Let $K_{\alpha }$ is kinetic energy of $\alpha$ particle and $K_d$ is the kinetic energy of the daughter nucleus.

Let $A$ be the mass number of mother nucleus then the mass number of the daughter nucleus is $\left(A-4\right)$.

From the conservation of momentum

$\frac{K_{\alpha }}{K_d}=\frac{p^2}{2\left(4\right)}\frac{2\left(A-4\right)}{p^2}$

$\frac{K_{\alpha }}{K_d}=\frac{A-4}{4}$

$K_d=\frac{4}{A-4}{K}_{\alpha }$

$Q$  value $=K_{\alpha }+K_d$

$50=48+\frac{4}{A-4}\left(48\right)$

$2=\frac{4}{A-4}\left(48\right)$

$A-4=96$

$A=100$