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

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Department of Physics

University of Lucknow, Lucknow - 226007

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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.