A nucleus of uranium decays at rest into nuclei of thorium and helium. Then

$\text{ If }{\overrightarrow{p}}_{\text{Th and }}{\overrightarrow{p}}_{\text{He }}\text{ are the momenta of thorium and helium }$nuclei respectively, then according to law of conservation of linear momentum

$0={\overrightarrow{p}}_{\mathrm{Th}}+{\overrightarrow{p}}_{\mathrm{He}}\text{ or }{\overrightarrow{p}}_{\mathrm{Th}}=-{\overrightarrow{p}}_{\mathrm{He}}$

-ve sign shows that both are moving in opposite directions. But in magnitude

$p_{\mathrm{Th}}=p_{\mathrm{He}}$

If $m_{Th}$ and $m_{He}$ are the masses of thorium and helium nuclei respectively, then

$\begin{array}{l}\text{ Kinetic energy of thorium nucleus is }{K}_{\mathrm{Th}}=\frac{p_{\mathrm{Th}}^2}{2m_{\mathrm{Th}}}\\ \text{ and that of helium nucleus is }\end{array}$

$K_{\mathrm{He}}=\frac{p_{\mathrm{He}}^2}{2m_{\mathrm{He}}}$

$\therefore \frac{K_{\mathrm{Th}}}{K_{\mathrm{He}}}={\left(\frac{p_{\mathrm{Th}}}{p_{\mathrm{He}}}\right)}^2\left(\frac{m_{\mathrm{He}}}{m_{\mathrm{Th}}}\right)$

$\begin{array}{l}\text{ But }{p}_{\text{Th }}=p_{\mathrm{He}}\text{ and }{m}_{\mathrm{He}}<m_{\mathrm{Th}}\\ \therefore K_{\mathrm{Th}}<K_{\mathrm{He}}\text{ or }{K}_{\mathrm{He}}>K_{\mathrm{Th}}\end{array}$

Thus the helium nucleus has more kinetic energy than the thorium nucleus.