What does the binding energy per nucleon show?

$A$ 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 $........MeV$.

The radionuclide $^{11}_{6}C$ decays by $\beta^+$ emission. Given that $m(^{11}_{6}C) = 11.011434 \ u$,$m(^{11}_{5}B) = 11.009305 \ u$,$m_e = 0.000548 \ u$,and $1 \ u = 931.5 \ MeV/c^2$. The $Q$-value of this decay process is:

The binding energy for the following nuclear reactions are expressed in $MeV$.
${ }_2 He ^3+{ }_0 n ^1 \rightarrow{ }_2 He ^4+20 \ MeV$
${ }_2 He ^4+{ }_0 n ^1 \rightarrow{ }_2 He ^5-0.9 \ MeV$
If $X_3, X_4, X_5$ denote the stability of ${ }_2 He ^3, { }_2 He ^4$ and ${ }_2 He ^5$, respectively, then the correct order is:

$A$ nuclide $1$ is said to be the mirror isobar of nuclide $2$ if $Z_1 = N_2$ and $Z_2 = N_1$. $(a)$ What nuclide is a mirror isobar of $_{11}^{23}Na$? $(b)$ Which nuclide out of the two mirror isobars has greater binding energy and why?

Obtain the binding energy (in $MeV$) of a nitrogen nucleus $\left(^{14}_{7} N \right)$ given $m\left(^{14}_{7} N \right)=14.00307 \; u$. (Given: $m_{H} = 1.007825 \; u$,$m_{n} = 1.008665 \; u$) (in $; MeV$)