Consider the following nuclear reaction : $^{238}\text{U} \rightarrow ^{234}\text{Th} + ^4\text{He}$. Take masses of $^{238}\text{U}$, $^{234}\text{Th}$ and $^4\text{He}$ as $238.050 \text{ u}$, $234.043 \text{ u}$ and $4.003 \text{ u}$, respectively. The $Q$ value for the reaction, in $\text{keV}$, is : [Given : $1 \text{ u} = 931.5 \text{ MeV/c}^2$]

The binding energy per nucleon of $^{209}_{83}Bi$ is . . . . . . MeV. [Take $m(^{209}_{83}Bi) = 208.980388 \text{ u}$, $m_p = 1.007825 \text{ u}$, $m_n = 1.008665 \text{ u}$, $1 \text{ u} = 931 \text{ MeV}/c^2$]

The positions of four different nuclei in the binding energy curve are shown in the figure. The process that might not proceed spontaneously is

$A$ nucleus of mass $M + \Delta m$ is at rest and decays into two daughter nuclei of equal mass $\frac{M}{2}$ each. The speed of light is $c$. The binding energy per nucleon for the parent nucleus is $E_1$ and that for the daughter nuclei is $E_2$. Then:

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

Einstein's mass-energy relation emerging out of his famous theory of relativity relates mass $(m)$ to energy $(E)$ as $E = mc^2$,where $c$ is the speed of light in vacuum. At the nuclear level,the magnitudes of energy are very small. The energy at the nuclear level is usually measured in $MeV$,where $1\,MeV = 1.6 \times 10^{-13}\,J$; the masses are measured in unified atomic mass unit $(u)$,where $1\,u = 1.6605 \times 10^{-27}\,kg$.
$(a)$ Show that the energy equivalent of $1\,u$ is approximately $931.5\,MeV$.
$(b)$ $A$ student writes the relation as $1\,u = 931.5\,MeV$. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.