Here, when deuteron nucleus is given energy $\mathrm{E}=\mathrm{B}$, proton and neutron (say particle $1$ and particle $2$) become free from mutual binding but since energy supplied is getting consumed completely, no energy remains in balance and so they can not be emitted out. This fact can be proved as follows.

Suppose after spending energy $B$ out of supplied energy $E$, proton and neutron (after getting freed from mutual binding) move with momenta, say $p_{1}$ and $p_{2}$ respectively and possess kinetic energy say $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ respectively. If this would happen then,

$\mathrm{E}-\mathrm{B}=\mathrm{K}_{1}+\mathrm{K}_{2}$

… $(1)$

$\therefore \quad \mathrm{E}-\mathrm{B}=\frac{p_{1}^{2}}{2 m}+\frac{p_{2}^{2}}{2 m}$$….(2)$

(Where $m_{p} \approx m_{n} \approx m$ )

According to law of conservation of momentum, $p_{1}+p_{2}=p^{\prime} \quad\left(\right.$ As per the statement $\left.\overrightarrow{p^{\prime}}\left\|\overrightarrow{p_{1}}\right\| \overrightarrow{p_{2}}\right)$ Here momentum of photon,

Here momentum of photon,

$p^{\prime}=\frac{h}{\lambda}=\frac{h}{(c / f)}=\frac{h f}{c}=\frac{\mathrm{E}}{c}$$…(3)$

and so from above equation $p_{1}+p_{2}=\frac{\mathrm{E}}{c}$$…(4)$

If $E=B$ then from equation $(1)$,

$\frac{p_{1}^{2}}{2 m}+\frac{p_{2}^{2}}{2 m}=0$ $\therefore p_{1}^{2}+p_{2}^{2}=0$ $\Rightarrow p_{1}=0 \text { and } p_{2}=0$ $\Rightarrow\text { Proton and neutron can not be emitted, after being freed from mutual binding. }$

Now suppose, $\mathrm{E}=\mathrm{B}+\Delta \mathrm{E}$ (Where $\Delta \mathrm{E}<<\mathrm{E}$ )

$\therefore \mathrm{E}-\mathrm{B}=\Delta \mathrm{E}$$….(5)$