question_answer

A nucleus of mass number A originally at rest, emits alpha particle with speed v, the recoil speed of the daughter nucleus is:

A) \[\frac{4v}{A-4}\]                                            

B) \[\frac{4v}{A+4}\]

C) \[\frac{v}{A-4}\]                                              

D) \[\frac{v}{A+4}\]

Correct Answer: A

Solution :

From law of conservation of momentum, we have momentum before collision = momentum after collision The daughter nucleus has mass\[A-4\] \[\therefore \]  \[{{m}_{\alpha }}{{v}_{\alpha }}+{{m}_{N}}{{v}_{N}}=0\] \[\Rightarrow \]               \[4v+(A-4){{v}_{N}}=0\] \[{{v}_{N}}=\frac{-4v}{A-4}\]