
Introduction

In 1808, John Dalton on the basis of experimental results first assumed that atoms were
the smallest part of the element identical in mass. In 1815, Prout suggested that the
atoms of all the elements were made up of hydrogen atoms. At the beginning of the
20th century, the picture that has come out of the atomic model was the one given by
Thomson. According to him the atom was a sphere of positively charged electricity of
radius of about 10-10 m in which electrons, sufficient in number to balance the charge so
as to make the atom neutral, were distributed. In 1911, Rutherford, by his α-particle
scattering showed that the positive charge of the atom is concentrated in a small region
(about 10-14 m in diameter) surrounded by electrons. This small positively charged
region at the centre of an atom is called nucleus. Nearly whole mass of the atom is
concentrated in the nucleus. Its density is very high. In this chapter, we introduce some
of its most basic properties, its mass, size, shape, the nuclear force, binding energy,
mass defect and packing fraction.
1.1 Basic Properties of Nucleus
1.1.1 Composition of Nucleus



All nuclei are made up of fundamental particles called protons and neutrons.
A proton has a positive charge of the same magnitude as that of electron. Neutron is
electrically neutral. Protons and neutrons in all are called nucleons. The mass of proton is
1.007825 a.m.u. and mass of neutron is 1.008665 a.m.u. The total number of protons in the
nucleus is called atomic number and denoted by Z. The total number of protons and neutrons in
the nucleus is called mass number and denoted by A. A species of nucleus, called nuclide, is
represented by symbol (X) of the atom with the atomic number Z as the left subscript and mass
number A as the right superscript for example, 27Co60 , 6C 12. X is called chemical symbol of
the species. The number of neutrons is given as N = A – Z
As an example, the cobalt nucleus 27Co60 has Z = 27 protons, A = 60 nucleons and
N = 60 – 27 = 33 neutrons.

The nuclei are arranged in the periodic table according to their atomic number Z. The
chemical properties of the atom depend on this number Z.

1.1.2 Nuclear Size and Density


One of the most direct ways of determining the size of nucleus is Rutherford’s scattering
experiment. In principle, Rutherford’s theory of scattering of α particles should predict
scattering as long as the nucleus may be considered as point charge. In Rutherford’s scattering
experiment, the beam of α particles is bombarded on a thin foil of the metal. As the α particles
moved through the foil, they often passed near a nucleus of the metal. Both the α particle and
the nuclei have positive charge. The α particles of the beam are deflected by the nucleus in
consistent with Coulomb’s law in electrostatics. In fact, some particles were even deflected
backward, through an angle of 180° from the incident direction. This holds good until the
bombarding α particle touches or penetrates the nucleus. For smaller separation between the
α particle and the nucleus, Coulomb’s law is not obeyed because the nucleus is no longer the
point charge of alpha particles.
Rutherford employed energy calculation and found an expression for the distance, R, at which
a particle approaching a nucleus is turned around by Coulomb’s repulsion. The closest distance
of approach for the α particle is given by
R = Zze2 \4πεo Emini
where Z = atomic number of the target nucleus.
z = atomic number of the α nucleus.
Emini = minimum energy of the α particle.



When the E min is increased, the closest distance of approach decreases. If we make energy of
the α particle such that it just touches the nucleus, then the distance of closest approach in this
case will be radius of the nucleus.
Rutherford’s experiment on the scattering of α particles had firmly established the planetary
model of the atom i.e. nucleus is at the centre and electrons are orbiting around it. According
to the Rutherford’s work, the mean radius of the nucleus is of the order of 10–14 m to 10–15
m; while that of atom is 10-10 m. Thus, the nucleus is 1000 times smaller than the atom i.e.
nucleus is concentrated in very small at the centre of the atom.
There are number of methods to determine the size and the charge distribution for the
nucleThe methods are classified into :
1) Nuclear force methods :
These methods are based on the study of the range of nuclear force. A nucleon or light nucleus
is used for this purpose.

2) Electrical methods :
These methods are based on the electric field and charge distribution of the nucleus. An
electron or a muon is used for this purpose.


Empirically, formula for the radius of nucleus R is
R = RoA1\3
where A is the mass number and Ro is the constant having value 1.2 × 10-15 m .
Nuclei are so small that the unit fermi (fm) is appropriate unit and it is given by
1 fm = 10-15 m
Ro = 1.2 fm

The volume of the nucleus is
V = 4 \3 πR 3
Where R = Ro

A The mass of nucleus is, approximately M = Zmp + (A – Z)mn
where mn and mp are the masses of neutron and proton respectively. Taking mn ≈ mp, we get
M = Amp
Therefore, density of the nucleus is
Ρ = M\V
= Amp 4\ 3 πRo 3 A
= 3mp\ 4πRo 3
Where

mp = 1.67 × 10 –27 kg and
Ro = 1.2 fm = 1.2 × 10–15 m
∴ ρ = 3 × 1.67 × 10 –27\ 4 × 3.14 × (1.2 × 10 –15 ) 3
Or
ρ = 2.308 × 1017 kg/m3

Thus, the density of the nucleus is independent of atomic mass number A. Thus, all nuclei have
nearly the same density. The nuclear density is approximately 2.3 × 1014 times as great as the
density of water.

1.1.3 Nuclear Charge

Nucleus is made up of protons and neutrons. As neutrons are electrically neutral, the charge on
the nucleus is due to protons only. Each proton carries a single positive charge, equal in
magnitude to the electron charge (i.e. 1.6 × 10 –19 C ). If Z is the atomic number of the nucleus,
then the charge on the nucleus is +Ze.
1.1.4 Atomic Mass and Atomic Mass Unit ( A.m.u)


Atomic mass (the mass of atom containing the nucleus and Z electrons) can be measured with
great precision with the help of mass spectrographs and the analysis of nuclear reactions. The
proton is 1836 times heavier than the electron, and the masses of protons and neutrons are
nearly equal. It is convenient to define, for atomic masses a unit called atomic mass unit. It is
frequently used in atomic and nuclear physics. One atomic mass unit (1 a.m.u.) is defined as 1 12
th mass of an atom of carbon (6C 12) isotope. This isotope of carbon with mass number 12 and
atomic number 6 has a mass equal to 12 a.m.u. From this we get
1 a.m.u. = 1.660540 × 10 –27 kg.
In terms of a.m.u.,
Mass of proton = 1.007825 a.m.u.
Mass of nutron = 1.008665 a.m.u.

The energy equivalent of 1 a.m.u can be obtained by Einstein’s mass-energy relation E = mc2 .
Thus, energy equivalent to 1 a.m.u. is
E = mc2
E = 1.66 × 10 –27 × (3 × 108)2
E = 1.494 × 10-10 joules
Since 1 eV = 1.6 × 10-19 joules
E = 1.494× 10-10 \ 1.6 × 10-19
E≈ 931 × 106 Ev
∴ E = 931 MeV.

This energy is equivalent to 1 a.m.u., which is used as a standard conversion. Physicists often
express mass in terms of the unit MeV/c2.

1
a.m.u. = 931 MeV/c2
1.15 NUCLEAR SPIN



i)
In 1924, Pauli suggested that a nucleus like an electron possesses an intrinsic angular
momentum commonly known as spin. A nucleus consisting of nucleons moving in certain orbits
about the centre of massof nucleus.
Each nucleon has intrinsic spin quantum number 1\ 2 exactly equal to that of an electron and
integral orbital quantum number just as in atomic physics.
The resultant total angular momentum quantum number denoted by I is the vector sum of the
orbital angular momentum L and spin angular momentum S of the nucleus. ∴ I = L + S (I or J) Due
to nuclear spin
the nuclear force or potential becomes spin dependent.
ii)
the nuclear force or potential no more remain central but becomes tensor (non-central).
iii)
The nucleus assumes slightly spheroidal shape.
1.1.6 Nuclear Magnetic Moment


The nuclei have magnetic dipole moments. They arise from the intrinsic dipole
moments of protons and neutrons in the nuclei, and from the current circulating in the
nuclei due to motion of protons. Fig. 1.2 shows the magnetic dipole moment µ
associated with a current loop of area A and carrying current ‘i’.
From an elementary course in electricity and magnetism it is well known that the
magnetic dipole moment is given as
µ = current × area of loop
µ=i×A

The direction of the magnetic dipole moment is perpendicular to the plane of the
current as shown in fig.

The current of an electron revolving around a heavy nucleus is given as:
µ = evr\2
⸪
L = mv
and
µ
= e(L\mr)r\2
µ
= eL\m\2
L\mr = v
𝑒
µ = - (2𝑚)LZ

⸫
Negative sign shows that angular momentum of electron and magnetic dipole
moment has same axis but opposite in direction.
LZ = mL h
µ = (
𝑒ℎ
2𝑚
Where
so
)mL
𝑒ℎ
2𝑚
is called Bohars magneton which is denoted by µB . So
µ = µB.mL
This is magnetic dipole moment of electron.

Like an electron, a nucleon also possesses a nuclear magneton given by
µ = eh\ 4πm
µ = 5.05 × 10−27 J/tesla


Where mp = mass of proton = 1836 times than mass of electron.
Hence µN is 1836 times smaller than µB and is called Rabi-magneton.





The magnetic moment of proton is found to be + 2.79 µN instead of 1 µN.
Positive sign shows that the direction of magnetic moment coincides with → I or → J
when there were circulation of positive charge.
The neutron has no charge still it posseses magnetic moment of -1.9 µN.
The net magnetic moment µ of a nucleus is
µ = g ⋅ µN ⋅ I

where g is gyromagnetic ratio. The value of nuclear magnetic moment can be
estimated by nuclear magnetic resonance (NMR) spectrometer, microwave
spectrometer e.t.c
1.1.7 Nuclear Charge

Nucleus is made up of protons and neutrons. As neutrons are electrically neutral, the
charge on the nucleus is due to protons only. Each proton carries a single positive
charge, equal in magnitude to the electron charge (i.e. 1.6 × 10 –19 C ). If Z is the atomic
number of the nucleus, then the charge on the nucleus is +Ze. 1.1.4 Atomic Mass and
1.1.8 Atomic Mass Unit (a.m.u.)



Atomic mass (the mass of atom containing the nucleus and Z electrons) can be
measured with great precision with the help of mass spectrographs and the analysis of
nuclear reactions. The proton is 1836 times heavier than the electron, and the masses
of protons and neutrons are nearly equal. It is convenient to define, for atomic masses a
unit called atomic mass unit. It is frequently used in atomic and nuclear physics. One
atomic mass
unit (1 a.m.u.) is defined as 1\ 12 th mass of an atom of carbon (6C12) isotope. This
isotope of carbon with mass number 12 and atomic number 6 has a mass equal to 12
a.m.u. From this we get 1 a.m.u. = 1.660540 × 10 –27 kg.
In terms of a.m.u.,
Mass of proton = 1.007825 a.m.u.
Mass of neutron = 1.008665 a.m.u.

The energy equivalent of 1 a.m.u can be obtained by Einstein’s mass-energy relation
Thus, energy equivalent to 1 a.m.u. is
E = mc2
E = 1.66 × 10 –27 × (3 × 108)2
E = mc2

E = 1.494 × 10-10 joules
Since 1 eV = 1.6 × 10-10 joules
E = 1.49× 10-10 \ 1.6 ×-19
≈ 931× 10-6
∴ E = 931 MeV.

This energy is equivalent to 1 a.m.u., which is used as a standard conversion. Physicists
often express mass in terms of the unit MeV/c2. So, here the mass of 1 a.m.u. is
1 a.m.u. = 931M\eV2
1.1.9 Electric quadrupole moment


One of the most direct ways of determining the size and shape of nucleus is
Rutherford’s scattering experiment. In principle, Rutherford’s theory of
scattering of α particles should predict scattering as long as the nucleus may be
considered as point charge. In Rutherford’s scattering experiment, the beam of α
particles is bombarded on a thin foil of the metal. As the α particles moved
through the foil, they often passed near a nucleus of the metal. Both the α
particle and the nuclei have positive charge. The α particles of the beam are
deflected by the nucleus in consistent with Coulomb’s law in electrostatics. In
fact, some particles were even deflected backward, through an angle of 180°
from the incident direction. This holds good until the bombarding α particle
touches or penetrates the nucleus. For smaller separation between the α particle
and the nucleus, Coulomb’s law is not obeyed because the nucleus is no longer
the point charge of alpha particles.
Rutherford employed energy calculation and found an expression for the
distance, R, at which a particle approaching a nucleus is turned around by
Coulomb’s repulsion. The closest distance of approach for the α particle is given
by
R = Zze2 \4πεo Emini
Where
Z = atomic number of the target nucleus.
Z= atomic number of the α nucleus.
Emini = minimum energy of the α particle


When the E min is increased, the closest distance of approach decreases. If we make
energy of the α particle such that it just touches the nucleus, then the distance of
closest approach in this case will be radius of the nucleus.

Rutherford’s experiment on the scattering of α particles had firmly established the planetary
model of the atom i.e. nucleus is at the centre and electrons are orbiting around it. According to
the Rutherford’s work, the mean radius of the nucleus is of the order of 10-14 m to 10-15 m; while
that of atom is 10-10 m. Thus, the nucleus is 1000 times smaller than the atom i.e. nucleus is
concentrated in very small at the centre of the atom so he gave a formula which explain the
radius of nucleus which is given by
R = RoA1\3





Where R is radius of nucleus assumed to be spherical in shape but this formula was
only valid for lighter nuclei or nuclei having atomic number less than 20 . If we increase
the mass number of nucleus greater than 20 then shape of nucleus is not spherical but it
start to deviate.Electric quadrupole moment actually a parameter which explain that
deviation.
The nuclear electric quadrupole moment is a parameter which describes the effective
shape of the of nuclear charge distribution in classical physics .In quantum physics ,it
tells about nuclear spin. One of the expectations of the shell model for the nucleus is
that for closed shells the nuclear charge is spherically symmetric. If a nucleus is not
spherically symmetric, it will have a non-zero electric quadrupole moment, so the
measurement of the quadrupole moment is a test of the shell theory. Since the
quadupole moment depends upon the size and charge of the nucleus, a better
comparison is obtained by normalizing for those factors, giving what is called a "reduced
quadrupole moment".
A non-zero quadrupole moment Q indicates that the charge distribution is not
spherically symmetric. By convention, the value of Q is taken to be positive if the
ellipsoid is prolate and negative if it is oblate.
Electric quadrupole moments of nuclei can be measured from hyperfine splitting of
atomic spectral lines, from quadrupole hyperfine splitting of molecular rotational
spectra, and other spectroscopic techniques.
The electric quadrupole moment has the units of Coulomb x meter2. It is sometimes
tabulated in units of e x 10-24 cm2 where e is the electron charge. In nuclear crosssection measurements, the quantity 10-24 cm2 is called a "barn" and is represented by b.
So sometimes the quadrupole unit is written "eb" for electron-barns. Usually the
quadrupole moments will be a few tenths in these units until you reach mass numbers
around 150.



1.10. Meson theory of nuclear forces


We know that nuclear force exist between protons and protons or neutrons and neutrons. The
charges of protons repel each other but the nuclear force hold these charges together to
overcome that resistance at short range. The range of nuclear force is very short which is 1
Fermi. At this range ,nuclear force is much stronger than repulsive. If distance is more than 2.5
Fermi, nuclear force is not practically exist. At distance less than 0.7 Fermi , the force becomes
repulsive . This is the reason , nucleus is hold together without collapsing. In 1932, Heisenberg
gave his famous hypothesis. The hypothesis was nuclear forces was “Nuclear forces possess
exchange character’’. The origin of these exchange forces was not known at that time. The only
interaction known at that time to change the charge of nucleons is beta decay . In beta decay,
Aprotons converted into neutrons and neutrons is converted into protons by emitting positron
and electron with antineutrinos. As beta decay is a weak interaction and nuclear force is a
strong interaction so beta decay theory is not valid for nuclear forces.
In 1935, Yukawa proposed a theory for this strong interaction on the basis of quantum field
theory .








According to Yukawa theory , nuclear forces between nucleons is due to constant exchange of
some particles between nucleons. These particles were called mesons. Later on these particles
mesons were detected.
Mesons have mass 270 times of mass of electrons .
The mesons with positive , negative and neutral charges have been detected.
Nucleons consist of some sort of common core surrounded by cloud of 𝜋 mesons.
These mesons rapidly jumped forth and back between the nucleons ,changes their identity fast
and at the same time bind together.
The force between neutron , neutron and proton , proton is due to exchange of 𝜋o meson. The
force between neutron and proton is due to exchange of charged mesons (𝜋- or 𝜋+ )
Neutron by ejecting 𝜋- meson is converted into proton. While proton by absorbing 𝜋+ meson is
converted into neutron .
i)



→
p + 𝜋-
ii)
p + 𝜋 -→ n
Proton by ejecting 𝜋+ meson is converted into neutron.While neutron by absorbing 𝜋+
meson is converted into proton.
i)
P → n + π+
ii)
n + π+ → p
Proton by ejecting πo meson is converted into proton. While proton by absorbing πo meson is
converted into proton.
P1 → p + πo
i)

n
ii)
p + πo → p1
Neutron by ejecting πo meson is converted into neutron. While neutron by absorbing πo meson
is converted into neutron.
i)
N1→ n + πo
ii)
n + π o → n1
In this manner, nucleons are held together by due to exchange of π meson.
Conservation of energy

It can be noted that each one of exchange reaction violates law of conservation of energy for
very short time between absorption and emission of pions. This temporary failure of energy
conservation can be used to obtain a theoretical estimate of pion’s mass.
According to Heisenberg uncertainty principle
E.∆ T = h
𝑅
T =𝑐
⸫ R = ct
The range ‘’R’’ of nuclear force is 15 Fermi which is nearly equal to 1.5 x 10-13cm so

T = 1.5 X 10-13\3 X108
E =
ℎ
𝑇
= 6.5 x 10-22\1.5 x 10-13
=
=
0.5 X 10 -23sec
132MeV
This value of energy gives the mass of pion exchanged between nucleons which is much closed
to mass of pion ( 140 MeV) determined experimentally. This proved that exchange particle in
exchange reaction is pion which is predicted by Yukawa.
1.11 Nuclear parity




In classical physics, we usually used two terms to describe the behavior of particle
(i) position (ii) momentum. In Quantum physics, according to Heisenberg uncertainty
principle, it is impossible to find the position and momentum at the same time.
According to de Broglie hypothesis , a wave associated with a moving particle which is
describe by a new term named as wave function ‘Ψ’ (psi). Parity is a very important
concept in physics because it helps us understand the behavior of particles and forces.
For example, the strong force, which is responsible for holding together the nucleus of
an atom, is a parity-conserving force. This means that it does not change under parity
transformations. However, the weak force, which is responsible for radioactive decay, is
a parity-violating force. This means that it does change under parity transformations.
If we reflect the coordinate system at origin i.e. change from ( x , y ,z) to (-x, -y ,-z) then
the wave function of a physical system changes from Ψ ( x ,y ,z ) to ( -x , -y ,-z ) so. parity
is function of space coordinate.
Parity is also exhibited by the behavior of subatomic particles. In general, subatomic
particles exhibit two kinds of behavior: they can either be fermions or bosons. Fermions
are the building blocks of matter, while bosons are the force-carrying particles. The
best-known example of a fermion is the electron. Electrons are the subatomic particles
that orbit the nucleus of an atom. They are also the particles that make up the atomic
bombs that were used in World War II. Bosons include the photons that make up light
and the gluons that hold together the quarks that make up protons and neutrons.
In terms of physical reality, the existence of either matter or antimatter is equally
possible, but the two forms of matter cannot coexist for long because when they come
into contact they annihilate each other. If there were equal numbers of matter and
antimatter particles in the universe, they would have completely annihilated each other
long ago, leaving behind only energy. This fact suggests that there is an asymmetry
between matter and antimatter in the universe, with a predominance of matter. An
asymmetry is a situation in which two things are unequal. An asymmetry between
matter and antimatter suggests that there are more matter particles than antimatter
particles in the universe. In other words, there is a matter-antimatter imbalance in the

universe. The existence of such an imbalance raises the question of the origin of this
asymmetry.


This parity can be even or odd.
For even parity ,
Ψ( -x, -y , -z ) = Ψ ( x , y, z)
i.e Cos ( -x ) = Cos x
For odd parity ,
Ψ( -x , -y –z ) = - Ψ( -x , -y , -z)
i.e sin ( -x) = - sin x
It is represented by azimuthal quantum number .