The Nucleus
335
E m (n) =
2/3
n
C A
 
 
 
(9.44)
where
C =
2/3
2
2
2
0
9
32
2
h
mr




π


≈ 52 MeV
(9.45)
for r 0 ≈ 1.2 fm. Similarly the total kinetic energy is given by
E r =
2
2
p dn
m
∫
=
3
5
m
n E
(9.46)
For a nucleus with Z protons and (A – Z) neutrons, the total kinetic energy
is given by
E =
5/3
5/3
2/3
3 [
(
) ]
5
C Z
A Z
A
+ −
(9.47)
If Z =
1 ,
2
A the kinetic energy of the last nucleon is obtained from Eq. (9.44)
to be E ≈ 33 MeV. Since the binding energy of the last nucleon is about 8 MeV,
the average depth of the potential is of the order of 41 MeV. This is in agreement
with the earlier estimation is Eq. (9.31) based on the uncertainty principle. It is
also seen that for a given A,
E
Z
∂
∂
= 0 for Z = 2
A
(9.48)
i.e. the most stable nucleus has equal number of protons and neutrons. Expanding
E about Z =
1 ,
2
A
E =
2
1/3
2/3
3
2 (2 )
1
...
3
2
5(2)
C
C
A
Z
A
A


+
−
+




(9.49)
The second term, which gives the increase in the energy because of the
imbalance of the protons and neutrons, is
δE ≈
2
1
2
43.7
MeV
Z
A
A


−




(9.50)