456
8 Nuclear Physics – II
8.22 The mass-energy equation for the positron decay gives
M Na = M Ne + 2m e +
T max + T γ
931.5
= 21.991385 + 2 × 0.511 +
0.542 + 1.277
931.5
= 23.015338 amu
8.23 For electron capture
7 Be + e
−
→
7 Li + ν e ,
Q = (m Be − m Li ) × 931.5 = (7.016929 − 7.016004) × 931.5 = 0.8616 MeV
which is positive
For positron emission the Q-value must be atleast 1.02 meV which is not
available. Therefore, positron emission is not possible.
8.3.6 Fermi Gas Model
8.24 U =
3
5
AE F
(1)
p = −
∂U
∂ V
=
3
5
A
∂ E F
∂ V
(2)
From Fermi gas model
A = KVE
3/2
F
(3)
Differentiating (3) with respect to V
3
2
V
E F
∂ E F
∂ V
+ E
3/2
F = 0
whence
∂ E F
∂ V
= −
2
3
E F
V
(4)
Using (4) in (2)
p =
2
5
A
V
E F =
2
5
ρ N E F
where ρ N = A/V , is the nucleon density.
8.25 The Fermi momentum for N = Z = A/2, is
p F (n) = p F ( p) = (/r 0 )(9π/8)
1/3
cp F = (c/r 0 )(9π/8)
1/3
= (197.3 MeV.fm/1.3 fm)(9π/8)
1/3
= 231 MeV
p F = 231 MeV/c
E F = p F
2
/2M = (231)
2
/(2 × 940) = 28 MeV
If B is the binding energy of a nucleon
V = E F + B = 28 + 8 = 36 MeV