Nuclear Physics

Nuclear Models

Liquid drop Model

Semi Empirical Mass Formula

Shell Model

Concept of Magic number

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

Nuclear models:

1. the “water-drop” model

2. the “shell” model

Models describe aspects

of the structure of nuclei

and how they behave.

The water-drop model

Or Liquid drop model

Assumptions1. 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.

Volume Energy As a first approximation, we can think of

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

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

Also we know that if nucleus is spherical then

Volume energy

AR3

3

4

EV = a1 A

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

Coulomb Energy

It is important for lighter nuclei since a greater fraction of their nucleons

are on the surface.

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.

Since there are Z(Z-1)/2 pairs of protons,

r

eV

0

2

4

av

Cr

eZZV

ZZE

1

8

)1(

2

)1(

0

2

avr

1Where is the value of 1/r averaged over all proton pairs.

If the protons are uniformly distributed throughout a nucleus of

radius R, is proportional to 1/R and hence 1/A1/3

Therefore

avr

1

3

13

)1(a

A

ZZEC

The Coulomb energy is negative because it arises from an effect

that opposes nuclear stability.

The total binding energy Eb of a nucleus

3

133

2

21

)1(

A

ZZaAaAaEEEE CSvb

Volume energies

Surface energies

Coulomb energies

The binding energy per nucleon is

3

433/1

21

)1(

A

ZZa

A

aa

A

EBE

Empirical binding energy

per nucleon curve

theoretical binding energy

per nucleon curve (using

Liquid drop model concept)

Correction to the formulaCorrection 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.

Asymmetry Energy

• 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.

Correction to the formulaCorrection 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)

2

8

22

1

2

1

ZN

ZNZNE Because N=A-Z, (N-Z)2=(A-2Z)2

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

A

ZAaEEa

2

4

2

Note: The asymmetry energy is negative because it

reduces the binding energy of the nucleus

Correction to the formula

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

6C, 168O etc. Appear as peak on the empirical curve of BE/A.

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

Pairing Term

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 e-e.

• p and n have different energy levels small overlap of wave functions. Two p(n) in same level with opposite values of jz

have AS spin state sym spatial w.f. maximum overlap maximum binding energy because of short range attraction.

Nuclei Pairing

term

e-e +ive

e-o 0

o-o -ive

Correction to the formula ……

The pairing energy Ep is

given by the relation

4/3

50,A

aEP

Semi Empirical Mass Formula (SEMF)

4/3

5

2

4

3

13

3/2

21 )0,()2()1(

),(A

a

A

ZAa

A

ZZaAaAaZAEb

Volume Term

Surface Term

Coulomb Term Asymmetry Term

Pairing Term

a5=33.5 MeVa4=19.0 MeV

a3=0.595 MeV

a2=13.0 MeV

a1=14.1 MeV

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]

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

13

1

3

1

1

43

1

3

43

1

3

4

3

1

3

15219.1

76595.0

82

4

024

12

AA

A

AaAa

aAaZ

ZAA

aZ

A

a

dZ

dEb

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.

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 model1. Nuclear fission

(very large nuclei break up)

2. Nuclear fusion

(very small nuclei fuse together)

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.

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.

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

Shell Model ……

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 verylightest nuclei in which the l values are necessarily small in their normal configuration.

5. The heavier nuclei exhibit j-j coupling.

……

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.

……

Problem: Using shell model calculate the

spin and parities of the following nucleus1. 27

13Al

2. 3316Ar

3. 168O

Parity: (-1)l

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.

Where l is orbital quantum number

If l is even parity is even, If l is odd parity is odd

),,(),,(

),,(),,(

zyxzyxP

zyxzyxP Even

Odd

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.

Quadrupole moment

Prolate (b > a)Spheroid

Q is +ve

Oblate (a > b)Spheroid

Q is -ve

Sphere (b = a)

Q is 0

22

5

2abzQ

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 16

8O 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: 168O (N, Z = 8),

4020Ca (N, Z = 20), All are abundant in nature

20682Pb ( Z = 82,N=126)

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, 16

8O, 4020Ca and

20682Pb are particularly tightly bound.

6. The binding energy of the next neutron and proton after magic

number is very small.

……

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.

Main Limitation of Shell Model (SM)

1. It fail to explain the stability of four stable nuclei 21H, 63Li, 105B

147N.

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.