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DISTANCE EDUCATION SELF LEARNING MATERIAL
PROGRAMME : M.Sc.
YEAR : Final
PAPER : III
TITLE OF PAPER : Nuclear & Particle Physics
MADHYA PRADESH BHOJ (OPEN) UNIVERSITY BHOPAL (M.P.)
Course Name – M.Sc. (Final)
PAPER- III (Nuclear & Particle Physics)
DISTANCE EDUCATION SELF LEARNING MATERIAL
MADHYA PRADESH BHOJ (OPEN) UNIVERSITY
BHOPAL (M.P.)
FIRST EDITION - September/2012
UNIVERSITY - M. P. Bhoj (Open) University, Bhopal
PROGRAMME - M. Sc. (Final)
TITLE OF PAPER - Nuclear & Particle Physics
BLOCK NO. - 1, 2
UNIT WRITER - Dr. Kaushlendra Chaturvedi
Department of Physics
Bundelkhand University, Jhansi - 284128
EDITOR - Dr. Rajiv Manohar
Department of Physics
University of Lucknow, Lucknow - 226007
COORDINATION - Dr. (Mrs.) Abha Swarup,
Director (Printing & Translation)
COMMITTEE - Maj. Pradeep Khare (Rtd)
Consultant, M. P. Bhoj (Open) University, Bhopal.
M.P. Bhoj (Open) University
ALL RIGHTS RESERVED
No part of this publication may be reproduced in any form, by mimeograph or any other means,
without permission in writing from M.P. Bhoj (Open) University.
The views expressed in this SLM are that of the author (s) & not that of the MPBOU.
The cost of preparation and printing of Self-Learning Materials is met out of DEC grant.
Further information on the MPBOU courses may be obtained from the University’s office at Raja
Bhoj Marg, Kolar Road, Bhopal (M.P.) 462016
Publisher: Registrar, M.P. Bhoj (Open) University, Bhopal (M.P.)
Phone: 0755-2492093
Website: www.bhojvirtualuniversity.com.
Course Name – M.Sc. (Final)
DISTANCE EDUCATION SELF LEARNING MATERIAL BLOCK: 1 UNIT 1 – Basic Properties of Nuclei UNIT 2 – Nuclear Force and Two Body Problem BLOCK: 2 UNIT 3 – Nuclear Models and Nuclear Reactions UNIT 4 – Nuclear Decay UNIT 5 – Elements of Particle Physics- Elementary Particles
Paper (III) Title - Nuclear & Particle Physics Block (1, 2) Block Introduction
Block 1 contains two units, namely- Unit 1 - Basic Properties of Nuclei Unit 2 - Nuclear Force and Two Body Problem Block 2 contains three units, namely- Unit 3 - Nuclear Models and Nuclear Reactions Unit 4 - Nuclear Decay Unit 5 - Elements of Particle Physics- Elementary Particles
Unit –I – Basic Properties of Nuclei
1.0 Introduction
The atomic nucleus was discovered by Lord Rutherford in 1911 through his alpha particle
experiment. He concluded that more than 99.9% of the atomic mass is concentrated within
the small volume of nucleus. In this unit, we will study about the basic properties of atomic
nucleus like its charge, mass, size, magnetic dipole moment and quadrupole moment. From
the Schmidt lines, we can get magnetic moment values of odd proton and odd neutron nuclei
while from quadrupole moment, the information about nuclear shape can be obtained.
Nuclear size can be estimated by various electrical and nuclear methods. We will study some
of these methods in this unit. The information about the stability of nuclei can be obtained
from the binding energy curve and semi empirical mass formula. From this formula, equation
of mass parabolas can be deduced explaining the stability of even or odd mass number
nuclides. These topics will also be covered in this unit.
1.1 objectives
This unit gives the description of basic properties of nucleus. After going through this unit,
you will be able to:
Know about constituents of nuclei, nuclear mass, density, nuclear size and various
methods to determine the size of nucleus.
Explain binding energy curve, its related semi empirical mass formula and know
about the various applications of this formula.
Get an understanding of magnetic dipole moment and electric quadrupole moment of
nuclei along with their experimental determination.
Know about mirror nuclei and isotopic spin of nucleus.
1.2 Constituents of nuclei
With the discovery of neutron by Chadwick in 1932, it was recognized that the atomic nuclei
are composed of two different type of elementary particles-protons and neutrons.
Collectively, protons and neutrons are known as nucleons. The proton is identified as the
nucleus of the lightest isotope of Hydrogen. It carries one electronic charge, +e and has a
mass of about 1836 times the electronic mass me. The neutron on the other hand, possesses
no charge i.e. it is electrically neutral. The mass is slightly more than mass of proton. Thus
electronic mass is negligibly small in comparison to proton and neutron mass.
According to Coulomb‟s law, the positively charged protons, closely spaced within the
nucleus, should repel each other strongly and they should fly apart. It is therefore difficult to
explain the stability of the nucleus unless it is assumed that nucleons are held together under
the influence of very short range attractive force. This force is different from other commonly
known forces like gravitational or electromagnetic forces and is known as strong force.
The atomic number of the nucleus is determined by number of protons in the nucleus and is
called Z value or the proton number. The sum of protons (Z) and neutrons (N) inside the
nucleus is called mass number A
(1)
A nuclide is symbolically represented by .
1.3 Nuclear mass and binding energy
The nuclear mass Mnu is obtained from the atomic mass M (A, Z) by subtracting the masses
of Z orbital electrons
(2)
The nuclei are very strongly bound and energies of few MeV are needed to break away a
nucleon from the nucleus. In contrast, only a few eV is needed to detach an orbital electron
from the atom. So, to break up a nucleus of Z protons and N neutrons completely into
separate particles, a minimum energy is to be supplied to the nucleus. This energy is called
Binding Energy, EB of the nucleus. Conversely, to build up a nucleus of mass number A and
nuclear charge Z from Z protons and N neutrons at rest, an amount of energy EB is evolved.
Binding energy, (3)
where, is the amount of mass disappeared in forming a nucleus out of the constituent
particles. If MH and Mn be the mass of hydrogen atom and neutron respectively, then
(4)
Hence,
(5)
In energy unit,
(6)
dropping c2 from the above equation.
1.4 Atomic mass unit
The unit of atomic mass is defined as one twelfth of the mass of the atom of carbon isotope
12C taken to be exactly 12 units, and is denoted by „u‟, abbreviated for „unified atomic mass
unit‟.
The unit of atomic mass in 12
C scale is
where NA is Avogadro number kg.
The energy equivalents of rest mass of electron, proton and neutron are:
Electron (me) = 9.10953 × 10-31
kg =5.48580 × 10-4
u
Proton (mp) = 1.677265 × 10-27
kg = 1.0072765 u
Neutron (mn) = 1.67495 × 10-27
kg = 1.0086650 u
1.5 Binding energy and nuclear stability
The EB value is the measure of the stability of the nucleus.
(i) If EB >0, i.e. for positive EB, the nucleus is stable and energy from outside is to be
supplied to break the nucleus into its constituents.
(ii) If EB <0, i.e. for negative EB, the nucleus is unstable and will disintegrate of itself.
The EB values range from 100MeV for Carbon to 1800MeV for Uranium.
Example: Computation of binding energy of α particle or nucleus.
The helium nucleus is made up of 2 protons and 2 neutrons.
Total = 4.031882 u
Atomic mass of = 4.002603 u
Difference = +0.029279 u
+ sign indicates that nucleus is stable.
Binding energy in MeV= 0.029279 ×931= 27.16 MeV. This energy divided by the number of
nucleons is called binding energy per nucleon EB/A.
Q.1 Calculate the binding energy in MeV of 4He from the following data: Mass of
4He =
4.003875 amu, Mass of 1H = 1.008145 amu, mass of neutron = 1.008986 amu.
Q.2 Calculate the binding energy per nucleon of 10
B (mass=10.01612 amu) and 29
Si
(mass=28.98571amu).
Q.3 Find the energy release if two 1H2 nuclei fuse together to form 2He4 nucleus. The
binding energy per nucleon of 1H2 and 2He4 is 1.1MeV and 7 MeV respectively.
1.6 Mass defect and packing fraction
1.6.1 Mass defect: The difference between the measured atomic mass M (A,Z) expressed
in u, and the mass number A of a nuclide is called mass defect and is given by
(7)
The mass defect can be both positive and negative. Mass defect is positive for very light and
very heavy atoms and negative in the region between the two.
1.6.2 Packing fraction: The packing fraction f is defined as the mass defect per nucleon
in the nucleus, i.e. the mass defect of an atom divided by its mass number. So
or
(8)
The „f‟ has same sign as and thus positive
for very light and very heavy atoms and
negative for intermediate range atoms.
Packing fraction f varies with mass number A
as shown in fig. 1. It is observed that f is
positive for very light nuclei and as A
increases, f decreases rapidly, being negative
for A>20. f attains minimum (negative) value
at A60 and then starts increasing slowly. For A 180, it becomes positive again.
Q.3A The mass of hydrogen atom and neutron are 1.008142 and 1.008982 amu
respectively. Calculate the mass defect, packing fraction and binding energy per nucleon of
16O nucleus. (Ans: 0.005085 amu, 3.178x10
-4, 8.26 MeV)
1.7 Binding energy curve
The binding energy curve is the variation of binding energy per nucleon EB/A with A for
different nuclei and is shown in fig. 2. From the study of the curve, following observations
are noteworthy.
(i) EB/A is very small for light nuclei and goes on
increasing with increasing A and reaches a value
8.8 MeV/nucleon for mass number A 20. The
reason for low value of EB/A for light nuclei is that
in these nuclei, most of the nucleons lie on the
surface of the nucleus rather than inside. For heavy
nucleus, most nucleons lie inside rather than on
surface. Since surface nucleons are surrounded by
fewer number of nucleons, they are not tightly
bound. This fact reduces the binding energy in
light nuclei.
(ii) For A>20, the rise of curve is much slower, reaching a maximum value of 8.7 MeV per
nucleon for A=56. If A is increased further, the curve decreases slowly. For A> 180, i.e. for
heavy nuclei, EB/A value decreases monotonically with increasing A and it is7.5
MeV/nucleon for heaviest nuclei. EB/A decreases for large values of A due to the coulomb
effect. Between every pair of protons, coulomb repulsion increases as Z2. For naturally
occurring nuclei, Z2 increases faster than A, so coulomb effect is not adequately compensated
by increase in A and hence binding energy decreases due to this coulomb repulsion between
protons in heavy nuclei.
(iii) In the mass range, 20<A<180, the variation in EB/A is very slight and in this region,
EB/A may be considered to be virtually constant with a mean value 8.5 MeV/nucleon. The
near constant value of EB/A implies that binding energy is independent of the size of the
nucleus in this region. This behavior tells that nucleons interact with only their nearest
neighbors and not all the nucleons of the nucleus. If nucleons interact with all other nucleons
present in the nucleus, then binding energy should increase in proportion to A2 (where A is
the mass number) whereas EB/A remains constant for most of the time in this region. Hence
nucleons interact with only limited number of their neighboring nucleons and this property is
called saturation property of nuclear forces.
(iv) B/A increases to a maximum of about 8.8 MeV per nucleon around mass number 56, and
then decreases slowly to 7.6 MeV around mass number 238. This indicates the energy release
in the fission of heavy elements and fusion of light elements.
(v) There are notable peaks at A=4n (4He,
8Be,
12C,
16O,
20Ne,
24Mg). This reflects the
peculiar stability of alpha particle structure. This greater stability is due to the pairing off of
the two protons with opposite spins.
(vi) Peaks are also seen at Z or N= 20,28,50,82, 126. These are called magic numbers. These
peaks indicate that corresponding nuclei are more stable relative to those in the
neighbourhood.
1.8 Nuclear Size
From the experiments, it is observed that majority of the nuclei are spherical or nearly so in
shape. The radius R of various nuclei is approximately given by
where A is mass number and r0 is a constant called nuclear radius parameter. The value of r0
ranges from (1.1-1.5) F and can be evaluated by different methods. To obtain the above
equation, we consider that nuclear charge is uniformly distributed or the nuclear charge
density is nearly constant. Also from the experiments, it is observed that nuclear matter
density m can be taken as approximately constant i.e.
Assuming spherical nucleus
Where R is the nuclear radius and mN is the mass of the nucleon.
Hence
or
i.e.
The nuclear radius is the radius of nuclear mass distribution.
Experimental methods to investigate the size of nucleus are classified into two main groups:
(A) Electrical methods
(1) Mesonic X rays (2) Electron scattering
(3)Coulomb energies of mirror nuclei
(B) Nuclear methods
(1) Neutron scattering (2) α – decay
(3)α particle scattering (4) Isotopic shift in line spectra
Some of these methods are discussed below:
1.8.1 Mesonic X rays
In this method, the probing particle is meson (μ). The advantage of the μ meson over electron
is its larger mass (207 times me) which allows the μ meson to penetrate the nucleus long
before it decays or captured by the nucleus.
Fitch and Rainwater conducted the experiment using a beam of 385 MeV protons to produce
negative pions of kinetic energy 110 MeV and the beam of pions of required energy and
charge was selected by magnetic analyzer. If negative pions are not captured by nucleus, they
decay into negative muons as follows:
(9)
the negative μ meson after being slowed down to thermal energies by ionizing collision may
be captured in Bohr type orbits around the nucleus, forming mesonic atoms. The muon in
these Bohr like orbits makes transition from one orbit to another and emitted X rays are
studied using ray spectrometer. The energies of such X rays depend on the value of R and
can be used to estimate the size of nucleus. The average value of R estimated by this method
is (1.20.03)A1/3
x 10-15
m.
1.8.2 Electron scattering method
In this method, elastic scattering of high energy electrons is studied and from the knowledge
of scattered electrons the radius constant r0 is computed. The value of r0 comes out to be
1.8.3 Mirror nuclei method
The mirror nuclei are pair of nuclei obtained from each other by the interchange of neutron
with proton e.g. ( , , ). Taking the pair
is unstable and is converted to by positron emission.
(10)
where is a neutral particle, called neutrino. If Z be the atomic number of daughter nucleus,
then difference in Coulomb energies of mirror nuclei is given by
Where R is the radius of daughter nucleus. The above energy is used in providing
(i) rest mass energy mec2 to produce positron
(ii) kinetic energy Eβ+ of positron
(iii)rest mass (mn-mp)c2 required for converting neutron into proton
Thus
From this equation R can be calculated. The average value of R estimated by this method is
1.23A1/3
x10-15
m.
1.8.4 Neutron scattering method
When the energy of neutrons is between 10 to 50 MeV, the total scattering cross section is
given by
where λ is the de-Broglie wavelength of incident particle. The value of r0 calculated from the
above expression comes out to be 1.2 fm.
1.8.5 Alpha (α) decay and α scattering
From Gamow theory of α decay and Rutherford theory of α particle scattering, nuclear radius
R can be estimated. We discuss the Rutherford α particle scattering from gold nucleus to
estimate the nuclear size. Scattering of alpha particle by gold nucleus is shown in fig. 3.
Ze is the charge of target (gold) nucleus, R is its radius and „b‟ is the impact parameter. The
deflection is less for b>R distances due to less Coulomb force experienced by alpha particle.
For b=R, deflection is maximum. The Coulomb repulsion experienced by alpha particle in the
presence of gold nucleus is given by
where 2e is the alpha particle charge. This force acts approximately at perpendicular direction
to the incident direction of alpha particle over distance „b‟. If particle velocity is „v‟, then the
time during which this force operates is . This force produces a momentum p in a
direction perpendicular to incident direction. By Newton‟s law
The deflection suffered by alpha particle is
where . Thus
In Rutherford scattering experiment, radian when . Taking the velocity of alpha
particle as 107 m/sec, we get
Thus , which may be considered as radius of nucleus.
Q.4 Determine the radii of 16
O and 206
Pb nucleus, given that r0=1.2 fm.
Q.5 Determine the mass number A of a stable nucleus whose radius is one third of that of Os
189 nucleus.
1.9 Nuclear density
The nuclear density is given by
Where A is mass number, mN is the mass of nucleon and R is the nuclear radius.
Using
We get
Using mN = 1.67 ×10-27
kg and r0 = 1.2 fm, we get ρ ≈ 1017
kg/m3
In terms of number of nucleons,
=1044
nucleons/m3
Thus nucleus is a very tightly bound system of particles with a large potential energy.
1.10 Nuclear spin
The spin of nucleus is the resultant of the spins of its constituent particles – neutrons and
protons. Both of these are spin 1/2 particles and have angular momentum . In addition to
spin angular momentum, neutrons and protons have orbital angular momentum with their
magnitude being integral multiple of . Thus intrinsic angular momentum of nucleus is a
vector such that
where summation stands for the nucleons inside the nucleus. The magnitude of total angular
momentum vector is then
where is the maximum value of in the given Z direction.
For even A type nuclei having either odd Z, odd N or even Z, even N nucleons, will be zero
or integral multiple of . For odd A nuclei having either odd Z, even N or even Z, odd N
nucleons, will be an odd half-integral multiple of .
The total angular momentum of a nucleus is sometimes loosely called „spin‟ of the nucleus
but it is different from the spin angular quantum number.
1.11 Semi Empirical Mass Formula
Semi Empirical Mass Formula was set up by Weizsacker in 1935. This is formula for the
atomic mass of a nuclide in terms of binding energy correction terms. This formula can be
used to predict the stability of nuclei against particle emission, energy release and stability for
fission.
The mass of a nucleus is given by
(13)
where B is the binding energy in terms of mass units. Weizsacker and others developed this
formula assuming the liquid drop model of the nucleus regarding B as latent heat of
condensation. Thus nucleus is analogous to a drop of incompressible fluid of very high
density 1017
kg/m3. The value of B was calculated empirically as made up of number of
correction terms given as
(14)
These correction terms are volume energy correction, surface energy, coulomb energy,
asymmetry energy and pairing energy correction terms discussed below:
1.11.1 Volume energy correction (B1)
The first correction term tells that since the nuclei are bound, the nuclear binding energy is
proportional to the volume of the nucleus or to the total number of nucleons A and hence
(15)
where is a positive constant.
1.11.2 Surface energy correction (B2)
The nucleons which are situated in the surface region of the nucleus are more weakly bound
than those in the nuclear interior because they have fewer immediate neighbours. The number
of such nucleons are proportional to surface area of nucleus and therefore to R2 and so
proportional to A2/3
(as R α A1/3
). Thus
(16)
where is a positive constant. Sign of B2 is opposite to that of B1 since this effect
corresponding to the surface tension of liquid drop represents weakening in binding energy.
This weakening is least or stability is greatest when droplet is spherical in shape since then
surface area is minimum for a given volume.
1.11.3 Coulomb energy correction (B3)
Assuming the nuclear charge Ze to be uniformly distributed throughout the nuclear volume,
the coulomb energy of the nucleus comes out as
Protons inside the nucleus obey Pauli Exclusion Principle and two of them cannot occupy the
same place. Thus the assumption that protons are uniformly distributed is far from correct.
The effect of coulomb self energy on binding energy is that it reduces the binding energy due
to repulsive effect. The correction term B3 is given as
For small value of Z in case of light nuclei, above formula modifies to
where is positive constant.
1.11.4 Asymmetry energy correction (B4)
In light nuclei, condition N=Z corresponds to stability and is called symmetry effect. Any
deviation from N=Z reduces the stability and hence the binding energy of nuclei. This
reduction in binding energy depends on neutron excess (N-Z) and is proportional to .
The correction term is given as
is a positive constant.
1.11.5 Pairing energy correction (B5)
The stability of nucleus also depends whether proton number Z and neutron number N in the
nucleus is even or odd. Nucleus with even Z, even N (even even nuclei) are most stable, even
odd and odd even nuclei are less stable, odd odd nuclei are least stable. This pairing effect is
incorporated in mass formula as
Where δ is given by
(19)
Collecting all the above correction terms, the semi empirical mass formula is given by
(20)
The empirical formula for binding energy per nucleon is
1.12 Mass parabolas for isobaric nuclei
The mass formula shows that M (A,Z) is a quadratic function of Z for a given mass number
A. Thus the graph of M (A,Z) versus Z will be a parabola, the minimum vertex of which
represents the most stable isobar. For odd A nuclei, there is only one stable isobar, for even A
nuclei, there two or three stable isobars. Isobaric nuclides have same mass number A but
different atomic numbers Z. To observe their behavior, we write eqn. (20) as
where
δ has zero value for odd A nuclei.
1.12.1 For odd A isobaric nuclei
For odd A isobaric nuclei
For constant A, this equation represents a parabola.
Differentiating eqn. (21) and equating to zero, the
condition for most stable Z is obtained.
The mass of stable isobar is then
where Zs=Zstable
From eqn. (22) (24)
Subtracting eqn. (23) from eqn. (24), we get,
(25)
This is the parabolic mass relationship for odd A isobaric nuclei displayed in fig. (4).
1.12.2 For even A isobaric nuclei
For odd A nuclei, δ =0 and so they are represented by
single parabola whereas for even-even nuclei, δ is positive
and for odd odd nuclei, δ is negative, so that masses of
even even isobars fall on a separate lower parabola than
that for odd odd isobars. The vertical separation between
the two parabolas is 2δ as shown in fig. 5.
Q.6 Using semi empirical mass formula, find the most stable isobar for A=25. Given a3=
0.58 MeV, a4=19.3 MeV
1.13 Nuclear magnetic dipole moment
Proton inside the nucleus moving in orbits behaves like
charged particle moving in closed path producing a
magnetic field which at large distances acts as a
magnetic dipole located at the current loop. If a particle
having charge q and mass m moves around a force centre
with frequency ν, then the equivalent current is i= qν. From Kepler‟s third law of areas, area
dA swept out in time dt by the particle is related to its angular momentum I is
On integrating over one period T
Hence magnetic moment of ring of current around area A (fig. 6) is
Experimentally it is found that spin is also a source of magnetic moment. Since nucleons
posses spins, hence using q=e, and a dimensionless factor „ ‟, eqn. (26) can be written as
Factor „ ‟ is different for protons and neutrons. Similarly using factor we can write
The total magnetic dipole moment μ is given by
For the nucleus of mass number A, magnetic dipole moment is
In terms of total angular momentum of the nucleus
where is the gyromagnetic ratio ( factor) of the nucleus. The magnetic moment is
measured in terms of nuclear magneton defined as
where μβ is the Bohr magneton given by
1.13.1 Determination of magnetic dipole moment μ
We can determine the dipole moment through the Zeeman splitting of a nuclear level of
energy E0 when the magnetic field B is applied. The particle magnetic moment μ interacts
with the magnetic field B, the interaction energy being – . Using the Schrodinger equation
and
we have
The spin independent Hamiltonian H0 corresponds to energy E0. If Z axis is chosen as
direction of magnetic field, then
The eigen values of Iz are M and hence energy eigen values are given by
(32)
The quantum number M takes values from – to + .
Zeeman splitting of a nucleus with nuclear spin 3/2 is shown in fig. 7 as example.
Experimentally the splitting ΔE is determined. Then from the relation , can be
obtained if B is known. Again from the relation
Dipole moment μ can be determined if and are known. Factor is obtained as earlier and
total number of Zeeman levels being (2 +1), can be found. Apart of Zeeman effect studies,
other methods of determining μ are NMR and perturbed angular correlation techniques.
1.13.2 Schmidt values of magnetic moments
Schmidt values are magnetic dipole moment values of single particle like proton and neutron
for two cases of orbital and spin angular momentum being parallel or anti parallel to each
other. For a single nucleon, the orbital angular momentum with eigen value and
spin s with eigen value couple to total angular momentum j with eigen value
in units of . The magnetic moment associated with orbital angular momentum is
, magnetic moment associated with spin angular momentum s is and the
magnetic moment associated with total angular momentum j is . To calculate the
Schmidt values we need a relationship between and . This in terms of and is
given by
For a single spin 1/2 particle, there are two cases:
(i) parallel to s (stretch case),
(ii) antiparallel to s (jackknife case),
Hence,
So that for case
And for case
These values of magnetic moments for proton and neutron are called Schmidt values and the
curves plotted between μ and j for the two cases are called Schmidt lines. These are displayed
below in fig. 8.
1.14 Electric quadrupole moment of nuclei
Let the nucleus has a charge density ρ(x,y,z) with its charge centre at the origin. As nucleus is
surrounded by the orbital electrons, so the electrostatic potential originating from these
electrons produces an electrostatic interaction energy resulting from interaction between ρ
and . This energy is given by
(35)
Using the Taylor series expansion of potential about origin, this energy can be expressed in
terms of electric moments of distribution, given as
The first term gives interaction energy of a point charge (monopole). Second term gives the
energy of dipole. Third term is quadrupole energy term. This third term in expanded form is
written as
For an ellipsoid of rotation due to symmetry, integrals involving cross products xy, yz, xz
vanish. When z axis is symmetry axis, integral over y2 and x
2 give same result. Hence the
quadrupole interaction energy is
Using the Laplace equation
We get
Where quadrupole moment Q is defined as
from this relation it is evident that:
(I) Q = 0 for spherically symmetric charge distribution (<x2> = <y
2> = <z
2> = r
2).
(II) Q is positive when 3z2
> r2 and charge
distribution is stretched in z direction
(prolate shape).
(III) Q is negative when 3z2
< r2 and charge
distribution is stretched perpendicular to z
direction (oblate shape).
The prolate and oblate shape of the nucleus is shown
in fig. 9. Quadrupole moment has dimension of area
measured in barns (1 barn = 10-28
m2).
1.15 Isotopic spin of nucleus
Neutrons and protons are same particles except their charge and hence these can be regarded
as different manifestations of the same inherent particle called the nucleon. To describe their
quantum state, quantum number used was termed isotopic spin quantum number by Wigner.
Now it is named as isospin or T-spin. A nucleon is assigned an isospin of 1/2, similar to the
spin quantum number and in electromagnetic field, two charged states with isospin quantum
numbers 1/2 and -1/2 can be distinguished as proton and neutron respectively. Similarly, π
particle is assigned an isospin of 1and three charged states of π viz. π+, π
0 and π
- can be
identified from three isospin quantum numbers +1, 0 and –1 respectively. The isospin is a
vector in three dimensional space called isospin space which is virtual one and nothing to do
with physical spin space. The third component of isospin is called T3 along the direction of
the third axis in isospin space.
1.16 Summary
Now we summarize what we have discussed so far:
We have learnt that protons and neutrons are basic constituents of nucleus. Sum of
proton and neutron masses determine the mass of nucleus.
Nuclear size is determined by the expression , where r0 is constant about
1.2-1.4 fermi. This constant can be estimated from various electrical and nuclear
methods.
Binding energy curve tells about stability of nuclei and binding energy itself can be
regarded as made up of number of terms. These terms like surface energy term,
volume energy term etc. give important information about stability of nuclei.
Semi empirical formula contains the various binding energy terms and this can be
changed to equation of mass parabolas telling about the stability of isobars.
Proton and neutron magnetic moment values are known as Schmidt values and curve
between magnetic moment values & angular momentum j is called Schmidt line.
Electric quadrupole moment (Q) of nucleus tells whether nucleus is spherical, prolate
or oblate type. Q=0 means nucleus is spherical whereas non zero positive and
negative value of Q tells about the prolate and oblate shape of nucleus respectively.
The isotopic spin of particle gives information about its various charge states. Isospin
is not real but a fictious quantity.
1.17 Answer to questions
Ans. 1: 28.29 MeV
Ans. 2: 6.47 MeV, 8.44 MeV
Ans. 3: 23.6 MeV
Ans 4:3.02 fm, 7.08 fm
Ans 5: 7
Ans 6: Z=12, 12Mg25
1.18 References
Concepts of Nuclear Physics; Bernard L. Cohen, Tata McGraw Hill Publishing Ltd.
Elements of Nuclear Physics; M.L. Pandya & R.P.S. Yadav, Kedar Nath Ram Nath
Publication.
Introductory Nuclear Physics; Samuel S.M. Wong, Prentice Hall of India Ltd.
Atomic and Nuclear Physics; A.B. Gupta and D. Ghosh, Books and Allied Pvt. Ltd.
Basic Ideas and Concepts in Nuclear Physics; K. Heyde, Institute of Physics (IOP)
Publishing.
Nuclear Physics; S.B. Patel, New Age International (P) Ltd.
Nuclear Interactions, Sergio De Benedetti, John Wiley & Sons.
1.19 Questions
1. Name the various methods for determining the size of nucleus. Describe one of them in
detail.
2. Write Von Weizsacker mass formula explaining each term. Obtain parabolic mass
relationship for odd A isobaric nuclei from it.
3. Show that nuclear density is always constant irrespective of size of the nucleus.
4. Discuss the quadrupole moment, isotopic spin and parity properties of nuclei.
5. Obtain the Schmidt values of neutron and proton magnetic moment.
6. Draw a curve showing the variation of binding energy per nucleon with mass number A.
Explain its main features.
Unit II – Nuclear Force and Two Body
Problem
2.0 Introduction
What holds the nucleons together in the nucleus? Systems are held together by forces. The
gravitational force and electromagnetic force cannot hold the nucleons together. The principle
electromagnetic force between protons is Coulomb repulsion which tends to tear the nucleus
apart. The gravitational force is attractive force but it is smaller by a factor of about 1039
than
the electrical force between protons. Then for existence of nuclei, one must recognize the
third force in nature called, nuclear force. This force is very strong at distances of the order
of nuclear size since it has to dominate the coulomb repulsion between protons. On the other
hand, nuclear force is negligible at distances of the order of spacing between nuclei in
molecules because molecular structure can be adequately accounted by
electromagnetic forces alone. Hence nuclear force is short range force falling off rapidly with
distance. The nuclear force is also called strong interaction because it is strongest of all the
four basic forces or interactions namely gravitational, electromagnetic, weak and strong
interactions, found in nature.
The more information about nuclear force can be obtained by considering two body problem.
The only bound system of two nucleons is the deuteron which consists of one proton and one
neutron. This system is weakly bound and no excited states exist. Apart of the study of
deuteron, other way to gather information about nuclear force is the study of neutron proton
(n-p) and proton proton (p-p) scattering. In this unit, we will basically study the two problem
like deuteron and scattering of one particle by the other like n-p and p-p scattering at low and
high energies. Some other aspects like, exchange nuclear forces and Yukawa theory of
nuclear forces will also be discussed in the present study.
2.1 objectives
The main aim of this unit is to study about the nuclear forces, their types and properties and
the study of two body problem like neutron proton bound state viz deuteron and its basic
properties like binding energy, its size, spin, magnetic and quadrupole moments etc. This unit
also gives the description of n-p scattering at low energies, scattering range and analysis of
scattering cross section through partial waves. After going through this unit you should be
able to:
Understand the various properties of nuclear forces and 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 of the results of low energy n-p and p-p scattering.
Know the spin dependence and scattering length.
Learn about the exchange forces between nucleons.
2.2 Basic properties of Deuteron
Nuclei are quantum mechanical systems composed of nucleons. These nucleons are hold
together by nuclear forces. The simplest case in which nuclear force is effective is when there
are only two nucleons present and interacting. Of the three possible bound states of two
nucleon system, di-neutron (nn), di-proton (pp) and deuteron (np), di-neutron and di-proton
bound states do not exist in nature due to Pauli exclusion principle and so only bound state of
two nucleons existing in nature is deuteron. The experimentally measured properties of
deuteron are:
(i) Charge: +e
(ii) Mass: 2.014735 amu = 3.34245 x 10-27
kg
(iii)Total angular momentum (called spin) = 1 (in unit of )
(iv) Parity (π) = positive
(v) Statistics: Bose Einstein statistics
(vi) Radius: rms value of deuteron radius is 2.1 fermi
(vii) Binding energy: 2.225 0.003 MeV (experimental value)
(viii) Electric quadrupole moment (Qd) = 2.82 x 10-31
m2 = 0.0028 barn
(ix) Magnetic dipole moment (μd) = 0.85736 0.0003
Among the above given properties, some of them like binding energy, magnetic dipole
moment and electric quadrupole moment are briefly explained below.
(a) Binding energy: The binding energy of deuteron can be determined from number of
experiments. One of these experiments allows slow neutrons to be captured by
protons in paraffin (containing hydrogen) and measuring the energy of emerging γ
rays. The resulting reaction called (np) capture reaction is given by
From the above reaction, binding energy comes out to be 2.225 MeV.
(b) Parity: The parity of a state describes the behavior of its wave function under
reflection of coordinate system through origin
Parity is positive for ground state of deuteron. This experimental fact can determine
the orbital angular momentum value L of deuteron ground state. To find it, we
separate the wave function of deuteron in three parts:
Wave function of deuteron ground state = Intrinsic wave function of proton (p) +
intrinsic wave function of neutron (n) + orbital wave
function for relative motion between p and n.
Parity (P) of deuteron ground state = (Parity of intrinsic wave function of p) ×
(Parity of intrinsic wave function of n) × (Parity of
orbital wave function for relative motion between p and
n)
Since intrinsic wave function of proton and neutron has same parity, hence product of
parity is positive for both intrinsic wave functions so that parity of deuteron is
determined by relative motion between two nucleons. For state of given angular
momentum L, the angular dependence of wave function is given by spherical
harmonics .
Thus parity of spherical harmonics = (-1)L. Since Parity of deuteron ground
state is positive, hence value of L should be even. Again ground state total angular
momentum (J) of deuteron is 1, so J=1where J = L + S. For two spin 1/2 nucleons S
may have value 0 or 1. For S = 0, any even value of L can not correspond to J=1,
hence S =1 for even value of L. Again J=1 is possible only for L =2 and L =0 because
L >2 will not result for J=1 when S =1. Hence L =0 and L=2 are only possible values
of orbital angular momentum of deuteron ground state.
(c) Electric Quadrupole moment: The observed value of Qd is 2.82 x 10-31
m2, which is
nonzero. However, if deuteron ground state is considered to be spherically symmetric
i.e. it has no dependence on and depends only on the relative coordinate „r‟ of
nucleons then Qd must be zero. This discrepancy tells that ground state is actually not
L=0 state but is a mixture of higher states.
Since deuteron is a bound state of n and p, so wave function need not to be anti-
symmetric. Wave function comprising of space part and spin part only is said to be
symmetric if space and spin both are symmetric and wave function is anti-symmetric
if either space part or spin part is anti-symmetric. Space part is symmetric for even
values of L (L = 0, 2, 4...) and anti-symmetric for odd values of L (L = 1, 3, 5…).
Spin part is symmetric for parallel spin of two nucleons (S=1) and anti-symmetric for
anti parallel spins (S = 0). Thus
Anti-symmetric wave function combination corresponds to identical systems like (n-
n) and (p-p) systems and symmetric wave function combination belongs to (n-p) or
deuteron. Again from parity consideration, the only possibility for ground state is S=1
and L=0, L=2, hence states are 3S1,
3D1 i.e. ground state is a mixture of
3S1 and
3D1
states. It comes out that deuteron spends 96% of its time in 3S1 state and 4% of its
time in 3D1 state.
(d) Magnetic dipole moment: The observed value of dipole moment of deuteron is
0.85736 0.0003 . Also, the experimental value of proton and neutron magnetic
moments are:
μp = 2.79281 0.00004
μn = -1.913148 0.000066
Sum of the two moments (μp + μn) = 0.879662 0.00005
If deuteron ground state is spherically symmetric (L=0) with parallel spins of nucleons
(S =1) then observed dipole moment of deuteron should be equal to the sum of two
nucleon magnetic moment value 0.879662 0.00005 . However the observed value
differs from the sum value by 0.0223 0.0002 and this difference is attributed to the
fact that ground state of deuteron is a mixture of 3S1 and
3D1 states and difference of the
above value is contributed by 3D1 state.
2.3 Existence of excited states of deuteron
Experimentally there is no evidence for the existence of excited states of deuteron. Apart of
ground state ( =0), no excited bound state for which orbital angular momentum 0, exists
for deuteron. To prove this fact mathematically, we start by writing Schrodinger equation for
the two body problem as
where , is the reduced mass of n-p system.
M: average nucleon mass
V(r): potential energy as a function of separation between neutron and proton
E: total energy of the system.
Using the spherical polar coordinates , the Schrodinger equation (1) takes the form
When the potential is spherically symmetric, above equation can be separated into angular
and radial parts. The radial part of the wave function is given by
The last term is called centrifugal potential.
Substituting u(r) = r ψ(r), we get
Here =0 substitution gives the ground state calculation. Now setting =1 for first excited
state, we get
Now putting E= - EB, the binding energy of deuteron in the p state ( =1) and using a square
well potential for r < r0 for the p state, above equation changes to
Putting
above equations may be written as
The least well depth, just required to produce this bound state, is the one for which the
binding energy EB is just equal to zero, i.e., when γ = 0 and =k0 (let).
If we put the wave equation become
The solution of equation (7b) is
Equation (7a) is solved by putting so that it reduces to
Differentiating this equation with respect to x and dividing by x throughout, we get
Now since , must vanish for x = 0, the solution of above equation comes out as
On integrating it, we get
To satisfy continuity condition at the boundary r = r0 or x= k0r0, these solutions yield,
or
The smallest positive root of this equation is . Hence a bound state of the deuteron
for l 0 can exist only if and this contradicts the fact that k0r0 must certainly be
less than π for all positive values of binding energy or to say for bound states. Therefore it is
concluded that no bound states exist for deuteron when l 0, i.e., deuteron does not possess
any excited state.
2.4 Neutron-proton (n-p) scattering at low energies
2.4.1 Scattering: When an intense and collimated beam of nucleons is bombarded on
target nuclei, the interaction between incident nucleus and target nuclei takes place. As a
result we may observe the following two possibilities:
(i) The interaction does not change the incident particles, i.e., incoming and outgoing
particles are 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 as 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.
Among the nucleon-nucleon scattering, neutron proton (n-p) scattering is the simplest one,
because here the complication due to coulomb forces is not present.
2.4.2 Partial wave analysis of (n-p) scattering: 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. 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 low energy range, most of the
measurements of scattering cross section are due to Melkonian and Rainwater et.al. A
beryllium target bombarded by deuterons accelerated in a cyclotron, provided the neutron
beam which was shot at a target containing free protons.
These results (Fig. 2) 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
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 system and 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
where M is the reduce mass of the n-p system. The scattered particles outside the interaction
region are represented by spherical waves eikr
/r radiated outward from centre of interaction.
The wave function in the asymptotic region (at large r) consists of (i) unscattered particles
and incident plane wave represented by eikz
(ii) spherical wave representing scattered particles
i.e.
where = scattering amplitude which measures the fraction of incident wave scattered in
the direction with polar angle θ.
For a central potential, the relative angular momentum l between the two nucleons is a
conserved quantity in the reaction. Under such condition, the wave function is expanded as
sum over the contributions from different partial waves,
where are the expansion coefficients. Only spherical harmonics with m=0
appears since in the absence of polarization, wave function is independent of azimuthal angle
.
The radial wave function for partial wave l satisfies the equation
Using modified radial wave function , above equation
simplifies to
For the short range potentials, V(r) goes to zero as r, hence in the asymptotic region
above equation changes to
At large r, the function takes the form
where constants Al, Bl or Cl ( ) are to be determined from boundary conditions.
2.4.2.1 Phase shift: The angle is known as phase shift introduced for the potential term.
The effect of potential on the scattering is seen by introducing phase shift in the asymptotic
form of wave function. Potential distorts the wave function in the potential region but at large
distances far away from the potential region, the wave is undistorted but with only change in
phase. A repulsive potential pulls out the wave function, so causes negative phase shift where
as attractive potential pulls in the wave function, introducing positive phase shift.
For a plane wave in the asymptotic region,
where asymptotically, Bessel function has the form
where . For a real potential only elastic scattering takes place. If potential is also
central one, orbital angular momentum l is a good quantum number and probability current
density in each l partial wave channel is conserved. Hence wave function due to the presence
of V (r) can only be changed by phase δ known as phase shift.
2.4.2.2 Elastic scattering cross section: The scattering wave function from equation (15) and
(17) can be written as
Since equation (14) and (20) represent the same wave functions, hence
where and k in the numerator is absorbed in . Using equation (19) and (21)
From the coefficient of we obtain,
Substituting this value back in equation (22) and comparing the coefficient of , we get
In terms of phase shift, differential scattering cross section is
and total elastic scattering cross section is
Using orthogonality properties of spherical harmonics
Equation (24) gives the total elastic cross section.
2.5 n-p scattering at low energies with square well potential
The theory for scattering cross section is a theory for phase shift which in turn depends on
the nature of scattering potential V(r). Using an attractive square well potential of depth V0
given as
The radial Schrodinger equation for l= 0 using equation (3) can be written as
Using equation (25a) and (25b) becomes
here ui and u0 are the wave functions inside and outside the well respectively.
Equation (26a) has solution
and equation (26b) has solution
this can be written as
where is the phase shift. At the boundary i.e. r = ro, the logarithmic derivative of solutions
(27a) and (27b) must be continuous viz.
which gives
It is assumed that inside the potential well, the logarithmic derivative of the inside
wave function for scattering can be approximated by the value of logarithmic derivative of
ground state wave function of deuteron viz. – γ where . This approximation holds
good as, inside the well, the n-p scattering energy and deuteron binding energy both are small
in comparison to well depth. The only difference is that here E is positive whereas deuteron
binding energy is negative. Hence, we can write
Again r0 is very small in comparison to , hence neglecting in the above
expression
From equation (24), the total elastic cross section for l = 0 is given by
Putting the value of and on solving, we finally get
This relation was first derived by E.P. Wigner which although agrees with experimental
results at higher energies but fails badly at low energies.
2.6 Spin dependence of n-p force: The total scattering cross section relation
derived above fails at low energies. To get an estimate of this, we calculate ζ0 under zero
energy approximation (E << EB)
where as experimental value is 20.36 barns as measured by Melkonion at low energies. This
discrepancy was sorted out by suggesting that inter nuclear forces are spin dependent.
Neutrons and protons are spin 1/2 particles, therefore in n-p system, their spins may be
parallel or anti parallel and possibly, the equation (28) holds good for parallel spins of
neutron and proton. The state of parallel spins is a spin triplet state and has statistical weight
3 corresponding to three allowed orientations of angular momentum vector under an external
magnetic field. The state of anti parallel spin is a singlet state and has statistical weight 1. In a
scattering experiment, in general neutron and proton spins are randomly oriented and hence
singlet and triplet states in the n-p system occur in proportion to their statistical weights. The
statistical weights for these states are 1/4 and 3/4 respectively. Therefore total scattering cross
section is considered to be composed of two parts, ζl,0- the cross section for scattering in
triplet state and ζs,0- the cross section for scattering in singlet state i.e.
Experimentally, the weighted average of the cross section is measured. To get an idea of
relative magnitude of and we take the triplet cross section value of 2.33 barn as
calculated above and experimental value of 20.36 barn and find the magnitude of singlet state
cross section as
or
which means is very large in comparison to for slow neutrons and contributes for
most of the scattering cross section.
2.7 Scattering length: The elastic scattering cross section is given by
For l=0 partial wave (S wave), elastic scattering cross section is
Now if k0, E 0 and , which means at E = 0 there is an infinite cross section.
Infinite cross section means particle will not scatter at all and there will be a bound state.
Physically, is an absurd solution, hence scattering cross section must be finite if E=0.
Therefore, a length parameter „a‟ is defined in a form
such that at
From equation (29) and (30)
Sign convention is such that for attractive potential, the scattering length is positive if a
bound state exists and scattering length is negative if there is no bound state. For a repulsive
potential, scattering length is always positive.
Scattering length „a‟ can also be defined using equation (27c). From this equation (27c)
taking B ,
In case of very low incident neutron energies or in zero energy approximation
Using equation (31)
In the limit of zero energy,
Equation (32) gives a
graphical representation of
scattering length and is
displayed in fig. 4. The
wave function is a straight
line which intercepts r axis
at r = a. Hence scattering
length can also be defined
as distance of point of
intersection of modified
radial function with r axis. Scattering length can be positive or negative depending on bound
or unbound state for attractive potential but is always positive for repulsive potential.
2.8 Effective range theory of n-p scattering: The variation of scattering
cross section with energy depends upon two parameters, one the scattering length „a‟ and
other the effective range „re‟, which has dimension of length. This effective range theory
expresses the phase shift as a function of energy. To derive the expression for effective range
re of nuclear potential, we proceed by writing the radial wave function for partial wave l.
From equation (16)
for l = 0 or S wave, above equation changes to
For two states of energies and , let the two solutions are and
respectively, then they satisfy the differential equations
Multiplying equation (34a) by and equation (34b) by and integrating the
difference over variable „r‟, we get
Integrating first integral, by parts
Cancelling the two integrals in L.H.S., we are left with
Above equation (35) is potential independent and holds good for any potential including V(r)
= 0. Let another wave function satisfies equation (33) when V(r) = 0, then
Proceeding the same steps as for , we arrive at a relation similar to equation (35)
If the range of the potential is short in equation (33) then potential vanishes as r , so that
equation (33) and equation (36) are identical to each other in the asymptotic region. Hence
the solutions must also have same form at r = i.e.
where A is the amplitude to be determined. Using
we write the L.H.S. of equation (35) in terms of viz.
(40)
Subtracting equation (35) from (37) and using equations (39) and (40), we get
We fix to get the normalization constant A i.e. from equation (38)
Substituting equation (42) and the condition in equation (41), we get
From equation (42)
Hence equation (43) reduces to
Now considering the special case of k10 i.e. for zero energy neutrons
Hence, equation (44) becomes
In the low energy region up to 10MeV, V(r) >>E, so that k2 and k1, which correspond to
energy E may be considered to be equal to some value k, hence
where is defined as effective range. Hence
In terms of re, the scattering cross section is given by
Thus scattering cross section ζ not only depends on energy E through k but depends on
scattering length „a‟ and effective range „re‟ as well. Therefore ζ can be determined by
measuring a and re at various low energies. It is noted that scattering cross section is
independent of shape of potential, hence also named shape independent approximation.
2.9 Low energy proton-proton (p-p) scattering: In proton-proton
scattering, in addition to nuclear forces, coulomb repulsive forces are also present. At low
energies (<100 KeV), the coulombian repulsion prevents any close contact of nucleons so
that nuclear force does not operate. Hence, p-p scattering is predominantly due to Coulomb
repulsive forces. As energy increases, nucleons approach closer and closer so that beside the
coulomb repulsion, nuclear forces also become effective and hence interaction potential
consists of nuclear plus coulomb potential.
Also in p-p scattering,
incident particle and target
or scatterer and scattered
both particles are identical
(fig. 5). Hence, wave
function describing p-p
system should be anti-
symmetric with respect to
interchange of protons
because of the Pauli
Exclusion Principle as they
are fermions. This means if spatial part of the wave function is symmetric then spin part
should be anti-symmetric and vice-versa. At low incident energies only l=0 partial wave (S
wave) contributes to scattering process, hence spatial part is symmetric and for total wave
function to be anti-symmetric, wave function remains in singlet (S= 0) state. Therefore, the
information about nuclear interaction is limited to singlet state only.
As energy increases, higher l partial waves also contribute to scattering process, hence triplet
state (S=1) also contributes for nuclear interaction fulfilling the exclusion principle
requirement for total wave function to be anti-symmetric.
2.9.1 p-p scattering cross section: The p-p scattering cross section is related to
scattering amplitude f(θ) through the relation . Since detector cannot
make a distinction between scatterer or scattered particle as particles are identical as shown in
fig. 5, the scattering cross section is given by
but quantum mechanically the waves of incident particles interfere. The interference will
mostly be destructive because Coulomb forces are repulsive. The interference terms are
different for scattering in singlet and triplet states. Due to this interference, the quantum
mechanical differential cross section is obtained by
where positive sign refers to singlet state and negative sign refers to triplet state. At very low
energies, the scattering is mostly due to coulomb interaction because protons do not reach
sufficiently close to feel nuclear forces. The pure Coulomb scattering cross section calculated
by Rutherford is
where θ is the scattering angle in centre of mass (C.M.) system and M is the proton mass. The
Rutherford formula for p-p scattering in C.M. system is (Z1=Z2=1 for protons)
If E0 is the incident proton energy and , then Rutherford formula in Lab system,
reduces to
where is the angle in Lab system. The term is not present in equation (48) but it is
added in equation (50) because each proton scattered through an angle in the Lab system is
accompanied by a recoil proton at an angle . However, equation (50) does not
reproduce correct p-p scattering at low energies because of neglecting the symmetry
consideration of quantum mechanics.
Mott calculated the quantum mechanical expression for p-p scattering cross section taking
symmetry into account. When unpolarized beam of protons were scattered from unpolarized
targets then singlet state occurs one quarter of time and triplet state occurs three quarter of
time, therefore
This expression is due to Mott. Now is calculated as
where and v is the relative velocity of two particles. Using equation (52),
equation (51) can be written as
This equation remains unchanged if θ is replaced by π-θ, hence explains the
indistinguishability of particles scattered by angle θ or π-θ. On comparing equation (49) and
(53), we see that classical scattering formula differs from equation (53) by the presence of
third term. This is called quantum mechanical interference term and arises from the identity
of incident and target particles. For non identical or distinguishable particles, this term
vanishes. The negative sign in the third term shows that protons are fermions.
2.9.2 Effect of nuclear potential: Since nuclear forces are charge independent, they do
not differentiate between n-p and p-p scattering. Hence in p-p scattering the scattering
amplitude due to nuclear potential should have same expression as that in the n-p scattering.
Thus, presence of Coulomb interaction and nuclear force modify the scattering amplitude as
In case of p-p scattering, the expression for Coulomb scattering is corrected by the
requirement of anti-symmetry under space spin exchange.
For (S=0) singlet state, there is a symmetric combination given by
& for (S=1) triplet state, there is an anti-symmetric combination
Again it is noticed that nuclear part of p-p scattering is present in singlet (S=0) 1S0 state only
(presence in the triplet state is ruled out due to Pauli Exclusion Principle) whereas coulomb
scattering comes from all angular momentum states. Thus for singlet state, both coulomb and
nuclear scattering is considered while for triplet state only Coulomb scattering is considered.
Now
And
The singlet and triplet scattering adds incoherently, therefore
Using equation (55), (56) and (57), we finally write the differential scattering cross section
formula as
The first bracket of this equation is just the Coulomb scattering term given by Mott in
equation (53), second bracket gives the interference term between the coulomb and S-wave
nuclear interaction and third term of the equation is specific nuclear scattering. This equation
reduces to Mott cross section formula when phase shift is set to zero. Thus p-p scattering
is satisfactorily explained by the above equation.
2.10 Exchange forces: Meson theory of nuclear force: From the study
of deuteron, it is known that nuclear force depends on whether the spin of deuteron system is
0 or 1 and whether the orbital angular momentum l is even or odd but why should the nuclear
forces depend on these? The answer is provided by the famous Yukawa theory of exchange
of mesons by nucleons. This meson exchange leads to exchange forces.
In 1935, the Japanese physicist, Hideki Yukawa, conceived the idea of treating the nuclear
force originating from the exchange of a particle with non zero rest mass. Yukawa postulated
that each nucleon is surrounded by a meson field. As a result, it continuously emits and
absorbs particles called mesons. This exchange of mesons gives rise to nuclear force.
The emission of meson should reduce the mass of nucleon but it is not observed. Thus, the
exchange of meson must take place in such a short time that uncertainty in energy is
consistent with Heisenberg uncertainty principle This makes the detection of
meson not possible and hence mesons in this exchange process are referred to as virtual
mesons.
2.10.1 Estimation of mass of meson through Heisenberg uncertainty
principle: If a nucleon emits a virtual meson π of rest mass Mπ, it loses an amount of
energy,
For example if a proton emits a π0 meson of rest mass Mπ, we have
Clearly the proton mass must decrease but it is not observed in careful experiments. The rest
mass of proton is constant and always found to be 1836 times rest mass of an electron. This
„violation‟ of conservation of energy is allowed if emitted meson is again absorbed in a short
time Δt such that The complete reaction would be then,
Thus emission and absorption of a particle of rest mass energy is allowed
through uncertainty principle and consistent with conservation of energy if the particle
encounters the same or other nucleon within a time Δt given by
If it is assumed that meson travels at maximum possible speed viz. speed of light then the
maximum distance it can cover in the time is
From this equation, one can estimate the rest mass of meson by putting the
experimentally observed value of R. If we put R=1.4 fermi, we get
In terms of electron mass me,
The theoretically predicted mass of the meson of 270 me with short range of nuclear force of
1.4 Fermi agrees well with experimental results. A particle of mass about 206 me called μ
meson was discovered in cosmic rays just after one year of Yukawa prediction and about 12
years later, π mesons (mass of 273 m0) were discovered using high energy accelerators at
Berkeley, University of California.
2.10.2 Yukawa potential: Yukawa potential has been found to be successful in the
deuteron problem and also in understanding the low energy nucleon scattering data. To derive
its form, we start from writing the relativistic relationship between the total energy E,
momentum p and the rest mass m0 of a particle as
Replacing E by the quantum mechanical operator and p by , we obtain
Introducing a potential function (r,t) and writing the above equation as
This is known as Klein Gordan equation for a spinless relativistic particle. For π mesons, this
equation changes to
The time independent part of above equation is obtained by substituting or
setting , which gives
or
where K . Equation (61) valid in case of a meson field is an analogue of
Laplace‟s equation for electromagnetic field which is valid in the absence of electric charges.
In the presence of charges, we require an analogue of Poisson‟s equation viz.
where e is the electronic charge and ρ is the electron particle density. The analogue of
Poisson‟s equation is
where „g‟ is the nucleon charge which measures the strength of interaction between nucleon
and meson field. If the only nucleon present is the point nucleon present at origin (r = 0), then
is replaced by where is the Dirac delta function. Hence,
The solution of equation (62) is
where „g‟ is an undetermined constant analogue to charge e in electromagnetic theory. The
interaction potential V (r) of a nucleon of charge g with the meson field is then
or
This is the required Yukawa potential and is shown below in fig. 6.
2.10.3 Exchange forces: Scattering experiments indicate that nuclear forces are strong
and short range and that they depend not only on „r‟ the separation between the two nucleons
but also on the orientation of „r‟ in space and the spins of nucleons. If r1, ζ1 are space and
spin coordinate of first nucleon and r2, ζ2 are space and spin coordinate of second nucleon
then wave function for nucleon pair is given by ψ(r1 ζ1, r2 ζ2). There are three possibilities
for exchange of space and spin coordinates to one another-
(i) exchange of position vectors r1 and r2 of the two nucleons (space exchange) only
(ii) exchange of spin coordinates ζ1 and ζ2 of the two nucleons (spin exchange) only
(iii) exchange of space as well as spin coordinates of the two nucleons (space spin exchange)
These give rise to following types of exchange forces-
(A) Majorana force: An exchange of coordinates changes the sign of wave function ψ if
parity is odd and leaves ψ unchanged if parity is even. The odd or even parity is
determined by the orbital angular momentum L. Thus if Majorana potential is denoted
by VM after operating on ψ we have
Here α (r) is a function of r alone.
(B) Barlett force: If the spins of two nucleons are parallel (S=1), then interchange of
spins does not affect the wave function while for S=0 or anti parallel spins, it changes
the sign of wave function. Thus if Barlett potential is denoted by VB after operating on
ψ we have
Here is a function of r alone.
(C) Heisenberg force: The third type of force gives the combined effect of space and spin
coordinate exchange. Thus if Heisenberg potential is denoted by VH after operating on
ψ we have
Here γ(r) is a function of r alone. In terms of isobaric spin T, the above condition can be
written as
(D) Wigner force: In addition to three exchange forces, there can be an ordinary or no
exchange force called Wigner force. It can be written as
Thus the resultant nuclear potential can be written as
The meson theory provides a physical basis for these exchange potentials. All these potentials
depend on r and so are central in character. To take into account the non-central character or
tensor behavior of nuclear force, a term VT is added to above equation and then the resultant
nuclear potential is given by
The spin dependence of nuclear forces can be visualized by the study of n-p scattering cross
sections in singlet and triplet spin states.
2.11 Summary
Now we recall what we have discussed so far:
We have learnt the basic properties of deuteron, 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 deuteron
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
deuteron suggests that at least a part of the neutron proton force acting in deuteron is
non-central.
For low energy incident neutrons and proton targets, elastic scattering cross section is
calculated and is given by where is the phase shift
showing the effect of potential on scattering.
Scattering length „a‟ is a length parameter defined as such
that in the zero energy approximation, the scattering cross section does not become
infinite and attains a finite value.
p-p scattering is different from n-p scattering in that here apart of nuclear forces,
Coulomb repulsive forces are also present. At low energies (<100 KeV), the
coulombian repulsion prevents any close contact of nucleons so that nuclear force
does not operate. Hence, p-p scattering is predominantly due to Coulomb repulsive
force at low energies.
As incident proton energy increases, nuclear forces also become effective and hence
interaction potential consists of nuclear plus coulomb potential. This results in the
inclusion of nuclear term and interference term apart from coulomb term in scattering
cross section.
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.
Yukawa theory of exchange forces explains the interaction between nucleons. The
nucleons exchange mesons and interact among themselves. These forces include the
exchange of spin and/or position coordinates of two nucleons involved in the
interaction.
2.12 References
Nuclear Interactions; Sergio De Benedetti, John Wiley & Sons.
Theory of Nuclear Structure; M.K. Pal, East West Press Pvt. Ltd.
Elements of Nuclear Physics; M.L. Pandya & R.P.S. Yadav, Kedar Nath Ram Nath
Publication.
Concepts of Nuclear Physics; B.L. Cohen, Tata McGraw Hill Publishing Ltd.
Nuclear Physics; V. Devanathan, Narosa Publishing House.
Nuclear Physics; S.B. Patel, New Age International (P) Ltd.
Introductory Nuclear Physics; S.M. Wong, Prentice Hall of India Ltd.
2.13 Questions
1. Write the basic properties of deuteron.
2. Show that ground state of deuteron is a mixture of L=0 and L=2 angular momentum states,
considering the parity property.
3. 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.
4. Show that there exist no excited states of deuteron. Assume square well potential for
solution.
5. Explain clearly how the properties of the deuteron indicate the presence of spin dependent
force or tensor force between two nucleons.
6. Discuss n-p scattering at low energies. How it explains the nature of nuclear force?
7. What are phase shifts? Obtain the elastic scattering cross section in case of n-p scattering
through partial wave analysis.
8. Write short note on scattering length.
9. Differentiate between n-p and p-p scattering.
10. Obtain the scattering cross section in terms of scattering length „a‟ and effective range
„re‟.
11. Estimate the mass of meson through Heisenberg uncertainty principle.
12. Discuss the Yukawa theory of nuclear forces and obtain the Yukawa potential.
13. Write short note on various types of exchange forces.
Unit III – Nuclear Models and Nuclear
Reactions
3.0 Introduction
The interaction between two nucleons has been studied on the basis of two body system as
deuteron and scattering theory but the results cannot be applied to many body problem.
Hence, to study the complex interrelationship between the nucleons, a number of nuclear
models have been developed on the basis of which, nuclear structure and nuclear interactions
are studied. Although none of the proposed models explain all the observed nuclear
properties satisfactorily, the nuclear shell model has been quite successful in predicting the
static properties of the ground state as well as indicating the broad structure of nuclear level
spectrum. Of the various models suggested, the important ones are: Fermi Gas model, Liquid
Drop model, Shell model, Collective model and Optical model. Out of these models, we will
study about Liquid drop model, Shell model and Collective model in this unit.
The other topic which is to be covered in this unit is nuclear reactions. Nuclear reactions are
the processes in which incident and emergent particles are not same but are different. Thus
nuclear reaction is a process in which a change in the composition or energy of target nucleus
is brought through bombardment with a nuclear projectile. As the energy of the projectile
increases, the type and variety of possible type of nuclear reactions also increases. For
example, up to 10 MeV of projectile energy, only one particle is emitted while if incident
particle energy is increased up to 20-30 MeV, two particles instead of one are emitted. If
projectile energy is increased up to 200 MeV which is threshold energy for pion production,
then emission of π mesons takes place. At low incident energies, nuclear reaction may
proceed via an unstable intermediate state called compound nucleus state. In this case, the
incident particle loses so much energy that it cannot escape the nucleus. In the present unit,
we will study about direct and compound nuclear reactions, will calculate the cross sections
of nuclear reactions and finally discuss the Breit Weigner formula for resonances.
3.1 objectives
The present unit involves the study of nuclear models and nuclear reactions. After going
through this unit, you will become familiar with:
Liquid drop model and its role for explaining the theory of nuclear fission
Nuclear shell model and its predictions for ground state properties of nuclei like spin
and parity, magnetic moment and quadrupole moment etc.
The collective model of nucleus which comprises of vibrational and rotational states.
Various types of nuclear reactions, disintegration energy Q and its standard form.
Theory of compound nucleus and concept of level width.
Calculation of elastic and reaction cross sections along with Breit Wigner formula for
isolated resonances.
3.2 Liquid Drop Model: The liquid drop model was proposed by N. Bohr and
Kalcker which provided the reasonable explanation of many nuclear phenomena, not
explained on the basis of other nuclear models. Some of these important phenomena are
given below.
3.2.1 Phenomena explained by liquid drop model:
(1) The constant density of nuclei with radius, just as density of liquid drop is independent of
the size of liquid drop.
(2) Systematic dependence of neutron excess (N-Z) on A5/3
for stable nuclei.
(3) The approximate constant value of binding energy per nucleon (B/A) which is analogues
to latent heat of vaporization
(4) Fission by thermal neutrons of U235
and other odd N nuclides
(5) Systematic variation of decay energies with N and Z
Bohr suggested that the properties of the nucleus can very well be compared with that of
liquid in which the molecules of liquid drop correspond to nucleons in the nucleus. The
similarities between liquid drop and nucleus are given below.
3.2.2 Similarities between nucleus and liquid drop:
(1) The density of liquid is almost independent of its size, so that radius R of the liquid is
proportional to cube root of number A of molecules in the drop.
(2) Both the molecules of the liquid drop and nucleons in the nucleus interact only with their
immediate neighbours.
(3) The energy necessary to completely evaporate the liquid drop into its molecules is
approximately proportional to mass number A just as the binding energy of nucleus is
proportional to mass number A.
Attempts were made to describe the nuclear dynamics in terms of motion of liquid drop. The
most important motions are surface vibrations. A deformation of spherical liquid drop gives
rise to periodic oscillations of the surface.
3.2.3 Surface vibrations of liquid drop: Let there exists a spherical drop of radius R.
Any deformation of its surface can be described by a function ) which is the distance
of deformed surface from centre.
The difference of the two radii is given by
can be expressed in terms of spherical harmonics as
Taking the cylindrically symmetric deformation for which m = 0, so that
The restoring force is supplied by the surface tension which opposes the deformation of
surface. The surface energy is
where α is coefficient of surface tension. In Weizsacker-Bethe mass formula, the surface
energy term is given as
Equating the two values of Es, we get
Rayleigh gave the formula for frequency as
where μ is the mass of liquid drop. Taking μ = MA, where M is the mass of single nucleon
and putting the value of α, we get
This gives too high value of excitation energy. The frequency ω is reduced by Coulomb
effect as
where γ is the ratio between the coulomb energy and surface energy
. This equation leads to somewhat smaller frequencies for heavier nuclei.
However even this correction is not able to explain the low lying energy levels. The liquid
drop model is however able to explain the stability of nuclei against the breakup into two
fragments i.e. nuclear fission, explained later.
3.3 Energy release in symmetric fission
If a nucleus (A,Z) breaks into two equal halves (A/2,Z/2), then the fission is called
symmetric fission. The energy release in this symmetric fission is
From the semi empirical mass formula, binding energy for nucleus (A,Z) is
and binding energy for two fission fragments (A/2,Z/2) is
From the mass formula
and using equation (7), (8) and (9) we finally get
Pairing energy difference is small and can be neglected. Taking =13MeV and =0.6 MeV
From this equation Q > 0 for
Thus for all nuclei having Z > 35, A > 80, fission is possible and will release energy.
However the slow neutron fission does not take place even with many of heavy nuclei. This
discrepancy was explained by Bohr and Wheeler by considering the Coulomb potential
barrier of the two fragments at the instant of separation.
3.3.1 Potential barrier for symmetric fission: The existence of potential barrier
prevents the immediate breaking of two fission fragments. Let the height of potential barrier
is Ec then the nucleus will be unstable and will break in two parts if Q > Ec. The barrier height
of coulomb potential between the two symmetric fragments when they just touch each other,
is
Thus the condition for stability is or
may have value 50 for a nucleus of mass number 250, hence nuclei (A>250) would be
too unstable to exist.
3.3.2 Nuclear fission and deformation of liquid drop: The fission process can be
explained with the help of liquid drop model as shown in fig. 1. A highly energetic compound
nucleus is formed when an incident neutron combines with the nucleus. This energy starts a
series of rapid oscillations in the drop which distorts the spherical shape of liquid drop and
drop becomes ellipsoidal in shape. The surface tension force makes the drop to return to its
original shape while the excitation energy tends to distort the shape further. For sufficiently
large excitation energy, the drop may acquire the dumb bell shape. Again if oscillations are
violent enough to produce the critical stage (stage IV) then nucleus cannot return to first stage
and finally fission takes place. The energy required to produce fourth stage is called threshold
energy or critical energy.
The potential energy curve for fission is
plotted in fig 2. On X axis, is the distance
r which is the separation of centers of
two fission fragments. The curve is
divided in three regions. In region I
fragments are completely separated and
their potential energy is the electrostatic
Coulomb energy resulting from their
mutual repulsion. When distance r = 2R, the drops just touch each other. The energy E at this
point is less than the corresponding Coulomb energy by an amount equal to CD. In region II
when we reach the critical distance rc, the potential energy curve has its maximum value
corresponding to the barrier height and explains why fission does not take place when Q>0.
A critical energy called threshold or activation energy given by Ea =Ec- Q is required to
overcome the potential barrier and fission to take place. In region III, the fragments coalesce
and short range attractive forces become dominant.
For the mathematical analysis, drop is considered incompressible, the volume of sphere of
radius R is same as that of ellipsoid of semi major axis „a‟ and semi minor axis „b‟ i.e.
If denotes the eccentricity then,
Then surface energy of ellipsoid is, Es= (area of ellipsoid) × surface tension
And the Coulomb energy of ellipsoid is
Taking the terms up to ε2 only, the stability condition against spontaneous fission is
This equation gives much more experimentally realistic stability limit against the
spontaneous fission.
3.4 Nuclear Shell Model: The liquid drop model only discusses the properties of
nuclear matter but does not tell anything about single nucleon. For certain numbers of
neutrons and protons called magic numbers, nuclei exhibit special stability. This stability is
not explained by liquid drop model. Also the other properties of nucleons like spin, magnetic
moment and quadrupole moments are unexplained in liquid drop model.
There are several evidences for the existence of magic numbers. Nuclei 2He4 (neutron =2,
proton =2) and 8O16
(n=8, p=8) are particularly stable. Isotopic abundance of Sr88
(n=50),
Ba138
(n=82) and Ce140
(n=82) are exceeding 60%. Sn (Z=50) has ten stable isotopes whereas
Ca (Z=20) has six stable isotopes. The doubly magic nuclei 2He4, 8O
16, 20Ca
40 and 82Pb
208
(n=126, p=82) are particularly tightly bound. All these examples show that nuclei having
neutron or proton number equal to 2,8,20, 50, 82, 126 known as magic numbers are more
than usual stable. The magic numbers and other properties of nuclei are explained by nuclear
shell model.
In the single particle shell model, it is assumed that nucleons move independently in a
common (mean) potential determined by the average motion of all other nucleons. Most of
the nucleons are paired. A pair of nucleons contributes zero spin and magnetic moment. The
paired nucleons form an inert core. The properties of odd A nuclei are determined by
unpaired nucleon and of odd-odd nuclei by unpaired proton and neutron. The initial magic
numbers are reproduced by considering the harmonic oscillator potential and rest are
produced by spin orbit interaction.
Let a particle moves in a spherically symmetric central field of force. The eigen states
available to nucleon of mass „m‟ moving in spherically symmetric potential are determined
by solution of the Schrodinger equation
where E is the energy eigen value. We find the solution in spherical polar coordinates. Since
potential is spherically symmetric, we have
The radial function R(r) satisfies the usual radial Schrodinger equation
where and using
We define where „b‟ has dimension of length called oscillation parameter
and a dimensionless radial constant . Writing equation (14) in terms of ρ,
The solution of above equation is given by
Substituting R(r) in equation (15), we get
The solution of this equation is,
This solution satisfies R(ρ) in equation (16) when ρ only when
n,l are two new quantum numbers. is called total oscillator quantum number and is given by
. Corresponding to , quantum numbers (n,l) can acquire any integer value. The
following table 1 gives various energy levels in order of increasing energy. A given n,l state
has (2l+1) degenerate sub states corresponding to ml = -l…+l. Each such (nlml) substate can
accommodate two nucleons of each kind corresponding to two possible alignment of spins
(ms=1/2). Thus by Pauli principle, there are 2(2l+1) nucleons of each kind to go to each n,l
state.
Table 1: Shell closure of isotropic 3D Harmonic Oscillator
(n l) Nucleons needed to fill the
shell
Total nucleons at shell closure
0 (0 0) 2 2
1 (0 1) 6 8
2 (1 0) (0 2) 12 20
3 (1 1) (0 3) 20 40
4 (2 0) (1 2) (0 4) 30 70
5 (2 1) (1 3) (0 5) 42 112
6 (3 0) (2 2) (1 4) (0 6) 56 168
Here only first three numbers 2,8,20 corresponding to He4, O
16, and Ca
40 agree with known
magic numbers. The rest numbers differ from observed magic numbers 50, 82, 126. To
explain these higher magic numbers, Mayer and Haxel in 1949 suggested that a non central
term must be added to the force acting on nucleons in the nucleus. This is the interaction
between orbital angular momentum and intrinsic angular momentum (spin) of the particle.
The magnetic moment μs is associated with the spin of the particle while magnetic field B is
induced by orbital motion. The interaction energy is
or potential function
This potential causes a splitting of j = l ½ levels. The energy of the single particle state
is expressed as
Here is the radial integral
. The state
is pushed up due to its
positive sign and state is
pushed down due to its negative sign
in the energy expression. The energy
splitting of two components of given
„l‟ is found to be
Fig.3 shows the energy level scheme
obtained by shell model. The lowest
shell 1s1/2 contains two nucleons
making the shell closure at 2. The next shell closure occurs when all six p states are filled, at
2+6=8. The next shell closure occurs at 8+12 =20 where 12 nucleons are filled 1d5/2, 1d3/2
and 2s1/2 states forming a single shell. For the state 1f7/2, l=3 value is high enough to lower
the state below all other N=3 states but does not form the group with N=2 shell. Hence next
shell closure occurs at 20+8=28. The next shell contains 1f5/2, 2p3/2, 2p1/2 and 1g9/2 states. The
last one comes down from N=4 level to N=3 due to spin orbit coupling. Thus shell closure
occurs at 28+22=50. The next shell consists of 32 sublevels of 1g7/2, 2d5/2, 2d3/2, 3s1/2 and
1h11/2, thus forming the closure at 50+32=82. The next shell contains 44 sub levels of 1h9/2,
2f7/2, 2f5/2, 3p3/2, 3p1/2 and 1f13/2 and closes at 82+44=126. Thus shell closure occurs at magic
numbers 2,8,20,28,50,82,126 as required by experiments.
3.4.1 Intruder states: With increasing l, increases such that 1g9/2 level from 4+
oscillator shell comes down to 3- shell and couples to 1f and 2p levels so that shell closure
occurs at N=50 not on 40. Similar is the case with 1h11/2 level. These opposite parity states
which couple to states of one group of same parity are called intruder states. Due to this, the
individuality of oscillator quantum number does not remain anymore and spin orbit
interaction mixes the states of different shells, thus making shell closure at magic numbers.
3.4.2 Predictions of shell model:
(a) Spin and parity of ground state: There are following rules for the angular momentum
and parity of ground state:
(i) The even even nuclei has total ground state angular momentum
(ii) With an odd number of neutrons or protons, resulting orbital angular momentum and spin
direction are just of that single odd particle. Parity of ground state will be the parity of odd
particle state, given by (-1)l. Rest of nucleons pair off as far as possible.
(iii) An odd odd nucleus will have total angular momentum which is the vector sum of odd
neutron and odd proton levels. L.W. Nordheim proposed the following rules to calculate the
total angular momentum of ground state for odd odd nucleus:
If for two odd nucleons, j1+ j2 + l1+ l2 is an even number, then and parity of the
state and if j1+ j2 + l1+ l2 is an odd number, then and parity of
the state . Here j and l are total angular momentum quantum number and
orbital angular momentum quantum number of odd single particle state.
Example 1: Calculate the ground state angular momentum and parity for 6C12
and 5B11
, as
predicted by shell model.
Sol. In 6C12
, there are 6 protons and 6 neutrons which are even numbers, hence by (a)(i)
prediction, Iπ=0
+. In 5B
11, there are 5 protons and 6 neutrons. We will concentrate only odd
number of particles i.e. protons. These 5 protons are filled in shell model scheme as (1s1/2)2,
(1p3/2)3. Thus last proton is filled in 1p3/2 level which means, ground state angular momentum
is I=3/2 and parity of ground state = (-1)l = (-1)
1 = negative, i.e. .
Exceptions: Exceptions to the above rule are found among heavy nuclei for 33As75
(Z=33),
28Ni61
(N=33) and 81Tl207
(Z=81). For Z=33 or N=33, shell model predicts the filling of
energy levels as (1s1/2)2 (1p3/2)
4 (1p1/2)
2 (1d5/2)
6 (2s1/2)
2 (1d3/2)
4 (1f7/2)
8 (2p3/2)
4 (1f5/2)
1 so that
ground state spin of the nuclei is predicted as 5/2 corresponding to odd nucleon occupied
orbit but the experimentally observed value of spin is 3/2. This discrepancy was solved by
modifying the above rule as if high spin shell comes after low spin shell, then high spin shell
fills faster than low spin shell pairing its particles before low spin shell can be filled. Thus
modified rule fills the levels as (1s1/2)2 (1p3/2)
4 (1p1/2)
2 (1d5/2)
6 (2s1/2)
2 (1d3/2)
4 (1f7/2)
8 (2p3/2)
3
(1f5/2)2 so that ground state spin is 3/2 which agrees with the experiment.
(b) Magnetic moments of nuclei: In an odd nucleus, the total angular momentum I of the
nucleus is equal to angular momentum j of the last unpaired nucleon. Thus magnetic moment
of the nucleus is determined by odd nucleon only. We have studied about magnetic moments
of nuclei in Unit I. Thus for odd proton and odd neutron nuclei, magnetic moments are
Odd proton μ = I + 2.29 for I =l+1/2
μ = I - 2.29 I/(I+1) for I =l- 1/2
Odd neutron μ = -1.91 for I =l+1/2
μ = 1.91 I/(I+1) for I =l-1/2
(c) Electric quadrupole moments of nuclei:
(i) For odd proton nuclei: For protons, the electric quadrupole moment operator is given by
where is the renormalized spherical harmonics. If the
state of last proton is given by then quadrupole moment of odd proton nucleus
is given by expectation value of this operator for maximum projection m = j i.e.
where rrms is mean square radius of the orbit. The value of radial integral is positive definite
calculated for harmonic oscillator potential. Hence q0 for one particle state is negative. For
uniform charge distribution, where Coulomb radius Rc =1.2×A1/3
fermi.
(ii) For odd neutron nuclei: The quadrupole moment of odd neutron nuclei is non zero and
found to be negative quantity. This anomaly was resolved by considering the core
contribution for odd neutron nuclei. The protons in the core give their contribution for the
observed magnetic moment of odd neutron nuclei due to the interaction of odd neutron with
the proton. The core contribution to the quadrupole moment is due to the effect of core
polarization.
Q. 1 What angular momentum (I) and parities (π) are predicted by shell model for the ground
state of 8O17
30Zn67
and 7N16
?
Q. 2 Calculate the magnetic dipole moment and quadrupole moment of 8O17
and 16S33
nuclei
according to shell model.
3.5 Collective nuclear model: J. Rainwater in 1950 suggested that the
shortcomings of nuclear shell model like the deviation of magnetic moments and quadrupole
moments from observed values and improper description of excited states of nuclei can be
overcome in odd A nuclei by considering the polarization of the even-even core by the
motion of odd nucleon. The nuclear core consisting of even number of nucleons thus has
spheroidal shape instead of spherical shape. The idea of deformed nuclear core was suggested
by Bohr and Mottelson. The entire shell configuration undergoes periodic oscillations in
shape which affects the individual particle orbits because it changes the potential of the
region in which these particles move.
The collective motion of nucleons in the core and motion of loosely bound surface nucleons
results in rotational, vibrational and nucleonic energy states in the nuclei. The rotational and
vibrational states arise due to the motion of nuclear core and nucleonic states arise from the
motion of loosely bound nucleons.
3.5.1 Vibrational states: Before discussing the vibrational states, we briefly outline the
parameters used to describe the deformed surfaces.
A moving nuclear surface may be described by an expansion in spherical harmonics with
time dependent shape parameters
Here denotes the nuclear radius in direction ( ) at any time t and R0 is the radius
of spherical nucleus. are time dependent amplitudes describing the vibrations of
nucleus and thus serving as collective coordinates. The various vibrational modes are
characterized by the value of as shown in fig 4. i.e.
=0: corresponds to monopole deformation which involves variation in size without
changing overall shape of the
nucleus.
=1: describes dipole
deformation. As an isoscalar
dipole, nucleus moves as a
whole and in isovector dipole
deformation, neutrons and
protons oscillate in opposite
phase.
=2: gives information about
quadrupole deformation.
Here nucleus changes shape
from prolate to oblate shape
and vice versa.
=3: describes octupole deformation. Here elongations along all the three axes are unequal
and nucleus changes from spherical to spheroidal shape.
For the case of pure quadrupole deformation in lab frame, the nuclear surface is given by
The parameters are called deformation parameters and define the stretching or
contraction of the nucleus in appropriate direction. The parameter α20 describes the stretching
of z axis with respect to x and y axis. α22 describes the relative length of x axis compared to
y axis as well as oblique deformation in xy plane. α21 indicate an oblique deformation of z
axis in lab frame.
3.5.1.1 Vibrational frequency for multipole λ: In a nucleus, the shape change involves
kinetic energy in transporting nucleons from one location to another and potential energy, in
forcing the nucleus away from its equilibrium shape. For small amplitude vibrations, kinetic
energy may be expressed in terms of rate of change of shape parameters i.e.
where coefficient D plays the same role as mass in kinetic energy expression (K= mv2).
For a classical vibrational flow, D is related to mass density ρ and equilibrium radius R0 of
nucleus in liquid drop model as and potential energy V is given by
For small amplitude vibrations, terms depending on higher powers of are neglected and
only quadratic terms are retained. Quantity C is related to surface and Coulomb energies of
the fluid in liquid drop model of the nucleus
where α2 and α3 are surface and Coulomb energy parameters. In terms of C and , the
Hamiltonian for vibrational mode can be written as, i.e.
Differentiating Hλ with respect to t and equating to zero we get equation of motion as
On comparing this equation with that for a harmonic oscillator i.e. , we get
the frequency of vibration as
The quantum of vibrational energy for
multipole λ is called phonon and is
given by . The frequency =0 for
λ=0 and λ=1.
Example: Quadrupole vibrations of
114Cd nuclei
In 114
Cd, quadrupole vibrations are observed. As in fig. 5, the first excited state is one phonon
state with Jπ=2
+. For about twice the excitation energy, we get triplet of states 0
+, 2
+, 4
+
corresponding to two phonon excitations but these three states are non-degenerate. The
splitting of three J levels is not possible in vibrational model.
3.5.1.2 Shortcomings of vibrational model: Spectra predicted by spherical vibrator model
are approximated for a small number of nuclei near closed shells which are assumed to be
spherical in their ground states. The predicted triplet of 0+, 2
+, 4
+ states in
114Cd nuclei are
non-degenerate. Also the quadrupole moments and B(E2) values are not predicted well by
this nuclei. All these drawbacks imply that harmonic oscillator with its high degree of
mathematical symmetries is simply too idealized to explain the observed properties of
deformed nuclei.
3.5.2 Rotational states: There may be two types of rotational motion shown in fig 6a,
one is rigid in which particles actually move
in circles around the axis of rotation and
other is wavelike in which case particles
perform oscillatory motions and only the
geometrical shape of the nucleus changes.
The wave like rotation is observed only in
deformed nuclei. In general, the non closed
nuclei are deformed. The nuclear shape is prolate at the beginning of major shell and oblate
towards the end of major shell. The rotational motion of a deformed object such as ellipsoid
can be detected by observing the changes in the orientation of the axis of symmetry with
time.
In quantum mechanics, three quantum numbers
are required to describe the rotational state.
First is the total angular momentum J, second
is its projection M along the quantization axis
in the lab frame and third is K, the projection
of J on the 3 axis in the body fixed frame as
seen in fig. 6b. In the body fixed system, K
plays the same role as M in the laboratory
system and there are (2J+1) possible values of
K ranging from –J to +J.
3.5.2.1 Rotational Hamiltonian and energy states: Classically, the angular momentum J of a
rotating object is proportional to angular velocity ω or where I is the moment of
inertia associated with the body. The rotational energy is given by the square of the angular
frequency and is proportional to J2 so that
Quantum mechanically, the rotational Hamiltonian may be written as
where Ii is the moment of inertia along the ith
axis. For an axially symmetric object, the
moment of inertia along a body fixed or intrinsic set of coordinate axes 1, 2 &3 has the
property ( or else it is spherical). The Hamiltonian in this case can be
written in the form
The expectation value of Hamiltonian in the body fixed system is therefore a function of J
(J+1), the expectation value of J2 and K, the eigen value of J3. The eigen functions are D
functions also called rotation matrices so that
The energy of the rotational state is then,
I3 is quite small, so that for low lying rotational states, energy is,
For K=0 band,
Example of 72Hf170
: To plot the theoretical values of the states, we take experimental value
of 2+ state and calculate the value of value back in equation (31), we calculate
rest of the energy states (fig. 7). The
symmetric energy difference with increasing
J is because of increasing I with increasing
J. Moment of inertia I increases due to
centrifugal force which increases with high
angular velocity at high J. When this
correction is included, the energy expression
becomes so
that agreement between theory and
experiment becomes good.
3.5.2.2 Rotational band: A nucleus in a given intrinsic state describing a particular shape can
rotate with different angular velocities in the laboratory. A group of states, each with different
angular momentum J but sharing same intrinsic state forms a rotational band. Since only
difference between these states is in their rotational motion, members of a band are related to
each other in energy, static moments and electromagnetic transition rates. The relative
position of members of rotational band is given by J(J+1) with constant of proportionality
related to moment of inertia I.
3.5.2.3 Value of J for different K’s: The D function transforms under an inversion of
coordinate system as, . Then the rotational wave
function of definite parity can be written as
where sign is for positive and negative parity respectively.
From equation (32), for K=0, the wave function for positive parity state vanishes if J is odd,
as a result, only even J values are allowed for K=0+ band. Similarly for K= 0
- band, only odd
J values are allowed for negative parity state. The results are
For K > 0, the only restriction on the allowed spin in a band is , arising from the fact
that K is the projection of J on body fixed quantization axis or 3 axes. The possible spins are
then
3.5.2.4 Quadrupole moment: The static moments of members of rotational band is given by
rotational model. For an axially deformed object, the quadrupole moment is given by the
integral
where is nuclear density distribution. is called intrinsic quadrupole moment since it
is related to shape of the intrinsic state. The observed quadrupole moment is related to Q0
by the relation
To calculate , knowledge of intrinsic quadrupole moment Q0 and the value of K is
required. K is known from minimum value of J for the band and the value of for only
one state gives the value of Q0 itself. In practice, direct measurements of quadrupole
moments are possible only for the ground state of nuclei. For excited states, the quadrupole
moments can be deduced indirectly through reactions such as Coulomb excitation.
Q.3 The first excited state of W182
is 2+ and is 0.1MeV above the ground state. Estimate the
energies of lowest lying 4+ and 6
+ states of W
182. (Hint: Use equation (31)).
3.6: Nuclear reactions and their types: When a nucleus is bombarded with
different projectiles, then either elastic or inelastic scattering may take place or one or more
particles which are all together different may be knocked out of the nucleus. When the mass
number and/or atomic number of target nuclei changes, we say a nuclear reaction takes place.
Typically a nuclear reaction is written as
where x is incident projectile, X is target nucleus, Y is new nucleus and y is outgoing particle.
The above reaction can also be written is short form as . Nuclear reactions are
classified on the basis of projectiles used, particles detected and residual nucleus. There are
two types of nuclear reactions:
(A) Direct reaction: In this type of reaction, projectile nucleons enter or leave the target
nucleus without disturbing the other nucleons. These are further classified as-
(i) Scattering: In the scattering reaction, the projectile and outgoing particles are same. The
scattering is elastic when residual nucleus is left in ground state. The scattering is called
inelastic when residual nucleus is in excited state.
(ii) Pickup reactions: When the projectile gains nucleons from the target, the nuclear reaction
is called pick up reaction. i.e.
(iii) Stripping reaction: In this type of nuclear reaction, the projectile loses nucleons to the
target nucleus viz.
(B) Compound nuclear reaction: Here projectile and target form a compound nucleus which
has a life time of 10-16
sec. The compound nucleus does not „remember‟ how it was formed
as life span of 10-16
sec is very large in comparison to nuclear time of 10-22
sec. The same
compound nucleus can be formed by number of nuclear reactions and can decay in a number
of ways or channels.
3.7 Conservation laws for nuclear reactions: Nuclear reactions are
governed by certain conservation laws which provide very valuable information about them
necessary in analyzing the experimental data. Some of these conservation laws are:
(i) Conservation of electric charge: The total electric charge is conserved in all nuclear
reactions.
(ii) Conservation of nucleon number: In all nuclear reactions, nucleon number is always
conserved.
(iii) Conservation of energy and momentum: In all nuclear reactions, at least to an order of
accuracy beyond experimental interest, the total energy (sum of rest mass energy and kinetic
energy), the linear and angular momentum, spin angular momentum and isotopic spin are
conserved. The conservation of energy helps to calculate Q value of the reaction.
(iv) Conservation of statistics: The conservation of statistics is also followed by all nuclear
reactions.
3.8 Nuclear disintegration energy (Q): The nuclear disintegration energy is
defined as change in total kinetic energy of a nuclear reaction. For a reaction,
mass energy conservation gives,
where are kinetic energy and rest energy of projectile, is rest energy of
target and so on. The target nucleus X is assumed to be at rest.
The Q value is expressed as,
Using equation (34), (35) becomes
Thus Q is also the change in total rest mass. A
nuclear reaction is exoergic if Q value is positive
and endoergic if Q value is negative.
3.8.1 The Q equation: The relationship between
kinetic energy of projectile and outgoing particle and
nuclear disintegration energy Q is called Q equation.
Once again considering the reaction,
From equation (35)
The kinetic energy of residual nucleus EY is hard to measure and can be eliminated.
Conservation of linear momentum along the direction of projectile gives
Using the relation , we get
(37)
as target is at rest. Conservation of linear momentum normal to the direction of
projectile gives,
(38)
On squaring and adding equation (37) and (38),
Putting the value of EY from equation (39) in equation (35), we get
This is the standard form of Q equation. From this equation, the energy of outgoing particle
can be calculated in terms of energy of projectile for fixed Q. For this, the equation can
be regarded as quadratic in and solution can be written as
when is real and positive, the reaction is energetically possible.
(A) Exoergic reactions (Q > 0):
(i) Very low energy projectiles: for thermal neutrons, we have , then
(ii) Finite energy projectiles: For most of the reactions, and so for all
Hence .
(B) Endoergic reactions (Q < 0): For every reaction with positive Q, there is an inverse
reaction with negative Q of same magnitude.
(i) Very low energy projectiles: , so that , and hence is
imaginary, which means no reaction occurs.
(ii) Threshold energy In an endoergic reaction, the energy –Q is needed to excite
the compound nucleus sufficiently so that it will break up. The bombarding particle must
supply this energy in the form of kinetic energy. Thus the smallest value of projectile energy
at which an endoergic reaction can take place is called threshold energy for that reaction. The
threshold energy is given as,
Q. 4 Calculate the Q value of the reaction; occurring in Rutherford α range
nitrogen experiment, given mN=14.0031amu, mα=4.0026amu, mO=16.9994amu,
mp=1.0078amu.
Q. 5 Calculate the lowest neutron energy or the threshold energy for the following reaction
to occur; , having Q value -3.9MeV.
Q. 6 Obtain the threshold energy of the reaction; . Given, m(209
Bi) =
208.98039 amu, md = 2.0141amu, m(208
Bi)=207.9797 amu.
3.9 Compound nucleus theory: In 1936, Bohr proposed his theory of compound
nucleus. According to Bohr, the nuclear reaction proceeds through two distinct steps: (i)
formation of compound nucleus C and (ii) the disintegration of compound nucleus into the
products of reaction. The compound nucleus is formed by the amalgamation of an incident
particle x and target nucleus X i.e.
The incident particle gives up its energy to the nucleons of target nucleus and this energy is
quickly distributed among them. The new nucleus thus formed is in excited state. The
excitation energy of compound nucleus is given by
where is kinetic energy and is the binding energy of projectile in the compound
nucleus. The mode of disintegration of compound nucleus ( ) is independent of its
mode of formation and depends only on its energy, angular momentum and parity. A
compound nucleus once formed can decay in a number of different ways each with its own
probability. For example
3.9.1 Level width: The excited state of the compound nucleus has a definite mean life
time before it decays by one of the possible modes of de-excitation. The mean life time of
energy state is , where λ is the disintegration constant. λ is connected to level width
by the relation
Comparing the above equation with Heisenberg uncertainty relationship, , if
corresponds to mean life time of excited state then, level width stands for uncertainty in
the energy of excited state. Hence the level width of excited state is a spread in the energy of
excited level due to uncertainty in that level. A long life time means a very fine and narrow
energy level and short life time means a broad energy level of larger level width. For each
individual mode of decay, there is a different probability of decay and therefore a different
partial width for each decay product. Hence, total width of an energy level is the sum of
individual partial widths, i.e.
3.9.2 Elastic and reaction cross section: Let the cross section for forming a
compound nucleus N through incident channel be represented by The decay of N to a
particular exit channel with final state consisting of particle b and B is characterized by
transition probability or partial width .
N a + A (decay product) (exit channel)
b + B ( ) ( )
c + C ( ) ( )
……… ………..
The total width of decay is given by sum of all partial widths, and
is the probability for decaying through channel then reaction cross section from
entrance channel α to exit channel β is given by the product of probability to form compound
nucleus N and for N to decay through β i.e. .
Let in each reaction channel, there is channel radius Rc, outside which there is no reaction
between scattered particle and the residual nucleus. Then in the outside region (r > Rc) the
particle may be considered as free and wave functions are given by plane waves. In the inside
region (r < Rc), wave function is given by interaction between nucleons. At the boundary r =
Rc, the logarithmic derivative of modified radial wave function of each channel must be
continuous i.e.
Let at low energy, only S wave scattering takes place. The modified radial wave function for
l=0 (S wave) has the asymptotic form
Here is inelasticity parameter and is complex phase shift for l=0 (S wave)
channel. Now
On simplification we get,
If is real, then , which means only elastic scattering takes place and there is no
absorption of particles. In general, is a complex quantity i.e. and for
to be less than 1( ) we must have as negative quantity. If is positive
then which does not belong to scattering, generally considered. Now the elastic
cross section
and reaction cross section
Putting and simplifying the above expression we get
From equation (43)
For elastic scattering, is real and so is zero, hence
This elastic cross section can be studied under two cases:
(i) if , then
If then for low energy neutrons,
This is hard sphere scattering formula and represents scattering from an impenetrable sphere
of radius Rc. This approximation works well at low energies.
(ii) if , then
which is resonance scattering formula. It has maximum value when
.
3.10 Breit Wigner formula for isolated resonances: Resonance means
reaction cross section i.e. should be maximum. Now from equation (44), is
maximum when Let Ec is the energy where resonance takes place i.e. is
maximum for energy Ec of incident particle. Then real part of i.e. can be expanded
in terms of Taylor series in E around Ec.
Similarly can be expanded as , where a and b are some parameters.
Then near the resonance energy, we may write elastic cross section and reaction cross section
from equation (43) and (44) in terms of a and b as
where is total width and is the partial width for entrance channel. The reaction width is
then
In expression, the term is called potential scattering term. Elastic cross section
is dominated by the term containing (E-Ec) so that
This formula is called compound elastic cross section. Now i.e. cross section for forming
the compound nucleus through entrance channel α can be calculated as
Hence reaction cross section from entrance channel α to
exit channel β is given by
This is known as Breit Wigner one level resonance formula.
This gives information that resonance occurs at energy E=Ec and width of resonance is given
by
3.11 Summary
Now we summarize what we have learnt so far in this unit.
In liquid drop model, the molecules of liquid drop correspond to nucleons in the
nucleus.
This liquid drop model explains the stability of nuclei against the breakup into two
fragments i.e. nuclear fission.
For certain numbers of neutrons and protons called magic numbers, nuclei exhibit
special stability. This stability is not explained by liquid drop model. Also the other
properties of nucleons like spin, magnetic moment and quadrupole moments are
unexplained in liquid drop model.
The magic numbers and other properties of nuclei are explained by nuclear shell
model.
In shell model, all magic numbers are reproduced when spin orbit interaction term is
included in simple harmonic oscillator potential.
Shell model predicts that ground state spin of even even nuclei is always 0+. Spin of
odd even and even odd nuclei is decided by the angular momentum of last single part
state occupied by odd nucleon. Spin of odd odd nuclei is governed by Nordeim rules.
The quadrupole moment of nucleus in shell model is given by Q= where
Rc is nuclear radius.
In collective nuclear model, the collective motion of nucleons in the core is also taken
into account so that motion of core nucleons and motion of loosely bound surface
nucleons results in rotational, vibrational and nucleonic energy states in the nuclei.
The rotational and vibrational states arise due to the motion of nuclear core and
nucleonic states arise from the motion of loosely bound nucleons.
A moving nuclear surface may be described by deformation parameter λ which tells
about the monopole, dipole, quadrupole and octupole vibrations.
The quantum of vibrational energy for multipole λ is called phonon and is given by
.
A rotational state is described by three quantum numbers. First, the total angular
momentum J, second, its projection M along the quantization axis in the lab frame and
third is K, the projection of J on the 3 axis in the body fixed frame.
For low lying rotational states, energy is given by . For K=0
band,
When the mass number and/or atomic number of target nuclei changes due to the
bombardment of different projectiles, a nuclear reaction takes place. Typically a
nuclear reaction is written as , also written as .
The nuclear disintegration energy is defined as change in total kinetic energy of a
nuclear reaction and is expressed as,
A nuclear reaction is called exoergic when Q > 0 and endoergic when Q < 0.
The threshold energy for an endoergic reaction is the smallest value of projectile
energy at which that reaction takes place.
A compound nucleus is formed by the amalgamation of an incident particle x and
target nucleus X i.e. .
Breit Wigner one level resonance formula represents reaction cross section in terms of
level widths . This gives information that resonance occurs at energy E=Ec.
3.12 Answer to questions
Ans 1: O17
: ; Zn67
: , N
16:
Ans 2: O17
: = -1.91, Q= -0.0326 barn; S33
: =1.146, Q=-0.0355barn
Ans 3: E(4+)=0.333MeV, E(6
+)=0.7MeV
Ans 4: -1.418MeV
Ans 5: 4.105MeV
Ans 6: 5.26MeV
3.13 References
Nuclear Models; Walter Greiner, Springer Publication.
Introductory Nuclear Physics; Samuel S.M. Wong, Prentice Hall of India Ltd.
Elements of Nuclear Physics; M.L. Pandya & R.P.S. Yadav, Kedar Nath Ram Nath
Publication.
Concepts of Nuclear Physics; B.L. Cohen, Tata McGraw Hill Publishing Ltd.
Nuclear Physics; S.B. Patel, New Age International (P) Ltd.
Basic Ideas and Concepts in Nuclear Physics; K. Heyde, Part C, Institute of Physics
(IOP) Publishing.
Nuclear Physics; D.C. Tayal, Himalaya Publishing House.
Theory of Nuclear Structure; M.K. Pal, East West Press Pvt. Ltd.
Nuclear Interactions, Sergio De Benedetti, John Wiley & Sons.
3.14 Questions
1. Discuss the phenomenon explained by liquid drop model. What are the similarities
between nucleus and liquid drop?
2. Explain symmetric fission of nucleus using liquid drop model.
3. Draw and discuss the potential energy curve obtained for nuclear fission using liquid drop
model.
4. State the assumptions of shell model of a nucleus. Obtain the sequence of magic numbers
using harmonic oscillator potential and spin orbit interaction.
5. Discuss the predictions of shell model in relation to experimental results.
6. Discuss the various vibrational modes described by the moving nuclear surface.
7. Obtain the expression for vibrational frequency of the nucleus.
8. Show that energy of low lying rotational states is given by where I is the
moment of inertia of nuclei.
9. Discuss various kinds of nuclear reactions. What are the associated conservation laws?
10. What is nuclear disintegration energy?
11. Obtain the solutions of standard Q equation for exoergic and endoergic cases.
12. Discuss the threshold energy in case of endoergic nuclear reaction.
13. Obtain the hard sphere scattering cross section and resonance scattering cross section for
compound nucleus.
14. Discuss Breit Wigner one level resonance formula for compound nucleus.
Unit IV – Nuclear Decay
4.0 Introduction
The radiations from a natural radioactive substance are classified as alpha (α) rays, beta (β)
rays and gamma (γ) rays. Beta rays are identical with electrons. Bucherer in 1909 determined
the charge to mass ratio (e/m) for β particles and found them to be nearly same as that of
electrons. The radioactive nuclide exhibit three kinds of β radioactivity viz. electron
emission, positron emission and electron capture. The most common of three is electron
emission. The energy spectrum of beta particles when studied by magnetic spectrographs was
found to be continuous. This continuous nature of spectrum was explained by assuming the
existence of a hypothetical particle called neutrino. In this unit, we will study about some of
these interesting aspects like existence of neutrinos, theoretical explanation of beta decay,
parity violation phenomenon etc. The other type of radiation called gamma rays is emitted
when nucleus from an excited state comes down to the ground state. The transition from an
excited state of nucleus to the lower level of same nucleus can also be achieved without
emission of γ ray photon. This process is called internal conversion. We will study about
these processes and others like nuclear isomerism, allowed gamma ray transitions in nuclei
etc. in detail in this unit.
4.1 Objectives
This unit presents a brief discussion on beta and gamma rays. After going through this unit,
you will be able to:
Understand beta decay energy spectrum, its nature and associated problems.
Get an idea, how Pauli explained the existence of neutrinos.
Know about neutrino and its antiparticle called antineutrino.
Get acquainted with Fermi theory of beta decay, about Fermi and Gamow Teller
transitions in beta decay.
Learn about the phenomenon of parity violation and experimental evidences for
existence of neutrinos.
Get knowledge about multipolarity and multipole transitions in gamma ray emission.
Learn about internal conversion process and nuclear isomerism, and about angular
correlations in gamma ray emission.
4.2 Beta ray spectrum
When beta ray energies are studied by mass spectrographs, a continuous nature of energy
spectrum is found as shown in fig.1. This continuous spectrum is characterized by following
properties:
(i) Every continuous β ray
spectrum has a definite maximum
height and position of which
depends on nucleus emitting the
particles.
(ii) The particle β+ or β
- possesses
some upper limit of energy called
the end point energy. This energy
is different for different nuclides.
4.2.1 Difficulties associated with continuous β ray spectrum: The continuous
nature of β ray energy spectrum gave rise to serious difficulties in understanding β decay.
The main problems are as follows:
(i) Nature of spectrum: β decay is an energy transition between two definite energy states as
shown in fig. 2. The parent nucleus
emits a β particle and form the
daughter nucleus . Thus
monoenergic beta rays are expected
and spectrum should be a straight
line, but observed spectrum is
continuous, implying that electrons
emitted in β decay process do not
have the same kinetic energy.
(ii) Conservation of energy: From fig 1, it is concluded that most of the electrons
(corresponding to peak of curve) are emitted with only (1/3)rd
the maximum energy Emax.
Then where the missing energy (about (2/3)rd
of Emax) goes? This missing energy must be
accounted to follow the principle of conservation of energy.
(iii) Conservation of Spin: Taking the example of tritium β decay,
the charge and mass number are conserved. However, spin angular momentum of is half
integral while that of + is either 0 or 1. Thus L.H.S. has half integral spin while
R.H.S. has integral spin angular momentum and so spin angular momentum is not conserved.
(iv) Statistics: Since L.H.S. of above equation is fermion , so obeys the Fermi Dirac
statistics while R.H.S. is boson ( + ), so follows Bose Einstein statistics.
All these problems were resolved by considering neutrino hypothesis.
4.3 Pauli neutrino hypothesis
Pauli in 1930 suggested that a particle called „neutrino‟ may accompany β particle to resolve
the above problems. This particle has zero charge and almost zero mass ( ). This has
spin 1/2, so it is a fermion. These properties of neutrino supported Pauli hypothesis in the
following way:
(i) Energy conservation: This particle carries away an amount of energy equal to difference
between the observed energy for a particular β particle and the maximum energy Emax of the
continuous β spectrum. Thus principle of conservation of energy is satisfied.
(ii) Conservation of spin: Neutrino is spin 1/2 particle, so that R.H.S of equation (1) with
emission of neutrino and β particle also has half integral spin. Hence, L.H.S and R.H.S both
have spin half integral, which follows spin angular momentum conservation.
(iii) Statistics: Since, L.H.S and R.H.S of equation (1) are now fermions, both sides follow
Fermi Dirac statistics.
4.3.1 Neutrino and antineutrino: The particle ν has an antiparticle called antineutrino, . An
antineutrino has zero charge, mass as that of neutrino and spin 1/2. Then how can we make
distinction between neutrino and antineutrino? This is done by their „handedness‟ property.
4.3.2 Handedness property of neutrino:
Neutrino is described as left handed particle which
means its spin Sν is always antiparallel to its
momentum Pν while antineutrino is described as
right handed particle having its spin parallel to its
momentum as illustrated in fig. 3.
Mathematically, the idea of handedness is described
by the quantity called „helicity‟ defined as
where . If angle between and is 1800 i.e. they are anti parallel to each
other, then , implying that . Hence for neutrinos, helicity .
Similarly, if angle between and is 00 i.e. they are parallel then , implying that
. Hence for antineutrinos, helicity . Thus neutrino remains in negative
helicity state whereas antineutrino is in positive helicity state. If a neutrino is massless, it will
have fixed helicity states but if it possesses some mass then helicities are not fixed. A left
handed particle may be viewed as right handed particle and vice-versa.
The neutrino ν is a particle emitted in positron emission (β+ decay) where as anti-neutrino
is emitted in electron emission (β- decay). Thus the equations for beta decay can be written as
for β- decay
for β+
decay
for electron capture
Thus equation (1) and decay of Bismuth can be correctly can be written as
and
A simple example of β decay is the disintegration of free neutron with a half life of about 12
minutes. The decay process is represented as
In fact, there are three types of antineutrinos say, electron antineutrino ( ), muon
antineutrino and tau antineutrino and in the above equation (2), should be
replaced or electron antineutrino. We will study about these particles in unit V.
4.4 Fermi theory of beta decay
Enrico Fermi in 1934 postulated that electron and neutrino are created at the time of β particle
emission. He assumed that interaction responsible for β decay process is weak and so
perturbation theory can be used. The probability that an electron of momentum between pe
and pe+dpe emitted per unit time is given by
This is also called Fermi golden rule. Here is the Hamiltonian operator that operates
between initial states i and final states f. Quantity is called matrix element of β
decay and its computation involves nuclear wave functions of initial and final states. is the
wave function of daughter nucleus while is initial state wave function of parent nucleus.
is the density of state factor which gives the number of available final states per unit
energy. Thus
where N is the number of energy states available to final decay products that can be put in a
given volume and E is the total energy of particles.
4.4.1 Computation of : Let a particle is moving along X axis with momentum px,
then x-px plot is known as phase space. Classically, the state of a particle can be represented
by a point in phase space but according to quantum mechanics, the particle in phase space has
to be represented by a cell and not a point. The volume of cell is determined by Heisenberg
uncertainty principle and is equal to h. Then from fig. 4, it is evident that
number of cells that can be put in volume px is
N represents number of states in the volume px in
phase space. In three dimensional space, equation
(5) becomes,
Here is the number of states for one particle. It
is convenient to choose unit volume for the particle so that . Hence
Therefore, density of state factor for one particle is
From the relativistic expression,
So that
Using equation (7) and (9) in equation (8), we get
The β decay problem ( ) involves
three particles in the final state. Thus final state
consists of three particles. Two of these three
particles can change the momentum
independently but conservation of momentum
fixes the momentum of third particle i.e. if and
are the momenta of two particles, then , the
momentum of third particle gets fixed by the
relation , so that only two particles appear free and independent. Hence
number of states that can be put in unit volume in phase space for three particles are
Therefore, density of final states is
Fig. 5 shows the recoil of daughter nucleus, emitted electron and the neutrino. If Emax is the
total energy of disintegration, then this energy is shared by e- and only as recoil nucleus is
much more massive than these two particles and hence energy carried by recoil nucleus is
negligible. Thus equation (11) can be written as
is solid angle in which an electron having momentum between and is
emitted. Similarly is the solid angle corresponding to antineutrino. To calculate ,
electron energy and momentum is kept constant. Hence
Neglecting recoil energy of daughter nucleus, we have,
If rest mass of neutrino is considered zero, then,
(15)
Using equation (14) and (15) in equation (13), we get
Substituting this value in equation (13), we find
Putting , we get
Substituting this density of state factor from equation (16) back in equation (3), the transition
probability for a transition between initial state i and final state f is,
If is the average value of matrix element, averaged over the angle between
electron and antineutrino, then we can put and , so that
The emitted β particles experience Coulomb force. The nucleus with charge Ze will
accelerate positrons and retard electrons. This means we shall have more electrons of low
energy and positrons of high energy
than predicted by equation (18). This
Coulomb correction is taken into
account by inserting a Fermi function
in equation (18) such that
This equation represents transition
probability of electron with
momentum between pe and pe+dpe. Above equation can be verified experimentally if we
rewrite this as
The L.H.S of above equation is called Kurie function K and is determined experimentally.
When this function is plotted against measured electron energy Ee, a straight line is obtained
over a wide energy range for momentum independent matrix element. This plot is called
Fermi plot or Kurie plot (fig.6). Experimentally is the number of beta particles per
energy interval and a G.M. counter is used to detect and count the number of beta particles.
Kurie plots are straight lines and the intercept on energy axis gives the end point energy,
Emax. Experimental data leading to the Kurie plot provide an upper limit on neutrino rest mass
(fig. 6). In the figure, the solid lines are for the case of finite neutrino mass mν and dashed
lines are for mν= 0. The two solid lines correspond to an upper limit of 1keV and 0.25keV
neutrino mass.
4.4.2 Evaluation of matrix element: The total transition probability is obtained by
integrating equation (19) over momentum i.e.
Matrix element being momentum independent is taken out of integral. The integral can be
evaluated and comes out as
where f is function of Emax i.e. . Constant is used to make f dimensionless.
Thus
here η is the mean life time of beta emitter, measured experimentally. Thus matrix element
can be evaluated by the knowledge of .
4.4.3 Allowed and forbidden transitions
(A) Allowed transitions: There are two selection rules in practice:
(i) Electron and antineutrino are emitted in beta decay without orbital angular momentum (s
wave electron and antineutrino); hence if they are emitted with their spins anti-parallel, then
there is no change in total nuclear angular momentum. This corresponds to selection rule
, called Fermi selection rule.
(ii) If electron and antineutrino are emitted with their spins parallel, then there is a change of
one unit in total nuclear angular momentum. This corresponds to selection rule
, known as Gamow Teller (GT) selection rule. The examples are
; this is allowed GT transition as
.
; this is allowed Fermi transition as
allowed by both transitions as and
.
These allowed transitions are also categorized on the basis of parity of initial and final state.
(i) The allowed transition is called favoured (superallowed) if there is no change in the
parity of initial and final state.
(ii) The allowed transition is called unfavoured (allowed) if the parity of initial and final
state changes.
(B) Forbidden transitions: If transition from initial to final state takes place with the
emission of p wave electron and antineutrino then, there is also a change in orbital angular
momentum i.e. . Hence may be 1, 2, 3… as p, d, f wave electron and antineutrino
are emitted. Thus total change in momentum may be and so on. Also, the
parity change takes place for odd value and no parity change for even value.
The overall transitions are summarized in the following table 1.
Table 1: Allowed and forbidden transitions
Selection rule Fermi transition (F) GT transition (change in
parity)
Allowed
No
Ist forbidden
Yes
IInd
forbidden
No
IIIrd
forbidden
Yes
Q.1 Classify the following transitions on the basis of selection rules:
(a)
(b)
(c)
(d)
4.5 Detection of Neutrino: Cowan and Reines experiment
Cowan and Reines first observed the direct interaction of free neutrino in 1953. The basic
idea behind the experiment is inverse beta decay reaction in which anti neutrinos are
bombarded over proton target. These protons emit positrons to become neutrons.
Cowan and reins utilized the large flux of antineutrinos existing in the neighborhood of a
nuclear reactor to observe the above reaction. The schematic diagram of experimental
arrangement is shown in figure 7.
The protons for the reaction are provided by large plastic tank containing about 200 litres of
water. I and II are liquid scintillator detectors containing 400 litres of liquid scintillator
solution. CdCl2 was dissolved in water to provide Cd nuclei for neutron detection. To
accomplish the above reaction, the following sequence of events should take place.
(i) An antineutrino from the reactor interacts with a proton in the water tank. A neutron and a
positron are produced as a result of this interaction.
(ii) The positron annihilates with an electron resulting in two gamma rays emitted in opposite
directions. Each of these gamma rays carries energy equal to electron rest mass of 0.511
MeV. These γ rays can be detected by two liquid scintillators I and II. It was observed that
this annihilation radiation follows very quickly in an interval of second after the
emission of positron.
(iii) The neutron diffuses in water (with dissolved CdCl2) and after many collisions with
protons slows down to thermal energies to get captured by a Cd nucleus. This neutron capture
results in the emission of three γ rays with the time gap of about seconds. One of
these γ rays is detected in the upper detector and rest two rays are detected in lower detector.
Cowan and Reines observed the above three sequences and used an oscilloscope to
photograph the sequence of annihilation γ ray and neutron pulse. To check the result, they
used water without any cadmium and found that neutron pulse due to capture γ rays
disappeared. Also when reactor was shut down, they found that no reaction occurs. Thus
Cowan and Reines experiment proved the existence of neutrinos.
4.6 Parity non conservation in beta decay
The parity is conserved if the probability of finding a particle between a certain range of
coordinates is not affected by coordinate reflection. We define a parity operator P such that
for even parity wave function,
and for odd parity wave function
Parity is related to the mirror image of an object or process and conservation of parity means
the mirror image of an object or process is
indistinguishable from object or process
itself.
4.6.1 Parity non conservation in
neutrinos: We consider the mirror image of
neutrino as in fig. 8. Let a neutrino travels
with momentum Pν and spin Sν towards the mirror. This neutrino has oppositely directed
momentum and spin but the mirror image of this neutrino has spin and momentum in the
same direction i.e. its mirror image is not same to it. In fact the mirror image is its anti
particle called antineutrino. This is known as non conservation of parity by neutrino.
Before discussing parity non conservation in beta decay, we define pseudoscalar and vector
quantities.
(i) Polar vector: Quantities that change sign under coordinate reflection are called polar
vectors. Examples are position vector, linear momentum of particle etc.
(ii) Axial vector or pseudo vector: Quantities that do not change sign under coordinate
reflection are called axial vectors. Examples are angular momentum, spin, magnetic field,
magnetic moments of particles etc. Angular momentum does not change sign
under reflection of coordinates because and both are polar vectors and change sign under
coordinate reflection. Hence product vector remains invariant under coordinate reflection.
(iii) Pseudo scalar: Quantities that are product of one polar vector and one axial vector are
called pseudo scalars. These quantities change sign under coordinate reflection, hence
named pseudo scalar (a scalar quantity does not change sign under coordinate reflection).
Example is product of total angular momentum and linear momentum i.e.
. As both are axial vectors that do not change sign under coordinate
reflection, hence is an axial vector while is a polar vector. Thus product consists of one
polar vector and one axial vector and is pseudo scalar.
4.6.2 Experimental verification of parity non conservation: In an experiment, if
one gets a pseudo scalar quantity, it is concluded that parity is not conserved in the
experiment. The confirmation of parity violation came in 1957 from observation of beta
decay of 60
Co nuclei by Wu and his fellow workers when they found a pseudo scalar in the
experiment. 60
Co nuclei decays as a
Ground state of odd odd nucleus has spin-parity . A non zero spin makes it
possible for the nucleus to be polarized along a magnetic field applied externally. The ground
state decays predominantly to state of at excitation energy 2.5 MeV (fig. 9). If
the spin of all cobalt (Co) nuclei are aligned, then this direction may be indicated by unit
vector σ parallel to the alignment of 60
Co ground state spin J. The angular distribution of
emitted electron with momentum p and energy E is expressed as
Where θ is the angle with respect to J through which electron is emitted. Parameter
represents the intensity of angular
dependence. Now is an axial vector
which does not change sign under
parity operation and p is a polar vector
that changes sign under parity
operation. This in turn implies that
changes sign under parity operation and
hence it is a pseudoscalar quantity. The
first term in equation (23) being scalar
remains invariant under parity
operation. Now if the parity is conserved in beta decay experiment then second term
containing term must vanish because it is a pseudoscalar term and its nonzero value will
show the non conservation of parity in the experiment. Now the second term will vanish if
is equal to zero i.e. there should be no angular dependency of emitted electrons or the angular
distribution of emitted electrons must be isotropic. In the experiment, it is found that
i.e. beta particles are emitted preferentially in the direction opposite to
alignment of 60
Co ground state spin J which makes pseudoscalar term nonzero and gives rise
to maximum parity violation.
4.6.3 Explanation of anisotropy of electrons: Cobalt nuclei (60
Co) with ground state
spin aligned in the direction of magnetic field (fig 10b) decays to 60
Ni with spin
Hence spins of electron and antineutrino must be in the same direction (of nuclear
spin) to make up this decrement. As antineutrino is a right handed particle, so electron should
be emitted in left helicity state to conserve the momentum (fig. 10a). In other words,
electrons are emitted preferentially with momentum opposite to direction of nuclear spin,
thus causing anisotropy.
4.7 Gamma decay
When a nucleus decays by α emission or β emission, it is left in an excited state. The nucleus
in an excited state comes down to ground state by emission of gamma (γ) rays. Let Ei and Ef
be the energy of excited state and lower (ground) state of nucleus, then energy release is
. This energy may be emitted by (i) γ ray emission (ii) internal conversion and
(iii) internal pair production. For internal pair production, must be greater than or equal to
1.02 MeV. Hence, first two processes occur more frequently than the third.
4.7.1 Selection rules for γ ray emission: If the spin of the initial state of nucleus is
and of final state is then angular momentum carried away by gamma (γ) ray photon is
In other words,
There are following selection rules for γ decay:
(i) the γ photon must carry away at least one unit of angular momentum.
(ii) parity is always conserved in γ decay.
4.7.2 Multipolarity in γ transitions
The γ emission is classified according to multipole order. There can be dipole, quadrupole,
octupole or still higher order transitions. For any order, the transition may be electric or
magnetic type. These multipole orders have following properties:
(i) Multipole order of a transition is said to be 2L.
(ii) If , multipole order is 2 and there is dipole transition. Similarly for
, multipole order is 4 which means there is quadrupole transition and so on.
(iii) If parity changes in nuclear transition, then we have electric type multipole transition and
if there is no parity change in transition we have magnetic type multipole transition.
If parity of initial state is and that of final state is , then the two selection rules for
transition are given by
where π is the parity of emitted radiation. Further γ transitions are classified as electric 2L
pole or magnetic 2L pole according as:
π = (-1)L for electric 2
L pole radiation and π = (-1)
L+1 for magnetic 2
L pole radiation. The
electric and magnetic multipole transitions are written as (EL) and (ML) respectively.
The selection rules for multipole transitions are summarized in the table 2.
Example: Calculate EL and ML for the nuclear transition, .
Sol. Parity of emitted radiation = (+).(+) = positive. Also angular momentum
carried away by γ ray photon is;
Using these L values for the rule; π = (-1)L for electric 2
L pole radiation and π = (-1)
L+1 for
magnetic 2L pole radiation, we get following electric multipole (EL) and magnetic multipole
(ML) transitions: M1, E2, M3, E4.
Table 2: Selection rules for multipole radiation
Multipole Symbol L Parity change
Electric dipole E1 1 Yes
Magnetic dipole M1 1 No
Electric quadrupole E2 2 No
Magnetic quadrupole M2 2 Yes
Electric octupole E3 3 Yes
Magnetic octupole M3 3 No
Electric pole EL L No for L even
Yes for L odd
Magnetic pole ML L No for L odd
Yes for L even
Q.2 What are the expected gamma rays transitions between following states of odd A nuclei?
(i) (ii) (iii)
4.7.3 Multipole radiation and transition probability: When the nucleus interacts
with electromagnetic radiation field, gamma rays are emitted. In the interaction, energy is
transferred from nucleus to the field and this excitation of field appears as γ ray. An
electromagnetic wave is an oscillating electric and magnetic field. A changing electric field
gives rise to magnetic field and changing magnetic field gives rise to electric field. Such an
electromagnetic wave can be produced by an oscillating electric charge which produces
oscillating electric field or by an oscillating electric current which sets up a magnetic field.
These oscillating charges and current elements radiate energy and angular momentum. The
angular momentum is as such not radiated but the angular momentum equal to can be
assigned to the photon constituting the radiation. For L= 0,1,2,3 etc, angular momentum
associated with photon in the Z direction will be The radiation emitted from
oscillating electric field is called electric multipole radiation and is represented by EL where
L is the angular momentum carried away by radiation while radiation emitted from oscillating
magnetic field is called magnetic multipole radiation and is represented by ML.
The knowledge about the angular momentum carried away by γ radiation can be inferred
from the mean life time of excited state which gives rise to γ transition. This mean life time η
depends on L and is reciprocal of transition probability λ(L). The expression for transition
probability for EL multipole is obtained as
where
E is transition energy and R is nuclear radius given by . The estimates obtained
for some of the lower order electric multipoles are
;
;
Similarly, the expression for transition probability for ML multipole is given by
The estimates for some of the lower order magnetic multipoles are
;
;
4.8 Internal conversion
A nucleus from an excited state can perform a transition to lower state not only by γ ray
emission but also by transmitting the energy directly to the electrons surrounding the nucleus.
Due to this, an atomic electron is ejected from a bound orbit. The kinetic energy of emitted
electron is equal to transition energy minus the binding energy of the orbital electron. This
process is called internal conversion. These electrons produce a series of monoenergetic
lines. The lowest energy line corresponds to emission of K shell electron i.e.
Similarly conversion of L shell electrons gives
These conversion electrons appear as discrete lines in the continuous beta decay energy
spectrum as shown in fig.
11. The total transition
probability λ from
nuclear state a to state b
is the sum of two terms
where are
partial decay constants
for conversion electron
emission and for gamma
emission respectively. The ratio of two decay constants is called conversion coefficient (α)
and is defined as
Thus,
As a result of internal conversion, a number of groups of electrons are emitted and for each
gamma ray, there are several conversion lines corresponding to ejected electrons from
different atomic shells. Therefore, total conversion coefficient is
Here … are partial conversion coefficients. The approximate relations for K
conversion coefficient in transitions for which the binding energy of K electron is small
compared with transition energy are given by
Thus conversion coefficients have following properties:
(i) Conversion coefficients are proportional to Z3, hence conversion process dominates in
heavy nuclei and γ ray emission is favored in light nuclei.
(ii)These increase as power of , transitions corresponding to small energies are
preferentially converted.
(iii) They increase with multipole order L of the transition.
(iv) In general αm
are larger than αe.
4.9 Nuclear isomerism
The long lived, low lying excited states are often found among intermediate and large mass
nuclei. The life times of these states vary from 10-10
sec. to 108 years. These states are called
isomeric states and transitions from these states are called isomeric transitions. Nuclei which
have same atomic and mass number but different radioactive properties are called nuclear
isomers and their existence is referred as nuclear isomerism. The phenomenon of nuclear
isomerism was discovered by O. Hawn in 1921. For example, we consider the decay
the decay scheme is shown in fig. 12. Here (protactenium) decays by emission of β
particles with two different half lives, one of 1.18 minute and other of 6.66 hours. These two
half lives are due to two different excited energy states of the same nucleus. These states are
known as isomeric states. Sometimes they are also called metastable states, denoted by
.
Isomeric transitions are
characterized by a
sufficiently large change
in angular momentum and
very small change in
energy. Both these factors
favour high internal
conversion. The nuclear
isomers have been
successfully separated
chemically by the use of large internal conversions.
4.10 Angular correlation in gamma emission
When two gamma rays are emitted in rapid succession by the same nucleus, it is found that
directions of two emitted
gamma rays are correlated.
This means if direction of
emission of first gamma ray
is taken along Z axis, then
probability of falling second
ray into the solid angle dΩ
is not constant but depends
upon the angle θ between
the two directions. This
type of correlation is known as γ-γ correlation as shown in fig. 13. Such a correlation is
observed when spin of excited state differs from ground state by several units of while spin
of intermediate state lies between the two.
Consider two successive γ rays in the sequence as in fig. 13. The nuclear excited state (spin
) decays to intermediate state (spin ) with the emission of ray and finally to ground
state of (spin ) through the emission of ray. The multipolarity of rays and is L1
and L2 respectively. The directional correlation W(θ) is defined as the relative probability
that ray is emitted into the solid angle dΩ at an angle θ with respect to ray and is given
by
where are the coefficients which depend on spins of levels and and on L1 and L2, and
are the even Legendre polynomials. An equivalent and more common form of above
expression is
where coefficients are the functions Ia, Ib, Ic, L1and L2.
Since no monopole (L =0) radiation exists, hence transition between states of angular
momentum Ia =Ic =0 do not occur. The transition probability diminishes rapidly with
increasing order L. In fact the main contribution comes from dipole (L=1) and quadrupole
(L=2) transitions.
4.10.1 Dipole-dipole angular correlation: The γ-γ correlation-ship can be compared
with dipole antenna radiation. A dipole antenna
radiates power which is proportional to sin2θ where
θ is the angle relative to antenna rod. There is no
radiation of power in the direction parallel to rod.
In case of gamma emission when a photon is
emitted carrying angular momentum L with z
component m, then (L,m) determines the antenna
pattern. L = 0, m = 0 gives sin2θ antenna pattern
while L=1, m= +1 and L=1, m= -1 gives
antenna pattern. When all the three
m states are present in equal amount then the sum of three antenna patterns are isotropic and
each gives an equal amount of radiation. As an example, we consider a nuclear transition
by successive E1 (electric dipole) γ ray transition as shown in fig. 14. Z axis
is chosen as the direction of emission of first γ ray. In this case, there is no m =0 radiation
since antenna pattern for m=0 (sin2θ) gives no intensity in the Z direction. From conservation
of angular momentum, this implies that m=0 component of 1- state is not populated in the
transition. In the emission of second γ ray, angular momentum conservation requires that γ2
comes off with m=1 or m= -1 both of which have a antenna pattern relative to
Z axis which is the direction of first gamma ray. Thus in the transition, , the
angular correlation of γ2 relative to γ1 is given by . These angular correlations are
very specific to the transitions.
4.10.2 Experimental determination of angular correlation: The experimental
arrangement for determination of
angular correlation is shown in fig.
15. The gamma rays are emitted from
source S and detected by scintillation
detectors D1 and D2. Detector D1 is
fixed while D2 is movable. Pulses
from these detectors are amplified by
amplifiers AMP and then passed into
single channel analyzer SCA1 and
SCA2. The output from these two
SCA are fed into a coincidence
analyzer COINC which produces an output pulse when two inputs arrive at the same time
indicating that two γ rays are emitted from same decay. These output pulses are counted in a
scalar. Counts are taken for different values of angle θ between the two detectors.
4.11 Summary
Now we recall what we have discussed so far:
We have learnt about the nature of beta decay spectrum and how neutrino hypothesis
resolved the problems associated with continuous beta spectrum.
The handedness property differentiates neutrino from antineutrino.
Enrico Fermi in 1934 presented the theory of beta decay which can be verified
experimentally using Kurie plots.
There are two selection rules called Fermi and Gamow teller selection rules which
allow or forbid beta decay transitions.
Cowan and Reines first observed the direct interaction of free neutrinos in 1953 in the
inverse beta decay experiment.
Wu and his coworkers in 1957 confirmed the violation of parity in beta decay from
observation of beta decay of 60
Co nuclei.
Gamma rays are emitted when a nucleus from excited state comes down to ground
state.
The emitted γ ray photon must carry at least one unit of angular momentum and parity
must be conserved in γ transition.
The parity of emitted radiation decides electric or magnetic type multipole transition.
An internal conversion is a process in which the nucleus from an excited state can
perform a transition to lower state by transmitting the energy directly to the electrons
surrounding the nucleus. The emitted electron is called conversion electron.
Nuclei which have same atomic and mass number but different radioactive properties
are called nuclear isomers and their existence is referred as nuclear isomerism.
Nuclear isomers often possess long lived, low lying excited states. These states are
called isomeric states and transitions from these states are called isomeric transitions.
Finally, when two gamma rays are emitted in rapid succession by the same nucleus, it
is found that directions of two emitted gamma rays are correlated. This means if
direction of emission of first gamma ray is taken along Z axis, then probability of
falling second ray into the solid angle dΩ is not constant but depends upon the angle θ
between the two directions. This type of correlation is known as γ-γ correlation.
4.12 Answer to questions
Ans 1: (a) Allowed GT transition, (b) Ist forbidden GT transition (c) Allowed mixed, both F
and GT transition (d) IInd forbidden, mixed decay, F and GT transition
Ans 2: (i) M4,E5 (ii) E3, E5, E7, M4, M6, M8 (iii) E5,E7, M4, M6
4.13 References
Introductory Nuclear Physics; Samuel S.M. Wong, Prentice Hall of India Ltd.
Basic Ideas and Concepts in Nuclear Physics; K. Heyde, Part B, Institute of Physics
(IOP) Publishing.
Elements of Nuclear Physics; M.L. Pandya & R.P.S. Yadav, Kedar Nath Ram Nath
Publication.
Nuclear Physics; S.B. Patel, New Age International (P) Ltd.
Concepts of Nuclear Physics; Bernard L. Cohen, Tata McGraw Hill Publishing Ltd.
Nuclear Physics, D.C. Tayal, Himalaya Publishing House.
Nuclear Physics, V. Devanathan, Narosa Publishing House.
Nuclear Interactions, Sergio DeBenedetti, John Wiley & Sons.
4.14 Questions
1. Discuss the silent features of β ray spectrum and explain how Pauli neutrino hypothesis
solved the anomalies in β ray spectra.
2. How will you differentiate between neutrino and antineutrino?
3. What is Fermi golden rule?
4. Give Fermi theory of beta decay.
5. Define parity. Explain violation of parity in β decay process.
6. Discuss Cowan and Reines experiment which has determined the existence of neutrinos.
7. Discuss the allowed and forbidden transitions in case of beta decay.
8. Define multipolarity in gamma ray transitions.
9. What are the selection rules for multiple radiations?
10. Write short notes on (i) internal conversion (ii) nuclear isomerism.
11. Discuss the angular correlation in gamma emission. How it is experimentally determined?
Unit V – Elements of Particle Physics
Elementary Particles
5.0 Introduction
The search for elementary particles from which all matter is composed has started from the
time of discovery of negatively charged particle called electron by Sir J.J. Thomson in 1898.
The scattering experiment of Lord Rutherford in 1914 indicated the presence of positively
charged protons inside the nucleus. With the discovery of neutron in 1932, the number of
elementary particles became three-electron, proton and neutron. In 1935, Yukawa postulated
the meson as the particle which is exchanged between the nucleons to generate nuclear force.
Such a particle called μ meson was discovered by Neddermeyer et al in 1937 but it interacted
very weakly with nucleons. After 10 years, Powell et al discovered the Yukawa particle
called now π meson. With the development of high energy accelerators, it had become
possible to produce high energy protons and observe their collision on selected targets. The
result was the discovery of newer and newer particles. Antiproton was first produced in
Berkeley by using protons of energy 6 GeV from Bevatron and antineutrons were discovered
in 1958. A host of other short lived particles known as strange particles were produced in
pairs and their decay observed. About hundred of particles have now been discovered but
most of them are the excited states or resonance states of other stable particles. In this unit,
we will study about the classification scheme of these elementary particles, about their
interactions and quantum numbers which are assigned to them. Some basic ideas about
quarks and their role in constructing baryon and meson multiplets will be presented at the
end.
5.1 Objectives
In this unit, a brief discussion on various elementary particles is presented. After going
through this unit, you will be able to:
Know about the different classification schemes of elementary particles and about the
interactions in which they participate.
Get an idea of different quantum numbers assigned to particles for their identification.
Understand the various conservation laws that allow the production and decay of
interacting particles.
Learn how particles are arranged in groups of octet, nonet and decuplet using group
theory.
Know about the mass formulas in which masses of particles of one group are related
to each other.
Get acquainted with Quark model in which all the elementary particles except leptons
are made of quarks.
Know about gluons and basic ideas of QCD.
5.2 Classification of elementary particles
Elementary particles may be classified according to their mass, spin and their interaction as
follows-
(a) Classification on the basis of mass: Elementary particles having mass up to 134
MeV are called leptons, while those having the mass greater than 134 MeV but less than 938
MeV are called mesons. Also there are particles which have mass about the nucleon mass of
938MeV are known as baryons while particles having mass greater than nucleon mass are
called hyperons. This particle classification can be summarized as:
Light mass particle (lepton) – mass: 0-134MeV
Intermediate mass particle (meson):
Heavy mass particle (baryons and hyperons) – mass
(b) Classification on the basis of spin: Elementary particles are broadly classified as
fermions, having half integral spin and bosons, having integral spin. Bosons consist of
mesons and gravitons while leptons and baryons come under fermions. Electrons, muons and
tau particles and their corresponding neutrinos are called leptons while baryons are spin half
particles (proton, neutron and hyperons) and spin 3/2 particles (resonance states). This
classification is summarized as follows-
(c) Classification on the basis of interaction: Those particles which take part only in
electromagnetic and weak interactions are called leptons while those which participate in
strong interactions are known as hadrons.
Leptons Possible interaction
Electron Electron neutrino
( )
Electromagnetic and weak interaction
Muon Muon neutrino
Tau Tau neutrino
Hadrons Strong interaction
Mesons:
Baryons: (proton (p), neutron (n), hyperons
(0,
,0,
0,-,
-))
5.3 Fundamental interactions
The four fundamental interactions occurring in nature are described as follows-
(i) Gravitational interaction: The gravitational force between two nucleons separated
by a nuclear diameter is
Thus we see that it is extremely weak force and plays no role in nuclear interactions. In
nineteenth century, forces were thought to be propagated by fields and in twentieth century,
these fields are explained in terms of messengers that actually propagate the effect.
Gravitation is thus explained in terms of gravitons. Graviton has mass zero and they move
with velocity of light. Gravitational fields are extremely weak.
(ii) Electromagnetic interaction: For interaction of point charges, Coulomb law is
given by
For two protons, separated by nuclear diameter of 10-15
m, the repulsive force is
about 1035
times greater than the gravitational force. When charge is accelerated, the energy
is radiated in the form of electric and magnetic pulses. This pulse is called photon which
travels with velocity of light. Thus, interaction between charged particles is due to the
exchange of these photons. is called fine structure constant and is due
to photon exchange. The mutual annihilation of particles and antiparticles is an example of
electromagnetic interaction.
(iii) Strong interaction: Strong nuclear interaction is independent of electric charge.
This force is same for proton-proton and neutron-neutron interaction. Strong interaction can
be characterized by the life time of the particles produced. If particles decay in 10-23
second
then interaction is called strong. These interactions involve mesons and baryons. The mesons
and baryons together are called hadrons. The range of strong interaction is very short in
comparison to electromagnetic and gravitational interaction. Yukawa in 1935 predicted that
like photons, pions are exchange particles for nucleons. The mass of these pions is about 200-
300 times that of electron mass. The strength of nuclear interaction is represented by
magnitude of dimensionless coupling constant It is about 1000 times
greater than electromagnetic coupling constant . Strong interactions are responsible for
kaon production whereas decay of mesons, nucleons and hyperons proceed through
electromagnetic or weak interaction.
(iv)Weak interaction: Weak interaction is responsible for decay of strange and non
strange particles. The dimensionless weak coupling constant is of the magnitude
. Weak interaction involves the change in strangeness of baryons or mesons
and this change in strangeness must be equal to change in charge. The change in isospin
must not be zero. Like strong, electromagnetic and gravitational interaction, weak interaction
is also mediated by charged particles called „intermediate charged particles‟ or „vector
bosons‟ and symbolically represented as W. These particles are most massive. Weak
interaction range is minimum among all the interactions. Neutrinos and antineutrinos are also
associated with weak interaction as they are most weakly interacting particles.
All these interactions are summarized in following table 1.
Table 1: Four fundamental interactions
Interaction Particle
affected
Range
(meter)
Relative
strength
Particle
exchanged/mass
Role in universe
Strong Quarks,
hadrons
1 Gluons, mesons
)
Holds quarks together to
form nucleons, holds
nucleons together to form
atomic nuclei
Electromagnetic Charged
particles
Photons
(mass zero)
Determines structure of
atoms, molecules and
solids
Weak Quarks,
leptons
Intermediate
bosons, W, Z
(mass
)
Mediate transformations
of quarks and leptons
Gravitational All Graviton
(mass zero)
Assemble matter into
planets, stars and
galaxies
5.4 Quantum numbers of elementary particles
The various quantum numbers associated with individual particles are as follows:
(i) Orbital quantum number ‘l’: This gives the value of orbital angular momentum with
respect to centre of reference. The numerical value of orbital angular momentum is
where l is a positive integer.
(ii) Magnetic orbital quantum number: This gives the component of orbital angular
momentum l in the direction of externally applied magnetic field. Quantum number ml takes
values from –l to +l, i.e. (2l+1) values.
(iii) Spin quantum number (s): The intrinsic angular momentum of particle is called spin
and is represented by quantum number s. The numerical value of spin angular momentum is
. Quantum number s takes half integer values for particles obeying Fermi Dirac
statistics called fermions while s is integer for bosons obeying Bose Einstein statistics.
(iv) Magnetic spin quantum number (ms): This is the component of s in the quantization
direction. For s = 1/2, ms = +1/2 and -1/2 corresponding to parallel and anti-parallel spin
alignment.
(v) Total angular momentum quantum number (j): This is given by . The
magnitude of angular momentum corresponding to given j value is .
(vi) Isospin (T): Since nuclear forces are charge independent, they cannot distinguish
between neutron and proton as they appear only different charge states of same particle called
nucleon. Hence idea of isotopic spin T and its projection T3 along certain axis was adopted to
differentiate the various charge states of the same particle. For example, nucleons have
isospin quantum number T = 1/2, so that 2T+1= 2 and hence there are two possible values of
T3. Proton has been assigned as T3 = +1/2 and neutron has T3 = -1/2. Similarly for pions, T =
1, so that 2T+1=3 and hence the three charge states of pions viz. π+, π
0, π
- have been assigned
T3 values as +1, 0, -1 respectively. If Q is the charge on the particle, B is its baryon number
and T3 be the third component of isospin, then we have the relation,
(vii) Strangeness: Kaons and hyperons are never produced separately but always produced in
association with each other via strong interaction called associate production. Examples of
associate production are:
; ;
If these are produced via strong interaction, they should decay also by strong interaction with
the life time of the order of 10-23
second, but it is found that decay of these particles takes
place with the life time of 10-10
second. Thus their unusual stability against decay termed
them as strange particles and then a new quantum number called strangeness quantum
number was assigned to them. The decay of strange particles (kaons and hyperons) is
governed by weak interaction.
(viii) Parity: In quantum mechanics, particle is described by means of wave function
which is a function of all its position and spin coordinates. Parity is related to the symmetry
of the wave function. If the wave function remains unchanged under space inversion r-r,
then parity is conserved and if wave function changes then parity is not conserved. If P is the
parity operator, then
which means eigen values of P = 1. Hence parity is conserved when P = +1 and parity is
violated when P = -1. Nucleons and electrons are assigned positive or even intrinsic parity.
All the hyperons also have positive parity. π meson, K meson and meson have negative
intrinsic parity. All spin 1/2 particles (fermions) have positive parity and their antiparticles
have opposite (negative) parity. For bosons, particles and antiparticles have same parity.
(ix) Charge conjugation (cc): Charge conjugation is the symmetry operation in which every
particle of the system is replaced by its antiparticle. If antimatter or anti-system exhibits the
same phenomenon, we say that charge conjugation symmetry is preserved. Let and
are the wave function of proton and neutron. Using the creation operators (t3 is third
component of isospin) for a particle, proton and neutron wave function can be written as
and where is wave function for
vacuum.
The wave function for an antiproton and antineutron can be written in the same way using
creation operator such that
and
using the fact that third component of isospin changes sign for particle and its antiparticle.
The two creation operators are related by a phase factor given by
Using the above relation, the proton and neutron wave function change to antiproton and
antineutron wave function under charge conjugation as
5.5 Conservation laws
There are two types of conservation laws:
(A) Exact conservation laws (B) Approximate conservation laws
(A) Exact conservation laws: These are the conservation laws which are obeyed in
every reaction. These laws consist of conservation of momentum, energy, charge, baryon
number and lepton number. Below, we briefly describe these laws.
(i) Conservation of momentum: Conservation of angular momentum involves both types
(orbital and spin) of momentum together. First one is motion of object about external chosen
axis and other one is due to its intrinsic motion about centre of mass. Fermions have half
integral spin. Strongly interacting bosons (, K, ) are zero spin particles. Leptons (e, , , e,
μ, ) are spin half particles. Massless boson (photon) has spin 1. Graviton has spin 2.
(ii) Conservation of energy: In the reactions, sum of the rest energy associated with mass,
kinetic energy or potential energy always remains constant. Thus
because rest energy of K0
is not enough to produce four pions, however
is allowed. Other example is the decay of neutron into proton with electron emission,
which is possible, whereas decay
of proton into neutron with positron emission ( ) is not possible in free
space.
(iii) Conservation of charge: Charge is conserved in all nuclear reactions. All elementary
charges are 0 or 1. Fractional charges are carried by quarks.
(iv) Conservation of baryon number: The number of baryons minus number of anti-
baryons is always conserved i.e. net baryon number remains unchanged. All normal baryons
have B = +1 and anti-baryons etc. have B = -1. All
mesons have B =0.
Example (i)
Reaction is allowed as B = 1-1 = 0
(ii)
Reaction is forbidden as B = -1-1= -20
(v) Conservation of lepton number: The number of leptons minus number of anti-leptons is
always conserved i.e. net lepton number remains unchanged. Lepton number is further
divided into three categories Le, L, L.
For , . Similarly, for , and for ,
Through lepton number conservation it is possible to say that neutrino and
antineutrino are two different particles. For this we consider the reaction,
This reaction does not occur as = +1+1 = +20.
While for the reaction;
= +1-1 = 0, so this reaction is allowed which clearly shows that and are two
different particles. In any reaction, Le, L, L numbers are separately conserved. For
example-
Le =0; Lμ =0
(B) Approximate conservation laws: Conservation of isospin, hypercharge,
strangeness, parity, charge conjugation and time reversal operation come under the
approximate conservation laws. These quantities are not conserved in every interaction. For
example, isospin and strangeness is conserved in strong interaction but not in weak
interaction. Parity is also violated in weak interaction. These approximate conservation laws
are discussed below-
(i) Conservation of isospin: Isospin is a quantum number used to differentiate the various
charge states of the same particle. Isospin is conserved in strong and electromagnetic
interaction but not in weak interaction.
(ii) Conservation of hypercharge: Hypercharge is a quantity which is defined as twice the
average charge on the group of particles. For pion group (π+, π
0, π
-), average charge is zero
so that hypercharge is also zero. For kaons (K+, K
0), average charge is 1/2 so hypercharge is
+1. For other kaon group , average charge is -1/2 so hypercharge is -1. Hypercharge
„Y‟ can also be defined as
The hypercharge, Y= B + S (sum of baryon number and strangeness quantum number).
(iii) Conservation of strangeness: Strangeness is conserved in strong and electroweak
interaction but not in weak interaction. In fact S=0 for strong interaction and S= 1 for
weak interaction. The decay of kaon and hyperon is governed by weak interaction.
Example:
S = 0, so this reaction is allowed by strong interaction.
S = 0- (-1) = +1, strangeness is not conserved and this reaction corresponds to weak
interaction.
(iv) Conservation of parity: Parity is conserved in strong and electromagnetic interaction
but is violated in weak interaction.
(v) Conservation of charge conjugation (cc): Charge conjugation is the symmetry operation
in which every particle of the system is replaced by its antiparticle. Gravitational and
electromagnetic interactions are charge conjugation invariant. The gravitational interaction
given by remains charge conjugation invariant.
The electrostatic interaction remains charge conjugation (c.c.) invariant as under c.c.
operation, . Similarly, for electromagnetic interaction given by
we have under c.c. operation, and so that R.H.S.
remains same and interaction remain invariant.
(vi) CP invariance: CP operation is the combined effect of charge conjugation and parity
operations. There are reactions which are not allowed by the application of these operations
separately but the combined operation allows the reaction. Parity operation changes the
handedness of the particle i.e. a left handed particle gets converted to right handed particle
and vice versa under parity operation and charge conjugation operation replaces particle by
its antiparticle. To see the effect of CP operation we take the example of decay of positive
pion,
(L means left handed particle)
Charge conjugation: (This is an impossible process and not
allowed)
Parity operation: (This is an impossible process and not
allowed)
Combined operation: (This is possible process and allowed)
Thus under CP operation, a left handed neutrino gets converted to right handed antineutrino
and decay of negative pion becomes possible.
(vii) Time reversal: Time reversal is the operation which reverses the direction of time or
direction of all motions. Under this operation, displacement, acceleration and electric field
remains invariant while momenta, angular momenta and magnetic fields invert their signs. In
quantum mechanics *(x, -t) is the time reversal wave function of (x, t) so that under time
reversal operation, . An example of time reversal process is the
creation of e+, e
- pair by the collision of two photons. Strong, electromagnetic and
gravitational interactions are time reversal invariant while in K0 decay, time reversal
symmetry is not obeyed.
(viii) CPT theorem: CPT means combined operation of charge conjugation, parity and time
reversal. An important theorem in relativistic quantum mechanics is that if all the physical
processes could be described in terms of relativistic field equations, the physical laws must
be invariant under combined operation of C.P.T. This is known as C.P.T. theorem. No
violation of invariance of C.P.T. has been observed so far.
Below we present a table in which all the baryons, mesons and leptons are given along with
their various quantum numbers.
Table 2: Elementary particles with their quantum numbers (T3: Third comp. of isospin)
Particle Lepton number
(L)
Baryon
number (B)
Strangeness
(S)
Hypercharge
(Y=B+S)
Spin
(s)
Isospin
(T3)
Baryons
p+ 0 +1 0 +1
+1/2
p- 0 -1 0 -1 -1/2
n0 0 +1 0 +1
-1/2
0 -1 0 -1
+1/2
0 +1 -1 0
0
0 +1 -1 0
+1
0 +1 -1 0 0
0 +1 -1 0
-1
0 +1 -2 -1 +1/2
0 +1 -2 -1 -1/2
0 +1 -3 -2 3/2 0
Leptons
e-, Le = +1, Lμ = L = 0 0 0 0 1/2 0
e+, Le = -1, Lμ = L = 0 0 0 0 1/2 0
Lμ = +1, Le = L =0 0 0 0 1/2 0
, Lμ = -1, Le = L = 0 0 0 0 1/2 0
, L= +1, Le = Lμ = 0 0 0 0 1/2 0
, L= -1, Le = Lμ = 0 0 0 0 1/2 0
Mesons
0 0 0 0 0 0
K+ 0 0 +1 +1 0 +1/2
K0 0 0 +1 +1 0 -1/2
0 0 -1 -1 0 +1/2
0 0 -1 -1 0 -1/2
0 0 0 0 0 +1
0 0 0 0 0 0
0 0 0 0 0 -1
Example: The nuclear reaction takes place through strong interaction.
Identify the unknown particle X.
Sol. For strong interaction, all conservation laws are obeyed. Applying following
conservation laws to above reaction-
(i) Charge conservation (Q): -1 + 1 = +1 + QX QX = -1
(ii) Baryon number conservation (B): 0 + 1 = 0 + BX BX = +1
(iii) Strangeness (S): -1 + 0 = +1 + SX SX = - 2
(iv) Third component of isospin (T3): -1/2 + 1/2 = +1/2 + T3X T3X = - 1/2
Thus the particle is negatively charged baryon with strangeness = -2 and T3 = -1/2. This is
hyperon.
Q.1 For the following reactions, identify the unknown particle X.
(i) (ii)
Q.2 Identify the following interactions as strong, electromagnetic or weak.
(i) (ii) (iii) (iv)
(v)
Q.3 Determine which of the following reactions are allowed or forbidden by (a) charge
conservation (b) baryon number (c) strangeness (d) Third component of isospin conservation.
(i) (ii) (iii) (iv)
5.6 SU(N) group and its properties
A unitary quadratic matrix with n rows and n columns can be written as where
is Hermitian quadratic matrix with n rows and n columns. All such matrices form a group
under matrix multiplication. This group is called U(N) which stands for „unitary group in n
dimensions‟. Since is Hermitian, the diagonal matrix elements are real, i.e. and
Thus and therefore allows for n2 independent parameters.
The trace of Hermitian matrix is real. For a Hermitian matrix , the unitary condition is
Now
Again , hence
Now if This group is called „special unitary group‟ in n
dimensions. Special stands for unit determinant value of . It depends on n2-1 real
parameters and is denoted by SU(N).
5.6.1 SU(2) group and spin matrices: The two dimensional matrices of SU(2) contain
parameters. The generators are three linearly independent traceless matrices. The
Pauli spin matrices given by
are linearly independent Hermitian matrices which span matrix space
completely. Since are traceless, so they can be treated as
generators of SU(2). The algebra or commutations of generators of the group is called Lie
algebra. The commutation relation of is
Also we can write, as generators, satisfying the commutation relations
where
These relations define Lie algebra of SU(2) group. Since no pair of generators
commute, hence maximum number of commuting generators is one which in turn implies that
rank of SU(2) is one.
5.6.2 Lie algebra of SU(3) group: The special unitary group in three dimension is
SU(3) having generators, given by . The three generators can be
formed from SU(2) generators extending to three dimension
The traces of vanish as required. The remaining five generators are
and .
All ‟s are Hermitian and traceless. Factor is added in so that the relation
holds. The generators and are derived from and generators and
are derived from . The commutation relation is identical to that of Pauli matrices and is
given by
where structure constants are totally anti-symmetric under the exchange of any two
indices,
The generators of SU(3) group can be redefined as and hence
The spherical representation of operators are given by
The maximum number of commuting generators of SU(3) Lie algebra are 2 represented as
Hence rank of SU(3) is 2.
5.6.3 Classification of particles under SU(3) symmetry: In SU(3) group, there is
a particular method of classifying the strongly interacting particles into groups of 1, 8, 10 and
27 numbers. This is often known as eight fold way because it covers the relations between
eight conserved quantities. The choice of name is attributed to the teachings of Buddha:
“Now this, O monks, is noble truth that leads to the cessation of pain: this is the noble eight
fold way: namely, right views, right intension, right speech, right action, right living, right
effort, right mindedness, right concentration”.
All the mesons form nonet and are constructed with quark-antiquark pair while baryons form
nonet and decuplet and are composed of three quarks. Quarks are basic building blocks from
which the strongly interacting particles or hadrons are made of. Quarks will be discussed in
detail later in this unit. At present it is sufficient to know that there are three types of quarks-
up (u), down (d) and strange (s).
5.6.3.1 Meson multiplets
(i) Pseudo-scalar mesons : With two quarks u, d and anti-quarks we can
construct a total of pseudoscalar mesons (three π mesons, one meson) but when
s is also included, a total of pseudoscalar mesons are constructed which are
separated in two groups as Eight of these mesons form an octet which transform into
each other under a rotation in flavor space i.e. interchanging among u, d, s quarks, the wave
function of eight mesons transform into each other as an irreducible representation of SU(3)
group. These eight mesons of spin zero and odd parity can be grouped to form a
meson octet. The singlet pseudo-scalar meson is with Y=0, T3= 0. Sometimes this
particle is joined to the point Y=0, T3= 0 in the octet. Then the diagram is called „a pseudo-
scalar meson nonet‟ as shown in below in fig 1. The properties and quark constituents of
these nine pseudo-scalar mesons are given in table 3.
Table 3. Pseudo-scalar meson nonet (8+1), their properties and quark constituents
Mesons Quantum numbers Mass
(MeV/c2)
Mean
lifetime(sec)
Main
Decay
modes
Quark contents
T T3 B S Y=B+S
1 +1 0 0 0 139.6
1 0 0 0 0 135.0
2
1 -1 0 0 0 139.6
0 1 1 493.8
0 1 1 497.8
0 -1 -1 497.8
0 -1 -1 493.8
0 0 0 0 0 548.6
2
0 0 0 0 0 958.0
2
(ii) Vector mesons :
Vector mesons have spin 1 and parity odd. This group consists of kaon resonances and
rho ( ), omega ( ) and phi ( ) meson. These resonances again form an octet similar to the
corresponding octet of pseudo-scalar mesons ( ). As is singlet, phi ( ) meson is
singlet too and a nonet is formed again if is mixed with octet of vector mesons. The vector
mesons, especially and mesons play an important role in the theory of nuclear forces
between two nucleons at small distances. They give rise to repulsive contribution to the
strong interaction between the nucleons if they come sufficiently close together. The
properties and quark constituents of this nonet of vector mesons are given in table 4. The
corresponding Y-T3 plot is shown in figure 2.
Table 4. Vector mesons nonet (8+1), their properties and quark constituents
Mesons Quantum numbers Mass
(MeV/c2)
Mean
lifetime(sec)
Main Decay
modes
Quark
contents T T3 B S Y=B+S
1 +1 0 0 0 773.0
1 0 0 0 0 773.0
1 -1 0 0 0 773.0
0 1 1 892.0
0 1 1 898.0
0 -1 -1 898.0
0 -1 -1 892.0
0 0 0 0 0 782.7
0 0 0 0 0 1019
5.6.3.2 Baryon octet:
Baryons are spin 1/2 particles with positive parity. This group consists of eight particles
which are neutrons, protons and hyperons. Hyperons consist of lambda
( particles. Sigma hyperon is spin 3/2 particle and
belongs to the group of baryon decuplet. All these baryons are constructed by taking three
quarks at a time viz. qqq type structure. The three quarks used for constructing baryon octet
are up (u), down (d) and strange (s). The reason for taking three quarks is that quarks are spin
1/2 particles and baryons are also spin half integer particles. The properties and quark
contents of the octet of spin 1/2 baryons are given in table 5. The corresponding Y-T3
diagram is plotted in figure 3.
Table 5. Baryons octet, their properties and quark constituents
Baryons Quantum numbers Mass
(MeV/c2)
Mean
lifetime(sec)
Main Decay
modes
Quark
contents T T3 B S Y=B+S
1 0 1 938.3
uud
N
1 0 1 939.6 (15 min) udd
0 0 1 -1 0 1116
uds
1 1 -1 0 1189
uus
1 -1 0 1192 uds
1 -1 0 1197 dds
1 -2 -1 1315
uss
1 -2 -1 1321
dss
5.6.3.3 Baryon Resonances decuplet:
In the pion nucleon scattering, , the intermediate state is formed
which is a short lived resonance state. The nature of these resonances is similar to that of
compound nuclei in nuclear physics. The above intermediate state with mass equal to 1232
MeV is actually a charge quartet, denoted by .
Thus resonance is a state. Baryon resonances may be supposed as excited baryons.
Other resonances are , which form an iso-doublet and , forming an iso-triplet. Below in
table 6, we outline the properties and quark constituents of this spin 3/2 baryon resonance
decuplet and the corresponding Y-T3 plot is shown in figure 4.
Table 6. Baryon resonances decuplet, their properties and quark constituents
Baryon
resonances
Quantum numbers Mass
(MeV/c2)
Mean
lifetime(sec)
Main Decay
modes
Quark
contents T T3 B S Y=B+S
1 0 1 1232.0
uuu
1 0 1 1232.0 uud
1 0 1 1232.0
udd
1 0 1 1232.0 ddd
1 1 1 -1 0 1382.3 uus
1 -1 0 1382.0 uds
1 -1 0 1387.4 dds
1 -2 -1 1531.8
uss
1 -2 -1 1535.0
dss
0 0 1 -3 -2 1672.0 sss
5.6.
4
Discovery of quark
Only the Delta, Sigma and Xi resonances, a total of nine particles were known up to 1963 in
the baryon decuplet and the tenth particle was missing. SU(3) symmetry predicted that this
particle should have S = -3,Y = -2, T = T3 = 0. The possible decay mode is .
The expected mass of baryon is 1677 MeV as mass of baryons are 1115 MeV
and 496 MeV respectively. The search for was soon undertaken at Brookheaven and it
was produced in 1964 with the mass of 1683 MeV without any ambiguity. This discovery
presented the remarkable success of SU(3) classification of elementary particles.
5.7 Gell-Mann Okuba mass formula
Gell-Mann and Okuba obtained a mass formula for the several charge multiplets observed in
baryon and meson SU(3) classifications. For baryons and mesons, formula is as follows:
(A) For Baryon multiplets
where are constants, Y is hypercharge and I is spin of the particle.
(i) Baryon octet
Using equation (1), the masses of spin 1/2 baryons can be expressed in terms of constants a
and b.
Nucleon: Y= 1, I = 1/2
(2)
Sigma: Y= 0, I = 1
(3)
Lambda: Y = 0, I = 0
(4)
Psi: Y = -1, I = 1/2
(5)
Adding equation (2) and (5), we have
(6)
and from equation (3) and (4), we have
(7)
On combining equations (6) and (7), we get the following relation between the masses of
various multiplets of baryon octet
(8)
(ii) Baryon decuplet
For resonance particles, (9)
And for the particle
All these values satisfy the relation . Putting this relation in equation (1), we get
(10)
Using Y=1, 0, -1 and -2 for , , particles in equation (10), we get
From these equations, we have the following relation for baryon decuplet,
(11)
(B) For meson octet
For mesons, the Gell-Mann Okuba mass formula is given by
(12)
For these particles
And for
So that
From these equations, we eliminate a and b and finally obtain
This is the mass relationship between mesons of 0- octet.
5.8 Quark model
Gell Mann and G. Zweig proposed quark model in 1964 according to which hadrons are
made up of more fundamental particles called quarks. Initially there were three fundamental
quarks, up (u), down (d) and strange (s) and they carry fractional charges and fractional
baryon numbers. The up (u) and down (d) quarks have same mass, around 0.39GeV/c2 while
strange (s) quark is more massive, around 0.51GeV/c2. It was proposed that baryons are made
up of three quarks. Since each baryon has got baryon number B=1, hence each quark has
been assigned a baryon quantum number B = 1/3. Also factional charges have been assigned
to quarks as each baryon has integral charge. Mesons are constructed with one quark and one
anti-quark pair. In 1974, at SLAC and Brookhaven National Laboratory, a new particle called
was discovered. was shown to have the structure which gave rise to the existence
of new quark called charm (c). Later the mesons with charm +1 and -1 and baryons with
charm +1, +2 were observed. Charm quark has rest mass greater than those of up, down and
strange quarks, around 1.65 GeV/c2. The birth of c quark prompted the search for even
heavier quarks. In this way, the presence of bottom or beauty (b) quark was found in the
narrow resonance Upsilon state ( ), observed at energies of 10GeV in 1977. This put the rest
mass of b quark at around 5GeV/c2. The top (t) quark postulated to be even heavier has so far
escaped experimental confirmation. The list of six quarks with their charge and quantum
numbers are given in table 7.
Table 7: Quantum numbers of quarks
Flavor Mass
(GeV/c2)
Charge
Q (e)
Quantum numbers
B T T3 S C
u (up)
+2/3 1/3 ½ 1/2 0 0 0 0
d (down)
-1/3 1/3 ½ -1/2 0 0 0 0
s (strange)
-1/3 1/3 0 0 -1 0 0 0
c (charm)
+2/3 1/3 0 0 0 1 0 0
b (bottom)
-1/3 1/3 0 0 0 0 -1 0
t (top)
+2/3 1/3 0 0 0 0 0 1
Abbreviations: B: baryon number, T: isospin, S: strangeness, C: charm, : bottom, : top
In the Standard Model of Particle Physics which is a highly successful theory in which
elementary building blocks of matter are divided into three generations of two kinds of
particle - quarks and leptons, may be summarized as:
GENERATION I II III
QUARKS
LEPTONS
Beside the six types of quarks as stated above, the family consists of three flavors of charged
leptons, the electron (e¯), muon (μ¯) and tau (η¯) together with three flavors of neutrinos - the
electron neutrino (νe), muon neutrino (νμ) and tau neutrino (νη).
5.8.1 Quark structure of mesons: To see the quark structure of mesons, especially for
pions, we start with π- meson. The mesons are created with one quark (q) and one anti-quark
( ) pair. π- meson has T=1 and T3 = -1. The only pair which has T=1 and T3 = -1 is .
Hence
(14)
For the π0 meson, T=1 and T3 = 0. Hence wave function of π
0 can be constructed from π
-
meson using the isospin raising operator which acting on d and quark gives
and
Using the normalization factor of , the final wave function for π0 meson comes out as
Again applying the isospin raising operator on , the wave function for can be
written as (16)
The wave function for meson (T=0, T3 =0) is orthogonal to in equation (15) and is
given by
(17)
The remaining K mesons are hadrons with nonzero strangeness S and can be constructed with
s quark. They come out as two isospin doublets, one doublet consisting of K+ ( ) and K
0
( ), and the other doublet and .
5.8.2 Quark structure of baryon resonances: For the wave function of , all the
three quarks have same flavor. The complete symmetric wave function in flavor is symmetric
product of the wave functions of three u quarks,
(18)
Wave function for can be found by using isospin lowering operator ( ) to equation (18).
Since consists of uud quark combination, hence any one of the three u quarks in can
be changed to d quark. This in turn implies that wave function of is a linear combination
of all three possibilities with equal weights. The normalized and symmetrized wave function
of is therefore
or simply
Similarly applying isospin lowering operator ( ) to once, we get and on twice we
get resonance state. The wave functions are
(21)
The baryon resonance has and can be obtained from by replacing one u
quark with s quark. The normalized wave function for is therefore
Now isospin lowering operator ( ) is applied to to get state and then operation of
( ) to gives resonance state. Since s quark is isospin 0 particle, hence isospin
lowering operator changes only u quark to d quark. The wave functions are
The wave function for strangeness members of the decuplet can be constructed easily
and found to be
For ( baryon, the only possibility is
since the baryon number of is 1, its charge is -1 and strangeness quantum number is -3.
5.8.3 Quark structure of baryon octet: The wave function of each member of this
group is anti-symmetric under the combined exchange of both flavor and intrinsic spin. The
combined symmetry of the spin and flavor parts of the wave function is determined after
assigning a flavor to each one of the three quark involved. For example, the wave function of
proton is first written by assigning the first two quarks with different flavors i.e.
The combination of spin and flavor is symmterized in two stages. First the process is carried
out for first two quarks to get
Next all the other terms are generated by making the permutation P31 and P32 on each of the
four terms, which gives twelve terms in all. Grouping the identical terms together we finally
get
In above expression, quark numbers 1, 2 and 3 are omitted but the order is retained. The
neutron wave function can be written from proton wave function by replacing all the u quarks
by d quarks and vice versa. Similarly the wave function of all the other members of octet can
be written from proton wave function.
5.8.4 Magnetic dipole moment of baryon octet: The magnetic dipole moment of
baryon comes from two sources: the intrinsic dipole moments of constituent quarks and
orbital motion of quarks. Here the three quarks involved are symmetric in spatial part of their
wave function with relative motion between them in l=0 state, hence there is no contribution
to magnetic dipole moment from orbital part of motion. The operator for magnetic dipole
moment is given by
where s is the operator for intrinsic spin. The dipole moment is measured in units of nuclear
magneton where , being the mass of quark. For a particle having
intrinsic spin 1/2, g is equal to 2. Since the masses of u and d quarks are not known exactly,
their masses may be assumed as equal. The ratio of their magnetic dipole moment is then
given by ratio of their charges i.e.
(30)
To determine the dipole moment of proton, the number of and are counted
explicitly from the proton wave function given in equation (29). This is done by summing the
squares of the coefficients in the wave function for each one of the four possible
combinations of flavor and spin orientations. The sum of squares of coefficients is 5/3 for
number of u quarks with spin up, 1/3 for number of u quarks with spin down, 1/3 for number
of d quarks with spin up and 2/3 for number of d quarks with spin down. The net u quark
contribution to proton dipole moment is then 5/3-1/3=4/3 and d quark contribution is 1/3-2/3
= -1/3, so that the final result is
The neutron magnetic dipole moment is obtained by interchanging u and d quarks i.e.
The baryon is made of two u quarks and one s quark. Comparing with proton constituent
, it is seen that only difference between and proton is that one d quark in proton
replaces s quark in . Hence to get the magnetic dipole moment of , d quark is replaced
by s in magnetic dipole expression of proton, with the result
The quark content of baryon is . If it is compared with neutron constituent , the
only difference is the replacement of s quark by u quark. Hence using neutron dipole moment
from equation (32), the magnetic dipole moment of is given by
Using similar method, the dipole moment of and baryons are
Finally, the remaining two baryons and have quark contents which have three
different flavors. Hence using isospin analysis, the dipole moments are calculated with the
final results
5.8.5 Colour degree of freedom: Every quark has an additional degree of freedom
called colour degree of freedom besides flavor. The need of this additional quantum number
comes from the quark structure of particle which is . The intrinsic parity of is
positive. Thus the spatial part of wave function for three quarks in is symmetric. The
intrinsic spin of is 3/2, hence intrinsic spin part of wave function is also symmetric.
Similarly the isospin part of wave function is also symmetric as isospin of is 3/2. Thus
the product of space, intrinsic spin and isospin part of the wave function of is symmetric
under a permutation among three quarks. On the other hand, quarks are fermions and the
Pauli Exclusion Principle requires that total wave function of three identical quarks must be
anti-symmetric with respect to the permutation of any two of the three quarks. Thus the wave
function of Pauli Exclusion Principle unless there is another degree of freedom
for quarks in which the wave function is anti-symmetric. This is the colour degree of
freedom. Thus a colour is assigned to each quark for example R (red), G (green) and B (blue).
The net colour in hadron must vanish as hadrons are colourless. For mesons it is easy to see
as quark of one colour is neutralized by antiquark of opposite colour. The antiquarks are
assigned three anticolours as corresponds to cyan where as and correspond
to magenta and yellow.
The other evidence for colour quantum number comes from the experiment at colliders.
The ratio of annihilation cross section is
where is the quark charge. If u, d, c, s, t and b quarks contribute to above ratio at energy
of 10 MeV, then for colourless quarks
If there exist several colours then this value must be multiplied by number of distinct colours.
In case of three colours,
Experimentally for energies about 10GeV, ratio R is about 4, thus satisfying the three colour
hypothesis.
5.9 Quantum Chromodynamics (QCD)
The name Quantum Chromodynamics (QCD) stands for the theory which deals with strong
interaction involving coloured quarks. The quark-quark interaction is mediated by particles
called gluons which themselves are coloured. There are eight gluons made of different
combinations of R, B and G. The three colours are expressed as
and their anticolours
Thus nine possible reactions are
, ,
,
The colour octet of gluons has following combinations:
; ; ; ;
;
This colour octet can be expressed in form of matrices as
,
‟
,
‟
, .
These are equivalent to eight generators of SU(3) group given earlier.
The fundamental difference between QCD and QED (Quantum Electrodynamics) is that in
QED, the mediating photons do not carry electrical charges whereas in QCD, gluons carry
colour charges. QCD is a non-abelian gauge field theory. Its two important features are
asymptotic freedom and infrared slavery. Asymptotic freedom means at short distances, the
interaction between quarks become weak and they behave as free particles where as at large
distances the interaction becomes stronger and stronger which leads to permanent
confinement of quarks within colour singlet objects. This latter feature is known as infrared
slavery. Due to this colour confinement, quarks are not seen as free particles. Several bag
models in which quarks are assumed to be inside the „bag‟ have been developed to account
these features.
5.10 Summary
Now we summarize what we have discussed so far:
We have learnt that elementary particles may be classified on the basis of their
masses, spin and interactions in which they participate.
There are four fundamental interactions known as gravitational, electromagnetic,
weak and strong interactions.
Mesons and baryons [proton (p), neutron (n), lambda (), sigma (), psi (),
omega ()] are known as hadrons and they take part in strong interactions.
There are various quantum numbers associated with elementary particles like, baryon
number, lepton number, isospin, strangeness, parity, hypercharge etc.
Kaons (K) and hyperons ( ) are called strange particles as they are produced
through strong interaction but decay through weak interaction. Strange particles are
always produced in pairs called associate production.
There are two types of conservation laws known as exact and approximate
conservation laws. Exact conservation laws like baryon number, lepton number and
charge conservation are obeyed in every nuclear interaction whereas approximate
conservation laws like isospin, parity and strangeness are not followed in weak
interaction.
SU(3) is a special unitary group in three dimension having eight generators.
In SU(3) group, there is a particular method of classifying the strongly interacting
particles into groups of 1, 8, 10 and 27 numbers. This is often known as eight fold
way.
In SU(3) scheme, pseudo scalar mesons and vector mesons form a nonet (8+1) of
particles where as baryons form an octet and baryon resonances form a decuplet.
Gell-Mann Okuba mass formula correlates the masses of various particles of a
multiplet among themselves.
Gell Mann and G. Zweig proposed quark model in 1964 according to which hadrons
are made up of more fundamental particles called quarks.
There are six types of quarks known as up (u), down (d), strange (s), charm (c),
bottom (b) and top (t).
Baryons are composed of three quarks u, d and s where as mesons are made of one
quark and one anti-quark pair.
Magnetic dipole moment of baryons can be expressed in terms of magnetic dipole
moments of u, d and s quarks.
Besides flavor, quarks are assigned another degree of freedom called „colour‟. There
are three colours known as R (red), G (green) and B (blue). Consequently, quarks are
described as coloured quarks or red quark, green quark and blue quark and the theory
which deals with strong interaction involving coloured quarks is known as Quantum
Chromodynamics (QCD).
5.11 Answer to questions
Ans 1: (i) (ii)
Ans 2: (i) strong (ii) electromagnetic (involves ) (iii) weak (involves antineutrino) (iv)
weak (decay of strange ) (v) weak (involves antineutrino)
Ans 3: (i) forbidden (S, T3 not conserved); (ii) Allowed; (iii) forbidden (Q not conserved);
(iv) Allowed (weak decay, T3 not applicable).
5.12 References
Quantum Mechanics: Symmetries; Walter Greiner & Berndt Muller, Springer
Publication.
Introductory Nuclear Physics; Samuel S.M. Wong, Prentice Hall of India Ltd.
Elements of Nuclear Physics; M.L. Pandya & R.P.S. Yadav, Kedar Nath Ram Nath
Publication.
Nuclear Physics; V. Devanathan, Narosa Publishing House.
Groups, Representations and Physics; H.F. Jones, Institute of Physics (IOP)
Publishing.
Nuclear Physics, D.C. Tayal, Himalaya Publishing House.
Quarks and Leptons; F. Halzen and Alen D. Martin, John Wiley & Sons.
5.13 Questions
1. What are elementary particles? Discuss their classification schemes.
2. Describe the different types of interactions that occur between the elementary particles.
Give their relative strengths.
3. What are strange particles? Discuss about their strangeness quantum number.
4. Discuss the isospin, parity and charge conjugation properties of elementary particles.
5. What are exact conservation laws? Discuss baryon number and lepton number
conservation with examples.
6. Discuss various exact conservation laws. What is CPT theorem?
7. Explain why reaction occurs but does not occur?
8. What do you mean by SU(3) group? Discuss its Lie algebra.
9. What are hadrons? Describe their SU(3) classification.
10. How SU(3) classification scheme helped in the discovery of omega () particle?
11. Give the members of meson 0- octet, baryon (1/2)
+ octet and baryon (3/2)
+ decuplet.
12. Draw the isospin (T3) versus hypercharge (Y) plot for baryon decuplet and give their
quark constituents.
13. Write the isospin (T3), strangeness (S) and hypercharge (Y) quantum numbers of baryon
octet and draw its T3 versus Y plot.
14. Write the Gell-Mann Okuba mass formula and correlate the masses of baryon and meson
multiplets among themselves.
15. How many types of quarks are there? Give their various quantum numbers.
16. What are the quark structures of π and mesons?
17. Write the quark structures of delta particles and sigma resonances.
18. Write the proton wave function in terms of its quark constituents. How the magnetic
moment of proton is obtained from its wave function?
19. Obtain the magnetic moments of sigma particles in terms of magnetic moments of their
quark constituents.
20. What is the need for colour quantum number for quarks? Give the experimental evidence
that shows that there are three colours.
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