The energy in $MeV$ released due to the complete transformation of $1\, kg$ of mass into energy is $(c = 3 \times 10^8\, m/s)$:

The atomic mass of ${ }_{6} C^{12}$ is $12.000000 \ u$ and that of ${ }_{6} C^{13}$ is $13.003354 \ u$. The required energy to remove a neutron from ${ }_{6} C^{13}$,if the mass of a neutron is $1.008665 \ u$,will be: (in $MeV$)

If $M_0$ is the mass of isotope ${ }_{5}^{12} B$,$M_p$ and $M_n$ are the masses of a proton and a neutron respectively,then the nuclear binding energy of the isotope is:

Find the binding energy $(BE)$ per nucleon of ${^{56}Fe}$,where $m({^{56}Fe}) = 55.936 \ u$,$m_{n} = 1.00866 \ u$,and $m_{p} = 1.00727 \ u$ (in $MeV$).

The positions of four different nuclei in the binding energy curve are shown in the figure. The process that might not proceed spontaneously is

Einstein's mass-energy relation emerging out of his famous theory of relativity relates mass $(m)$ to energy $(E)$ as $E = mc^2$,where $c$ is the speed of light in vacuum. At the nuclear level,the magnitudes of energy are very small. The energy at the nuclear level is usually measured in $MeV$,where $1\,MeV = 1.6 \times 10^{-13}\,J$; the masses are measured in unified atomic mass unit $(u)$,where $1\,u = 1.6605 \times 10^{-27}\,kg$.
$(a)$ Show that the energy equivalent of $1\,u$ is approximately $931.5\,MeV$.
$(b)$ $A$ student writes the relation as $1\,u = 931.5\,MeV$. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.