Q.

A heavy nucleus N, at rest, undergoes fission N→P+Q, where P and Q are two lighter nuclei. Let δ=MN-MP-MQ, where MP, MQ and MN are the masses of P, Q and N, respectively. EP and EQ are the kinetic energies of P and Q, respectively. The speeds of P and Q are vP and vQ, respectively. If c is the speed of light, which of the following statement(s) is(are) correct                          [2021]

1 EP+EQ=c2δ  
2 EP=(MPMP+MQ)c2δ  
3 vPvQ=MQMP  
4 The magnitude of momentum for P as well as Q is c2μδ, where μ=MPMQ(MP+MQ)  

Ans.

(1, 3, 4)

For nuclear fission reaction,

N→P+Q

Energy released =ΔMc2=(MN-MP-MQ)c2=δc2

This will be distributed as kinetic energy of P and Q

∴   EP+EQ=δc2         ⋯(i)

By conservation of momentum,

VPMP=VQMQ

∴  VPVQ=MQMP            ⋯(ii)

Kinetic energy can be written as KE=P22m

Hence, kinetic energies are divided in the inverse ratio of masses.

EP=MQMP+MQδc2                   ⋯(iii)

By equation (i),

⇒P22MP+P22MQ=δc2

⇒P22μ=δc2⇒P=c2μδ                  [∵μ=MPMQMP+MQ]