The deuteron is bound by nuclear forces just | vedclass

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(D) The binding energy of an $H$-atom is given by the formula:$E = \frac{m_e e^4}{8 \epsilon_0^2 h^2} = 13.6 	ext{ eV}$ ... $(1)$Now,considering the

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(D) The binding energy of an $H$-atom is given by the formula:$E = \frac{m_e e^4}{8 \epsilon_0^2 h^2} = 13.6 	ext{ eV}$ ... $(1)$Now,considering the deuteron nucleus as a binary system,we replace $m_e$ with the reduced mass $m'$ and $e$ with the effective charge $e'$. The binding energy of the deuteron is:$E' = \frac{m' e'^4}{8 \epsilon_0^2 h^2} = 2.2 	imes 10^6 	ext{ eV}$ ... $(2)$Taking the ratio of equation $(2)$ to equation $(1)$:$\frac{m'}{m_e} 	imes \left(\frac{e'}{e}ight)^4 = \frac{2.2 	imes 10^6}{13.6} \approx 1.617 	imes 10^5$ ... $(3)$The reduced mass $m'$ of the deuteron (proton-neutron system) is:$m' = \frac{m_p m_n}{m_p + m_n} = \frac{m 	imes m}{2m} = \frac{m}{2}$Given $m \approx 1836 m_e$,we have $m' = \frac{1836 m_e}{2} = 918 m_e$.Thus,$\frac{m'}{m_e} = 918$.Substituting this into equation $(3)$:$918 	imes \left(\frac{e'}{e}ight)^4 = 1.617 	imes 10^5$$\left(\frac{e'}{e}ight)^4 = \frac{1.617 	imes 10^5}{918} \approx 176.14$$\frac{e'}{e} = (176.14)^{1/4} \approx 3.65$.

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