Question

A radioactive nucleus undergoes $\alpha$- emission to form a stable element.What will be the recoil of the daughter nucleus if $v$ is the velocity of $\alpha$ emission?

• A. $\frac{4v}{A-4}$
• B. $\frac{2v}{A-4}$
• C. $\frac{4v}{A+4}$
• D. $\frac{2v}{A+4}$

Answer 1

The correct option is A $\frac{4v}{A-4}$

We assume that mass number of nucleus when it was at rest = $A$

$\because$ mass number of $\alpha$-particle = 4

$\therefore$ mass number of remaining nucleus = $A-4$

As there is no external force, so momentum of the system will remain conserved.

$\Rightarrow$ $0=(A-4)v^{\prime }+4v\Rightarrow v^{\prime }=-\frac{4v}{(A-4)}$

Negative sign represents that direction is opposite to the direction of motion of $\alpha$-particle.