The mass of a proton is $1.0073 \, u$ and that of a neutron is $1.0087 \, u$. The binding energy of $_2He^4$ is .... ($u = \text{atomic mass unit}$,mass of helium nucleus = $4.0015 \, u$)

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 for a deuteron $(_{1}^{2}H)$ and an $\alpha -$ particle $(_{2}^{4}He)$ are $x_1$ and $x_2$ respectively. The energy $(Q)$ released in the reaction $_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{4}He + Q$ 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?

If an electron and a positron annihilate,the energy released is ........

The deuteron is bound by nuclear forces just as the $H$-atom is made up of a proton and an electron bound by electrostatic forces. If we consider the force between the neutron and proton in a deuteron as given in the form of a Coulomb potential but with an effective charge $e'$: $F = \frac{1}{4\pi \epsilon_0} \frac{e'^2}{r^2}$,estimate the value of $(e'/e)$ given that the binding energy of a deuteron is $2.2 \text{ MeV}$.