The deuteron is bound by nuclear forces just as H -atom is made up of $p$ and $e$ bound by electrost

Question

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 in the form of a Coulomb potential but with an effective charge $e^{\prime}$, \ $ F=\frac{1}{4 \pi \varepsilon_0} \frac{e^{\prime 2}}{r} $ \ Estimate the value of $\left(e^{\prime} / eight)$ given that the binding energy of a deuteron is 2.2 MeV .

Professor PC Thomas Classes · Answer

(c) : We know that binding energy of hydrogen atom in ground state, 
$E=\frac{m e^4}{8 \varepsilon_0^2 h^2}=13.6 \mathrm{eV}$ ...(i)
 Replacing $e$ by $e^{\prime}$ and $m$ by $m^{\prime}$, reduced mass of neutron proton, 
$m^{\prime}=\frac{M 	imes M}{M+M}=\frac{M}{2}=\frac{1836 m}{2}=918 m$ ...(ii)
 ( $M=$ mass of neutron/proton) $	herefore \quad$ \ Binding energy, $E^{\prime}=\frac{918 m e^{\prime 4}}{8 \varepsilon_0^2 h^2}=2.2 \mathrm{MeV}$\ Dividing eq. (ii) by (i), we get $ 918\left(\frac{e^{\prime}}{e}ight)^4=\frac{2.2 	imes 10^6}{13.6} \quad 	herefore \quad \frac{e^{\prime}}{e}=(176.21)^{1 / 4}=3.64 $