The deuteron is bound by nuclear forces just as $H-$ atom is made up of $p$ and $e$ bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given 1 e 2 in the form of a Coulomb potential but with an effective charge $e' \,:\,F = \frac{1}{{4\pi { \in _0}}}\frac{{e{'^2}}}{r}$ . estimate the value of $(e'/e)$ given that the binding energy of a deuteron is $2.2\, MeV.$

Binding energy of H-atom is given by formula,

$\mathrm{E}=\frac{m_{e} e^{4}}{8 \epsilon_{0}^{2} h^{2}}=13.6 \mathrm{eV}$$....(1)$

Now, considering deuteron nucleus as a binary system, let us denote $m$ by $m^{\prime}$ and $e$ by $e^{\prime}$ in above formula. Now the binding energy of deuteron,

$\mathrm{E}^{\prime}=\frac{m^{\prime} e^{\prime 4}}{8 \epsilon_{0}^{2} h^{2}}=2.2 \times 10^{6} \mathrm{eV} \quad$ (as per statement)$....(2)$

Taking ratio of equation $(2)$ and $(1)$, $\frac{m^{\prime}}{m_{e}} \times\left(\frac{e^{\prime}}{e}\right)^{4}=\frac{2.2 \times 10^{6}}{13.6}$$...(3)$

Here, reduced mass of binary system like deuteron nucleus,

$m^{\prime}=\frac{m_{p} \times m_{n}}{m_{p}+m_{n}}$

$m^{\prime}=\frac{m \times m}{m+m}=\frac{m}{2}=\frac{1836 m_{e}}{2}$

$\therefore m^{\prime}=918 m_{e} \Rightarrow \frac{m^{\prime}}{m_{e}}=918$

$\ldots$ $(4)$

$\left(\because m_{p}=m_{n}=m=1836 m_{e}\right)$