Calculate the binding energy of a nitrogen nucleus $\left[{ }_7^{14} N\right]$,given that the mass of the nucleus $m\left[{ }_7^{14} N\right] = 14.00307 \ u$. (Take mass of proton $m_p = 1.00783 \ u$ and mass of neutron $m_n = 1.00867 \ u$) (in $MeV$)

$A$ plot of the number of neutrons $(N)$ against the number of protons $(Z)$ for stable nuclei exhibits upward deviation from linearity for atomic number $Z > 20$. For an unstable nucleus having an $N/Z$ ratio less than $1$,the possible mode$(s)$ of decay is(are):
$(A)$ $\beta^{-}$-decay ($\beta$ emission)
$(B)$ Orbital or $K$-electron capture
$(C)$ Neutron emission
$(D)$ $\beta^{+}$-decay (positron emission)

Draw conclusions from the two main features of the graph of binding energy per nucleon versus the atomic mass number $(A)$.

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:

The binding energy per nucleon of $^{16}O$ is $7.97 \ \text{MeV}$ and that of $^{17}O$ is $7.75 \ \text{MeV}$. The energy (in $\text{MeV}$) required to remove a neutron from $^{17}O$ is:

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.