The correct Solution is (2)
$
\begin{array}{l}
\mathrm{U} \rightarrow \mathrm{Th}+\alpha \\
\mathrm{KE}_{\mathrm{Th}}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}_{\mathrm{Th}}}, \mathrm{KE}_{\alpha}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}_{\alpha}}
\end{array}
$
Since $\mathrm{m}_{\mathrm{a}}$ is less so $\mathrm{KE}_{\mathrm{a}}$ will be mone.