A radioactive element of mass number 208 at rest disintegrates by emitting an $\alpha$-particle. If E is the energy of the emitted $\alpha$-particle, the energy of disintegration is:

Mass number of daughter nucleus (M) = 208 - 4 = 204. 
Now, total energy of disintegration = energy of daughter nucleus + energy of $\alpha$-particle 

 $E_t= \frac{p^2}{2m}+\frac{P^2}{2M}$

Since momentum is conserved, and there is no initial momentum, hence p = P. 

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

Energy of $\alpha$-particle = $\frac{p^2}{2m}= E$. Hence 

    $E_t= \left(\frac{m+M}{M}\right)E$

         = $\left(\frac{4+204}{204}\right)E=\frac{52}{51}E$