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  -particle are m and q respectively)
 

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

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

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

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