One atomic mass unit is equivalent to .............. $MeV$ energy.

$A$ nucleus of mass $M + \Delta m$ is at rest and decays into two daughter nuclei of equal mass. If $C$ is the speed of light,and $E_1$ is the binding energy per nucleon for the parent nucleus and $E_2$ is that for the daughter nuclei,then:

The mass of a nucleus $_Z^AX$ is denoted by $M(A, Z)$. If $M_p$ and $M_n$ are the masses of a proton and a neutron respectively, the binding energy of this nucleus is given by:

The binding energy per nucleon versus mass number curve for nuclei is shown in the figure. $W, X, Y$ and $Z$ are four nuclei indicated on the curve. The process that would release energy is

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 $fm$)

If the binding energy of the deuterium is $2.23 \, MeV$, the mass defect in $a.m.u.$ is: