464
8 Nuclear Physics – II
E α =
QM Li
M α + M Li
=
2.79 × 7.018
4.004 + 7.018
= 1.78 MeV
E Li = Q − E α = 2.79 − 1.78 = 1.01 MeV
8.51 p +
27 Al → n +
27 Si + Q
(1)
27 Si →
27 Al + β
+
+ ν
(2)
M si − M Al = E max + 2m e
where masses are atomic. In terms of nuclear masses
M Si − M Al = E max + 2m e = 3.5 + 0.51 = 4.01 MeV
In (1), Q = m p + m Al − m n − m si
where masses are nuclear
Q = −
m n − m p
− (m Si − m Al ) = −0.8 − 4.01 = −4.81 MeV
E Threshold = |Q|
1 +
m p
m Al
= 4.81 ×
1 +
1
27
= 5.0 MeV
8.52 (a) The threshold energy for appearance of neutron in the forward direction is
E p (threshold) = |Q|
1 +
m p
m H 3
= 0.764 ×
1 +
1
3
= 1.019 MeV
(b) The threshold for the appearance of neutrons in the 90
◦ direction is
E p (threshold) =
|Q|m He
m He − m p
= 0.764 ×
3
3 − 1
= 1.146 MeV
8.53 d +
30 Si →
31 Si + p
( 1 )
Q = M d + M Si 30 − M Si 31 − M p
(2)
Given
M Si 30 + M d = M p 31 + M n + 5.1
( 3 )
M Si 31 = M p 31 + M e + 1.51
(4)
Subtract (4) from (3)
M Si 30 + M d − M Si 31 = M n − M e + 3.59
(5)
Further
M n = M p + M e + 0 + 0.78
(6)
Add (5) and (6)
M Si 30 + M d − M Si 31 − M p = Q = 3.59 + 0.78 = 4.37 MeV