The nucleus of ${}_{90}{}^{230}\mathrm{Th}$ decays to  ${}_{88}{}^{226}\mathrm{Ra}$ and  ${}_2{}^4\mathrm{He}$ with the emission of energy. If the original nucleus was at rest, the ratio of kinetic energies of He and Ra nuclei will be very nearly equal to:

If the original nucleus Th is at rest, i.e. if the momentum of the system before $\alpha$-decay is zero, the total momentum after the decay will also be zero. Thus Ra and He will have equal and opposite momenta.

$\therefore$      $m_{He}{v}_{He}=-m_{Ra}{v}_{Ra}$

or       $m_{He}^2{v}_{He}^2=m_{Ra}^2{v}_{Ra}^2 .......................\left(i\right)$   

Now  $\frac{ K.E\left(He\right)}{K.E\left(Ra\right)}=\frac{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)m_{He}{v}_{He}^2}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)m_{Ra}{v}_{Ra}^2}$

                           = ${\left(\frac{m_{He}{v}_{He}^{ }}{m_{Ra}{v}_{Ra}}\right)}^2\times \left(\frac{m_{Ra}}{m_{He}}\right)$

                           = $\frac{m_{Ra}}{m_{He}}=\frac{226}{4}=\frac{113}{2}$               [use Eq.(i)]