1 Liquid drop model(Weizschher’s Semi-empirical Formula):

In 1935, Weizsher showed it is possible to achieve the correct explaination of thebinding energy formula using analogy with the liquid drop.The various termsinvolved in the expalination are given as follows:

1.1 Volume energy term, Bv:

The volume contribution of the energy in the liquid drop is due to the latentheat of the liquid drop. For a liquid drop

B = LMmA

where, L=latent heatMm=mass of a moleculeA=no.of moleculesTaking this analogy in nucleus we can write Volume term,

Bv = avA

where av is called volume coefficent.

1.2 Surface term:

As we know in a liqud drop the surface energy is contributed by the surfacetension of the liquid.

−Bs ∝ 4πR2

AndR=1.07 A

13

Then−Bs = asA

23

where asis surface coefficient and the negative sign indicates the reduction ofenergy

1.3 Columb Energy term, Bc:

If we consider a nucleus as uniformly charged sphere of radius R and havingcharge density

ρ =Ze

43πR

3

we can find the self-energy of this sphere as

W =3Z2e2

4πε0R

1

And the self energy of proton is given by

3e2

4πε0R

Since we are considering already formed protons in the nucleus we must deductthe self energy of the proton.So the true columb energy term is given by

−Bc =3Z2e2

4πε0R− Z 3e2

4πε0R

=3Z(Z − 1)e2

4πε0R

The negative sign indicates the disruptive natrue of the energy And on simpli-fication

−Bc = acZ(Z − 1)

A13

1.4 The asymmetry Energy Term Ba:

This term is depends on the excess neutron term i.e (A-Z) term For a nucleusthe mass term can be written as

M = ZMp +NMn −B = ZMp + (A− Z)MN −B

Substuting various terms of B we get

M = AMn − Z(Mn −Mp)− avA+ asA23 + ac

Z(Z − 1)

a13

On diff above wrt Z we get the condition for the stablity and determine the Zfor stable condition. But perfoming and calculating for A=216 we find Z=3 sosome terms is missing in binding energy. So far we have neglected the quantizedenergy levels of the nucleons. Now the excess neutron (A-Z) will occupy thehigher energy levels which are less tightly hold. To get Ba is simple way, wenote that the number of excess neutrons is N-Z and the neutron excess pernucleon is,

(N − Z)

A

Since the total number of excess neutron is (N-Z), the total deficit in nuclearbinding energy is proportional to the product of these.ie

−Ba = aa(NZ)2

A

−Ba = aa(A− 2Z)2

A

Thus,by incorporating a purely quantum mechanical effect in the binding energy,one can go beyond the liquid drop analogy.

2

1.5 The paring energy term Bp:

Since spins is not taking under consideration in liquid drop model and shelleffects, the effect can be corrected by adding a paring energy term . The energyterm is accounted by

−Bp = apA− 3

4

Typically, the values of the empirical coefficients are: av = 14Mevas = 13Mevac = 0.60Mevaa = 19Mevap = −33.5Mev for even-even= 0 for odd-even= 33.5Mev for odd -odd

2 Mass Parabola:

We can write the mass of nuclei as:

M(Z,A) = A(Mn−av)+asA23 +Z[(Mp−Mn)− ac

A13

−4aa]+Z2[ac

A13 + 4aa

A

]+aaA±δ

= αA+ βZ + γZ2 ± δ

where δ = 0, for odd and even combinationδ = 1, for odd and odd combinationδ = −1,for even and even combination

3 Stablity limits against spontaneous fission:

The success of the semi enpirical mass formula encourages Bohr and Wheelerto take idea of liquid drop model a little further and suggest an explaination forthe important procss of nuclear fission. Consider U92238 nucleus divided intotwo nuclei with

A =238

2

The B/A value increases from ∼ 7.6Mev to ∼ 8.5Mev.i.e 0.9 Mev per nucleon.The total increase in teh binding energy is

(0.9Mev)× 238 = 214Mev

3.1 Energetics of ’Symmetric Fission’:

The energy released in a symmetric fission is:

Q = M(Z,A)− 2M(Z/2, A/2)

3

From the semi-empirical mass formula, binding energy for a nucleus (Z,A) is:

B(Z,A) = avA− asA23 − ac

Z(Z − 1)

A13

− aa(A− 2Z)2

A− apA

−34

and binding energy for two fission fragments (Z/2,A/2) is:

B(Z,A) = av(A/2)− as(A/2)23 − ac

Z(Z − 1)

(A/2)13

− aa(A/2− 2Z)2

A/2− ap(A/2)

−34

Giving

Q = asA23 (1− 2

13 ) + ac

Z(Z − 1)

A13

(1− 1

223

)

+paring energy differenceIt can be approximated as

Q = asA23 (1− 2

13 ) + ac

Z(Z − 1)

A13

(1− 1

223

)

using the values of ac and as we get

Q ∼ A 23 (−3.38 + 0.22

Z(Z − 1)

A)

From this we see thta Q becomes greater than 0 for

Z2

A≥ 15

for the symmetric fission of U23892 ,

(238)23 (−3.38 + .22

92(92− 1)

238)Mev = 170Mev

4