The energy equivalent of $1 \text{ kg}$ of matter is about

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

Obtain the binding energy (in $MeV$) of a nitrogen nucleus $\left(^{14}_{7} N \right)$ given $m\left(^{14}_{7} N \right)=14.00307 \; u$. (Given: $m_{H} = 1.007825 \; u$,$m_{n} = 1.008665 \; u$) (in $; MeV$)

If the binding energy per nucleon for $\,_3^7\,Li\,\,$ and $\,_2^4\,\,He$ are $5.60 \,MeV$ and $7.06 \,MeV$ respectively, then the energy of the proton in the reaction $p\,\, + \,\,_3^7\,\,Li\,\, \to \,\,2\,_2^4\,\,He$ must be .......... $MeV$.

Derive the formula for the binding energy of a nucleus by explaining the mass defect of the nucleus and write the formula for the binding energy per nucleon.

The ratio of the orders of the spacings of nuclear energy levels and atomic energy levels is