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.

(D) Let the deuteron nucleus be at rest. When a $\gamma$-ray of energy $E$ is incident on it,the conservation of energy gives: $E = B + K_p + K_n$,where $K_p$ and $K_n$ are the kinetic energies of the proton and neutron.
From the conservation of linear momentum,the momentum of the photon must equal the sum of the momenta of the proton and neutron: $p_{\gamma} = p_p + p_n$. Since the photon's momentum is $p_{\gamma} = E/c$,we have $p_p + p_n = E/c$.
If $E = B$,then $K_p + K_n = 0$. Since kinetic energy cannot be negative,this implies $K_p = 0$ and $K_n = 0$,which means $p_p = 0$ and $p_n = 0$. However,this contradicts the momentum conservation equation $p_p + p_n = E/c$,as $E/c \neq 0$. Thus,$E = B$ is impossible.
For the process to occur,we must have $E > B$. Let $E = B + \Delta E$. The kinetic energies are $K_p = p_p^2 / 2m$ and $K_n = p_n^2 / 2m$. To minimize the energy required,we assume the particles move together with the same velocity,so $p_p = p_n = p/2$. Then $p = E/c$,so $p_p = p_n = E/2c$.
Substituting into the energy equation: $E - B = (E/2c)^2 / 2m + (E/2c)^2 / 2m = E^2 / 4mc^2$.
Since $E \approx B$,we have $\Delta E \approx B^2 / 4mc^2$.