Vikas dessertation

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CONTENTS
Page no.
1. INTRODUTION [1- 26]
1.1 Historical Background. [4-15]
1.1.1 Dalton’s Atomic Hypothesis… 4.
1.1.2 Thomson Model of Atom. 6.
1.1.3 Rutherford Model of Atom. 8.
1.1.4 Hypothesis for Structure of Nucleus. [12- 15]
(a)Proton-Neutron Hypothesis. 12.
(b)Proton-Electron Hypothesis. 14.
1.2 What is a Nucleus? 15.
1.3 Terms Associated with Nucleus. 16.
1.4 Quantitative Facts about the Nucleus. [17-26]
1.4.1 Size
17.
1.4.2 Mass 18.
1.4.3 Density 19.
1.4.4 Charge 20.
1.4.5 Binding Energy 21.
1.4.6 Nuclear Spin
25.
1.4.7 Magnetic Dipole Moment 26.
2. LITERATURE VIEW [27-40]
NUCLEAR MODELS 28.
2.1 Introduction 28.
2.2 Liquid Drop Model 29.
2.2.1 Semi Empirical Mass Formula. 30.
2.2.2 Achievements of Liquid Drop Model. 34.
2.2.3 Limitations of Liquid Drop Model. 35.
2.3 Shell Model 36.
2.3.1 The Square wave potential. 37.
2.3.2 The Harmonic Oscillator. 39.
2.3.3 The Spin-Orbit Interaction. 42.
2.3.4 Wood-Saxon Potential. 43.
2.3.5 Achievements of Shell Model. 44.

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3. Computer Program and Result. [45-55]
4. Conclusion. 56.
5. References. [57-60]

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1.1 Historic Development
The beginning of the nuclear physics may be traced back to the studies on atomic
structure started with the discovery of radioactivity in 1896 by Henry Becquerel. Further,
three different modes of radioactivity were observed emitting α-particles, β-particles and
γ-rays. It is well known that α-particles are Helium nuclei, β-particles are either electrons
or positrons and γ-rays are high-energy electromagnetic radiations. Scattering of α-
particles with matter revealed the existence of nucleus inside the atom. In the following
content, we give major models in establishing the structure of atom and nucleus.
1.1.1 Dalton’s Atomic Theory
Although the concept of the atom dates back to the ideas of Democritus, the
English meteorologist and chemist John Dalton formulated the first modern description
of it as the fundamental building block of chemical structures. Dalton developed the law
of multiple proportions (first presented in 1803) by studying and expanding upon
the works of Antoine Lavoisier and Joseph Proust[1].
Proust had studied tin oxides and found that their masses were either 88.1% tin and
11.9% oxygen or 78.7% tin and 21.3% oxygen (these were tin (II) oxide and tin dioxide
respectively). Dalton noted from these percentages that 100g of tin will combine either
with 13.5g or 27g of oxygen; 13.5 and 27 forms a ratio of 1:2. Dalton found an atomic
theory of matter could elegantly explain this common pattern in chemistry - in the case of
Proust's tin oxides, one tin atom will combine with either one or two oxygen atoms.
Dalton also believed atomic theory could explain why water absorbed different
gases in different proportions: for example, he found that water absorbed carbon dioxide
far better than it absorbed nitrogen[3]. Dalton hypothesized this was due to the
differences in the mass and complexity of the gases' respective particles. Indeed, carbon

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dioxide molecules (CO2) are heavier and larger than nitrogen molecules (N2).
Dalton proposed that each chemical element is composed of atoms of a single,
unique type, and though they cannot be altered or destroyed by chemical means, they can
combine to form more complex structures (chemical compounds). Since Dalton reached
his conclusions by experimentation and examination of the results in an empirical
fashion, this marked the first truly scientific theory of the atom.
Figure 1: - Dalton’s atomic model.

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Dalton's Atomic Theory [2]
The main points of Dalton's atomic theory are:
1. All atoms of an element are identical[4].
2. The atoms of different elements vary in size and mass.
3. Compounds are produced through different whole-number combinations of atoms.
4. A chemical reaction results in the rearrangement of atoms in
the reactant and product compounds.
Atomic theory has been revised over the years to incorporate the existence of
atomic isotopes and the inter conversion of mass and energy. In addition, the discovery
of sub atomic particles has shown that atoms can be divided into smaller parts. However,
Dalton's importance in the development of modern atomic theory has been recognized by
the designation of the atomic mass unit [5] as a Dalton.
1.1.2 Thomson Model Of Atom
In 1897, J.J. Thomson discovered a negatively charged particle known as an
electron. Thomson discovered electron by cathode ray tube experiment. Cathode ray tube
is a vacuum tube. Thomson assumed that an electron is two thousand times lighter than a
proton and believed that an atom is made up of thousands of electrons having the
negative charge. In this model, he considered atoms to have a cloud of negative charge
and the positive charges. He along with Rutherford was also the first to demonstrate the
ionization of air by X-rays. Thomson’s model of an atom is similar to plum pudding
model or a watermelon.[6]

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Postulates of Thomson’s atomic model
1. An atom consists of a positively charged sphere with electrons filled into it. The
negative and positive charge present inside an atom is equal and as a whole, an
atom is electrically neutral.
2. Thomson’s model of the atom was compared to plum pudding and watermelon.
He compared the red edible part of the watermelon to positively charged sphere
whereas the seeds of watermelon to negatively charged particles.

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Figure 2:- Thomson model.
Limitations of Thomson’s atomic model
1. This model of atom failed to explain how a positive charge holds the negatively
charged electrons in an atom. Therefore, it failed to explain the stability of an
atom.
2. This theory also failed to account for the position of the nucleus in an atom.
3. Thomson’s model failed to explain the scattering of alpha particles.
Although Thomson’s model was not an accurate model to account for the atomic
structure, it proved to be the base for the development of other atomic models. The study
of the atom and its structure has paved the way for numerous inventions that have played
a significant role in the development of humankind.
1.1.3 Rutherford Model Of Atom
Rutherford atomic model, also called nuclear atom or planetary model of the atom,
description of the structure of atoms proposed (1911) by the New Zealand-born
physicist Ernest Rutherford. The model described the atom as a tiny, dense, positively
charged core called a nucleus, in which nearly all the mass is concentrated, around which
the light, negative constituents, called electrons, circulate at some distance, much
like planets revolving around the Sun. The nucleus was postulated as small and dense to
account for the scattering of alpha particles from thin gold foil, as observed in a series of
experiments performed by undergraduate Ernest Marsden under the direction of
Rutherford and German physicist Hans Geiger in 1909[7].
Experimental Setup:
A radioactive source capable of emitting alpha particles (i.e., positively charged

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particles, identical to the nucleus of the helium atom and 7,000 times more massive than
electrons) was enclosed within a protective lead shield. The radiation was focused into a
narrow beam after passing through a slit in a lead screen. A thin section of gold foil was
placed in front of the slit, and a screen coated with zinc sulfide to render
it fluorescent served as a counter to detect alpha particles. As each alpha particle struck
the fluorescent screen, it would produce a burst of light called scintillation, which was
visible through a viewing microscope attached to the back of the screen. The screen itself
was movable, allowing Rutherford and his associates to determine whether or not any
alpha particles were being deflected by the gold foil.
Figure 3:- Experimental setup for α-scattering experiment.

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Observations and conclusions of Rutherford Model:
After performing the experiment of scattering of α-particles, Rutherford came to the
following Conclusion:-
1. Most alpha particles were observed to pass straight through the gold foil, which
implied that atoms are composed of large amounts of open space.
2. Some alpha particles were deflected slightly, suggesting interactions with other
positively charged particles within the atom.
3. Still other alpha particles were scattered at large angles, while a very few even
bounced back toward the source.

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Figure 4: Rutherford atomic model.
Only a positively charged and relatively heavy target particle, such as the proposed
nucleus, could account for such strong repulsion. The negative electrons that balanced
electrically the positive nuclear charge were regarded as traveling in circular orbits about
the nucleus. The electrostatic force of attraction between electrons and nucleus was
likened to the gravitational force of attraction between the revolving planets and the Sun.
Most of this planetary atom was open space and offered no resistance to the passage of
the alpha particles.
The Rutherford model supplanted the “plum-pudding” atomic model of English
physicist Sir J.J. Thomson, in which the electrons were embedded in a positively charged
atom like plums in a pudding. Based wholly on classical physics, the Rutherford model
itself was superseded in a few years by the Bohr atomic model, which incorporated some
early quantum theory.
Comparison to the nuclear radius (~10-15
) with the atomic radius (~10-10
) shows that
nuclear radius is about 105
times smaller than the latter. Further, the strong binding of the

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constituents of the nuclei results from forces which must have very short range of action,
less than at least 2*10-15
m[8].
Drawback:
Rutherford model of the atom has one serious drawback. Such an atom cannot be
a stable configuration. The electromagnetic theory of light predicts that the revolving
electrons, due to their centripetal acceleration, should continually emit electromagnetic
radiations so that they would move spirally inwards and ultimately plunge into the
nucleus. It was left to Niels Bohr of Denmark (1913) to suggest a way out of the
difficulty, which however involved entirely new concepts and that were at variance with
some of the fundamental concepts of classical mechanics and of Maxwell’s
electromagnetic theory of light. This is known as Bohr’s Quantum Theory, in a more
developed form at present, constitutes the theoretical basis of subatomic phenomena[9].
Figure 5:- Drawback of Rutherford atomic model.
1.1.4 Hypothesis On Structure of Atom
As we know the Nucleus of the atom has a very small size. This compact size of the
nucleus contains whole of the positive charge and practically the whole mass of the atom.
Following are the two hypothesis to explain the compact structure of the nucleus[10]:

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(A) Proton – Electron Hypothesis: In order to explain observed properties of the
nucleus. Firstly it was proposed that the nucleus consisting protons and electrons.
This model is known as Proton- Electron Model. The concept of the build-up of
the nucleus in terms of elementary constituents was based on the fact that certain
atoms emit α- and β-rays, which are corpuscular in nature. As proposed by Prout,
atomic weights A of the elements are closed to integers. The fractional parts are
contributed by the isotopes of the elements. The mass of the proton is
approximately equal to the mass of the Hydrogen atom. In fact, the hydrogen
nucleus was given the name of proton, which shows its importance as a
fundamental constituent of nuclei of all atoms.
To account for mass of the nucleus whose atomic weight is close to integer A,
called the mass number, it is necessary to assume that nucleus contains A protons.
But if this was the case the charge of the nucleus will be equal to A, nearly the
same as atomic weight and not equal to atomic number Z. As is well known,
value of Z is half or less than half of the atomic weight. To get over this
difficulty, it was assumed that in addition to protons the nuclei contains A-Z
electrons. The presence of electrons would not contribute to the mass of nucleus
but would make the charge Z as required. Thus, it was possible to consider atom
made up of a nucleus containing A protons and A-Z electrons surrounded by Z
extra-nuclear electrons. This hypothesis seems to be consistent with the emission
of α- and β-particles in radioactive elements. The presence of electrons directly
ensures the emission of β-particles, and emission of α-particles is assumed by the
combination of 4 protons and 2 electrons in the nucleus. These α-particles may
exist as such or may be formed at the instant of emission[11].
Failure of Proton-Electron Hypothesis:
1. Spin and Statistics: The statistical nature of nuclei can be built up from

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rotational spectra of diatomic molecules. If the nucleus (A, Z) contains A
protons and (A-Z) electrons, the spin of odd-odd nucleus or odd-even nucleus
would not agree with experimental results.
2. Nuclear Magnetic Moment: Magnetic moment of an electron is one Bohr
magnetron (μB = eħ/2me), while that of a proton is one nuclear magnetron
μN=eħ/2mp. Thus, μB= 1850*μN, where mN/mB = 1850. If the nucleus consisted
of protons and electrons, the nuclear magnetic moment should be of the order
of μB while experimentally nuclear magnetic moment is of the order of μN.
Thus, electrons do not exist inside the nucleus.
(B) Proton – Neutron Hypothesis:
After the discovery of neutrons by Chadwick in 1932 through research on
transmission of nuclei by α-particles, Heisenberg had earlier proposed that nuclei
might be composed of protons and neutrons, collectively called nucleons. The
neutron carried mass slightly greater than that of the protons, but is electrically
neutral. Due to no charge, neutron was hard to detect and several unsuccessful
efforts were made before it was finally observed in 1932. Thus, 8O16
nucleus
contains 8 protons and 16-8 = 8 neutrons; 16
O8 nucleus is surrounded by 8
electron to balance the nuclear charge. In general, A
XZ nucleus will contain Z
protons and (A-Z) neutrons. To balance the nuclear charge it will be surrounded
by Z electrons. This Model obviously avoids the failures of proton-electron
hypothesis[12].
Following facts supports the proton-electron hypothesis:
1. Spin: Both protons and neutrons have spin quantum number 1/2. According to
quantum mechanics, if the number of nucleons in a nucleons in a nucleus is
even, the resultant spin will be an integral multiple of ħ and if they are odd the

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spin will be half integral multiple of ħ. This observation is in agreement with
the experimental results.
2. Magnetic Moment: According to proton-neutron hypothesis, there are no
electrons inside the nucleus. Hence, we do not expect the magnetic moment of
the nucleus to be of the order of Bohr Magnetron. On the other hand, the
nuclear magnetic moment is of the order of nuclear magnetron.
3. Isotopic Masses: It is possible to explain the existence of isotopes of different
elements. Different isotopes of an element have same number of protons but
different number of neutrons in the nucleus.
1.2 WHAT IS A NUCLEUS ?
The nucleus is the center of an atom. It is made up
of nucleons (protons and neutrons) and is surrounded by the electron cloud. The size
(diameter) of the nucleus is between 1.6 fm (10−15
m) (for a proton in light hydrogen) to
about 15 fm (for the heaviest atoms, such as uranium)[13]. These sizes are much smaller
than the size of the atom itself by a factor of about 23,000 (uranium) to about 145,000
(hydrogen). The nucleus has most of the mass of an atom, though it is only a very small
part of it. Almost all of the mass in an atom is made up from the protons and neutrons in
the nucleus with a very small contribution from the orbiting electrons.

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Figure 6:- Atomic Nucleus
Neutrons have no charge and protons are positively charged[14]. Because the nucleus is
only made up of protons and neutrons it is positively charged. Things that have the same
charge repel each other: this repulsion is part of what is called electromagnetic force.
Unless there was something else holding the nucleus together it could not exist because
the protons would push away from each other. The nucleus is actually held together by
another force, the strong nuclear force[15].
The word nucleus is from 1704, meaning “kernel of a nut”. In 1844, Michael
Faraday used nucleus to describe the “central point of an atom”. The modern atomic
meaning was proposed by Ernest Rutherford in 1912. The use of the word nucleus in
atomic theory, however, did not happen immediately[16]. In 1916, for example, Gilbert
N. Lewis wrote in his famous article The Atom and the Molecule that "the atom is
composed of the kernel and an outer atom or shell"[17].
1.3 Terms Associated with the Nucleus
Atomic Number: It is the total number of Protons present in the nucleus. For example,
Nitrogen has 7 protons, so Z for Nitrogen is 7, Z for Uranium is 92 and for Hydrogen, Z
is 1[18].
Mass Number: It is the total number of protons and Neutrons present in the nucleus, for
example, carbon has 6 protons and 6 neutrons, so its mass number is 12, uranium has 92
protons and 143 neutrons, therefore, mass number of uranium is 235, ordinary Hydrogen
has only 1 proton in its nucleus, so its mass number is 1. It is obvious that A can never be
less than Z[19].
Neutron Number: It is the total number of Neutrons present in the nucleus and is equal to
A-Z. A nucleus X with atomic number Z, mass number A and neutron number N is

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represented as A
XZ[20].
Nucleons: The term nucleon refers to protons or neutrons present inside the nucleons.
Thus, the nucleus with mass number A having N neutrons and Z protons has A nucleons.
Isotopes: Nuclei of an element having the same atomic number but different mass
number are called isotopes of the element. There are two types of isotopes: (i) stable and
(ii) unstable. Stable isotopes are those which do not show radioactivity[21].
1.4 Quantitative Facts About Nucleus
Here we introduce some facts such as size, mass, density of the nucleus and charge on
the nucleus.
1.4.1 Size:
Rutherford in the gold foil experiment showed us that the atom was mainly empty
space with the nucleus at the centre and electrons revolving around it. When he fired
alpha particles towards the gold foil, he noticed that 1 in 20000 particles suffered a
change in direction of motion of more than 90 degrees. Rest of the 19999 article departed
from their trajectory by a very small margin. This meant that the atom consisted of an
empty space with most of the mass being concentrated in tiny volume in the centre. He
called this volume ‘the nucleus’; Latin for ‘little nut’.
Through the Rutherford experiment, it was possible to obtain the size of the nucleus. By
obtaining the point of closest approach of an alpha particle, we could calculate the size of
the nucleus. By firing alpha particles of kinetic energy 5.5 MeV, the point of closest
approach was calculated to be about 4×10-14
m. Since the repulsive force acting here is
Coulomb repulsion, there is no contact. This means that the size of the nucleus is smaller
than 4×10-14
m.
By firing alpha particles of greater energy and after many more iterations of the

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experiment, the sizes of the nuclei of various elements have been accurately measured.
Through this, we have obtained a formula to measure the size of the nucleus.
R = R0A1/3
Where R0 is a constant, known as the Nuclear Radius Parameter. Where R0= 1.2×10-15
m.
This means that the volume of the nucleus which is proportional to R3
is proportional to
A(mass number). One thing that should also be noticed is that there seems to be no
mention of density in the equation. That’s because the density of the nuclei does not vary
with elements. All nuclei have the same density. The density of the nucleus is
approximately 2.3×1017
kgm-3
. This is very high compared to the density of normal
things, ie water (1000 kgm-3
) or air (1.225 kgm-3
). This is because most of the atom is
empty and all the mass is concentrated in a very tiny space[22].
1.4.2 Mass:
An atom is extremely small and therefore its mass is also proportionally minute.
A regular unit of mass such as a Kilogram (Kg) cannot be used to weigh something as
small as an atom and to address this issue scientists have created a new unit of mass. It is
called the Atomic Mass Unit (u). Its reference is taken as Carbon-12 and 1 Atomic Mass
unit is equal to 1/12th
the weight of one atom of Carbon 12.
1 u = one atom of C-12/ 12
= 1.992647 10-26
/ 12 kg
1 u = 1.660539 10-27
kg
This is the mass of a hydrogen atom! Surprisingly except for a few elements, most of
them are whole multiples of the weight of the Hydrogen atom[22].

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The Nucleus of an atom consists of a tightly packed arrangement of protons and
neutrons. These are the two heavy particles in an atom and hence 99.9% of the mass is
concentrated in the nucleus. Of the two, the protons possess a net positive charge and
hence the nucleus of an atom is positively charged on the whole and the negatively
charged electrons revolve around the central nucleus. Since the mass concentration at the
nucleus of an atom is immense the nuclear forces holding the protons and the neutrons
together are also large. The protons are in such close vicinity to each other inside the tiny
nucleus and therefore the electrostatic forces of repulsion also act inside the nucleus.
Nuclear energy relies on nothing but releasing the energy trapped in the nucleus of an
atom. The total number of protons in a nucleus is equal to the number of electrons
revolving around the nucleus and hence the atom on the whole is electrically neutral.
1.4.3 Density:
Nuclear density is the density of the nucleus of an atom, averaging about
2.3×1017
kg/m3
. The descriptive term nuclear density is also applied to situations where
similarly high densities occur, such as within neutron stars.
The nuclear density for a typical nucleus can be approximately calculated from the size
of the nucleus, which itself can be approximated based on the number of protons and
neutrons in it. The radius of a typical nucleus, in terms of number of nucleons, is
R=R0A1/3
where A is the mass number and R0 is 1.25 fm, with typical deviations of up
to 0.2 fm from this value. The density of the nucleus is thus[24]:
The density for any typical nucleus, in terms of mass number, is thus constant, not
dependent on A or r, theoretically:

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The experimentally determined value for n is 0.16 fm−3
.
The mass density is the product of n by the nuclear mass. The calculated mass density,
using a nucleon mass of 1.67×10−27
kg, is thus:
1.4.4 Charge:
Rutherford as a result of α-particles scattering experiments concluded that all the
positive charge on atom is confined to a tiny central region called nucleus. Later on from
α-particles and X-ray scattering from the atoms, it was found that the number of unit
charges on the nucleus of any atom is approximately half of its atomic weight.
Rutherford also concluded that proton was identical with a hydrogen ion(electron
removed) that carried a single unit positive charge. Since hydrogen atom is neutral, so the
charge on proton must be equal to that of electron, but in opposite sign. Similarly, α-
particles is actually a Helium nucleus, that is a Helium atom minus its two electrons.
Therefore, Helium nucleus carries positive charge equal to 2e, where e is the charge
present on one electron. Thus, the charge on a nucleus carrying Z protons is Ze units[25].
1.4.5 Binding Energy:
Nuclear binding energy is the energy that would be required to disassemble
the nucleus of an atom into its component parts. These component parts
are neutrons and protons, which are collectively called nucleons. The binding energy of

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nuclei is due to the attractive forces that hold these nucleons together, and it is always a
positive number, since all nuclei would require the expenditure of energy to separate
them into individual protons and neutrons. The mass of an atomic nucleus is less than the
sum of the individual masses of the free constituent protons and neutrons (according to
Einstein's equation E=mc2
) and this 'missing mass' is known as the mass defect, and
represents the energy that was released when the nucleus was formed[26].
The term "nuclear binding energy" may also refer to the energy balance in
processes in which the nucleus splits into fragments composed of more than one nucleon.
If new binding energy is available when light nuclei fuse, or when heavy nuclei split,
either process can result in release of this binding energy. This energy may be made
available as nuclear energy and can be used to produce electricity as in (nuclear power)
or in a nuclear weapon. When a large nucleus splits into pieces, excess energy is emitted
as photons (gamma rays) and as the kinetic energy of a number of different ejected
particles (nuclear fission products).
The nuclear binding energies and forces are on the order of a million times greater
than the electron binding energies of light atoms like hydrogen. The mass defect of a
nucleus represents the mass of the energy of binding of the nucleus, and is the difference
between the mass of a nucleus and the sum of the masses of the nucleons of which it is
composed
Determining nuclear binding energy:
Calculation can be employed to determine the nuclear binding energy of nuclei. The
calculation involves determining the mass defect, converting it into energy, and
expressing the result as energy per mole of atoms, or as energy per nucleon.
Conversion of mass defect into energy:
Mass defect is defined as the difference between the mass of a nucleus, and the
sum of the masses of the nucleons of which it is composed. The mass defect is

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determined by calculating three quantities.[2]
These are: the actual mass of the nucleus,
the composition of the nucleus (number of protons and of neutrons), and the masses of a
proton and of a neutron. This is then followed by converting the mass defect into energy.
This quantity is the nuclear binding energy, however it must be expressed as energy per
mole of atoms or as energy per nucleon[27].
Nuclear energy is released by the splitting (fission) or merging (fusion) of
the nuclei of atom(s). The conversion of nuclear mass-energy to a form of energy, which
can remove some mass when the energy is removed, is consistent with the mass-energy
equivalence formula:
ΔE = Δ m c2
in which,
ΔE = energy release,
Δm = mass defect,
and c = the speed of light in a vacuum (a physical constant).
And the formula for Binding Energy is:
Binding Energy = ( ZMP + NMN - M(A,Z))
Now Binding Energy Per Nucleon is :
B.E. = (ZMP+ NMn – M (A,Z)) / A
Where
Z =atomic number
A = mass number
MP = mass of one proton
Mn =mass of one neutron
M(A,Z)= Atomic Mass.

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Figure 7 : Average Binding Energy per Nucleon.

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Nuclear binding energy curve:
In the periodic table of elements, the series of light elements from hydrogen up
to sodium is observed to exhibit generally increasing binding energy per nucleon as
the atomic mass increases. This increase is generated by increasing forces per nucleon in
the nucleus, as each additional nucleon is attracted by other nearby nucleons, and thus
more tightly bound to the whole.
The region of increasing binding energy is followed by a region of relative stability
(saturation) in the sequence from magnesium through xenon. In this region, the
nucleus has become large enough that nuclear forces no longer completely extend
efficiently across its width. Attractive nuclear forces in this region, as atomic mass
increases, are nearly balanced by repellent electromagnetic forces between protons,
as the atomic number increases[28].
Finally, in elements heavier than xenon, there is a decrease in binding energy per nucleon
as atomic number increases. In this region of nuclear size, electromagnetic
repulsive forces are beginning to overcome the strong nuclear force attraction.
At the peak of binding energy, nickel-62 is the most tightly bound nucleus (per nucleon),
followed by iron-58 and iron-56.[
This is the approximate basic reason why iron and
nickel are very common metals in planetary cores, since they are produced
profusely as end products in supernovae and in the final stages of silicon burning in
stars. However, it is not binding energy per defined nucleon (as defined above),
which controls which exact nuclei are made, because within stars, neutrons are free
to convert to protons to release even more energy, per generic nucleon, if the result
is a stable nucleus with a larger fraction of protons. In fact, it has been argued
that photodisintegration of 62
Ni to form 56
Fe may be energetically possible in an
extremely hot star core, due to this beta decay conversion of neutrons to protons.

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The conclusion is that at the pressure and temperature conditions in the cores of
large stars, energy is released by converting all matter into 56
Fe nuclei (ionized
atoms). (However, at high temperatures not all matter will be in the lowest energy
state.) This energetic maximum should also hold for ambient conditions, say T =
298 K and p = 1 atom, for neutral condensed matter consisting of 56
Fe atoms—
however, in these conditions nuclei of atoms are inhibited from fusing into the most
stable and low energy state of matter.
It is generally believed that iron-56 is more common than nickel isotopes in the
universe for mechanistic reasons, because its unstable progenitor nickel-56 is copiously
made by staged build-up of 14 helium nuclei inside supernovas, where it has no time to
decay to iron before being released into the interstellar medium in a matter of a few
minutes, as the supernova explodes. However, nickel-56 then decays to cobalt-56 within
a few weeks, then this radioisotope finally decays to iron-56 with a half life of about 77.3
days. The radioactive decay-powered light curve of such a process has been observed to
happen in type II supernovae, such as SN 1987A. In a star, there are no good ways to
create nickel-62 by alpha-addition processes, or else there would presumably be more of
this highly stable nuclide in the universe[29].
1.4.6 Nuclear Spin
Protons and neutrons have half integral spin i.e. plus +1/2 or -1/2. Spin, which
can be loosely associated with the picture of a particle spinning, is inherently quantum
mechanical in nature and related to the intrinsic angular momentum associated with the
sub-atomic particle. Spin is a vector quantity, with a total spin and a component of spin in
a specified direction. The total spin has a spin quantum number (symbol s) with value
equal to an integer for a boson, and a half-integer for a fermion and the word 'spin' is
often used to mean this quantum number.

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The overall spin of an atomic nucleus is by virtue of the spin of each nucleon
within it. The hydrogen nucleus, for example, contains one proton with a spin quantum
number of 1/2 - and this gives rise to a spin of 1/2 for a hydrogen atom (see fig. 3.12).
The spin produces a magnetic moment, and this forms the basis of the technique of
nuclear magnetic resonance[30].
Within a nucleus, nucleons (protons and neutrons) have a strong tendency to pair
i.e. neutron with neutron or proton with proton so that their spins cancel (spins pair anti-
parallel). Hence for all even-Z even N nuclei such as 12
C, 16
O, 32
S, the ground state spin is
always zero (see table 3.3). Nuclei with an odd number of protons, neutrons, or both, will
have an intrinsic nuclear spin. Although there is the tendency for nucleons to pair up
spins anti-parallel to become spin-0, the total spin is not necessary the lowest value after
pairing off - some nucleons remain unpaired and result in spins as high as 11/2[31].
1.4.7 Nuclear Magnetic Moment:
The nuclear magnetic moment is the magnetic moment of an atomic nucleus and
arises from the spin of the protons and neutrons. It is mainly a magnetic dipole moment;
the quadrupole moment does cause some small shifts in the hyperfine structure as well.
All nuclei that have nonzero spin also possess a nonzero magnetic moment and vice
versa, although the connection between the two quantities is not straightforward or easy
to calculate.
The nuclear magnetic moment varies from isotope to isotope of an element. For a nucleus
of which the numbers of protons and of neutrons are both even in its ground state (i.e.
lowest energy state), the nuclear spin and magnetic moment are both always zero. In
cases with odd numbers of either or both protons and neutrons, the nucleus often has
nonzero spin and magnetic moment[32].

27
Literature View
NUCLEAR MODELS

28
2.1 Introduction:
In order to understand the observed properties of the nucleus of an atom it is
necessary to have an adequate knowledge about the nature of the interaction. We know
that a very strong short range force acts between the nucleons. The exact mathematical
form of this interaction is still not known. Yukawa’s theory gives us some idea about it,
which is based on the exchange of a pion between two nucleons, when they are at a
distance less than the range of the interaction. However, there are alternative approaches
in which more than one pion exchange is also taken into account. None of the proposed
theories gives us a full understanding of the nature of the internucleon interaction.
It may be noted that even if the exact nature of the internucleon interaction were
known, it would have been extremely difficult to develop a satisfactory theory of the
structure of the nucleus made up of a large number of neutrons and protons, since it is
wellnigh impossible to solve the Schrodinger equation exactly for such a many body
system. Various methods have been developed for tackling the problem with different
degrees of approximation. However, the problem is still far from being solved
completely.
Because of the above difficulties in developing a satisfactory theory of nuclear
structure, different models have been proposed for the nucleus, each of which can explain
some of the different characteristics of the nucleus.
2.2 Liquid Drop Model:

29
The liquid drop model in nuclear physics treats the nucleus as a drop of
incompressible nuclear fluid of very high density. It was first proposed by George
Gamow and then developed by Niels Bohr and John Archibald Wheeler. The nucleus is
made of nucleons (protons and neutrons), which are held together by the nuclear force (a
residual effect of the strong force). This is very similar to the structure of a spherical
liquid drop made of microscopic molecules. This is a crude model that does not explain
all the properties of the nucleus, but does explain the spherical shape of most nuclei. It
also helps to predict the nuclear binding energy and to assess how much is available for
consumption[33].
According to this model, the atomic nucleus behaves like the molecules in a
drop of liquid. But in this nuclear scale, the fluid is made of nucleons (protons
and neutrons), which are held together by the strong nuclear force. The liquid drop model
of the nucleus takes into account the fact that the nuclear forces on the nucleons on the
surface are different from those on nucleons in the interior of the nucleus. The interior
nucleons are completely surrounded by other attracting nucleons. Here is the analogy
with the forces that form a drop of liquid.
In the ground state the nucleus is spherical. If the sufficient kinetic or binding energy is
added, this spherical nucleus may be distorted into a dumbbell shape and then may be
spitted into two fragments. Since these fragments are a more stable configuration, the
splitting of such heavy nuclei must be accompanied by energy release. This model does
not explain all the properties of the atomic nucleus, but does explain the predicted
nuclear binding energies[34].

30
Figure 8: Liquid Drop Model
2.2.1 Semi Empirical Formula:
In nuclear physics, the semi-empirical mass formula (SEMF) (sometimes also
called Weizsäcker's formula, or the Bethe–Weizsäcker formula) is used to approximate
the mass and various other properties of an atomic nucleus from its number
of protons and neutrons. As the name suggests, it is based partly on theory and partly on
empirical measurements. The theory is based on the liquid proposed by George Gamow,
which can account for most of the terms in the formula and gives rough estimates for the
values of the coefficients. It was first formulated in 1935 by German physicist Carl
Friedrich von Weizsäcker, and although refinements have been made to the coefficients
over the years, the structure of the formula remains the same today[35].
The SEMF gives a good approximation for atomic masses and several other
effects, but does not explain the appearance of magic numbers of protons and neutrons,
and the-energy and measure of stability that are associated with these numbers of
nucleons.

31
If we consider the sum of the following five types of energies, then the picture of a
nucleus as a drop of incompressible liquid roughly accounts for the observed variation of
binding energy of the nucleus:
Semi Empirical Formula without Correction Term:
1. Volume term:
The first two terms describe a spherical liquid drop of an incompressible fluid
with a contribution from the volume scaling with A and from the surface, scaling with
A2/3
. The first positive term aVA is known as the volume term and it is caused by the
attracting strong forces between the nucleons. The strong force has a very limited
range and a given nucleon may only interact with its direct neighbors. Therefore this term
is proportional to A, instead of A2
. The coefficient aV is usually about ~ 15.8 MeV.
B.E. = av A
2. Surface Energy:
The surface term is also based on the strong force; it is, in fact, a correction to the
volume term. The point is that particles at the surface of the nucleus are not completely
surrounded by other particles. In the volume term, it is suggested that each nucleon
interacts with a constant number of nucleons, independent of A. This assumption is very
nearly true for nucleons deep within the nucleus, but causes an overestimation of the
binding energy on the surface. By analogy with a liquid drop this effect is indicated
as the surface tension effect. If the volume of the nucleus is proportional to A, then the
geometrical radius should be proportional to A1/3
and therefore the surface term must be
proportional to the surface area i.e. proportional to A2/3
. The coefficient as is usually about
18.34 MeV
B.E. = -as A2/3

32
3. Coulomb Energy:
This term describes the Coulomb repulsion between the uniformly distributed
protons and is proportional to the number of proton pairs Z2
/R, whereby R is proportional
to A1/3
. This effect lowers the binding energy because of the repulsion between charges of
equal sign. The coefficient ac is usually about 0.71 MeV.
B.E = -aC.Z(Z-1).A-⅓
Total Binding Energy of Nucleons Prior to Correction is:
B.E. = av A - as A2/3
- aC.Z2
.A-⅓
And now Binding Energy per Nucleon is:
B.E./A = av- as A1/3
- ac Z(Z-1) A-4/3
.
Semi Empirical Formula with Correction Term:
4. Symmetry Energy:
This term cannot be described as ‘classically’ as the first three. This effect is not
based on any of the fundamental forces, this effect is based only on the Pauli exclusion
principle (no two fermions can occupy exactly the same quantum state in an atom). The
heavier nuclei contain more neutrons than protons. These extra neutrons are necessary for
stability of the heavier nuclei. They provide (via the attractive forces between the
neutrons and protons) some compensation for the repulsion between the protons. On the
other hand, if there are significantly more neutrons than protons in a nucleus, some of the
neutrons will be higher in energy level in the nucleus. This is the basis for a correction

33
factor, the so-called symmetry term. The coefficient aa is usually about 23.21MeV.
B.E = as(A-2Z)2
/A
5. Paring Energy:
The last term is the pairing term δ(A,Z). This term captures the effect of spin-
coupling. Nuclei with an even number of protons and an even number of neutrons are
(due to Pauli exclusion principle) very stable thanks to the occurrence of ‘paired spin’.
On the other hand, nuclei with an odd number of protons and neutrons are mostly
unstable. The coefficient ap is usually about 12 MeV.
B.E. = apA-3/4
The total Binding Energy after correction is:
B.E. = av A - asA2/3
– ac Z(Z-1) A-1/3
– as (A-2Z)2
/A - apA-3/4
Now the total Binding Energy per Nucleon is:
B.E./ A = av - asA-1/3
– ac Z(Z-1) A-4/3
– as (A-2Z)2
/A2
- apA-7/4
.

34
Figure 9: Semi Empirical Terms For Binding Energy.
2.2.2 Achievements of Liquid Drop Model[36]:
(i) Stable nucleus:
We can explain the stability of nuclei on the basis of liquid drop model .The
stability of liquid drop is due to cohesion between the molecules, similarly, the stability
of the nucleus is due to the binding energy of each nucleon .Just as to remove a molecule
from liquid drop energy has to supply to it in the form of heat ,energy will have to be
supplied to the nucleon equal to or greater than its binding energy removes it .Hence the
stability of nucleons is explained .
(ii) Radioactive nucleus:
Radioactive phenomenon can be explained by the liquid drop model as follows:

35
In a liquid, with the increase in temperature, thermal agitation of its molecules becomes
more and more rapid and at a particular stage, evaporation takes place .A molecule in a
liquid drop evaporates gaining energy from its neighboring molecules during the process
of collision, thus exhibiting the phenomenon of radioactivity.
(iii) Artificial radioactivity:
The liquid drop model explains the phenomenon of artificial radioactivity .It is
supposed that when a nucleus is bombarded by fast moving the particle, an incoming
particle enters the target nucleus forming a compounds nucleus. It quickly shares its
energy with the nucleons which already present so that no single particles has sufficient
energy for escape. The decay or disintegration of the compounds nucleus occur when the
energy is again accidentally concentrated on someone particles which escape giving rise
to the phenomenon of artificial radioactivity, Or energy may be lost by the emission of
a γγ-rays.
(iv) This model agrees with experimental graph B.E. except for the lighter nuclei .This
model also explains the nuclear stability against αα decay ββ- decay or disintegration.
Limitation of LDM[37]:
(i) The model fails to explain the high stability of nuclei with the magic number. i.e.
discontinuities of B/A curve can't be explained on the basis of this model .
(ii) This model fails to explain the concept of the pairing energy.
(iii) Naturally abundant even an isobar is not explained by this model.
(iv) This model cannot explain discontinuities in B/E nucleon
(v) The breaking of nuclei in lighter elements is not successfully interpreted by this
model.

36
2.3 Shell Model:
In nuclear physics the nuclear shell model is a model of the atomic nucleus which
uses the Pauli Exclusion Principle to describe the structure of the nucleus in terms of
energy levels.[
The first shell model was proposed by Dmitry Ivanenko in 1932. The
model was developed in 1949 following independent work by several physicists, most
notably Eugene Paul Wigner, Maria Goeppert Mayer and J. Hans D. Jensen, who shared
the 1963 Nobel Prize in Physics for their contributions.
The shell model is partly analogous to the atomic shell model which describes the
arrangement of electrons in an atom, in that a filled shell results in greater stability. When
adding nucleons (protons or neutrons) to a nucleus, there are certain points where
the binding energy of the next nucleon is significantly less than the last one. This
observation, that there are certain magic numbers of nucleons: 2, 8, 20, 28, 50, 82, 126
which are more tightly bound than the next higher number, is the origin of the shell
model.
The shells for protons and for neutrons are independent of each other. Therefore,
one can have "magic nuclei" where one nucleon type or the other is at a magic number,
and "doubly magic nuclei", where both are. Due to some variations in orbital filling, the
upper magic numbers are 126 and, speculatively, 184 for neutrons but only 114 for
protons, playing a role in the search for the so-called island of stability. Some semi magic
numbers have been found, notably Z=40 giving nuclear shell filling for the various
elements; 16 may also be a magic number[38].
In order to get these numbers, the nuclear shell model starts from an average
potential with a shape something between the square well and the harmonic oscillator. To

37
this potential a spin orbit term is added. Even so, the total perturbation does not coincide
with experiment, and an empirical spin orbit coupling must be added with at least two or
three different values of its coupling constant, depending on the nuclei being studied.
Nevertheless, the magic numbers of nucleons, as well as other properties, can be arrived
at by approximating the model with a three-dimensional harmonic oscillator plus a spin-
orbit interaction. A more realistic but also complicated potential is known as Woods
Saxon potential.
2.3.1. The Square Well Potential:
We have considered in some detail a particle trapped between infinitely high
walls a distance L apart, we have found the wave function solutions of the time
independent Schrödinger equation, and the corresponding energies. The essential point
was that the wave function had to go to zero at the walls, because there is zero probability
of finding the particle penetrating an infinitely high wall. This meant that the lowest
energy state couldn't have zero energy that would give a constant nonzero wave function.
Rather, the lowest energy state had to have the minimal amount of bending of the wave
function necessary for it to be zero at the two walls but nonzero in between-this
corresponds to half a period of a sine or cosine (depending on the choice of origin), these
functions being the solutions of Schrödinger's equation in the zero potential region
between the walls. The sequence of wave functions (Eigen states) as the energy increases
have 0, 1, 2, … zeros (nodes) in the well.
Let us now consider how this picture is changed if the potential at the walls is not
infinite. It will turn out to be convenient to have the origin at the center of the well, so we
take
V(x) = V0 for x < -L/2
V(x) = 0 for -L/2 < x < L/2

38
V(x) = V0 for L/2 < x.
Having the potential symmetric about the origin makes it easier to catalog the wave
functions. For a symmetric potential, the wave functions can always be taken to be
symmetric or antisymmetric. (If a wave function ψ(x) is a solution of Schrödinger's
equation with energy E, and the potential is symmetric, then ψ(-x) is a solution with the
same energy. This means that ψ(x)+ ψ(-x) and ψ(x)- ψ(-x) are also solutions, since the
equation is linear, and these are symmetric and antisymmetric respectively, and using
them is completely equivalent to using the original ψ(x) and its reflection ψ(-x).)
How is the lowest energy state wave function affected by having finite instead of
infinite walls? Inside the well, the solution to Schrödinger's equation is still of cosine
form (it's a state symmetric about the origin). However, since the walls are now
finite, ψ(x) cannot change slope discontinuously to a flat line at the walls. It must instead
connect smoothly with a function which is a solution to Schrödinger's equation inside the
wall.
The equation in the wall is
and has two exponential solutions (say, for x > L/2) one increasing to the right, the other
decreasing,
(We are assuming here that E < V0, so the particle is bound to the well. We shall find this
is always true for the lowest energy state.)
Let us try to construct the wave function for the energy E corresponding to this lowest
bound state. From the equation with V0 = 0, the wave function inside the well (let's
assume it's symmetric for now) is proportional to coskx, where

39
The wave function (and its derivative!) inside the well must match a sum of exponential
terms—the wave function in the wall—at x = L/2, so
(By writing just a cosine term inside the well, we have left out the overall normalization
constant. This can be put back in at the end.)
Solving these equations for the coefficients A, B in the usual way, we find that in general
the cosine solution inside the well goes smoothly into a linear combination of
exponentially increasing and decreasing terms in the wall. (By the symmetry of the
problem, the same thing must happen for x< -L/2.) However, this cannot in general
represent a bound state in the well. The increasing solution increases without
limit as x goes to infinity, so since the square of the wave function is proportional to the
probability of finding the particle at any point, the particle is infinitely more likely to be
found at infinity than anywhere else. It got away! This clearly makes no sense—we're
trying to find wave functions for particles that stay in, or at least close to, the well. We
are forced to conclude that the only exponential wave function that makes sense is the
one for which A is exactly zero, so that there is only a decreasing wave in the wall.
Requiring the decreasing wave function, A = 0, means that only a discrete set of values
of k, or E, satisfy the boundary condition equations above. They are most simply found
by taking A = 0 and dividing one equation by the other to give:
This cannot be solved analytically, but is easy to solve graphically by plotting the two
sides as functions of k (recall , and ) and finding
where the curves intersect[39].
2.3.2. The Harmonic Oscillator Potential:
Consider a three-dimensional harmonic oscillator. We can imagine ourselves
building a nucleus by adding protons and neutrons. These will always fill the lowest
available level. Thus the first two protons fill level zero, the next six protons fill level

40
one, and so on. As with electrons in the periodic table, protons in the outermost shell will
be relatively loosely bound to the nucleus if there are only few protons in that shell,
because they are farthest from the center of the nucleus. Therefore nuclei which have a
full outer proton shell will have a higher binding energy than other nuclei with a similar
total number of protons. All this is true for neutrons as well.
This means that the magic numbers are expected to be those in which all occupied shells
are full. We see that for the first two numbers we get 2 (level 0 full) and 8 (levels 0 and 1
full), in accord with experiment. However the full set of magic numbers does not turn out
correctly[40]. These can be computed as follows:
In a three-dimensional harmonic oscillator the total degeneracy at level n is
. Due to the spin, the degeneracy is doubled and
is .
Thus the magic numbers would be
for all integer k. This gives the following magic numbers: 2,8,20,40,70,112..., which
agree with experiment only in the first three entries. These numbers are twice the
tetrahedral numbers (1,4,10,20,35,56...) from the Pascal Triangle.

41
In particular, the first six shells are:
• level 0: 2 states (l = 0) = 2.
• level 1: 6 states (l = 1) = 6.
• level 2: 2 states (l = 0) + 10 states (l = 2) = 12.
• level 3: 6 states (l = 1) + 14 states (l = 3) = 20.
• level 4: 2 states (l = 0) + 10 states (l = 2) + 18 states (l = 4) = 30.
• level 5: 6 states (l = 1) + 14 states (l = 3) + 22 states (l = 5) = 42.
Where for every l there are 2l+1 different values of ml and 2 values of ms, giving a total
of 4l+2 states for every specific level.
These numbers are twice the values of triangular numbers from the Pascal Triangle: 1, 3,
6,10,15,21....

42
2.3.3.The Spin-Orbit Interaction:
We next include a spin-orbit interaction. First we have to describe the system by
the quantum numbers j, mj and parity instead of l, ml and ms, as in the hydrogen-like
atom. Since every even level includes only even values of l, it includes only states of
even (positive) parity; similarly every odd level includes only states of odd (negative)
parity. Thus we can ignore parity in counting states. The first six shells, described by the
new quantum numbers, are[41]:
• level 0 (n = 0): 2 states (j = 1
⁄2). Even parity.
• level 1 (n = 1): 2 states (j = 1
⁄2) + 4 states (j = 3
⁄2) = 6. Odd parity.
• level 2 (n = 2): 2 states (j = 1
⁄2) + 4 states (j = 3
⁄2) + 6 states (j = 5
⁄2) = 12. Even
parity.
• level 3 (n = 3): 2 states (j = 1
⁄2) + 4 states (j = 3
⁄2) + 6 states (j = 5
⁄2) + 8 states (j =
7
• level 4 (n = 4): 2 states (j = 1
⁄2) + 4 states (j = 3
⁄2) + 6 states (j = 5
⁄2) + 8 states (j =
7
⁄2) + 10 states (j = 9
⁄2) = 30. Even parity.
• level 5 (n = 5): 2 states (j = 1
⁄2) + 4 states (j = 3
⁄2) + 6 states (j = 5
⁄2) + 8 states (j =
7
⁄2) + 10 states (j = 9
⁄2) + 12 states (j = 11
Where for every j there are 2j+1 different states from different values of mj.
Due to the spin-orbit interaction the energies of states of the same level but with
different j will no longer be identical. This is because in the original quantum numbers,
when s is parallel to l , the interaction energy is positive; and in this case j = l + s = l + 1
⁄2.
When s is anti-parallel to l (i.e. aligned oppositely), the interaction energy is negative,

43
and in this case j=l−s=l− 1
⁄2. Furthermore, the strength of the interaction is roughly
proportional to l.
For example, consider the states at level 4:
1. The 10 states with j = 9
⁄2 come from l = 4 and s parallel to l. Thus they have a
positive spin-orbit interaction energy.
⁄2 came from l = 4 and s anti-parallel to l. Thus they have a
negative spin-orbit interaction energy.
⁄2 came from l = 2 and s parallel to l. Thus they have a
positive spin-orbit interaction energy. However its magnitude is half compared to
the states with j = 9
⁄2.
⁄2 came from l = 2 and s anti-parallel to l. Thus they have a
negative spin-orbit interaction energy. However its magnitude is half compared to
the states with j = 7
⁄2.
⁄2 came from l = 0 and thus have zero spin-orbit interaction
energy.
2.3.4 Wood Saxon Potential:
The Woods–Saxon potential is a mean field potential for
the nucleons (protons and neutrons) inside the atomic nucleus, which is used to describe
approximately the forces applied on each nucleon, in the nuclear shell model for the
structure of the nucleus.
The form of the potential, as a function of the distance r from the center of nucleus, is:

44
where V0 (having dimension of energy) represents the potential well depth, a is a length
representing the "surface thickness" of the nucleus, and is the nuclear
radius where r0 = 1.25 fm and A is the mass number.
Typical values for the parameters are: V0 ≈ 50 MeV, a ≈ 0.5 fm.
For large atomic number A this potential is similar to a potential well. It has the following
desired properties
• It is monotonically increasing with distance, i.e. attracting.
• For large A, it is approximately flat in the center.
• Nucleons near the surface of the nucleus (i.e. having r ≈ R within a distance of
order a) experience a large force towards the center.
• It rapidly approaches zero as r goes to infinity (r − R >> a), reflecting the short-
distance nature of the strong nuclear force.
When using the Schrödinger equation to find the energy levels of nucleons subjected to
the Woods–Saxon potential, it cannot be solved analytically, and must be treated
numerically[42].
2.3.5 Achievements of Shell Model:
This model predicts very closely spaced energy levels in nuclei which is contrary

45
to observation at low energies. The low lying excited states in nuclei ate actually quite
widely spaced, which cannot be explained by the liquid drop model. This and certain
other properties of the nucleus would require us to consider the motion of the individual
nucleons in a potential well which would give rise to the existence of a nuclear shell
structure, similar to the electronic shells in the atoms[43].
Computer Program and Result

46
Binding Energy:
Nuclear binding energy is the energy that would be required to disassemble
the nucleus of an atom into its component parts. These component parts
are neutrons and protons, which are collectively called nucleons. The binding energy of
nuclei is due to the attractive forces that hold these nucleons together, and it is always a
positive number, since all nuclei would require the expenditure of energy to separate
them into individual protons and neutrons. The mass of an atomic nucleus is less than the
sum of the individual masses of the free constituent protons and neutrons (according to
Einstein's equation E=mc2
) and this 'missing mass' is known as the mass defect, and
represents the energy that was released when the nucleus was formed.
The term "nuclear binding energy" may also refer to the energy balance in
processes in which the nucleus splits into fragments composed of more than one nucleon.
If new binding energy is available when light nuclei fuse, or when heavy nuclei split,
either process can result in release of this binding energy. This energy may be made
available as nuclear energy and can be used to produce electricity as in (nuclear power)
or in a nuclear weapon. When a large nucleus splits into pieces, excess energy is emitted
as photons (gamma rays) and as the kinetic energy of a number of different ejected
particles (nuclear fission products).
The nuclear binding energies and forces are on the order of a million times greater
than the electron binding energies of light atoms like hydrogen. The mass defect of a
nucleus represents the mass of the energy of binding of the nucleus, and is the difference
between the mass of a nucleus and the sum of the masses of the nucleons of which it is
composed.
Now Binding Energy Per Nucleon is :
B.E. = (ZMP+ NMn – M (A,Z)) / A
Where

47
Z =atomic number M(A,Z)= Atomic Mass.
A = mass number Mn =mass of one neutron
MP = mass of one proton
Program 1: For Binding energy Per Nucleon With help of Mass Defect:
C Binding Energy Per Nucleon
C D= Atomic Mass number
C A=Atomic Mass
C Z=Atomic Mass
write(*,*)'enter the value of D, A,and Z'
read(*,*) D,A,Z
write(*,*)'enter the vaulue of N'
read(*,*)N
B=(Z*1.007825+N*1.008665-A)*931.47
write(*,*)B
U=B/D
write(*,*)'Binding energy per nucleon is=',U
stop
end

48
Semi Empirical Formula:
In nuclear physics, the semi-empirical mass formula (SEMF) (sometimes also
called Weizsäcker's formula, or the Bethe–Weizsäcker formula) is used to approximate
the mass and various other properties of an atomic nucleus from its number
of protons and neutrons. As the name suggests, it is based partly on theory and partly on
empirical measurements. The theory is based on the liquid proposed by George Gamow,
which can account for most of the terms in the formula and gives rough estimates for the
values of the coefficients. It was first formulated in 1935 by German physicist Carl
Friedrich von Weizsäcker, and although refinements have been made to the coefficients
over the years, the structure of the formula remains the same toda
Binding Energy per Nucleon is:
- ac Z(Z-1) A-4/3
.
Program 2.: Binding Energy per Nucleon With Help of Semi Empirical Formula.
c Binding enegry per nucleon
write(*,*)'enter 1 for volume energy enter 2 for surface energy enter 3 for coulomb
energy’
read(*,*) n
c volume energy
if(n.EQ.1.)then
write(*,*)'Enter the value for how many of nuclei you want to find volume enegry'
read(*,*)n
do 10 i= 1,n
V =(15.8*A)
B=V/A
write(*,*)B
A=A+1

49
10 continue
stop
end if
c surface energy
if(n.EQ.2.)then
write(*,*)'Enter the value for how many of nuclei you want to find surface energy '
read(*,*)n
do 20 j= 1,n
S=(18.34*(A**(2/3)))
B=S/A
write(*,*)B
A=A+1
20 continue
stop
end if
c coulomb energy
if(n.EQ.3.)then
write(*,*)'enter the value of Z and A'
read(*,*) Z,A
M= (0.7*(Z**(2)))/(A**(1/3))
B= M/A
write(*,*)B
stop
end if
stop
end
Program description :

50
The program is made on the bases of FORTRAN. In this program we basically use read,
write, if else ect. Command use. Read statement use for storing the input data given by
user, write statement use for showing statement in the upper columns, if else condition
use for checking the condition is true for given input or not. Here we use if else ladder
condition.
The notation used:
D is used for Atomic Mass Number
A is used for Atomic Mass
Z is used for Atomic Number
N is used for Neutron Number
B is used for Binding energy
S used for surface energy
V is used for Volume energy
M is used for Coulomb energy.
Result And Graphs:

51
1.) Binding Energy graph with mass defect
0 20 40 60 80
0
1
2
3
4
5
6
7
8
9
Binding EnegryBindingEnegry(MeV)
Atomic Mass
Graph 1: This Graph is showing the Nuclear Binding Energy per nucleon i.e. atomic
mass number.
Firstly it increases with the increasing the atomic mass and then it starts slowly varying
with the atomic mass. For the small nuclei is very small and rises rapidly with A attaining
a value of 8 MeV per nucleon for A~20. It is then rises A~56. For higher A, it decreases
slowly.
2.) Volume Energy Graph:

52
0 10 20 30 40
-20
-15
-10
-5
0
5
10
15
20
Volumeenergy/nucleon(Mev)
Atomic Mass
Volume energy / nucleon
Graph 2: This Graph show the variation of Binding energy per nucleon with the atomic
mass of volume term. It is constant because the strong force between the nucleons has a
limited range and a given nucleon may only interact with its direct neighbours.
3.) Surface Energy Graph:

53
0 10 20 30 40
-20
-15
-10
-5
0
5
10
15
20
BindingEnergy(Mev)
Atomic Mass
Surface Energy
Graph 3: This Graph shows the binding energy per nucleon with atomic mass for Surface
Energy. The surface term is also based on the strong force; it is, in fact, a correction to
the volume term. It is negative because it is decreases the total binding energy per
nucleon.
4.) Coulomb Energy Graph:

54
0 10 20 30 40
-20
-15
-10
-5
0
5
10
15
20
Coulombenergy/Nucleon(Mev)
Atomic Mass
Coulomb energy / Nucleon
Graph 4: This graph show the variation of binding energy per nucleon with the atomic
mass for coulomb energy. This term describes the Coulomb repulsion between the uniformly
distributed protons and is proportional to the number of proton pairs Z2
/R, whereby R is
proportional to A1/3
. This effect lowers the binding energy because of the repulsion between
charges of equal sign.
5.) Total Energy Graph:

55
-20
-15
-10
-5
0
5
10
15
20
-20
-15
-10
-5
0
5
10
15
20
-20
-15
-10
-5
0
5
10
15
20
0 10 20 30 40
-20
-15
-10
-5
0
5
10
15
20
ATOMIC MASS
bindingenergry/A[MeV]
SURFACE ENERGY
COULOMB
TOTAL
VOLUME
Graph 5: This Graph show the total binding energy per nucleon with atomic mass. The
total cure gives the very accurate value of the binding energies of the atoms. This is
better than the binding energy given by the mass defect formula.
Conclusion
We have review a detailed study of the Nuclear Structure Models. Starting with the study

56
of history of the nucleus then basics of the nucleus and its properties we came across the
nuclear models. We have studied the liquid drop model and shell model. Firstly we
calculate the binding energy per nucleon form the Mass Defect formula which is:
B.E. = (ZMP+ NMn – M (A,Z)) / A
Then with the help of Semi-Empirical formula that gives the more accurate curve of
binding energy. Hence give the more accurate binding energy per nucleon. The Semi-
Empirical Formula without corrections is:
- ac Z(Z-1) A-4/3
.
We have made a Fortran program to calculate the binding energy per nucleon of the
atoms. The program generated calculates all aspects of binding energy formula. In the
program we give the data for A,Z & N The program easily calculate all the other aspects
also by given required data. The programming made calculation very easy and fast. So
that we easily calculate any value of B.E. Although we use computer programming on
very high scale and it makes our life more easier. This program is a small example of that
easiness.
The Graphs shows the variation of the Binding Energy per nucleon with the Atomic
Mass. There the four separate graphs for the different forms of energy like volume,
surface, coulomb and binding energy with the mass defect formula and fifth graph show
the total binding energy collectively of the terms volume, surface and coulomb which
gives the best fit curve for binding energy.
References:
[1] Boundless."http://www.boundless.com/ Boundless Learning CC BY-SA 3.0.
[2] Atomic theory. "http://en.wikipedia.org/wiki/Atomic_theory Wikipedia CC BY-SA
3.0.’

57
[3] John Dalton."http://en.wikipedia.org/wiki/John_DaltonWikipedia CC BY-SA 3.0.
[4] atom- ‘http://en.wiktionary.org/wiki/atomWiktionary CC BY-SA 3.0.
[5]http://www.boundless.com//chemistry/definition/atomic-mass-unitBoundless
Learning CC BY-SA 3.0.
[6] Thomson model “ http://byjus.com/chemistry/thomsons-model.”
[7] Rutherford Model”https://www.britannica.com/science/atom/Discovery-of-
radioactivity#ref496652”
[8] Nuclear Physics By S.N. Ghoshal ISBN: 81-219-0413-7 S.Chand Publications.
[9] Nicholas Giordano (1 January 2012). College Physics: Reasoning and Relationships.
Cengage Learning. pp. 1051–. ISBN 1-285-22534-1.
[10] Introduction to Nuclear and Particle Physics 2nd
Edition by V.K. Mittal, R.C. Verma
and S.C. Gupta, ISBN -978-81-203-4311-5 PHI Learning Privet Limited.
[13] wanenko, D.D. (1932). "The neutron hypothesis". Nature. 129 (3265):
798. Bibcode:1932Natur.129..798I. doi:10.1038/129798d0.
[14] Heisenberg, W. (1932). "Über den Bau der Atomkerne. I". Z. Phys. 77: 1–
11. Bibcode:1932ZPhy...77....1H. doi:10.1007/BF01342433.
[15] Heisenberg, W. (1932). "Über den Bau der Atomkerne. II". Z. Phys. 78 (3–4): 156–

58
164. Bibcode:1932ZPhy...78..156H. doi:10.1007/BF01337585.
[16] Heisenberg, W. (1933). "Über den Bau der Atomkerne. III". Z. Phys. 80 (9–10): 587–
596. Bibcode:1933ZPhy...80..587H. doi:10.1007/BF01335696.
[17] Miller A. I. Early Quantum Electrodynamics: A Sourcebook, Cambridge University Press,
Cambridge, 1995, ISBN 0521568919, pp. 84–88.
Edition by V.K. Mittal, R.C.
Verma and S.C. Gupta, ISBN -978-81-203-4311-5 PHI Learning Privet Limited.
[22] Size of Nucleus http://byjus.com/physics/size-of-the-nucleus.
[23] Nuclear Mass http://www.nature.com/news/pear-shaped-nucleus-boosts-search-for-
new-physics-1.12952
[24] Nuclear Density "The Atomic Nucleus". Retrieved 2014-11-18. (derivation of equations
and other mathematical descriptions)
[25] Charge: Introduction to Nuclear and Particle Physics 2nd
Edition by V.K. Mittal,
R.C. Verma and S.C. Gupta, ISBN -978-81-203-4311-5 PHI Learning Privet Limited.

59
[26] r. Rod Nave of the Department of Physics and Astronomy, Dr. Rod Nave (July
2010). "Nuclear Binding Energy". Hyperphysics - a free web resource from GSU. Georgia State
University. Retrieved 2010-07-11.
[27] Nuclear binding energy". How to solve for nuclear binding energy. Guides to
solving many of the types of quantitative problems found in Chemistry 116. Purdue
University. July 2010. Retrieved 2010-07-10. Guides
[28]"Nuclear Energy". Energy Education is an interactive curriculum supplement for secondary-
school science students, funded by the U. S. Department of Energy and the Texas State Energy
Conservation Office (SECO). U. S. Department of Energy.
[29] Stern, Dr. David P. (September 23, 2004). "Nuclear Physics". "From Stargazers to
Starships" Public domain content. NASA website. Retrieved 2010-07-11.
[30] Nuclear Spin: https://www.nucleonica.com/Application/Help/Helpfiles/ Nuclear
Spin.html
[31] Nuclear Spin: https://www.nucleonica.com/Application/Help/Helpfiles/ Nuclear
Spin.html
[32] Nuclear Magnetic Moment: K. Kirk Shung; Michael Smith; Benjamin M.W. Tsui
(2012-12-02). "Principles of Medical Imaging”. Academic Press. p. 216.
[33] Liquid Drop Model: http://www.nuclear-power.net/nuclear-power/fission/liquid-
drop-model.
drop-model.

60
drop-model.
[36] https://www.kullabs.com/classes/subjects/units/lessons/notes/note-detail/7965
[37] https://www.kullabs.com/classes/subjects/units/lessons/notes/note-detail/7965
[38] "New Magic Number, N=16, near the Neutron Drip Line". Physical Review
Letters. 84 (24): 549 Bibcode:2000PhRvL..84.5493O.
PMID 10990977 . doi:10.1103/PhysRevLett.84.5493.
[39] Caurier, E.; Martínez-Pinedo, G.; Nowacki, F.; Poves, A.; Zuker, A. P. (2005). "The
shell model as a unified view of nuclear structure". Rev. Mod. Phys. 77 (2): 427–
488. doi:10.1103/RevModPhys.77.427.
[40] Coraggio, L.; Covello, A.; Gargano, A.; Itaco, N.; Kuo, T.T.S. (2009). "Shell-model
calculations and realistic effective interactions". Progress in Particle and Nuclear
Physics. 62 (1): 135–182. doi:10.1016/j.ppnp.2008.06.001.
[41] B.R.; Navrátil, P.; Vary, J.P. (2013). "Ab initio no core shell model". Progress in
Particle and Nuclear Physics. 69: 131–181. doi:10.1016/j.ppnp.2012.10.003.
[42] Skyrme, T. H. R. (7 February 1961). "A Non-Linear Field Theory". Proceedings of the Royal
Society A: Mathematical, Physical and Engineering Sciences. 260 (1300): 127–
138. doi:10.1098/rspa.1961.0018.

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