The binding energy of a nucleus is a measure of its

If $E_e$ is the energy required to remove an electron from an atom and $E_n$ is the energy required to remove a nucleon from a nucleus,then:

One requires energy ${E_n}$ to remove a nucleon from a nucleus and an energy ${E_e}$ to remove an electron from the orbit of an atom. Then:

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 $; MeV$)

$A$ nucleus of mass $ 20 u $ emits a $ \gamma $ photon of energy $ 6 MeV $. If the emission is assumed to occur when the nucleus is free and at rest, then the nucleus will have a kinetic energy nearest to (take $ 1 u = 1.6 \times 10^{-27} kg $): (in $keV$)

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