In a nuclide, one $a.m.u.$ of mass is dissipated into energy to bind its nucleons. The energy equivalent of this mass is

What is $X$ in the following nuclear reaction?
$_{7}N^{14} + _{1}H^{1} \xrightarrow{} _{8}O^{15} + X$

The nuclear binding energy for $Ar$ $(39.962384 \, amu)$ is ......... $MeV.$ (Given: mass of proton $= 1.007825 \, amu$,mass of neutron $= 1.008665 \, amu$)

The isotopic mass of ${}_{3}^{7}Li$ is $7.016005 \ u$ and those of the $H$ atom and neutron are respectively $1.007825 \ u$ and $1.008665 \ u$. Then,the binding energy of the ${}_{3}^{7}Li$ nucleus is: (in $MeV$)

Calculate the mass defect in the following nuclear reaction:
$_1H^2 + _1H^3 \to _2He^4 + _0n^1$
(Given: mass of $_1H^2 = 2.014 \ amu$,$_1H^3 = 3.016 \ amu$,$_2He^4 = 4.004 \ amu$,$_0n^1 = 1.008 \ amu$) (in $amu$)

If the mass defect of a nuclide is $3.32 \times 10^{-26} \ g$, its binding energy in $MeV$ is: