1000 Solved Problems in Modern Physics

464 8 Nuclear Physics – II

Eα=

QMLi

Mα+MLi

=

2. 79 × 7. 018

4. 004 + 7. 018

= 1 .78 MeV

ELi=Q−Eα= 2. 79 − 1. 78 = 1 .01 MeV

8.51 p+^27 Al→n+^27 Si+Q (1)

(^27) Si→ (^27) Al+β++ν (2)
Msi−MAl=Emax+ 2 me
where masses are atomic. In terms of nuclear masses
MSi−MAl=Emax+ 2 me= 3. 5 + 0. 51 = 4 .01 MeV
In (1),Q=mp+mAl−mn−msi
where masses are nuclear
Q=−

(

mn−mp

)

−(mSi−mAl)=− 0. 8 − 4. 01 =− 4 .81 MeV

EThreshold=|Q|

(

1 +

mp

mAl

)

= 4. 81 ×

(

1 +

1

27

)

= 5 .0MeV

8.52 (a) The threshold energy for appearance of neutron in the forward direction is

Ep(threshold)=|Q|

(

1 +

mp

mH 3

)

= 0. 764 ×

(

1 +

1

3

)

= 1 .019 MeV

(b) The threshold for the appearance of neutrons in the 90◦direction is

Ep(threshold)=

|Q|mHe

mHe−mp

= 0. 764 ×

(

3

3 − 1

)

= 1 .146 MeV

8.53d+^30 Si→^31 Si+p(1)

Q=Md+MSi^30 −MSi^31 −Mp (2)

Given

MSi^30 +Md=Mp^31 +Mn+ 5 .1(3)

MSi 31 =Mp 31 +Me+ 1. 51 (4)

Subtract (4) from (3)

MSi 30 +Md−MSi 31 =Mn−Me+ 3. 59 (5)

Further

Mn=Mp+Me+ 0 + 0. 78 (6)

Add (5) and (6)

MSi 30 +Md−MSi 31 −Mp=Q= 3. 59 + 0. 78 = 4 .37 MeV