Nuclear Models

3.1 Introduction

The small size of the nucleus and the fact that the forces do not appear else where , make the

theoretical approach to the nuclear model more difficult than the electronic structure. There are many

isolated facts which will require explanation when we adopt any nuclear model. For example (i) Why is

the binding energy per nucleon almost constant? (ii) why do nuclei emit α particles and β particles when

these are known to contain only protons and neutrons ? (iii) How do we interpret the special properties

of nuclei such as stability, spin, magnetic moment ,etc.

The macroscopic properties, such as constant density and constant binding energy per nucleon,

resemble those of a drop of liquid. On the other hand, the microscopic properties, such as nuclear wave

functions and particle motions, resemble those of a weakly interacting gas. The resemble to a drop of

liquid serves as the basis for the liquid drop model and collective model and the resemblance to a

weakly interacting gas serves as the basis for the fermi gas model and the shell model.

3.2. Liquid drop model

A nuclear model was suggested by Bohr in 1937. In this model, the forces acting in the nucleus are

assumed to be analogical to the molecular forces in a droplet of some liquid. Bohr observed that there

are certain similarities between an atomic nucleus and a liquid drop. The similarities between the

nucleus and a liquid drop are as follow:

(i) The nucleus is supposed to be spherical in shape in the stable state, as a liquid drop which is

spherical due to the symmetrical surface tension forces. The nuclear forces are analogous to

the surface tension force of the liquid.

(ii) The force of surface tension acts on the surface of the liquid drop. Similarly, there is a

potential barrier at the surface of the nucleus.

(iii) The density of liquid drop is independent of its volume. Similarly, the density of the nucleus

is independent of its volume.

(iv) The intermolecular forces in a liquid are short range forces. The molecules in a liquid drop

interact only with their immediate neighbors. Similarly, the nuclear forces are short range

forces. Nucleons in the nucleus also interact only with their immediate neighbors. This leads

to the saturation in the nuclear forces and a constant binding energy per nucleon.

(v) The constant binding energy per nucleon is analogous to the latent heat of vaporization.

(vi) The molecules evaporate from a liquid drop on raising the temperature of the liquid due to

their increased energy of thermal vibration. Similarly, when energy is given to a nucleus by

bombarding it with nuclear projectiles, a compound nucleus is formed which emits nuclear

radiation almost immediately.

(vii) When a small drop of liquid is allowed to oscillate, it breaks up into two smaller drops of

equal size. The process of nuclear fission is similar and the nucleus breaks up into two

smaller nuclei.

In spite of these similarities we can see following differences : (1) Molecules attract one another at

distances larger that the dimensions of the electron shells and repel strongly when the distance is

smaller than the size of the electron orbits. Nuclear forces are attractive within the smaller range,

the range of nuclear forces. (2)Average kinetic energy of the molecules in the liquid is of the order of

0.1 eV , corresponding de Broglie wavelength is 5 × 10−11 𝑚 which is very smaller than the

intermolecular distances. The average K.E. of nucleons in nuclei is of the order of 10 MeV, the

corresponding de-Broglie wavelength is nearly 6 × 10−15 m, which is of the order of inter-nucleon

distances (Tayal, 2003). Hence the motion of the molecules in the liquid is of classical character

whereas in nuclei, the motion of nucleons is of quantum character.

Semi-empirical mass formula

Binding energy of a nucleus is the total energy needed to separate completely it’s individual

nucleons (of total number A). In this section, we derive a formula that gives the binding energy of nuclei

as a function of the numbers Z, N and A(= 𝑍 + 𝑁). All of the terms in this formula have at least partial

theoretical justifications, but the values of the coefficients that appear are generally found by fitting the

formula to the measured binding energies. For this reason this formula is called semi empirical binding-

energy formula, or the Weizsacker semi empirical formula, after its inventor.

The liquid drop model can be used to obtain an expression for the binding energy of the nucleus. Some

of the properties of the nuclear forces such as saturation, short range, have been deduced from the

approximately linear dependence of the binding energy and the volume on the number of particles in

the nucleus, are analogous to the properties of the forces which hold a liquid drop together. Hence, a

nucleus may be considered to be analogous to a drop of liquid of very high density. This idea has been

used, together with other classical ideas such as electrostatic repulsion and surface tension , to set up a

semi-empirical formula for the binding energy of a nucleus in tis ground state. The formula has been

developed by considering the different factors which affect the nuclear binding. The factors are

discussed as follows:

1. To an excellent approximation, the binding energy of a nucleus is found to be proportional to the

number A, of its nucleons. Sine the range of nuclear force is only about 2 fm, hence for most nuclei,

each nucleon is bound to only a fraction of the other nucleons , namely, those neighbors within a

sphere of radius about 2 fm. Also the density of nucleons is about the same inside all nuclei. It

follows that the number of neighbors to which each nucleon is bound is approximately the same for

all nuclei. Therefore, the average binding energy of each nucleon is the same in all nuclei (i.e. B/A is

constant) hence total binding energy is proportional to A. Since A is proportional to the volume of

the nucleus, this approximation is called the volume term and is written as

𝐵 ≈ 𝐵vol = 𝑎vol𝐴…………. (3.1)

where 𝑎vol is a constant. This is energy is called volume energy of the nucleus.

It follows from equation (3.1) that larger the total number of nucleons A, the more difficult it will

be to remove the individual protons and neutrons from the nucleus.

2. There is an immediate correction to the approximation (3.1). Those nucleons near the nuclear

surface have fewer neighbors to attract them and are less tightly bound than those in the interior.

The number of these surface nucleons is proportional to the surface area of the nucleus, i.e.

4𝜋𝑅2 = 4𝜋𝑅02𝐴

2

3. Therefore, equation (3.1) must be reduced by an amount proportional to 𝐴2

3, and

our next approximation is 𝐵 ≈ 𝐵vol + 𝐵surf ,where the surface correction 𝐵surf is negative and has

the form

𝐵surf = −𝑎surf 𝐴2

3 …………. (3.2)

This negative energy is called surface energy of a nucleus.

3. A further correction is needed for the electrostatic repulsion of the protons. This repulsion reduces

the binding energy by an amount equal to the potential energy of the total nuclear charge. The

potential energy of a uniform sphere of charge Ze is

𝑈Coul =3

5 𝑘(𝑍𝑒)2

𝑅 ……………. (3.3), where 𝑘 = 1/4𝜋𝜀0

Since 𝑅 = 𝑅0𝐴1

3, this implies a correction of the form

𝐵Coul = −𝑎Coul 𝑍2

𝐴13

………… (3.4)

Where

𝑎Coul ≈3

5

𝑘𝑒2

𝑅0

4. The next correction arises from the symmetry effect i.e. lack of symmetry between the number of

protons(Z) and number of neutrons (N) in the nucleus. The maximum stability of a nucleus occurs

when 𝑁 = 𝑍. Any departure from this introduces an symmetry 𝑍 − 𝑁, which results in decreases in

the stability i.e. in the binding energy (B). Since B gets less as 𝑍 − 𝑁 increases in either direction,

this suggests a negative term proportional to (𝑍 − 𝑁)2. The observed binding energies of all nuclei

are best fitted by a correction of the form

𝐵Sym = −𝑎sym(𝑍−𝑁)2

𝐴 …………… (3.5)

where A comes in the denominator because from the detailed study of the symmetry effect shows

that B is also inversely proportional to A.

5. The final term in our formula for the binding energy comes from the property called pairing effect.

It is seen that majority of stable nuclei (it is 148, in fact) have both Z and N even; there are 100

nuclei with Z even but N odd or vice versa; and there are only 4 stable nuclei in which both Z and N

are odd. This preference for Z and N to be even is the property of the nuclear force called paring

effect. Due to this effect, a correction term to the binding energy be

𝐵pair = 𝜀𝑎𝑝𝑎𝑖𝑟

𝐴12

…………….. (3.6)

where 𝑎pair is some constant and

𝜀 = { 1 if Z and N are both even

0 if Z or N is even and other is odd −1 if Z and N are both odd

} ……..(3.7)

Putting all five of these contribution together, we obtain the semi-empirical binding-energy formula

𝐵 = 𝐵vol + 𝐵surf + 𝐵Coul + 𝐵sys + 𝐵pair

= 𝑎vol 𝐴 − −𝑎surf 𝐴2

3 − 𝑎Coul 𝑍2

𝐴13

−𝑎sym(𝑍−𝑁)2

𝐴+ 𝜀

𝑎𝑝𝑎𝑖𝑟

𝐴12

……. (3.8)

where 𝜀 is defined in (3.8). The five coefficients are generally adjusted to fit all the measured binding

energies as well as possible. One set of values that gives a good fit is as follows:

𝑎vol = 15.75 𝑀𝑒𝑉 𝑎surf = 17.8 𝑀𝑒𝑉 𝑎Coul = 0.711 𝑀𝑒𝑉 𝑎sym = 23.7 𝑀𝑒𝑉 𝑎pair = 11.2 𝑀𝑒𝑉

Merits

1. The liquid drop model explains many characteristics of the nuclear matter, such as observed

binding energies of the nuclei and their stability against 𝛼 and 𝛽 disintegration as well as nuclear

fission.

2. From the liquid drop model, the calculation of atomic masses and binding energies can be done

with good accuracy.

3.3 The IPA Potential Energy for Nucleons

IPA is often called the independent particle model, abbreviated IPM, or the mean field approximation.

In study of the atomic structure, we relied heavily on this model. In this approximation each of the Z

electrons was assumed to move independently in the field produced by the average distribution of the

other Z-1 electrons. In this approximation, it is found to have possible states of each separate electron

and , hence those of the atoms as a whole. In the same way, in the IPA , each of A nucleons assumed to

move independently in the field produced by the average distribution of the other A-1 nucleons.

For the real nucleus in three dimension, it is seen that the nucleons are distributed nearly

uniformly through a sphere of radius

R (nuclear surface)= (1.07 fm)𝐴1

3 ………. (3.9)

Therefore, the average potential energy seen by each

nucleon is found by averaging the nucleon-nucleon

potential energy over this sphere. The resulting IPA

potential energy, U(r), is spherically symmetric and is

sketched in figure 3.1. The radius of the potential well is

given approximately by

R (force range)= (1.17 fm)𝐴1

3 ……………. (3.10)

Comparing equation (3.10) with (3.9), we can see that, as

one would except, the force range is a little larger than

the nuclear radius since the force can “reach out” a little

beyond the distribution of nucleons. The depth of IPA well

is about 50 MeV and is approximately the same for most

nuclei.

The potential-energy function in figure 3.1 is the averaged

nuclear potential energy seen by each nucleon. Because

of the charge independence of the nuclear force, this IPA potential energy is the same for neutrons as

for protons.

To a good approximation, the electrostatic potential energy of a proton in the electric field of a uniform

sphere of charge (𝑍 − 1)e. Outside the nucleus (𝑟 > 𝑅) this is

Figure 3.1: The nuclear IPA potential well, U(r), seen by one nucleon under the influence of other A−𝟏 nucleons in a nucleus. The force range R can be defined as the radius at which the potential energy is half its maximum depth and is found to be somewhat larger than the radius of the nucleus.

UCoul(𝑟) = (𝑍 − 1)𝑘𝑒2

𝑟 …….. (3.11)

Inside the nucleus, it has the shape of the inverted

parabola and potential energy, 𝑈Coul(𝑟) is

maximum at the center, where it is 1.5 times its

value at the nuclear surface:

UCoul(0) = (𝑍 − 1)𝑘𝑒2

𝑟 ………. (3.12)

The coulomb potential varies from 1 or 2 MeV for

the lightest nuclei to about 30 MeV for the heaviest.

For protons in light nuclei, it is therefore a fair

approximation to ignore 𝑈Coul compared to the

nuclear potential energy of figure 3.1. For protons in

heavy nuclei, this is certainly not a reasonable

approximation, and the total potential energy is then

𝑈(𝑟) = 𝑈nuclear(𝑟) + 𝑈Coul(𝑟)

This function is plotted in figure 3.2, which shows the

IPA potential well for a proton in the middle mass

nucleus 𝑆𝑛50120 (tin).

Using the IPA potential-energy functions, one can calculate the energy levels of each nucleus. In this way

we can construct a shell model of nucleus, quite analogous to the atomic shell model.

3.3 The Shell Model

The shell model is an important part of our understanding of nuclei and is a beautiful application of

many of the ideas of the atomic shell model. Shell model assumes the energy levels for nucleons inside

the nucleus are similar to that of the electrons in the atom. According to this model, the protons and

neutrons are grouped in shells in the nucleus, similar to the electrons in various shells outside the

nucleus (Murugeshan, 2015).There is direct a direct analogy between the theoretical treatment of a

nucleons in a nucleus and an electron in an atom. The potential energy V(r) is analogous to the Coulomb

energy ,and the orbit (or quantum state) of a nucleon is analogous to an orbit (or quantum state) of an

electron. The theoretical treatment of the nucleon parallels that of the electron and closed nucleon

shells are found analogous to the closed-electron shells in the atom. Furthermore, two different kinds of

particles must be considered, protons and neutrons, rather than a single kind , electrons. The Pauli’s

exclusion principle must be applied to both protons and neutrons; it excludes two protons from

occupying the same quantum state, and two neutrons from having the same quantum numbers. (Arora

C. L., 2014)

Figure 3.2: The IPA potential energy of any one proton in the

nucleus 𝑺𝒏𝟓𝟎𝟏𝟐𝟎 . A neutron feels only 𝑼𝒏𝒖𝒄

The semi empirical binding-energy formula gives a remarkably accurate picture of general trends

of nuclear binding energies. Nevertheless, it is a smoothed average picture and shows none of the

fluctuations that one would expect if nucleons, like atomic electrons, occupy energy levels that

corresponds to some distinct shells (Taylor, 2004).

Recall that atomic binding energies fluctuate periodically with Z, being greatest when the electrons

exactly fill a closed shell and dropping steeply just beyond each closed shell. The closed shells are

formed when they contain specific number of protons or neutrons or both. These number are called

magic number and nuclei exhibit special characteristics of stability when neutrons or protons or both are

filled in the shell by such numbers. This is the reminiscent of the atomic shell characteristics shown by

noble gases among the atoms. The magic numbers are 2,8,20,50,82 and 126. The nuclei containing

2,8,20,50, 126 nucleons of the same kind form some sort of closed nuclear shell structures.

Evidence in favor of the existence of magic numbers

1. Mayer in 1948 suggested that nuclei with magic number of nucleons are especially abundant in

nature.

2. The inert gases with closed shells exhibit a high degree of chemical stability. Similarly, nuclides

whose nuclei contain a magic number of nucleons of the same kind exhibit more than average

stability.

3. 𝐻𝑒24 and 𝑂8

16 are particularly stable, can be seen from the binding energy curve. Here the

numbers 2,8, indicate stability.

4. Isotopes of elements having an isotopic abundance greater than 60% belongs to the magic

number category.

5. Tin(Sn) with Z=50 has ten stable isotopes, more than any other element, while Calcium (Ca) with

Z=20 has six isotopes. This indicates that elements with Z=50 and Z=20 are more than usually

stable.

6. The three main radioactive series, namely the uranium series, actinium series and thorium series

decay to lead ( 𝑃𝑏)82126 with Z=82 and N=126. Thus lead 𝑃𝑏82

126 is the most stable isotope. This

gain show the number 82 and 126 indicate stability.

7. It has been found that nuclei having a number of neutrons equal to the magic number can not

capture a neutron because the shells are closed and they can not contain an extra neutron.

8. It is found that some isotopes are spontaneous neutron emitters when excited above the

nucleon binding energy by preceding β-decay. These are 𝑂817 , 𝐾𝑟36

87 and 𝑋𝑒54137 for which N=9, 51

and 83 which can be written as 8+1, 50+1, and 82+1. If we interpret this loosely bound neutron,

as a valence neutron, the neutrons numbers 8,50, 82 represent the magic number and show

greater stability than other neutron numbers.

It is thus concluded that nuclear behavior is often determined by excess or deficiency of

nucleons with respect to closed shells of nucleons corresponding to the magic numbers. Hence,

it was thus suggested that nucleons revolve inside the nucleus just as electrons revolve outside

in specific permitted orbits. The protons and neutrons move in two separate systems of orbits

round the centre of mass of all the nucleons. The nucleons move in the orbits around a

common centre of gravity of all the constituents of the nucleus. However, electrons outside the

nucleus (also called extra-nuclear electrons) revolve in the Coulomb field of a relatively distant

heavy nucleus. Each nucleons shell has a specific maximum capacity, like that of the extra-

nuclear electrons. When the shells are filled to that capacity, they give rise to the particular

numbers, which are called magic numbers. The magic number nuclei gives rise to the unusual

stability like the closed electron shell structure of inert gases.

The shell model is able to account for several nuclear phenomena in addition to magic

numbers. For example even-even nuclei are ,in general, more stable than odd-odd nuclei which

can be explained from the shell model. According to Pauli’s principle, a single energy sublevel

can have a maximum of two nucleons , one with spin up and another with spin down. Therefore,

in an even-even nucleus only completed sublevels are present which means greater stability. On

the other hand, an odd-odd nucleus contains incomplete sublevels for both kinds of nucleons

which means lesser stability.

The shell-model also capable of predicting the total angular momenta of nuclei. In even-

even nuclei, all the protons and neutrons should pair off so as cancel out one another’s spin and

orbital angular momenta. Thus even-even nuclei correspond to zero angular momenta, as

observed experimentally. However, in even-odd and odd-even nuclei, the half-integral spin of

the single “extra” nucleon should be combined with the integral angular momentum of the rest

of nucleus for a half-integral total angular momentum. In odd-odd nuclei each have an extra

neutron and an extra proton whose half-integral spins should yield integral total angular

momenta. Both of these idea are experimentally confirmed.

Origin of the Magic Numbers

After introducing the existence of the nuclear magic numbers, physicists hoped to explain them

in terms of nucleon energy levels and the grouping of those levels into energy shells. The IPA

potential energy felt by any one proton or neutron was described §3.3 and sketched in figure

3.1.The shell model is often called the independent-particle model (IPA). According to this

model, although shows quite similarities between electron’s shell and nucleon’s shell, but in the

case of potential energy felt by nucleons, it is quite different from the coulomb potential,

because there is a nuclear attractive force. The potential has actually a form between the

square-well potential V= −𝑉0 and the so called oscillator potential 𝑉 = −𝑉0 + 𝑎𝑟2, where r is

the distance between the nucleon and center of force and a is a constant (Kaplan, 2002). By

treating the potential V as a harmonic oscillator potential ,we have Schrodinger wave equation

for the nucleus is now given by

∇2𝜓 +2m

ħ2(𝐸 − 𝑉)𝜓(𝑟) = 0 ………….(3.30)

where ∇2 is the Laplacian second order differential operator, V is the potential energy while E is

the kinetic energy of the nucleon, m is generally taken as the reduced mass i.e. m =mnmp

mn+mp≈

1

2M ( mass of proton or neutron).

The solution of equation (3.30) is

En = (N +3

2) ħω

where N is called oscillator quantum number; N=0,1,2, 3,…and 𝜔 is the angular velocity of the

oscillator (Arora C. , Refresher Course in Physics, 2013). Another quantum number similar to the

principle quantum number of the electronic orbit characterizes the radial part of the nuclear

wave function and is denoted by n. The value of 𝑛 = 1,2,3, … etc.

It can be shown that angular part of ψ requires that the oscillator quantum number N is

related to the orbital quantum number 𝑙 and quantum number 𝑛 by the relation (Arora C. ,

Refresher Course in Physics, 2013)

𝑁 = 2(𝑛 − 1) + 𝑙 = 2𝑛 + 𝑙 − 2

Like in atomic physics, for each n, there is a azimuthal quantum number 𝑙 which varies

from 0 to 𝑛 − 1. The number 𝑙 identifies the magnitude of the nucleon’s orbital angular

momentum L by the equation L= √𝑙(𝑙 + 1)ħ. The possible values 𝑙 = 0,1,2,3,4 … are indicated

by the code letters s, p, d, f, g, … For each value of 𝑙, there are 2𝑙 + 1 possible values of

𝑚𝑙 which go from 𝑙, 𝑙 − 1, … to − 𝑙 while 𝑚𝑠 has two values ±1

2 . The term

𝑚𝑙 and 𝑚𝑠 are called magnetic orbital quantum number and magnetic spin quantum number.

The degeneracy of each level is therefore 2(2𝑙 + 1). Total accumulation number for a shell will

be

∑ 2(2𝑙 + 1)

𝑙

Thus, all the parameters associated with angular

momentum are the same as in atomic physics. On

the other hand, nuclear physicists define the

quantum number n so that the levels of a given 𝑙 are

simply numbered 𝑛 = 1,2,3 … (Taylor, 2004).

Thus the levels with 𝑙 = 1 are labeled

1𝑝, 2𝑝, 3𝑝, … in order of increasing energy; the levels

with 𝑙 = 2 are 1𝑑, 2𝑑, 3𝑑, … , so on.

Table 1 Energy levels and their occupation of nucleons

Table(1) shows the different quantum states

with their degeneracy and accumulation number

and corresponding sketch of these levels is shown

in the figure 3.3. We can see that first three of the

closed-shell numbers i.e. 2,8 and 20 agree exactly with the observed magic numbers, but beyond this, all

of the predicted closed-shell numbers turn out to be wrong. There must be some lacking on this

assumption.

Level Degeneracy of the level {2(2𝑙 + 1)}

Accumulation number (closed-shell number)

1s 2 2

1p 6 8

1d,2s 10+2=12 20

1f, 2p 14+6=20 40

1g,2d,3s 18+10+2=30 70

1h,2f,3p 22+14+6=42 112

Figure 3.3: The filling order of the levels for a proton or neutron for the nuclear potential well for IPA. The number to the right of each level is its degeneracy. The levels fall into distinct energy shells, but the closed-shell numbers do not agree with the observed magic numbers. It follows that these levels are not correct.

The reasonable explanation was given independently in 1949 by Maria Goeppert-Mayer and by

Johannes Jensen and coworkers. It turns out that the nuclear force depends strongly on the relative

orientation of each nucleon’s spin and orbital angular momentum. The IPA potential well of figure 3.1 is

the correct potential-energy function for a nucleon with 𝑙 = 0, but if 𝑙 ≠ 0, the nucleon has an

additional potential energy that depends on the orbital angular momentum L and its orientation relative

to the spin S. The additional energy ∆𝑈 is found to have the form

∆𝑈 ∝ 𝑆𝐿 cos 𝜃

where θ is the angle between S and L, that is,

∆𝑈 ∝ 𝑺. 𝑳 ……………. (3.31)

Thus , the energy ∆𝑈 is called the spin-orbit energy.

Since the spin-orbit energy of each nucleon is proportional to 𝑺. 𝑳, each energy level of figure 3.3

(except 𝑙 = 0) is split into two levels according to the two possible orientations of 𝑺 relative to L, taking

the consideration of the fact that S has just two possible orientations relative to any direction. Like that

in the atomic physics magnetic moment(μ) is associated with the spin angular momentum and the

magnetic field(B) is induced due to the orbital angular momentum. Thus 𝝁 ∝ 𝑺 and 𝑩 ∝ 𝑳 and the spin

orbit energy is

∆𝑈 = −�⃗�. 𝑠……………. (3.32)

However, the nuclear spin-orbit splitting in not a magnetic effect. Rather, it is a new property of the

nuclear force. The nuclear splitting is much larger than the atomic effect and is in the opposite direction.

For an electron in an atom, the state with S and L parallel has slightly higher energy; for a nucleon in a

nucleus, the corresponding state has much lower energy. The first clear evidence for this nuclear spin-

orbit energy came from studies of nuclear shell structure, but scattering experiments with separate spin-

up and spin-down beams of nucleons have since confirmed its existence and properties.

The effect of the spin-orbit energy on the levels of figure 3.3 is shown in figure 3.4. The figure

shows the resulting levels (or subshells) with their degeneracies and their grouping into energy shells.

The effect of spin-orbit energy increases with increasing 𝑙. The 1s level is not split at all; 1 p level and d

level are split, but the resulting two levels are still in the same shell. However, 1 f level and all the levels

with 𝑙 ≥ 3, the splitting is so large that the resulting two levels belong to different shells. We can see

that as a result of these splitting, the closed-shell numbers agree perfectly with the observed magic

numbers.

To understand the labeling and degeneracy of the levels in figure 3.4, we must discuss briefly the

total angular momentum J of a nucleon,

𝑱 = 𝑳 + 𝑺

The magnitude of B of 𝐿 = √𝑙(𝑙 + 1) ħ with 𝑙 = 0,1,2, …, and that of S is 𝑆 = √𝑠(𝑠 + 1) ħ, with s is

equal to ½. It is found that the magnitude of J is given by a similar formula

𝑱 = √𝑗(𝑗 + 1) ħ ……….(3.33)

If its orbital angular momentum is zero (𝑙 = 0), the nucleon’s total angular momentum is just the spin,

and hence 𝑗 =1

2. If 𝑙 ≠ 0, it is found that there are two possible values of j, depending on the orientation

of S relative to L: If S is parallel to L, then

𝑗 = 𝑙 +1

2 (𝑺 and 𝑳 parallel )……… (3.34)

When S is antiparallel to L, then

𝑗 = 𝑙 −1

2 (𝑺 and 𝑳 antiparallel )…….. (3.35)

In figure 3.4 , the value of j in each level is shown by a subscript following the code letter for 𝑙.

Thus the original 1p level, shown on the left, is two levels with

𝑗 = 𝑙 ±1

2= 1 ±

1

2

Figure3.4: Filling order of the levels through Z or N= 𝟖𝟐 for a single nucleon in a nucleus, including the spin-orbit energy. The levels on the left are the corresponding levels in the absence of the spin-orbit energy. The numbers to the right of each level are the level’s degeneracy and in parenthesis, total of protons or neutrons needed to fill through that level. On the right are the closed shell numbers which agree perfectly with the observed magic numbers. Below 82, where the proton well is strongly distorted by Coulomb repulsion, the level ordering for protons and neutrons are significantly different.

=3

2 or

1

2

These are shown on the right of figure 3.4. as 1𝑝1

2

and 1𝑝3

2

where 1𝑝3

2

is at lower energy state.

For each magnitude of the total angular momentum, there are several possible orientations

of J, all with the same energy. These correspond to the different possible values of 𝐽𝑧 (Z-

component of total angular momentum), which are given by

𝑗𝑧 = 𝑚𝑗ħ ,where 𝑚𝑗 is called the magnetic total angular momentum

quantum number, which can have any of the value 2𝑗 + 1 values

𝑚𝑗 = 𝑗, 𝑗 − 1, … . . , −𝑗

Hence each of the levels of figure 3.4 has a degeneracy of (2𝑗 + 1) as indicated to the

right of each level.

Angular momenta of Nuclei

Any filled level of given j contributes zero total angular momentum. This is because for every

nucleon with 𝐽𝑧 = 𝑚𝑗ħ; there is a second with 𝐽𝑧 = −𝑚ħ. This implies that any nucleus in which

all the occupied levels are closed, i.e. no partial filled levels; should have zero total angular

momentum. That is,

𝐽𝑡𝑜𝑡 = √𝑗𝑡𝑜𝑡(𝑗𝑡𝑜𝑡 + 1) ħ = 0

For example, in 𝐶612 the protons and neutrons both completely fill their 1𝑠1

2

and 1𝑝3

2

levels.

Therefore, 𝐶612 should have, and does have, zero total angular momentum,𝑗𝑡𝑜𝑡 = 0. Referring to

figure 3.4, we can predict in the same way that the closed-subshell nuclei 𝑂816 , 𝑆𝑖14

30 and 𝐶𝑎2048

should all have 𝑗total = 0, and in every case this is found to be correct.

Let us consider next a nucleus with a single nucleon outside otherwise filled subshells. For

example, in 𝑂817 the right protons and eight of the nine neutrons fill the levels 1𝑠1

2

, 1𝑝3

2

and 1𝑝1

2

,

but the ninth neutron must occupy the 1𝑑5

2

level all by itself. Since the closed levels all have zero

angular momentum, the angular momentum of the whole nucleus is just that of the final, odd

neutron. Thus we predict that 𝑂817 should have 𝑗tot = 5/2, this proves to be correct.

Limitation of shell model

(i) This model does not predict correct values of isospin quantum number (𝐼) in case of

certain nuclei. For example, it predicts 𝑙 = 5/2 for 𝑁𝑎1123 whereas correct value is 3/2.

(ii) The four stable nuclei 𝐻12 , 𝐿𝑖3

6 , 𝐵510 , 𝑁7

14 do not fit in the scheme of this model.

(iii) This model can not explain the observed first excited states in even-even nuclei at

energies much lower than those expected from single particle excitation.

(iv) This model could explain the observed large quadruple moment of odd A nuclei,

particularly those which have their nucleon number far away from the magic numbers.

Workout Examples

1. Determine the spin parity of 𝐵𝑒49 .

Solution: For odd mass number nuclei, there is either proton number or neutron number is

odd. The spin parity in this case is given by

𝑃 = {𝑗, (−1)𝑙}

where j is the total angular momentum quantum number of last unpair nucleon, 𝑙 is the

orbital quantum number of that nucleon.

Now in our case, number of neutron is equal to 5 and hence last nucleon is a neutron

which lies on 1𝑝3/2 energy level. So for p level,

𝑙 = 1, 𝑗 = 3/2

∴ spin parity = [𝑗, (−1)𝑙]

= [3

2, (−1)𝑙]

∴ spin parity = [3

2, −]

2. Determine the spin parity of 𝑁714 .

Solution: Here number of proton is 7 and number of proton is 7. There is an unpair neutron

and an unpair proton. Hence both these nucleon give spin parity. Now this the even mass

number case of nuclei. So we use the concept of Nordheim number i.e.

𝑁 = 𝑗1 + 𝑗2 − 𝑙1 − 𝑙2 = 0, ±1

Case I

When N=0 then, we use strong rule

i.e. 𝑗 = |𝑗1 − 𝑗2| and

Spin parity = [𝑗, (−1)𝑙1+𝑙2]

Case II

When N= ±1 then we use soft rule

i.e. 𝑗 = (𝑗1 + 𝑗2) or (𝑗1 − 𝑗2)

Spin parity = [𝑗, (−1)𝑙1+𝑙2]

Here seventh proton lie on the level 1𝑝1/2 , where 𝑙1 = 1 and 𝑗1 = 1/2. This is also same for

the seventh neutron. Thus

𝑗2 = 1/2 and 𝑙2 = 1

Hence Northumb number,

N= 𝑗1 + 𝑗2 − 𝑙1 − 𝑙2

=1

2+

1

2− 1 − 1

= −1

Here 𝑁 = −1. So we use soft rule or weak rule

𝑗 = (𝑗1 + 𝑗2) or (𝑗1 − 𝑗2)

= (1

2+

1

2 )or (

1

2−

1

2)

= 1 or 0

∴ parity = 𝑗, (−1)𝑙1+𝑙2

= [−1, (−1)1+1 ] or [0, (−1)1+1 ]

= (−1, +) or (0, +)

Bibliography Arora, C. (2013). Refresher Course in Physics. New Delhi: S.Chanda.

Arora, C. (2014). Refresher course in Physics . New Delhi: S.Chanda.

Arora, C. L. (2014). Refreshor Course in physics. New Delhi: S.Chanda.

Kaplan, I. (2002). Nuclear Physics . New Delhi: Narosa Publishing House.

Murugeshan, R. (2015). Modern Physics . New Delhi: S. Chanda.

Tayal, D. (2003). Nuclear Physics . Mumbai: Himalaya Publishing House.

Taylor, J. R. (2004). Modern Physics. New Delhi: PHI Learning Private Limited .