A nucleus X of mass  M, initially at rest, undergoes alpha decay according the equation  ${ }_Z^{A}X{\to }_{Z-2}^{A-4}Y{+}_2^{4}He$
The alpha particle emitted in the above process is found to move in a circular track of radius r  in a uniform magnetic field $B.$  Then (mass and charge of  $\alpha -$ particle are m  and  q respectively)

The ratio of kinetic energies of the alpha particle and the daughter nucleus $Y$  approximately equals $\frac{M-m}{m}$

The ratio of kinetic energies of the alpha particle and the daughter nucleus  Y approximately equals $\frac{m}{M}$

The energy released in the process approximately equals  $\frac{M}{M-m}\left(\frac{r^2{q}^2{B}^2}{2m}\right)$

The energy released in the process approximately equals  $\frac{M-m}{M}\left(\frac{r^2{q}^2{B}^2}{2m}\right)$