A nucleus with mass number 220 initially at rest emits an $\mathrm{\alpha }$-particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the $\mathrm{\alpha }$-particle

Q-value of the reaction is 5.5 eV

$\mathrm{i}.\mathrm{e}., {\mathrm{k}}_1+ {\mathrm{k}}_2 =5.5 \mathrm{MeV} .......\left(\mathrm{i}\right)$

By conservation of linear momentum

$$\begin{gathered} {\mathrm{p}}_1={\mathrm{p}}_2\Rightarrow \sqrt{2\left(216\right){\mathrm{k}}_1} \\ =\sqrt{2\left(4\right){\mathrm{k}}_2} \end{gathered}$$

$\Rightarrow {\mathrm{k}}_2=54{\mathrm{k}}_1 ....\left(\mathrm{ii}\right)$

On solving equation (i) and (ii) we flet ${\mathrm{k}}_2= 5.4 \mathrm{MeV}$.