A proton accelerated through a potential difference of $100 \mathrm{V}$, has de-Broglie wavelength ${\lambda }_0$. The de-Broglie wavelength of an $\alpha$-particle, accelerated through $800 \mathrm{V}$ is

A nuclear reaction along with the masses of the particle taking part in it is as follows;

$\underset{1.002 \mathrm{amu}}{A}+\underset{1.004 \mathrm{amu}}{B}\to \underset{1.001 \mathrm{amu}}{C}+\underset{1.003 \mathrm{amu}}{D}+Q \mathrm{MeV}$

The energy $Q$ liberated in the reaction is