A nucleus of mass number 220, initially at rest, emits an $\alpha$-particle. If the  Q-value of the reaction is 5.5 MeV, the energy of the emitted $\alpha$-particle will be:

$\text{ Kinetic energy }=\frac{(\text{ momentum }{)}^2}{2\times \text{ mass }}$

Mass number of $\alpha$-particle (m) = 4 units. 

Mass number of daughter nucleus (M) = 220 –  4 = 216 units. 

If P and p denote the momenta of the daughter nucleus and the $\alpha$-particle respectively, then:

$Q=\frac{P^2}{2M}+\frac{p^2}{2m}$

Since momentum is conserved, P = p. Hence,

$Q=\frac{p^2}{2}\left(\frac{1}{M}+\frac{1}{m}\right)=\frac{p^2}{2m}\left(\frac{m}{M}+1\right)$

Now, $\frac{p^2}{2m}=\mathrm{KE} \mathrm{of} \mathrm{\alpha }-\mathrm{particle}={\mathrm{E}}_{\mathrm{\alpha }}.$  Thus,

$Q=E_{\alpha }\left(\frac{m+M}{M}\right)$

$\Rightarrow E_{\alpha }=\frac{QM}{(m+M)}=\frac{5.5 \mathrm{MeV}\times 216}{(4+216)} =5.4 \mathrm{MeV}$