Q. A radioactive nucleus undergoes $\alpha$-emission to form a stable element. What will be the recoil velocity of the daughter nucleus if $V$ is the velocity of $\alpha$-emission and $A$ is the atomic mass of radioactive nucleus
A) $\frac{4V}{A - 4}$
B) $\frac{2V}{A - 4}$
C) $\frac{4V}{A + 4}$
D) $\frac{2V}{A + 4}$
Explanation
(a) According to conservation of momentum $4v = (A - 4)v'$
$\Rightarrow v^{'} = \frac{4v}{A - 4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.