The radius of the first orbit of hydrogen is r_H, and the energy in the ground state is –13.6eV. Considering a µ^- -particle with a mass 207 m_e revolving around a proton as in Hydrogen atom. The energy and radius of proton and µ^- -combination respectively in the first orbit are (assume nucleus to be stationary)

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$\frac{- 13 .6}{207} e V, \frac{r_{H}}{207}$

$\frac{- 13 .6}{207} eV, 207 r_{H}$

$- 13 .6 \times 207 eV, \frac{r_{H}}{207}$

–207 × 13.6 eV, 207r_H

answer is A.

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Detailed Solution

i) $E_{n} = (- \frac{{me}^{4}}{8 h^{2} ϵ_{o}^{2}}) \frac{z^{2}}{n^{2}} ∴ Eαm$
$\begin{array}{l} \frac{E_{1}}{E_{2}} = \frac{m_{1}}{m_{2}}; \frac{- 13 .6 eV}{E_{2}} = \frac{m_{e}}{207 m_{e}} \\ ∴ E_{2} = - 13 .6 \times 207 eV \end{array}$
ii) $r_{n} = (\frac{h^{2} \in_{0}}{π {me}^{2}}) \frac{n^{2}}{z} ∴ rα \frac{1}{m}; ∴ \frac{r_{1}}{r_{2}} = \frac{m_{2}}{m_{1}}$
$∴ \frac{r_{H}}{r_{2}} = \frac{207 m_{e}}{m_{e}}; ∴ r_{2} = \frac{r_{H}}{207}$