1000 Solved Problems in Modern Physics

8.3 Solutions 463

8.48 Given reaction is
4
2 He+

14

7 N→

A

ZX+

1

1 H

As the atomic number (Z) and mass number (A) are conserved

2 + 7 =Z+ 1 ,orZ= 8

4 + 14 =A+1orA= 17

Therefore the productXis^178 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 minimumα- particle kinetic

energy required to initiate the above reaction is

Ethreshold=|Q|

(

1 +

MHe

MN

)

= 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
Mn+MHe 3 + 3. 27 =Mp+MH 3 + 4. 03
MH 3 −MHe 3 =

(

Mn−Mp

)

+ 3. 27 − 4. 03

= 1. 29 + 3. 27 − 4. 03 = 0 .53 MeV

Binding energies are given by

B(H^3 )=mp+ 2 mn−M(H^3 )

B(He^3 )= 2 mp+mn−M(He^3 )

∴B(H^3 )−B(He^3 )=mn−mp−

[

M(H^3 )−M(He^3 )

]

= 1. 29 − 0. 53 = 0 .76 MeV

The Coulomb energy of two protons

Ec=

1. 44 × 1 × 1

31

/

(^3) × 1. 3
= 0 .769 MeV
8.50 Given reaction can be written down as
n+^105 B→^115 B→α+^73 Li+Q
(a) Q=(Mn+MB 10 −Mα−MLi)×931 MeV
=(1. 008987 + 10. 01611 − 4 .003879–7.01822)× 931 .5MeV
= 2 .79 MeV
(b) The energy released is partitioned as follows