Liquid Drop Model

Our first model of nuclei.

The motivation is to describe the masses and binding energy of nuclei.

It is called the Liquid Drop Model because nuclei are assumed to behave in a similar way to a liquid (at least to first order).

The molecules in a liquid are held together by Van der Waals force that is only between near neighbors.

We have already seen that the nuclear force is a short range force that appears to act only between neighboring nucleons.

Liquid Drop Model From the definition of Binding Energy we can write

the mass of a nucleus X𝑍𝐴 as,

To first order we can write

The first term = Volume Term

This is due to the short range attraction between nucleons.

If we ignore the fact that the nucleus has a surface, each nucleon is attracted to the same number of nearest neighbors.

i.e. BE A

),()(),( 222 AZBcmZAcZmcAZM np

3/123/2 /),( AZaAaAaAZB CSV

Liquid Drop Model

But: some nucleons are on the surface – they

have fewer neighbors.

Therefore: BE will be reduced by an amount

number of nucleons on the surface.

Therefore: the second term = Surface Term

surface area

r2 and since r A1/3

A2/3

i.e. surface term = aSA2/3.

Liquid Drop Model

The Coulomb repulsion between the protons in the

nucleus also reduces the Binding Energy.

The Coulomb potential energy will be

Z2

and 1/x

x = average distance between proton pairs

x r A1/3

So the third term = Coulomb Term

= aC Z2/A1/3

Actually the PE of one proton is

So PE for a nucleus would be Z(Z 1)

but we will simplify to Z2

eZe )1)(1(

Liquid Drop Model There are other terms in the Liquid Drop Model.

These additional terms are quantum mechanical in origin.

The Asymmetry Term.

Imagine that the neutrons and protons are in a potential well (the nucleus).

We will assume that the well has approximately equally spaced energy levels.

Let the spacing be E.

E

Neutrons and protons are fermions

(spin ½ particles).

So the Pauli exclusion principle

allows only 2 p and 2 n in each level.

Liquid Drop Model For a nucleus with a given A, the lowest energy will

be when N = Z.

If we change a neutron to a proton

(e.g. via decay)

N Z

we increase the energy by 1E

from Z N = 0 case.

(Now Z N = 2)

(Z N = 0)

Liquid Drop Model If we change another neutron to a

proton…

we increase the energy by E

for a total change of 2E

from Z N = 0 case.

(Now Z N = 6)

(Now Z N = 4)

Another neutron to a proton…

we increase the energy by 3E

for a total change of 5E

from Z N = 0 case.

N Z

Liquid Drop Model Continuing this, we can find the energy shift

for a change in Z N or N Z.

Z N or N Z Total Energy Shift (E)

2 1

4 2

6 5

8 8

10 13

12 18

14 25

16 32

We can approximate this energy increase by

0.5

2

4.5

8

12.5

18

24.5

32

EZN

8

)( 2

8

)( 2ZN

Liquid Drop Model

It has been found that E decreases with A like A1.

(This can be justified by considering the phase space

available for fermions – but it is not worth the effort for

this phenomenological formula.)

Therefore we can write the asymmetry term as

(Some authors use (ZA/2)2 instead of (A2Z)2. In that

case aA will be different.)

A

ZAa

A

ZNa AA

22 )2()(

Liquid Drop Model

Pairing Term.

This term reflects the observation that the nucleus is in a lower energy state when neutrons are all paired with other neutrons.

We find a similar thing for protons.

e.g. A plot of the Neutron Separation Energy for Barium isotopes = energy needed to extract a neutron from a nucleus.

Neutron

separation

energy

N

N

N Ba56

56

74 76 78

Liquid Drop Model

Neutron Separation Energies

Liquid Drop Model

Proton Separation Energies

Liquid Drop Model We find that the energy

of the nucleus is changed

by an amount

pair = N even Z even

= 0 A odd (N or Z even)

= + N odd Z odd

i.e. Nuclei with both neutrons and protons paired are in a lower energy state than nuclei with just the neutrons or protons paired, and these are in a lower energy state than if neither are paired.

The best fit to data is with:

# of Stable Nuclei

165

105

4

2/1A

aP

Liquid Drop Model Finally we have the:

Semi-Empirical Mass Formula.

nucleieven -even /

nuclei odd 0

nuclei odd-odd /

/)2(

/

),(

2/1

2/1

2

3/12

3/2

Aa

A

Aa

AZAa

AZa

Aa

AaAZB

P

P

A

C

S

V

Best Fit Parameters

aV = 15.56 MeV aS = 17.23 MeV aC = 0.697 MeV aA = 23.285 MeV aP = 12.0 MeV

),()(),( 222 AZBcmZAcZmcAZM nHA

Liquid Drop Model

The value of the model is that we can predict the masses of nuclei that are far from stability, where masses are hard to measure. e.g. Nucleosynthesis models are highly dependent on

knowing such masses.

Many refinements, beyond what we have done, have been added to the model to improve accuracy.

The liquid drop model has several shortcomings. e.g. – it is not quantum mechanical

– assumes nuclei are spherical symmetric (which disagrees with the fact that nuclei have non-zero quadruple moments – more later.)

Nevertheless, the model does help us understand nuclear stability.

Liquid Drop Model Odd A Nuclei

(Paring term is zero.)

e.g. A = 101 isobars.

= decay by

either EC or +

Conclusion:

Odd A nuclei

have only one

stable nucleus

for a given A. (Stable, unless it

can decay by

emission.)

Liquid Drop Model Even A Nuclei

(Paring term alternates.)

e.g. A = 100 isobars.

= decay by

either EC or +

Conclusions:

o-o nuclei can

always decay. (There are 4

exceptions – mass

formula not accurate

enough to predict.)

e-e nuclei often

have two (or more)

stable isobars.

o-o nuclei

e-e nuclei

Liquid Drop Model

The semi-empirical mass formula does a reasonable job of describing the general observations of nuclear masses.

e.g. Binding energy per nucleon for stable nuclei

This indicates that our model of the nucleons being bound in the nucleus by a Nuclear Force with a short range is valid – at least to first order.

mass formula

measured for

stable nuclei