NUCLEAR MODELS

NUCLEAR MODELS
Inventory of Stable Nuclides:
1000 nuclides known to exist.
25% are stable.
Rest of the nuclei are radioactive and are mainly produced artificially.
Most of the heavy nuclei undergo 𝛽 – decay, few nuclei disintegrate by 𝛼 – decay.

Naturally occurring elements can be classified into two classes:
Elements[Natural]
Even values of Z Odd values of Z

 Elements with even value of Z have larger number of stable isotopes than with odd Z.
Number of stable isotopes associated with a nuclei having even number of Z can vary from 5 to
10.
Example:
 Calcium (Z = 20) , Selenium(Z = 34), Krypton (Z = 36) ------ Six stable isotopes
Zinc (Z = 30), Germanium (Z = 32), Zirconium (Z = 40) ------ Five stable isotopes
Cadmium (Z = 48) -------- Eight stable isotopes
Tin (Z = 50) -------- Ten stable isotopes.

Ratio of N/Z for stable nuclides is
confined within a narrow range.
For lighter nuclei:
number of protons and neutrons are
nearly equal N/Z = 1.
Stability line – equally inclined to Z
and N axes.

For heavier nuclei:
Number of neutrons is higher than
the number of proton.
N/Z > 1
Highest value is 1.6 for a very heavy
nuclei.
Stability line steeper at higher Z.

Isotopes of different elements (Z = constant) lie on different vertical lines.
Nuclides with different Z having same mass number (A = constant) lie along a line inclined at
1350 to the Z – axis known as isobars.
Segre chart:
Plot of N v/s Z.
Nuclides with same number of neutrons (N = constant) lie along different horizontal lines known
as isotones.

Isotopes of all elements can be classified into four groups:
Sr No Combination of protons and neutrons Abundance Number of
Stable isotopes
1 Even Z – Even N (e – e) 60 % 164
2 Even Z – Odd N (e – o)
comprise 20%
and
are equal in number
54
3 Odd Z – Even N (o – e) 50
4 Odd Z – Odd N (o – o) Smallest 4

Equality of Z and N for the lighter nuclei indicate:
Proton – proton and neutron – neutron forces are approximately equal with in a nuclei.
Known as charge – symmetry of the nuclear force.
For heavier nuclei:
Coulomb repulsion between the protons weaken the binding.
To compensate for this, the number of neutrons relatively larger than the number of protons –
increases the strength of binding.

𝛽−
active nuclei:
Nuclei lying to the left of the stability line –
number of neutrons increased keeping Z constant.
Neutron transforms into a proton spontaneously.
𝛽+
active nuclei:
Nuclei lying to the right of the stability line-
number of protons increased keeping N constant.
Protons transforms into a neutron spontaneously.

Nuclear Models:
To understand the observed properties of the nucleus of an atom it is
necessary –
Adequate knowledge about the nature of internucleon interaction.
Inside the nucleus –
Short range force exists – exact mathematical form of this interaction is still
unknown.
If the exact nature of the internucleon interaction were known….the
difficulties encountered –
Structure of nucleus consists of large number of protons and neutrons.
Impossible to solve the Schrodinger equation exactly for such a many body
problem.

Theory of atomic structure – What makes it easy to
understand????
Nature of interacting forces acting on the electrons in an atom
is electromagnetic force which is well understood.
Quantum mechanical theory of atomic structure is extensively
developed and agrees well with the experimental data.
To explain difficulties in developing a satisfactory theory of
nuclear structure –
 Different models proposed for the nucleus.
Each model explains some of the characteristics of the nuclei.

LIQUID DROP MODEL:
Liquid Drop model first proposed by:
N. Bohr and F. Kalckar in 1937
Later developed by C.F.von Weizsacker and H.A.Bethe to develop a semi – empirical formula for
the binding energy.

Similarity between a liquid drop and a nucleus:
Saturation of the force:
Each individual molecule within a liquid drop exerts an attractive force
upon group of molecules in its immediate neighbourhood.
Force of attraction does not extend to all molecules within the drop.
Calculating the potential energy:
Number of interacting pairs of molecules within a drop must be known.
If each molecule interacts with all the molecules in a drop, the number of
interacting pairs – N(N – 1)/2 where N is the total number of molecules.

If N is large, the number of pairs = 𝑁2
Potential energy ∝ 𝑁2
.
If each molecule interacts with a limited number of molecules in its immediate
vicinity –
Number of interacting pairs ∝ N
Potential energy ∝ N
Above relation supported by experimental facts.
E.g. total amount of heat required for evaporating a drop of liquid (latent heat) is
linearly proportional to the number of molecules within a liquid.
Amount of heat required to evaporate 2 g of a liquid is twice that required to
evaporate 1 g.

Binding energy of a nucleus:
Binding energy 𝐸𝐵 of a nucleus is proportional linearly to the number of
nucleons within it.
Mathematically: Binding energy 𝐸𝐵 ∝ Number of nucleons in the nucleus.
Binding fraction 𝑓𝐵 (binding energy per nucleon) ~ constant (8MeV) for
most nuclei.
Shows close resemblance of the nucleus with the liquid drop.
Conclusion:
Internucleon force within a nucleus attain a saturation value.
Each nucleon interacts with a limited number of nucleons in its close
vicinity.

Certain other points of resemblances between
the nucleus of an atom and a liquid drop:
Attractive force near the nuclear surface is
similar to the force of surface tension on a
surface of a liquid drop.

Figure: Attractive forces in nucleus and liquid drop at the surface

Density:
Density of a liquid drop is independent of the volume of the liquid.
Similarly, density of nuclear matter is independent of the volume.

Different types of particles, e.g. neutrons, protons, deuteron, 𝛼 – particles are emitted during
nuclear reactions.
Processes analogous to the emission of the molecules from a liquid drop during evaporation.

Internal energy of the nucleus is analogous to the heat energy within the liquid drop.

Formation of a short lived nuclide:
Formation of a short lived compound nucleus by the absorption of a nuclear particle in a
nucleus during a nuclear reaction is analogous to the process of condensation from the vapour
phase to the liquid phase in the case of a liquid drop.

BEITHE – WEIZSACKER FORMULA:
This semi – empirical formula for the nuclear masses (or nuclear binding energies) gives a
connection between the theory of the nuclear matter with the experimental information.
Based on the liquid drop model of the nucleus.
If M(A,Z) be the atomic mass of the isotope of an element X of atomic number Z and mass
number A, then we can write:
M(A,Z) = Z 𝑀𝐻 + N𝑀𝑛 - 𝐸𝐵
where 𝐸𝐵 is the nuclear binding energy
𝑀𝐻 and 𝑀𝑛 are the masses of the hydrogen atom and neutron.
N = A – Z is the number of neutrons in the nucleus.

The binding energy 𝐸𝐵can be expressed as the sum of a number of terms as given below:
1. Volume energy
2. Surface energy
3. Coulomb energy
4. Asymmetry energy
5. Pairing energy

Volume energy term:
𝐸𝑉 = 𝑎1𝐴

Surface energy term:
Nuclear force similar to surface tension.
Nucleons acted by attractive force due to
nucleons inside the sphere.
No forces acting from outside.
Existence of surface force tends to reduce
the binding energy of the nucleus by an
amount proportional to the surface area of
the latter.
𝐸𝑆 = -𝑎2𝐴2 3

Coulomb energy term:
𝑬𝑪 = −𝒂𝟑
𝒁𝟐
𝑨𝟏 𝟑
where
𝒂𝟑 =
𝟑
𝟓
𝒆𝟐
𝟒𝝅𝜺𝟎𝒓𝟎

Corrections in the coulomb energy
term:
1)non- uniformity of the nuclear
charge distribution.

the requirement of discrete arrangement of the charges on the proton.
Effect of uncertainty in the localization of the protons.
Non- sphericity of the nucleus.
Corrections of the positions of protons.

Asymmetry energy term:
𝑬𝒂 = −𝒂𝟒
(𝑨 −𝟐𝒁)𝟐
𝑨

Pairing term:
𝜹 = 𝒂𝟒 𝑨−𝟑 𝟒

Figure: Contribution of different
terms in semi – empirical mass
formula

Figure: Comparison between theoretical and empirical graph for Binding energy

Semi – Empirical Mass Formula:

Value of different constants in MeV in Semi – Empirical mass formula:
M(A,Z) = Z 𝑀𝐻 + N𝑀𝑛 - 𝐸𝐵
The pairing energy term (𝜹) is subtracted for e – e nuclei and is added for o – o nuclei.
M(A,Z) = Z 𝑀𝐻 + (A – Z)𝑀𝑛 - 𝑎1𝐴 + 𝑎2𝐴2 3
+ 𝒂𝟑
𝒁𝟐
𝑨𝟏 𝟑 + 𝒂𝟒
(𝑨 −𝟐𝒁)𝟐
𝑨
- 𝜹

APPLICATIONS OF THE SEMI – EMPIRICAL MASS FORMULA:
1. Alpha Decay
2. The Mass Parabolas and prediction of stability against beta activity
3. 𝛽 – disintegration energy of the mirror nuclei.

Alpha Decay:
If a nucleus ZXA undergoes 𝛼 – decay into the nucleus Z-2YA-4
ZXA
Z-2YA-4 + 2He4
The 𝛼 – disintegration energy –
𝑄𝛼 = M(A,Z) - M(A – 4, Z-2) – M[2He4]
written in terms of the binding energies 𝐸𝐵 of the nuclei involved:
𝑄𝛼 = 𝐸𝐵(A – 4, Z – 2) + 𝐸𝐵[2He4] - 𝐸𝐵 (A, Z)
= 𝑎1(𝐴 − 4) - 𝑎2(𝐴 − 4)2 3- 𝑎3
(𝑍−2)2
(𝐴 −4)1 3 - 𝑎4
(𝐴 −2𝑍)2
𝐴 −4
- 𝑎1𝐴 + 𝑎2𝐴2 3+ 𝑎3
𝑍2
𝐴1 3 + 𝑎4
(𝐴 −2𝑍)2
𝐴
+ 𝐸𝐵[2He4]

On simplification:
𝑄𝛼 = 28.3 - 4𝑎1 +
8
3
𝑎2 𝐴−1 3 +
4𝑎3𝑧
𝐴1 3 [1 -
𝑍
3𝐴
] -
4𝑎4
𝐴
(𝐴 −2𝑍)2
𝐴 −4
---------------------------------- [1]
Important note:
Pairing energy term has been neglected.
Binding energy of the 𝛼 – particle is taken to be 28.3 MeV.
Using the numerical values of 𝑎1, 𝑎2, 𝑎3, 𝑎4 expressed in MeV are used in equation [1] gives
𝑄𝛼 >0 for A > 160.
Nuclei with A > 160 should be 𝛼 – disintegrating according to equation [1]
Observation: Nuclei with A > 200 undergo 𝛼 – disintegration.
For light nuclei (A < 200) the energy release is so small that barrier penetration probability is very
small.

Mass Parabolas: Stability of Nuclei against 𝜷 – Decay:
M(A,Z) = Z 𝑀𝐻 + (A – Z)𝑀𝑛 - 𝑎1𝐴 + 𝑎2𝐴2 3+ 𝑎3
𝑍2
𝐴1 3 + 𝑎4
(𝐴 −2𝑍)2
𝐴
Above equation can be rewritten as:
M(A,Z) = 𝑓𝐴 + pZ + q 𝑍2
------------------------------------------------------[1]
where 𝑓𝐴 = A(𝑀𝑛 - 𝑎1+ 𝑎4) + 𝑎2 𝐴2 3
p = -4𝑎4 - (𝑀𝑛 - 𝑀𝐻)
q =
1
𝐴
(𝑎3 𝐴2 3
+ 4𝑎4)
Equation [1] is the equation to a parabola for a given A.

Differentiating equation (1) w.r.t Z for a given A and setting it equal to zero gives the lowest
point Z = 𝑍𝐴 :
𝜕𝑀
𝜕𝑍 𝐴
= p + 2qZ
p + 2q𝑍𝐴 = 0
𝑍𝐴 = -
𝑝
2𝑞
=
(𝑀𝑛 − 𝑀𝐻+4𝑎4)A
2(𝑎3 𝐴2 3 + 4𝑎4)

Putting Z = 𝑍𝐴 in equation M(A,Z) = 𝑓𝐴 + pZ + q 𝑍2 we get:
M(A, 𝑍𝐴) = 𝑓𝐴 + p 𝑍𝐴 + q 𝑍𝐴
2
------------------------(2)
On simplification:
M(A, 𝑍𝐴) = 𝑓𝐴 - 𝑝2/4𝑞 ---------------------------(3)
Subtracting (3) from (1) :
M(A,Z) - M(A, 𝑍𝐴) = 𝑝2/4𝑞 + pZ + q 𝑍2
M(A,Z) - M(A, 𝑍𝐴) = q (𝑍 − 𝑍𝐴)2

M(A,Z) - M(A, 𝑍𝐴) = q (𝑍 − 𝑍𝐴)2
Important note:
Above equation proves that the mass parabolas for a given isobar A( A = constant) has the
lowest point at Z = 𝑍𝐴, since R.H.S is positive.
M(A,Z) has the smallest value for a given A at Z = 𝑍𝐴 nucleus has the largest binging
energy amongst the isobars for a given A.
𝑍𝐴 gives the value of Z for the most stable isobar given by:
𝑍𝐴 =
𝐴
1.98+0.015𝐴2 3
Equation doesn’t yield integral value for 𝑍𝐴. Value of Z nearest to 𝑍𝐴 corresponds to the actual
stable nucleus for a given A.

Expression for the energies or the
amount of energy released in the 𝜷
– disintegration process

𝜷 – disintegration of mirror nuclei:
Mirror Nuclei: Mirror nuclei are pairs of isobaric nuclei in which the proton and neutron
numbers are interchanged and differ by one unit.
Examples: (1H3, 2He3), (3Li7, 4Be7), (5B11, 6C11).
Members of the pairs of higher Z are usually found to be 𝛽+
emitters such as:
6C11
5B11 + 𝛽+
+ 𝜈.
The (A – 2Z) value in the asymmetry term in the mass formula can be written as:
A – 2Z = N + Z – 2Z = N – Z
N – Z = ±1
A – 2Z = ±1

Semi – Empirical mass formula for Odd A nuclei:
M(A,Z) = Z 𝑀𝐻 + N𝑀𝑛 - 𝑎1𝐴 + 𝑎2𝐴2 3
+ 𝑎3
𝑍2
𝐴1 3 + 𝑎4
(𝐴 −2𝑍)2
𝐴
M(A,Z) = Z 𝑀𝐻 + (Z – 1)𝑀𝑛 - 𝑎1𝐴 + 𝑎2𝐴2 3+ 𝑎3
𝑍2
𝐴1 3 +
𝑎4
𝐴
--------------------------- (1)
For daughter nuclei:
M(A,Z - 1) = (Z – 1) 𝑀𝐻 + Z𝑀𝑛 - 𝑎1𝐴 + 𝑎2𝐴2 3
+ 𝑎3
(𝑍−1)2
𝐴1 3 +
𝑎4
𝐴
---------------------------(2)

From equation (1) and (2):
M(A,Z) - M(A,Z - 1) = 𝑀𝐻 - 𝑀𝑛 + 𝑎3
(2𝑍 −1)
𝐴1 3
𝑄
β
+ = [M(A,Z) - M(A,Z-1) - 2𝑚𝑒]
= 𝑀𝐻 - 𝑀𝑛 + 𝑎3
(2𝑍 −1)
𝐴1 3 - 2𝑚𝑒
= 𝑎3 𝐴2 3 - (𝑀𝑛 - 𝑀𝐻 + 2𝑚𝑒)
𝑸
β
+ = 𝒂𝟑 𝑨𝟐 𝟑 + 1.804 MeV

Figure shows plot of the disintegration energy
against 𝐴2 3 is a straight line with slope 𝑎3.
From 𝑎3 - determined from the graph the
nuclear radius parameter 𝑟0 can be found.
Value of 𝑟0 = 1.44 X 10−15m.
Above estimate is on the higher side because
of limitations imposed by the presence of
Coulomb energy term.
With corrections more precise value of the
Coulomb energy 𝐸𝑐 term can be deduced.

Energetics of Symmetric Fission:
Nuclear Fission: Process of breaking up of a nucleus into two fragment nuclei of comparable
masses.
It can be induced:
by an external agent
or
can occur spontaneously.
Fission of the heavy nuclei (e.g. uranium) induced by neutrons.

In spontaneous fission a nucleus ZXA undergoes the spontaneous transformation:
ZXA
𝑍1
𝐴1
𝑋1 + 𝑍2
𝐴2
𝑋2
where the two product nuclei have mass numbers and atomic numbers of comparable values.
𝐴1 + 𝐴2 = 𝐴 and 𝑍1 + 𝑍2 = 𝑍
𝐴1 = 𝐴2 = 𝐴/2 and 𝑍1 = 𝑍2 = 𝑍/2 -------------[Symmetric Fission]

Above process can occur if the Q – value of the transformation is positive:
𝑄𝑓 = M(A,Z) – M(𝐴1,𝑍1) - M(𝐴2,𝑍2) > 0
where all quantities are expressed in terms of atomic masses.
For a symmetric fission:
𝑄𝑓 = M(A,Z) – 2 X M(A/2,Z/2) > 0
In terms of binding energies:
𝑄𝑓 = 2 X B(A/2,Z/2) - B(A,Z)
= 2 X (A/2)𝑓𝐵
′
- 𝑓𝐵
𝑄𝑓 = A (𝑓𝐵
′
- 𝑓𝐵) = A . ∆𝑓𝐵

For 𝑄𝑓 to be positive 𝑓𝐵
′
> 𝑓𝐵
Binding fraction of the product nuclei > binding fraction of the parent nucleus.
In terms of the semi – empirical mass formula:
M(A,Z) = Z 𝑀𝐻 + (A – Z)𝑀𝑛 - 𝑎1𝐴 + 𝑎2𝐴2 3+ 𝑎3
𝑍2
𝐴1 3 + 𝑎4
(𝐴 −2𝑍)2
𝐴
---------------------------[1]
M(A/2,Z/2) = Z/2 𝑀𝐻 + N/2𝑀𝑛 - 𝑎1(𝐴/2) + 𝑎2(𝐴/2)2 3+ 𝑎3
(
𝑍
2
)2
(𝐴/2)1 3 + 𝑎4
(𝐴 −2𝑍)2
2𝐴
------- [2]

𝑄𝑓 = M(A,Z) – 2 X M(A/2,Z/2)
On simplifying:
𝑄𝑓 = - 0.26 𝑎2𝐴2 3 + 0.37𝑎3
𝑍2
𝐴1 3
Symmetric fission energetically possible (𝑄𝑓 > 0) if
𝑍2
𝐴
>
0.26 𝑎2
0.37𝑎3
Substituting the values of 𝑎2 = 0.019114 u and 𝑎3 = 0.0007626 u:
𝑍2
𝐴
> 17.6

𝑍2
𝐴
> 17.6
The condition is fulfilled for A > 90 and Z > 40.
For A = 90, Z = 40,
𝑍2
𝐴
> 17.8
Important note:
For nuclei for which A > 90, symmetric fission energetically possible.
Uncommon phenomena.
Rarely observed even amongst the nuclei of the heaviest atoms in the periodic table e.g.
uranium.
 For instance, only one S.F. per hour in 1 g of 235U corresponding to a half – life of 2 X 1017 yr.

Reason:
Quantum mechanical Barrier penetration.
Problem is much more acute in S.F., since the nuclei of the fission fragments carry much higher
charges than the 𝛼 – particle.

Stability Limit Against Spontaneous Fission:
[Neutron bombarded on a target nucleus and gets captured.
Distorts spherical shape and induces oscillations.]
[Deviation from spherical shape [ redistribution of electric charges gives
tendency to move far apart]
Activation energy not sufficient, Surface energy > Coulomb energy [nucleus
retains its shape]]
Dumbbell shape [Surface energy < Coulomb energy fission takes place]

Shape of the distorted nucleus is expressed in terms of
spherical coordinates:
R(𝜃) = 𝑅0 ( 1 + 𝑎2 cos 𝜃 + 𝑎3 cos 𝜃 +……….)
where a’s are small numbers that determine the amount of
distortion called distortion parameters.
P’s represent the Legendre’s polynomials.
𝑅0 is the radius of the undistorted spherical nucleus.

If 𝑎2 = 𝑎3 = ……….= 0, the nucleus is undistorted sphere for which R = 𝑅0.
UNDISTORTED NUCLEI DISTORTED NUCLEI
COULOMB ENERGY TERM [𝐸𝐶
0
] 𝑎3
𝑍2
𝐴1 3 =
0.71 𝑍2
𝐴1 3 (MeV) 𝐸𝐶
0
[ 1 -
𝑎2
2
5
- ……]
SURFACE ENERGY TERM [𝐸𝑆
0
] 𝑎2𝐴2 3
= 17.80 𝐴2 3
(MeV) 𝐸𝑆
0
[ 1 +
2𝑎2
2
5
+ …..]

Highers powers of the distortion parameters a are neglected.
Total deformation energy:
𝐸𝑇 = 𝐸𝐶 + 𝐸𝑆
Change in the energy:
∆𝐸 = (𝐸𝐶 + 𝐸𝑆) – (𝐸𝐶
0
+ 𝐸𝑆
0
)
∆𝐸 =
1
5
𝑎2
2
(2 𝐸𝑆
0
- 𝐸𝐶
0
)

𝐸𝑆
0
is positive ; 𝐸𝐶
0
is negative.
Stability of a nucleus against spontaneous fission is defined in terms of ∆𝐸.
Difference in energy Condition to be satisfied Nucleus
∆𝐸 > 0 𝐸𝐶
0
< 2 𝐸𝑆
0 Nucleus is Stable
∆𝐸 < 0 𝐸𝐶
0
> 2 𝐸𝑆
0 Nucleus is Unstable

For spontaneous fission ∆𝐸 < 0
𝐸𝐶
0
> 2 𝐸𝑆
0
Coulomb energy is greater than twice the surface energy.
0.71 𝑍2
𝐴1 3 ≥ 2 X 17.80 𝐴2 3
0.71
33.6
(𝑍)2
𝐴
≥ 1
(𝒁)𝟐
𝑨
≥ 50
 above is stability limit against spontaneous fission.
Nuclei stable against spontaneous fission have value of
(𝑍)2
𝐴
less than the limiting value.

Nuclear Shell Structure:
Different nuclear models proposed explain limited features of the nuclei.
Liquid Drop Model explains:
Observed variation of the nuclear binding energy with mass number,
Fission of the heavy nuclei.
Liquid Drop Model predicts:
Closed spacing of the energy levels in the nuclei [at low energies]

Observations:
Low lying excited states widely spaced.
Cannot be explained by Liquid Drop Model.
To explain certain properties of the nuclei following consideration to be taken into account:
Motion of individual nucleons in a potential well.
Potential well gives rise to existence of a nuclear shell structure.
Similar to the electronic shells in the atoms.

Extra – nuclear electrons in an atom are arranged in a number of shells.
Principal Quantum Number (n) Shell
1 K
2 L
3 M
4 N

Each shell has number of sub – shells characterized by different values of the azimuthal
quantum number (𝒍).
Principal Quantum Number (n) Azimuthal Quantum Number(𝒍)
1 0
2 0,1
3 0,1,2
4 0,1,2,3

A sub – shell of a given 𝒍 contains a maximum of 𝟐(𝟐𝒍 + 𝟏) electrons.
Azimuthal Quantum
Number(𝒍)
Notation Maximum number of
electrons in the given shell
2(2𝑙 + 1)
0 s 2
1 p 6
2 d 10
3 f 14

Inert Gas elements:
Ne (Z = 10), Ar (Z = 18), Kr (Z = 36), Xe (Z = 54), Rn (Z = 86) outermost p sub – shells are
completely filled up.
Lightest inert gas He (Z = 2) 1s sub shell is completely filled.
Electrons are tightly bound.
First ionization potentials are high.

Figure: First Ionization Potentials of atoms:

Electrons arranged in certain discrete levels Nucleons in the nuclei also arranged in
discrete levels.
Pointed out first by W.M. Elasser (1933)
Maria Gopert Meyer (1948) and independently O. Haxel, J.H.D Jensen and H.E.Suess (1949) -
showed nuclei containing the following numbers of protons and neutrons exhibit very high
stability.
Protons 2 8 20 28 50 82
Neutrons 2 8 20 28 50 82 126

Above numbers are known as magic numbers.
Analogous to the atomic number of the inert gases.

Evidence of the Existence of Shell Structure within the Nuclei:
Nuclei magic number of protons and neutrons show high stability compared to nuclei
with containing one or more nucleon of the same kind.
Measurement of the separation energy 𝑆𝑛 of a neutron shows –
𝑆𝑛 (nuclei with magic number of neutrons) > > nuclei (containing one more neutron)
Measurement of the separation energy 𝑆𝑝 of a proton shows –
𝑆𝑝 (nuclei with magic number of protons) > > nuclei (containing one more proton)

Nuclei with magic number of neutrons or protons have their first excited states at higher
energies than in cases of the neighbouring nuclei.

Neutron capture cross – section
of the nuclei with magic numbers
of neutrons are low.
Neutrons shells are filled up.
Probabilities of capturing
additional neutron is small.

Single Particle States in Nuclei:
Theoretical understanding of the origin of the nuclear structure based on assumption:
Existence of a dominant spherically symmetric central field of force.
Force governs the motion of individual nucleons in the nuclei.
Central field force in an atom: Electrostatic Force

Central field force in the nucleus:
Average field due to all nucleons in the nucleus.
No residual interaction exists in the nucleons.
Existence of potential energy of the form: 𝑉 = 𝑉(𝑟).
Possible to obtain a solution of the Schrodinger wave equation.

Assumption:
Presence of an infinite three dimensional
harmonic oscillator potential of the form:
𝑉(𝑟) = - 𝑉0 + ½ 𝑀𝜔2𝑟2
𝑀 = nucleon mass
𝑉0 = well – depth
𝜔 = circular frequency of the simple harmonic
oscillator

Three dimensional Schrodinger equation for the harmonic oscillator can be solved using
spherical polar coordinates.
Time independent Schrodinger equation:
-
ħ
2
2𝑚
𝜕2𝜓
𝜕𝑥2 + V𝜓 = E𝜓
Substituting the value of potential gives and using the variable separable form:
Radial equation:
1
𝑟2
𝑑
𝑑𝑟
(𝑟2 𝑑𝑅𝑙
𝑑𝑟
) +
2𝑀
ℏ2 {E - 𝑉(𝑟) -
𝑙(𝑙+1)ℏ2
2𝑀𝑟2 }𝑅𝑙 = 0
𝑅𝑙(𝑟) radial function.

Angular part of the wavefunction in the spherical harmonic: 𝑌𝑖
𝑚
(𝜃, 𝜑).
Total wavefunction:
𝜓𝑛𝑙𝑚 = 𝑅𝑙(𝑟) 𝑌𝑖
𝑚
(𝜃, 𝜑)
 Energy (harmonic oscillator):
E = (𝜆 +
3
2
) ℏ𝜔

1
𝑟2
𝑑
𝑑𝑟
(𝑟2 𝑑𝑅𝑙
𝑑𝑟
) +
2𝑀
ℏ2 {E - 𝑉(𝑟) -
𝑙(𝑙+1)ℏ2
2𝑀𝑟2 }𝑅𝑙 = 0;
E = (𝜆 +
3
2
) ℏ𝜔
Solution exists if 𝜆 = 2𝑛 + 𝑙 − 2
Allowed values of 𝑛 and 𝑙:
𝑛 = 1,2,3…and 𝑙 = 0,1,2,….𝜆 can assume values
0,1,2,3,…..

Discrete set of energy values, Degenerate energy levels, Eigen – functions classified according to 𝒏 and 𝒍
𝑛 𝑙 0 1 2 3 4 5
1 0 1 2 3 4 5
2 2 3 4 5 6 7
3 4 5 6 7 8 9
4 6 7 8 9 10 11
5 8 9 10 11 12 13
6 10 11 12 13 14 15

Diagonal lines connect the levels with different possible combinations of 𝑛 and 𝑙 values.
IMPORTANT NOTE:
Angular part of the wave – function: 𝑌𝑖
𝑚
(𝜃, 𝜑)
Degeneracy (2𝑙 + 1) for a given 𝑙 with magnetic quantum numbers 𝑚 = 𝑙, 𝑙 − 1, … . . −𝑙,
Each level with a given set of (𝑛, 𝑙) has a degeneracy (2𝑙 + 1).
Each level of a given energy (given 𝜆) contains several states of different (𝑛, 𝑙) values.
Degeneracy of a state: 2(2𝑙 + 1) over different values of 𝑙 values for a given 𝜆.
Factor 2 is due to the two possible spin orientations of the neutron and proton.

For a given 𝑙𝑚𝑎𝑥 = 𝜆.
Possible number of nucleons for a given 𝜆:
𝜆 – even: 𝑁𝑒 = 2(1 + 5 + 9+………. 2𝜆 + 1) = (𝜆 +1) (𝜆 +2)
𝜆 –odd: 𝑁𝑜 = 2(3 + 7 + 11+………. 2𝜆 + 1) = (𝜆 +1) (𝜆 +2)
Total number of nucleons up to a maximum value 𝜆𝑚 of 𝜆:
N = (𝜆𝑖+1)(𝜆𝑖+2) =
1
3
(𝜆𝑚+1)(𝜆𝑚+2)(𝜆𝑚+3)
 2(2𝑙 + 1) degenerate states for a given energy having different combinations of of
𝑛, 𝑙, 𝑚𝑙, 𝑚𝑠 values determine a sub level of a given energy.

𝜆 Energy in unit of
ℏ𝝎
Degenerate states
(𝑛, 𝑙)
No of nucleons
filling up the
shell:
2(2𝑙 + 1)
Total number of
nucleons for shell
closure
0 3/2 (1,0) 2 2
1 5/2 (1,1) 6 8
2 7/2 (2,0),(1,2) 12 20
3 9/2 (2,1),(1,3) 20 40
4 11/2 (3,0),(2,2),(1,4) 30 70
5 13/2 (3,1),(2,3),(1,5) 42 112
6 15/2 (4,0),(3,2),(2,4),(1,6) 56 168
7 17/2 (4,1),(3,3),(2,5),(1,7) 72 240

 Levels of different azimuthal quantum numbers 𝑙 are designated by symbols used in atomic
spectroscopy:
𝑙 Symbol
0 s
1 p
2 d
3 f
4 g
5 h
6 i

To explain the above discrepancies at higher magic numbers, Mayer and independently Haxel,
Jensen and Suess –
Suggested that a spin – orbit interaction term should be added to the central potential given by
𝑉(𝑟) = - 𝑉0 + ½ 𝑀𝜔2𝑟2
Spin – orbit potential, which is non – central given as:
𝑉𝑙𝑠 = - 𝜙 𝑟 𝑙. 𝑠
where
𝜙 𝑟 = 𝑏
1
𝑟
(
𝜕𝑓
𝜕𝑟
)
𝑙 and 𝑠 are azimuthal and spin angular momenta of the nucleon under consideration.

Strong coupling between the spin and orbital angular momenta of each individual nucleon
giving rise to a total angular momentum 𝑗 for each given by:
𝑗 = 𝑙 + 𝑠
𝑠 = ½ for each nucleon; the two possible values of 𝑗 = 𝑙 + ½ and 𝑙 − 1/2

Figure: Sequence for nuclear
levels according to shell model
taking into account spin orbit
interaction

NUCLEAR MODELS

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NUCLEAR MODELS