8.3 Solutions
463
8.48 Given reaction is
4
2 He +
14
7 N →
A
Z X +
1
1 H
As the atomic number (Z ) and mass number (A) are conserved
2 + 7 = Z + 1, or Z = 8
4 + 14 = A + 1 or A = 17
Therefore the product X is
17
8 O.
Q = [(4.0039 + 14.0075)–(17.0045 + 1.0081)] × 931.5
Q = −1.118 MeV
Thus it is an endoergic reaction for which the minimum α - particle kinetic
energy required to initiate the above reaction is
E threshold = |Q|
1 +
M He
M N
= 1.118 ×
1 +
4
14
= 1.437 MeV
8.49 Given reactions are
2 H +
2 H → n +
3 He + 3.27 MeV
2 H +
2 H → p +
3 H + 4.03 MeV
It follows that
M n + M He 3 + 3.27 = M p + M H 3 + 4.03
M H 3 − M He 3 =
M n − M p
+ 3.27 − 4.03
= 1.29 + 3.27 − 4.03 = 0.53 MeV
Binding energies are given by
B(H
3 ) = m p + 2m n − M(H
3 )
B(He
3 ) = 2m p + m n − M(He
3 )
∴ B(H
3 ) − B(He
3 ) = m n − m p −
M(H
3 ) − M(He
3 )
= 1.29 − 0.53 = 0.76 MeV
The Coulomb energy of two protons
E c =
1.44 × 1 × 1
3
1
3 × 1.3
= 0.769 MeV
8.50 Given reaction can be written down as
n +
10
5 B →
11
5 B → α +
7
3 Li + Q
(a) Q = (M n + M B 10 − M α − M Li ) × 931 MeV
= (1.008987 + 10.01611 − 4.003879–7.01822) × 931.5 MeV
= 2.79 MeV
(b) The energy released is partitioned as follows