$A$ nucleus with mass number $242$ and binding energy per nucleon as $7.6\,MeV$ breaks into two fragments,each with mass number $121$. If each fragment nucleus has a binding energy per nucleon of $8.1\,MeV$,the total gain in binding energy is $........MeV$.

For which mass number range is the binding energy per nucleon maximum?

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 mass of a $_3^7Li$ nucleus is $0.042 \, u$ less than the sum of the masses of its constituent nucleons. The binding energy per nucleon of the $_3^7Li$ nucleus is approximately ......... $MeV$.

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:

Assertion : The binding energy per nucleon,for nuclei with atomic mass number $A > 100$,decreases with $A$.
Reason : The nuclear forces are weak for heavier nuclei.