1. Nuclear models like the liquid drop model and shell model describe aspects of nuclear structure and behavior. The liquid drop model treats the nucleus like a liquid drop while the shell model treats nucleons as moving independently in nuclear orbits. 2. The shell model explains nuclear magic numbers and properties like spin and parity. Magic numbers correspond to nuclear stability when the number of protons or neutrons equals 2, 8, 20, 28, 50, 82, etc. The shell model accounts for magic numbers in terms of closed nuclear shells. 3. While insightful, nuclear models have limitations and do not fully describe all nuclear phenomena. The liquid drop model cannot explain magic numbers while the shell model fails to explain the stability of certain

1. Nuclear Physics

2.  Nuclear Models
 Liquid drop Model
 Semi Empirical Mass Formula
Shell Model
Concept of Magic number

3. Nuclear Models
Why is the binding energy per nucleon almost constant?
Why do certain nuclei emit α- and β-particles through these
particles do not exist inside the nucleus?
Why are the nuclei containing 2,8,20,28,50,82 nucleons most
stable?
Like any other quantum mechanical system, a nucleus also
exist in its excited states. The most stable state is the ground
state in which the nuclei are generally found.
Different models:
1. Liquid drop model
2. Shell model
3. Collective model
4. Optical model

4. Nuclear models:
1. the “water-drop” model
2. the “shell” model
Models describe aspects
of the structure of nuclei
and how they behave.

5. The water-drop model
Or Liquid drop model

6. Assumptions
1. The nuclei of all elements are considered to be behave like a liquid
drop of incompressible liquid of very high density.
2. In an equilibrium state the nuclei of atoms remain spherically
symmetric under the action of strong attractive nuclear forces just
like the drop of a liquid which is spherical due to surface tension.
3. The density of a nucleus is independent of its size just like the
density of liquid which is also independent of its size.
4. The nucleons of the nucleus move about within a spherical
enclosure called the nuclear potential barrier just like the movement
of the molecules of a liquid within a spherical drop of liquid.
5. The binding energy per nucleon of a nucleus is constant just like
the latent heat of vaporization of a liquid.

7. As a first approximation, we can think of
Volume Energy each nucleon in a nucleus as interacting
solely with its nearest neighbors.
Energy associated with each nucleon-
nucleon bond = U
Because each bond energy is shared by
two nucleons therefore
each has a binding energy of ½ U.
When an assembly of spheres of the
same size is packed together
into the smallest volume, each interior
sphere has 12 other spheres
In contact with it

8. Volume Energy
Hence each interior nucleon in a nucleus has a binding energy of
=12(1/2U) = 6U.
If all A nucleons in a nucleus were in the interior, the total binding
Energy would be
EV= 6AU
Or
EV = a1 A Volume energy
Also we know that if nucleus is spherical then
4
R3 A
3

9. Surface Energy
Actually, of course, some nucleons are on the surface of every
Nucleus and therefore have fewer than 12 neighbors.
Surface energy surface area
4πR2
If R = R0 A1/3 then
Surface energy 4πR0 A2/3
Or
Surface energy = - a2 A2/3
Surface energy A2/3

10. It is important for lighter nuclei since a greater fraction of their nucleons
are on the surface.
Coulomb Energy
The repulsion between each pair of protons in a nucleus also
contribute toward decreasing its binding energy.
The coulomb energy EC of a nucleus is the work that must be done
To bring together Z protons from infinity into a spherical aggregate
The size of the nucleus.
e2
V
4 0r
Since there are Z(Z-1)/2 pairs of protons,
Z (Z 1) Z (Z 1)e 2 1
EC V
2 8 0 r av

11. 1
Where r
is the value of 1/r averaged over all proton pairs.
av
If the protons are uniformly distributed throughout a nucleus of
radius R, 1 is proportional to 1/R and hence 1/A1/3
r av
Therefore Z ( Z 1)
EC a3 1
3
A
The Coulomb energy is negative because it arises from an effect
that opposes nuclear stability.

12. The total binding energy Eb of a nucleus
Surface energies
2
3
Z ( Z 1)
Eb Ev ES EC a1 A a2 A a3 1
3
A
Volume energies Coulomb energies
The binding energy per nucleon is
EBE a2 Z ( Z 1)
a1 1/ 3
a3 4
A A 3
A

13. Empirical binding energy
per nucleon curve
theoretical binding energy
per nucleon curve (using
Liquid drop model concept)

14. Correction to the formula Asymmetry Energy
Correction 1: When the neutrons in a nucleus outnumber the protons,
which means that higher energy levels to be occupied than would be
the case if N and Z were equal.
• Neutrons and protons are spin ½ fermions  obey
Pauli exclusion principle.
• If other factors were equal  ground state would
have equal numbers of n & p.
Illustration
Neutron and proton states with same spacing ε.
Crosses represent initially occupied states in ground
state.
If three protons were turned into neutrons the extra
energy required would be 3 3ε.
In general if there are N-Z excess neutrons over
protons the extra energy is
((Z-N)/2)2 ε. relative to Z = N.

15. Correction to the formula
Correction 1: When the neutrons in a nucleus outnumber the protons,
which means that higher energy levels to be occupied than would be
the case if N and Z were equal.
ΔE=(number of new neutrons)(energy increase/new neutron)
1 1 Because N=A-Z, (N-Z)2=(A-2Z)2
E N Z N Z
2 2 2
2
N Z
8
The greater the number of nucleon in a
nucleus the smaller is the energy level
spacing ε, with ε proportional to 1/A
Asymmetry energy due to difference
Between N and Z
2
A 2Z
Ea E a4
A

16. Correction to the formula Pairing Term
Note: The asymmetry energy is negative because it
reduces the binding energy of the nucleus
Correction 2: The correction term arises from the tendency of proton
pairs and neutron pairs to occur. Even –even nuclei are the most stable
And hence have higher binding energies. Therefore nuclei such as 42He,
12 C, 16 O etc. Appear as peak on the empirical curve of BE/A.
6 8
The pairing energy Ep is
positive for even-even
nuclei
The pairing energy Ep is
zero for odd-even and
even-odd nuclei
The pairing energy Epis
negative for odd-odd nuclei

17. Pairing Term
• Nuclei with even number of n or
even number of p more tightly
bound fig.
• Only 4 stable o-o nuclei cf 153 Nuclei Pairing
e-e.
• p and n have different energy term
levels  small overlap of wave
functions. Two p(n) in same e-e +ive
level with opposite values of jz e-o 0
have AS spin state  sym
spatial w.f. maximum overlap o-o -ive
maximum binding energy
because of short range
attraction.

18. Correction to the formula ……
The pairing energy Ep is
given by the relation
a5
EP ,0 3/ 4
A
Semi Empirical Mass Formula (SEMF)
a1=14.1 MeV Surface Term a3=0.595 MeV Pairing Term
Z ( Z 1) ( A 2Z ) 2 a5
Eb ( A, Z ) a1 A a2 A2 / 3 a3 1
a4 ( ,0) 3 / 4
3
A A
A
Volume Term Coulomb Term Asymmetry Term
a2=13.0 MeV a4=19.0 MeV a5=33.5 MeV

19. Numericals:
Numerical 1. The atomic mass of the zinc isotope 6430Zn is
63.929 a.m.u. calculate its binding energy using semi-empirical
mass formula and compare the results with direct formula.
[Ans: 561.7 MeV]

20. Numerical 2. Isobars are nuclides that have the same mass number
A. Derive a formula for the atomic number of the most stable
isobar of a given A and use its to find the most stable isobar of
A=25
Solution: To find the value of Z for which the binding energy Eb
is a maximum which correspond to maximum stability, we must
Solve dEb/dZ = 0 for Z. From the liquid drop mass formula
dEb a3 4 a4
1
2Z 1 A 2Z 0
dZ 3
A
A
1 1
3 3
a3 A 4a4 0.595 A 76
Z 1 1
3 1 3 1
2a3 A 8a4 A 1.19 A 152 A
For A =25 this formula gives Z = 11.7, from which we
conclude that Z = 12 should be the atomic number of
The most stable isobar of A=25. This nuclei is 2512Mg.

21. Main Achievements of liquid drop
model (LDM)
1.It explains binding energy of large number of nuclei.
2.It explains the fusion and fission processes nicely.
3. Explains energies of radioactive decays, fission and fusion.
Applications of the water-drop model
1. Nuclear fission
(very large nuclei break up)
2. Nuclear fusion
(very small nuclei fuse together)

22. Main drawbacks of liquid drop
model (LDM)
1. It is not able to explain the magic numbers.
2. It is not able to explain excited states.
3. It is not able to calculate the nuclear spin.

23. Shell Model
The basic assumption of the liquid drop model is that each nucleon in
a nucleus interact only with its nearest neighbors, like a molecule in a
liquid.
In shell model each nucleon interact chiefly with a general force field
produced by all the other nucleons.
The atoms with 2,10,18,36,54 and 86 electrons have all their
electron shell completely filled.
In the same way, nuclei that have 2,8,20,28,50,82 and 126 neutrons
and protons are more abundant than other nuclei of similar mass
numbers, suggesting their structures are more stable.

24. Shell Model ……
It has been observed that nuclei having either the
number of protons Z on number of neutrons N =A
- Z equal to one of the numbers 2, 8, 20, 50, 82
and 126 are more stable than their neighbours.
These numbers are called magic numbers

25. Main Assumptions ……
1. Nucleon forms subshells and shells with in the nucleus
1. The shell within the nucleus get closed with a suitable
number of nucleons
3. Each nucleon is supposed to possess a spin angular momentum
of ħ/2 and orbital angular momentum lħ.
4. This theory assumes that LS coupling holds only for the very
lightest nuclei in which the l values are necessarily small in their
normal configuration.
5. The heavier nuclei exhibit j-j coupling.

26. Shell model ……
It is assumed that the nucleons move in its orbit within the nucleus,
independently of all other nucleons.
The orbit is determined by a potential energy function V(r) which
represent the average effect of all interaction with other nucleons,
and is same for each particle.
Each nucleons is regarded as an independent particle and the
interaction between a nucleons is considered to be a small perturbation
On the interaction between a nucleon and the potential field.
There is a direct analogy between the theoretical treatment of a
Nucleus and an electron in an atom.

27. ……

28. Problem: Using shell model calculate the
spin and parities of the following nucleus
1. 27
13Al
2. 33
16Ar
3. 16
8O
Where l is orbital quantum number
Parity: (-1)l
If l is even parity is even, If l is odd parity is odd
The parity of a wave function refer to its behaviour under a
simultaneous reflection of the space coordinates i.e. x to –x,
y to –y and z to –z.
Parity of a nucleus refers to the behaviour of the wave function
as a result of the inversion of the coordinates.
P ( x, y , z ) ( x, y, z ) Even
P ( x, y , z ) ( x, y , z ) Odd

29. Electric quadrupole moment (Q): measure of how much nuclear charge
distribution depart from sphericity.
A spherical nucleus has no quadrupole moement, i.e. Q = 0
while one shaped like a prolate - speroid or egg shaped Q is +ve
In an oblate or disc like spheroid (pumpkin shaped) have Q is –ve.
Nuclei of magic N and Z are found to have zero quadrupole moment and
hence are spherical in nature.
Shell model is an attempt to account for the existence of magic
numbers and certain other nuclear properties in terms of nucleon
behavior in a common force field
Neutron and protons occupy separate sets of sates in a nucleus
Because proton interact electrically as well as through the specifically
nuclear charge.

30. Quadrupole moment
2 2 2
Q zb a
5
Prolate (b > a) Sphere (b = a) Oblate (a > b)
Spheroid
Spheroid
Q is 0
Q is -ve
Q is +ve

31. Experimental Evidence of Nuclear Magic
Numbers
1. It has been experimentally observed that the nuclei for which
N and Z are 2,8…….. 42He and 168O are more stable then their
neighbours.
2. Stability also related to the natural abundance. It has been observed
From the experimental data that the nuclei having number of nucleons
as magic number are in abundance as compared to other nuclei in
Nature.
For Example: 16
8O
(N, Z = 8),
40 Ca (N, Z = 20), All are abundant in nature
20
206 Pb ( Z = 82,N=126)
82

32. Experimental Evidence of Nuclear Magic
Numbers ……
3. Sn (Z = 50) has 10 stable isotopes, more than any other element ,
While Ca (Z = 20) has six isotopes. This indicates that elements with
Z=50 and Z=20 are more than usually stable.
4. No more than 5 isotones occur in nature for any N except N=50,
where there are 6 and N = 82, where there are 7. Neutron numbers of
82, 50 therefore, indicate particular stability.
5. The doubly magic nuclei (N &Z both magic) 42He, 168O, 4020Ca and
206 Pb are particularly tightly bound.
82
6. The binding energy of the next neutron and proton after magic
number is very small.

33. Main Achievements of Shell model
(SM)
1.It explains Magic numbers.
2. It explains the magnetic moment of some nuclei nicely.
3. It explains successfully the ground state spin.
4. It explains the great stability and high binding energy
5. It explains the phenomenon of nuclear isomerism.
Note: Atoms having the same mass number but different and
same atomic number, are distinguish by certain difference in the
internal structure of the nucleus are called isomers.

34. Main Limitation of Shell Model (SM)
1. It fail to explain the stability of four stable nuclei 21H, 63Li, 105B
14 N.
7
2. It does not predict correct values of nuclear spin for certain nuclei
3. The Quadrupole moment calculated using this model in also not in
good agreement.
4.The magnetic moment is also shows some deviation from
observed values.