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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{m e^4}{8 \varepsilon_0^2 n^2 h^2}$.Using the Bohr mod

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$R = {\left( {\frac{1}{{4\pi {\varepsilon _0}}}} \right)^2}.\frac{{2{\pi ^2}m{e^4}}}{{c{h^3}}}$

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{m e^4}{8 \varepsilon_0^2 n^2 h^2}$.Using the Bohr model,the wavenumber $\bar{
u}$ for a transition from $n_2$ to $n_1$ is $\bar{
u} = \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.From the energy difference $\Delta E = h c \bar{
u} = E_{n_2} - E_{n_1}$,we get $R = \frac{m e^4}{8 \varepsilon_0^2 c h^3}$.Substituting $k = \frac{1}{4 \pi \varepsilon_0}$,we have $\varepsilon_0 = \frac{1}{4 \pi k}$,so $\varepsilon_0^2 = \frac{1}{16 \pi^2 k^2}$.Thus,$R = \frac{m e^4}{8 (\frac{1}{16 \pi^2 k^2}) c h^3} = \frac{16 \pi^2 k^2 m e^4}{8 c h^3} = \frac{2 \pi^2 k^2 m e^4}{c h^3}$.Substituting $k = \frac{1}{4 \pi \varepsilon_0}$,we get $R = \left( \frac{1}{4 \pi \varepsilon_0} ight)^2 \frac{2 \pi^2 m e^4}{c h^3}$.

The Rydberg constant $R$ for hydrogen is

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