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(A) When the finite mass of the proton is taken into account,the electron mass $m_e$ in the energy expression is replaced by the reduced mass $\mu = \

Vedclass · Accepted Answer

$0.06$

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(A) When the finite mass of the proton is taken into account,the electron mass $m_e$ in the energy expression is replaced by the reduced mass $\mu = \frac{m_e m_p}{m_e + m_p}$.The energy levels are given by $E_n = -\frac{\mu e^4}{8 n^2 \epsilon_0^2 h^2} = \left( \frac{m_p}{m_e + m_p} ight) E_{n, 	ext{infinite mass}}$.The ratio of the reduced mass to the electron mass is $\frac{\mu}{m_e} = \frac{m_p}{m_e + m_p} = \frac{1}{1 + \frac{m_e}{m_p}}$.Using the binomial approximation $(1 + x)^{-1} \approx 1 - x$ for small $x = \frac{m_e}{m_p}$:$\frac{\mu}{m_e} \approx 1 - \frac{m_e}{m_p}$.The fractional change in energy is $\frac{\Delta E}{E} = \frac{m_e}{m_p} = \frac{9.10 	imes 10^{-31}}{1.60 	imes 10^{-27}} \approx 5.68 	imes 10^{-4}$.Converting this to percentage: $5.68 	imes 10^{-4} 	imes 100 \% \approx 0.0568 \% \approx 0.06 \%$.

If one takes into account the finite mass of the proton,then the correction to the binding energy of the hydrogen atom is approximately (take,mass of proton $= 1.60 \times 10^{-27} \, kg$ and mass of electron $= 9.10 \times 10^{-31} \, kg$) (in $\%$)

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