1000 Solved Problems in Modern Physics

462 8 Nuclear Physics – II

8.3.10 Nuclear Reactions (General).........................

8.45 The given decay is

(^13) N→ (^13) C+β++ν+ 1 .2MeV (1)
MN−MC− 2 me= 1 .2MeV (2)
where masses are atomic andc= 1
For the reaction
p+^13 C→^13 N+n
Q=Mp+MC−MN−Mn
where the masses are nuclear.
Add and subtract 7meto get Q in atomic masses
Q=MH+MC−MN−Mn=−[(Mn−MH)+(MN−MC)]
Q=−[0. 78 + 1. 2 + 2 me]=−3MeV
where we have used (2) and the mass differenceMn−MH= 0 .78 MeV and
me= 0 .51 MeV
Ethreshold=|Q|

(

1 +

mp

mc

)

= 3 ×

(

1 +

1

13

)

= 3 .23 MeV

8.46 Given reaction is

p+^7 Li→^7 Be+n− 1 .62 MeV (1)

MLi+MP−MBe−Mn=− 1 .62 MeV (2)

where the masses are nuclear.

It follows that

MBe−MLi= 1. 62 + 938. 23 − 939. 52 = 0 .33 MeV

Add and subtract 4meto get theQmass difference of Be and Li

MBe−MLi−me= 0. 33

orMBe−MLi= 0. 33 +me= 0 .84 MeV

whereme= 0 .51 MeV. Thus the total energy released in the electron capture

e−+^7 Be→^7 Li+νis 0.84 MeV

This energy is shared between the neutrino and the recoil nucleus. Energy

and momentum conservation give

EN+Eυ= 0 .84 MeV (1)

Pυ^2 =E^2 υ=PN^2 = 2 MNEN (2)

UsingMN∼=7amu= 6520 .5 MeV, (1) and (2) can be solved to obtainEN=

7 .5 keV andEυ∼= 0 .84 MeV.

8.47 The Q-value is− 1 .37 MeV. Minimum energy required is

Ea=|Q|

(

1 +

ma

mx

)

= 1. 37

(

1 +

1

24

)

= 1 .427 MeV