Let $m_p$ be the mass of a proton,$m_n$ the mass of a neutron,$M_1$ the mass of a ${}_{10}^{20}Ne$ nucleus,and $M_2$ the mass of a ${}_{20}^{40}Ca$ nucleus. Then:

In the nuclear process $n \to p + e^- + \bar{\nu}$,if the masses of proton,neutron,and electron are $1.6725 \times 10^{-27} \ kg$,$1.6747 \times 10^{-27} \ kg$,and $9 \times 10^{-31} \ kg$ respectively,then the energy released is ...... $MeV$.

For a nucleus ${ }_Z^A X$ having mass number $A$ and atomic number $Z$:
$A.$ The surface energy per nucleon $(b_s) = a_1 A^{2/3}$
$B.$ The Coulomb contribution to the binding energy $b_c = -a_2 \frac{Z(Z-1)}{A^{4/3}}$
$C.$ The volume energy $b_v = a_3 A$
$D.$ Decrease in the binding energy is proportional to surface area.
$E.$ While estimating the surface energy,it is assumed that each nucleon interacts with $12$ nucleons,($a_1, a_2$ and $a_3$ are constants)
Choose the most appropriate answer from the options given below:

The binding energies of $_1H^2$,$_2He^4$,$_{26}Fe^{56}$,and $_{92}U^{235}$ are $2.22 \ MeV$,$28.3 \ MeV$,$492 \ MeV$,and $1786 \ MeV$ respectively. Which nucleus is the most stable?

Which of the following statements is correct?

What is the energy produced in $MeV$ when the mass equivalent to $1$ proton is completely converted into energy?