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:

The mass of an $\alpha$-particle is

Binding energy per nucleon of a fixed nucleus $X^A$ is $8 \ MeV$. It absorbs a neutron moving with kinetic energy $2 \ MeV$ and converts into $Y$,emitting two photons of energy $1 \ MeV$ and $4 \ MeV$ respectively one after the other. The binding energy per nucleon of $Y$ (in $MeV$) is:

Assume that the nuclear binding energy per nucleon $(B/A)$ versus mass number $(A)$ is as shown in the figure. Use this plot to choose the correct choice$(s)$ given below.
Figure: $222706-q$
$(A)$ Fusion of two nuclei with mass numbers lying in the range of $1 < A < 50$ will release energy.
$(B)$ Fusion of two nuclei with mass numbers lying in the range of $51 < A < 100$ will release energy.
$(C)$ Fission of a nucleus lying in the mass range of $100 < A < 200$ will release energy when broken into two equal fragments.
$(D)$ Fission of a nucleus lying in the mass range of $200 < A < 260$ will release energy when broken into two equal fragments.

Deuteron is a bound state of a neutron and a proton with a binding energy $B = 2.2 \, MeV$. $A$ $\gamma$-ray of energy $E$ is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the $n$ and $p$ move in the direction of the incident $\gamma$-ray. If $E = B$,show that this cannot happen. Hence,calculate how much bigger than $B$ must $E$ be for such a process to happen.

$A$ plot of the number of neutrons $(N)$ against the number of protons $(Z)$ for stable nuclei exhibits upward deviation from linearity for atomic number $Z > 20$. For an unstable nucleus having an $N/Z$ ratio less than $1$,the possible mode$(s)$ of decay is(are):
$(A)$ $\beta^{-}$-decay ($\beta$ emission)
$(B)$ Orbital or $K$-electron capture
$(C)$ Neutron emission
$(D)$ $\beta^{+}$-decay (positron emission)