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 curve of binding energy per nucleon as a function of atomic mass number has a sharp peak for the helium nucleus. This implies that helium

The nucleus having the highest binding energy per nucleon is

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 mass and energy equivalent to $1\, a.m.u.$ are respectively:

The mass defect in a particular nuclear reaction is $0.3 \,g$. The amount of energy liberated in kilowatt-hour $(kWh)$ is: (Velocity of light $c = 3 \times 10^8 \,m/s$)