The binding energy per nucleon for ${O^{16}}$ and ${O^{17}}$ are $7.97 \, MeV$ and $7.75 \, MeV$ respectively. How much energy in $MeV$ is required to remove one neutron from ${O^{17}}$?

In the nuclear reaction $_1^2H + _1^3H \to _2^4He + _0^1n$, if the binding energies of $_1^2H$, $_1^3H$, and $_2^4He$ are respectively $a$, $b$, and $c$ (in $MeV$), then the energy (in $MeV$) released in this reaction is:

$M_p$ denotes the mass of a proton and $M_n$ that of a neutron. $A$ given nucleus,of binding energy $B$,contains $Z$ protons and $N$ neutrons. The mass $M(N, Z)$ of the nucleus is given by ($c$ is the velocity of light):

If $M(A, Z)$,$M_p$,and $M_n$ represent the masses of the nucleus ${}_{Z}^{A}X$,proton,and neutron in $u$ units respectively $(1u = 931.5 \, MeV/c^2)$,and $BE$ represents the binding energy in $MeV$,then which of the following relations is correct?

The electrostatic energy of $Z$ protons uniformly distributed throughout a spherical nucleus of radius $R$ is given by $E = \frac{3}{5} \frac{Z(Z-1) e^2}{4 \pi \varepsilon_0 R}$. The measured masses of the neutron,${ }_1^1 H$,${ }_7^{15} N$,and ${ }_8^{15} O$ are $1.008665 \ u$,$1.007825 \ u$,$15.000109 \ u$,and $15.003065 \ u$,respectively. Given that the radii of both the ${ }_7^{15} N$ and ${ }_8^{15} O$ nuclei are the same,$1 \ u = 931.5 \ MeV/c^2$ ($c$ is the speed of light),and $e^2 / (4 \pi \varepsilon_0) = 1.44 \ MeV \ fm$. Assuming that the difference between the binding energies of ${ }_7^{15} N$ and ${ }_8^{15} O$ is purely due to the electrostatic energy,the radius of either of the nuclei is $(1 \ fm = 10^{-15} \ m)$: (in $fm$)

Assertion : Binding energy (or mass defect) of hydrogen nucleus is zero.
Reason : Hydrogen nucleus contains only one nucleon.