Stationary nucleus ${}^{238}\mathrm{U}$ decays by a emission generating a total kinetic energy T

${}_{92}{}^{238}\mathrm{U}\to {}_{90}{}^{234}\mathrm{Th}+{}_2{}^4\mathrm{\alpha }$

What is the kinetic energy of the $\mathrm{\alpha }$ particle?

Let the kinetic energy of the $\mathrm{\alpha }$-particle be ${\mathrm{E}}_{\mathrm{\alpha }}$ and that of the thorium Th be ${\mathrm{E}}_{\mathrm{Th}.}$

The ratio of kinetic energies is 

$\frac{{\mathrm{E}}_{\mathrm{\alpha }}}{{\mathrm{E}}_{\mathrm{th}}}=\frac{{\displaystyle \frac{1}{2}}{\mathrm{m}}_{\mathrm{\alpha }}{{\mathrm{v}}_{\mathrm{\alpha }}}^2}{\frac{1}{2}{\mathrm{m}}_{\mathrm{th}}{{\mathrm{v}}_{\mathrm{th}}}^2}=\left(\frac{{\mathrm{m}}_{\mathrm{\alpha }}}{{\mathrm{m}}_{\mathrm{th}}}\right){\left(\frac{{\mathrm{v}}_{\mathrm{\alpha }}}{{\mathrm{v}}_{\mathrm{th}}}\right)}^2 .....\left(1\right)$

By conservation of momentum be momentum of  -particle and that of the recoiling thorium must be equal.Thus, 

$$\begin{gathered} {\mathrm{m}}_{\mathrm{\alpha }}{\mathrm{v}}_{\mathrm{th}}={\mathrm{m}}_{\mathrm{\alpha }}{\mathrm{v}}_{\mathrm{th}} \\ \mathrm{or} \frac{{\mathrm{v}}_{\mathrm{\alpha }}}{{\mathrm{v}}_{\mathrm{th}}}=\frac{{\mathrm{m}}_{\mathrm{th}}}{{\mathrm{m}}_{\mathrm{\alpha }}} .......\left(2\right) \end{gathered}$$

Subst, (2) into (1), we have 

$\frac{{\mathrm{E}}_{\mathrm{\alpha }}}{{\mathrm{E}}_{\mathrm{th}}}=\left(\frac{{\mathrm{m}}_{\mathrm{\alpha }}}{{\mathrm{m}}_{\mathrm{th}}}\right){\left(\frac{{\mathrm{m}}_{\mathrm{th}}}{{\mathrm{m}}_{\mathrm{\alpha }}}\right)}^2=\frac{{\mathrm{m}}_{\mathrm{th}}}{{\mathrm{m}}_{\mathrm{\alpha }}}=\frac{234 }{4}=58.5$

Thus, the kinetic energy of the  $\mathrm{\alpha }$-particle expressed as the fraction of the total kinetic energy T is given by

${\mathrm{E}}_{\mathrm{\alpha }}=\frac{58.5}{1+58.5}\mathrm{T}=\frac{58.5}{59.5}\mathrm{T}=0.98\mathrm{T}$

Which is slightly less than T.