuclear forces. It appears that
nuclear force between the protons is not sufficiently strong to bind the protons against
coulomb repulsion. (It is supported by the fact that no bound state with two protons
He3 exists). It will be seen later that p-p system remains unbound even in the absence
of coulomb repulsion.
The neutron neutron (n-n) scattering is not practically possible because of the non
availability of neutron target (because neutron decays into proton in a few minutes).
However, their are evidences to support if n-n forces are similar to p-p forces, a bound
state for two neutrons cannot exist.
Neutron – Proton Scattering at Low Energies
In the low energy range most of the measurements of scattering cross section
are due to Melkonian and Rainwater et.al. A beryllium target bombarded at by
deuterons accelerated in a cyclotron, provided the neutron beam which was shot at a
target containing free protons .
33
Fig. 2. n-p scattering cross section
Properties of Nuclei and Scattering
These results show that the scattering cross section depends very much on the energy
of the incident neutrons. At low energies below 10 Mev, the scattering is essentially
due to neutrons having zero angular momentum (l=0) and hence in the centre of mass
system, the angular distribution of scattered neutrons is isotropic. In order to avoid
complications due to Coulomb forces we shall consider the scattering of neutrons by
free protons viz. those not bound to molecules. However in practice the protons are of
course bound to molecules but the molecular binding energy is only about 0.1 ev.
Therefore if the incident neutrons have an energy greater than about 1ev. The protons
can be regarded as free.
In describing elastic scattering events like the scattering of neutrons by free protons it
is more convenient to use the center of mass system. The quantum mechanical
problem describing the interaction between two particles, in the center of mass
system, is equivalent to the problem of interaction between a reduced mass such as
the system. Although while wording out the following theory we shall think in terms
of a neutron being scattered by a proton but it applies equally well to spin less,
reduced mass particle which is being scattered by a fixed force center. Let us suppose
that the neutron and the proton interact via a spherically symmetric force field whose
potential function is V (r), where r is the distance between the particles .
The Schrodinger equation for a central potential V (r) in the center of mass system ,
for the n-p system is
 2 M

  h 2 E  V (r )  0
34
Where M is the reduce mass of the n-p system .
To analyze the scattering event, we have to solve this equation under proper boundary
conditions. In the immediate vicinity of the scattering center, the action will be violent
and its description is difficult. At a considerable distance from the scattering center
where the experimentalist lies in wait for the scattered particles, things will however
be simpler. For scattering the boundary condition is that at large distances from the
scattering center the wave should be made up of two parts: (i) an incident plane wave
that describes the un scattered particles and superimposed upon it, (ii) an outgoing
scattered spherical wave which emanates from the scattering center. To solve in
asymptotic form,
Nuclear Force and Scattering
   inc  sc
The wave function that describes an incident plane wave (a beam of particles )
moving in the positive z-direction is
 inc  eikz  eikz cos  ,
 ME 
Where k   2 
 h 
Which is a solution of the wave equation with V (r) set equation zero,
 2 ME 
  h 2  inc  0,
Setting V (r) equal to zero in this manner actually amounts to switching off the
scattering potential and thereby eliminate scattering so that the total wave
function becomes identical with the incident wave function
 inc . The wave
function represents one particle per unit volume since the square of the wave function
is equal to unity. Having known the form of the incident wave function, the next
problem is to devise a suitable form for the scattered wave function . This obviously is
 sc  f  
eikr
,
r
35
For large ‘r’ f (  ) in this expression indicate amplitude of the scattered wave in the
direction  . This wave function is a necessary consequence of the assumption that
the scatterer simply scatters the particles and does not absorb them at all.
The probability density and hence the number of scattered particles per unit volume
shall be proportional to  sc . If scattering is considered to be isotropic, the density
2
(number per unit volume) of scattered particles through a large spherical shell of
radius r is inversely proportional to r2 since the volume of the spherical shell, being
given by 4r 2 dr, is proportional to r2 and density therefore is proportional to 1/r2
which is also proportional to  sc . Hence 1/r2 dependence of  sc .
2
Therefore the wave function  , in a form we are actually interested viz. asymptotic,
may be written as .
Properties of Nuclei and Scattering
   inc  sc  eiks  f  
eikr
.
r
Now, in Fourier analysis we often expand as arbitrary function into a sense of
harmonic functions of various frequencies. So we expand the incident plane wave
function eikz in terms of Legendre Polynomials Pi (cos  ) and write.

 inc  eikrc0 s   B1 (r ) P1 cos  
l 0
Where l is an integer representing the various partial waves. This particular way of
writing the wave function is termed as the partial wave expansion.
The radial functions B1 (r) in this equation are given by
Bl (r )  i l (2l  1) jl (kr),
(1)
Where Jl(kr) is the Spherical Bassel function which is related to the ordinary Bessel
function through the formula
  
jl (kr)  

 2kr 
1/ 2
J l 1/ 2 (kr)
and can be represented as
l
 1 d   sin kr 
jl (kr)  (kr) 


 kr d (kr)   kr 
l
Whence asymptotically
36
jl (kr) r 
l 

sin  kr  
2

kr
Asymptotically, Bl(r) from is given by
Bl (r ) r 

l 

sin  kr  
2
 i l (2l  1) 
kr


i ( kr l / 2 )
1 l
i (2l  1). ei ( kr l / 2) e
.
2ikr
Nuclear Force and Scattering
The Spherical Bessel function Jl(kr) for various values of l are given below
 3
1
3 cos( kr)
j2 (kr)  
  sin( kr) 
.
3
kr2
 (kr) kr 
These functions are plotted in the Fig 3.
Similarly f (  ) may also be expanded in terms of the Legendre Polynomials as
follows
f ( ) 
i 
 f1 (2l  1) P1 (cos  ).
2k l  o
Substituting from equation in equation we obtain


l 0

   inc  sc   i l (2l  1) jl (kr)  f l
eikr 
P1 (cos  ).
r 
37
Fig.3. Variation of Bessel function with orbital angular momentum quantum
number.
Since each term in the sum [equation] with a specific value of the orbital angular
momentum quantum number ‘l’ , represents a solution of the wave equation in
spherical polar co-ordinates for constant potential energy. Therefore the expansion
classifies the particles in the beam according to their angular momenta which is of
great practical importance since at lower energies below10 Mev, most of the
scattering is due to l=0 particles , i.e. the number of partial waves is severely limited
in this case and it suffices to study the scattering only for l=0, i.e. S-wave.
Properties of Nuclei and Scattering
From above equation (1) for l=0
kr  .......
sin( kr)
B0 (r ) 
 1
kr
6
2
And for l=1
 sin( kr) cos( kr) 
B1 (r )  3i 

kr 
 kr
 kr kr 3

 3i  
 .......
30
3


B1 (r )
 (kr) 2 .
B0 (r )
We have found out the ratio of the square instead of just
B1 ( r )
, since the probability
B0 ( r )
density is determined by Bl2(r). To have an idea of the magnitude of this ratio, let us
38
consider a neutron of energy 1 Mev in the L-system, it will be 0.5 Mev in the C-M
system. Neutron momentum then is
1
 2 1.67 1024 1.6 106 
p  (2ME ) 2  

2


 1.63 1015 gm.cm. / sec .
and its wave number
k
p
1.63 1015

 1.55 1012 cm.1
 27
h 1.0545 10
If we assume the nuclear forces to have a range r = 2 Fermi, then
2


B1 (r )
2
 (kr) 2  1.55 1012  2 10 13  0.31 ,
B0 (r )
= 0.0961,
i.e. at an energy of 1 Mev only about 9% of the scattering is due to neutrons with l=1.
Similar calculations for a neutron of energy 10 Mev raises this percentage to about
49%. There fore in the energy range below 10 Mev. S-wave scattering (l=0) is
predominant.
Nuclear Force and Scattering
_____________________________________________________________________
2.5
RESULTS OF LOW ENERGY n- p AND p-p SCATTERING
The theory for the scattering cross-section developed in the previous section is in fact
a theory for the phase shift  l which in turn depends up on the assumptions regarding
the nature of the scattering potential V (r). by way of illustration, we now proceed to
carry out the calculations for the same rectangular potential well as was assumed in
sections for the deuteron ground state.
The radial Schrodinger equation for l=0, viz. equation inside and outside the nuclear
square potential well may be written as
d 2u ( r ) M
 2 E  V0 u (r )
dr 2
h
for r < r0,
39
d 2u ( r ) M
 2 Eu(r )  0
dr 2
h
for r > r0,
Since in the present case of n-p scattering, the negative binding energy is replaced by
a small positive energy E which is much smaller than the well-depth V0. These
equation may be written as
d 2 ui 2
K ui  0
dr 2
for r < r0,
d 2 ui 2
K ui  0
dr 2
for r > r0,
Where ui is the wave function inside the well and u0 that outside the well and
K2 
M ( E  V0)
h
,k2 
ME
.
h2
Equation has the solution
ui  A sin Kr
And equation has the solution
u0  C sin kr  D cos kr,
Which may be written as
u0  B sin( kr   0 ).
Properties of Nuclei and Scattering
Fig. 4. Two indistinguishable p-p collisions
40
In order to understand the significance of the phase shift  0 ,
the Schrodinger
equation would be equation V(r) set equal to zero, the solution of which would have to
be of the form
u (r )  sin kr.
Since it must vanish at r = o. The solution which holds good only outside the well.
Thus  0 is the phase shift at large distances introduced
by switching on
the
scattering potential .
We now require that the solution and join smoothly at r = r0 i.e. the logarithmic
derivative must be continuous at r=r0 viz,
1 dui
ui dr
r  r0

1 du0
u0 dr
r  r0
This condition, with the aid of equation gives
K cot Kro  k cot( kr0   0 )
This result may be compared with the continuity condition equation for the ground
state of the deuteron viz.
M (V0  EB )
h

cot  


M (V
 EB )  
 r0  
h
 
0
To simplify the matching condition in case of n-p scattering, we assume that inside
the well, The scattering wave function is not much different from the deuteron wave
function. This appears quite reasonable since the two situations differ only in that the
total energy E in this case although small, is positive whereas the deuteron binding
Nuclear Force and Scattering
energy EB is small, but negative. We therefore assume that the logarithmic derivative
K cot Kr0 of the inside wave function for scattering could be approximated by the
value of the logarithmic derivative of the ground state wave function of deuteron viz.
– y. Hence from
k cot( kr0   0 )   y.
At this point we introduce another approximation that r0 is very small (possibly zero)
compared to k  (MF / h) so that kr0 may be neglected in the above equation and
then
kcon 0   y or cot  0   y / k
Now the total scattering cross-section for l=0 from equation is given by
41
 sc,0 

4
4
1
4
1
sin 2  0  2 .
 2.
2
2
k
k (1  cot  0 ) k (1  y 2 / k 2 )
4
4h 2

.
k 2  y 2 M ( E  EB )
Where we have substituted the values of k2 and y2 from equations and respectively.
The relation was first arrived at by E.P. winger which although agrees with
experimental results at high energies but fails miserably at low energies.
Check Your Progress 2
Note: a) Write your answers in the space given below.
b) Compare your answers with the ones given at the end of the unit.
(1) Solve the Schrodinger equation for the deuteron in a S-state under
the
assumption of square well potential. Show that deuteron has
no excited state.
(2) Discusss n-p and p-p scattering at low energies. When light does it
through
on the nature of nuclear force ?.
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Properties of Nuclei and Scattering
_________________________________________________________
2.6 SPIN DEPENDENCE AND SCATTERING LENGTH
Spin dependence
E. P. Wigner suggested that the internucleon forces are spin-dependent. Since neutron
and proton are
1
spin particles, therefore in n-p scattering the neutron and proton
2
spins may either be parallel or anti parallel. In deuteron the bound state of the n-p
42
system, whose binding energy EB, the neutron and proton spins are parallel and
therefore this equation possibly holds good for parallel spin case.
The state of parallel spins, is a triple state and has a statistical weight 3 corresponding
to the three allowed orientations of the angular momentum vector under an external
magnetic field. The state of anti parallel spins is a singlet state on account of the non
orientability of a vector of zero length and has a statistical weight.
1-
In a scattering experiment in general neutron and proton spins are randomly
oriented and so are the spins of neutrons in the incident beam and therefore singlet
and triplet state of the n-p system will occur in proportion to the statistical weight
factors for these states which are
1
3
and
respectively. The total scattering cross2
4
section therefore shall be made up of two parts,  l , 0 - the cross-section for scattering in
the triplet state and  s , 0 the scattering cross-section in the singlet state, as follows
3
4
1
4
 0   t ,0   s ,0
From a naïve point of view, in a random distribution of spins as in n-p scattering, the
two spins are as often parallel as antiparallel giving equal statistical weights to the two
states. However quantum mechanically, the spin direction cannot be defined as
uniquely as a vector in space and hence the statement ‘spin pointing up’ simply tells
that the spin vector points somewhere along a cone around the vertical direction. The
following figure depicts schematically the four equally likely situations for the
relative spins of the two particles.
Nuclear Force and Scattering
43
(1)
(2)
(3)
(4)
Fig. 5 Spin dependence
Figures (1) and (4) correspond to a total spin unity corresponding to the magnetic
quantum number values +1and -1 respectively. In cases (2) and (3) the z-components
add up to zero but since the spins are not aligned along the z-direction, they may add
up to zero as in case (2) resulting in a singlet state or may add up to a total spin
perpendicular to the z-axis as in case (3) giving rise to a triplet state.
Scattering Length
Fermi and Marshall introduced a very useful concept the ‘scattering length a’ for the
discussion of nuclear scattering at very low incident neutron energy.
[i.e.E  0
 ME 
and hence k   2   0
 h 
Which is defined as follows:
Properties of Nuclei and Scattering
44
 sin  0 
a  Lim 
;
k 0
k 

With this definition, equation giving the total scattering cross section form S-wave
(l=0) may be written for very low incident neutron energy as
 4 sin 2  0 
  4a 2
Limsc, o   Lim
2
k 0
k 0
k


Equation then indicates that ‘a’ has the geometrical significance of being the radius of
a hard sphere surrounding the scattering center from which neutrons are scattered and
so has the dimensions of length, hence the name scattering length.
Now it is to be noted from equation that as k  0 (i.e.,) as the energy E of the
incident neutron approaches 0,  must approach either 0 or  other wise the crosssection at zero neutron energy would become infinite which is physically absurd.
There fore at very low incident neutron energies ( E  0 ), equation reduces to
a
0
k
Then at very low incident neutron energies, the wave function outside the range of
nuclear force as expressed by equation may be written as
sin( kr   0 

Lim U (r )  Lim (r 0 )  Lim ei 0

k 0
k 0
k 0
k


The equation then gives a simple graphical interpretation of the scattering length. This
equation represents a straight line for U (r) and the scattering length ‘a’ is the
intercept on the r-axis. This is indicated in Fig.6.
Having defined the scattering length by means of equations an inquisitive reader may
ask quite naturally as to Well, the significance of positive or negative scattering
length is that it tells us what is the significance of attaching a positive or a negative
sign with at the scattering length ? whether the system has a bound or an unbound
state.
Nuclear Force and Scattering
45
Fig. 6. Graphical interpretation of scattering length
From fig.6. it is clear that positive scattering length indicates a bound state and
negative scattering length indicates a virtual or unbound state. Since the deuteron
wave unction, i.e., the wave function for the bound state of n-p system, must curve
towards the r-axis in order to match the exponentially decaying solution (c.f. equation
i.e. r>r0 will give rise to a positive intercept on the r-axis indicating thereby a positive
scattering length. For unbound state the wave function has to match with an
increasing solution outside the range r0 and then extrapolation of U(r) shall produce a
negative intercept on the r-axis implying thereby a negative scattering length.
Check Your Progress 3
Note: a) Write your answers in the space given below.
b) Compare your answers with the ones given at the end of the unit.
(1) Write short note on scattering length.
(2) Explain clearly how the properties of the deuteron indicate the
presence of
spin dependent force and tensor force between two nucleons.
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46
Properties of Nuclei and Scattering
___________________________________________________________
2.7 LET US SUM UP
After going through this unit, you would be able to achieve the aforesaid objectives.
Now we recall what we have discussed so far.

We have learnt the basic properties of deutron, its charge (+e), mass (~2.014
amu), its radius (2.1 fermi), its binding energy (=2.225  .003 Mev), Spin (1  )
and statistics (Bose-Einstein) and the electric quadrupole moment Qd
=0.00282 barn .

The study of deuteron problem, although hopelessly limited in as much as
deutron possesses only the ground state and no-excited states exist for the
bound neutron-proton system, gives invaluable clues about the nature of the
nuclear force.

We learnt that neutron and proton can form stable combination (deuteron)
only in the triplet state means when the n & p spins are parallel. The singlet
state, i.e. a state of antiparallel n-p spins being unbound.

The existence of non-zero magnetic moment and electric quadrupole moment
for deutron suggests that at least a part of the neutron proton force acting in
deutron is non-central.

The nuclear forces are spin dependent i.e., nuclear forces not only depend
upon the separation distance but also upon the spin orientations of two
nucleons. They are independent of the shape of nuclear potential.
___________________________________________________________
2.8 CHECK YOUR PROGRESS: THE KEY
1. (1) For the deuteron properties see the section 2.2.
(2) The quadrupole moment is described in the last of section 2.2.
2. (1) For the excited states of deuteron, see the section 2.3.
(2) The n-p and p-p scattering at low energies is explained in section 2.4.1.
3. (1) The scattering length is described in the last of sub-section 2.4.2.
(2) For Spin dependence see the first part of section 2.4.2.
47
Nuclear Force and Scattering
REFERENCES AND SUGGESTED READINGS
1.”The Two Nucleon Problem” by M. Sugrwara and Hulthen, Encyclopedia of
Physics,
Berlin: Springer Ver.
2. “Nuclear Two Body Problems and Elements of Nuclear Forces” Experimental
Nuclear
Physics by N. F. Ramsey, Wiley: New York.
3. Lectures on Nuclear Theory (translated from the Russian) by Landau, Plenum
Press, New York.
4. Elementary Nuclear Theory, 2nd ed. by Bethe and Morrison, Wiley: New York.
5. The Atomic Nucleus by R D. Evans, McGraw-Hill Publications.
6. Atomic and Nuclear Physics by Brijlal and Subhraininyan.
7. Nuclear Physics by D. C Tayal.
8. Nuclear Physics by Irving Kaplan, Narosa Publishing House.
*************
48