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(A) The total energy of the $n$-th orbit is given by $E_n = -\frac{m e^4}{8 \varepsilon_0^2 h^2 n^2}$.Since $E_n \propto m$,the ratio of the energy of

Vedclass · Accepted Answer

$-13.6 	imes 207 	ext{ eV}, \frac{r_{H}}{207}$

Vedclass · Answer

(A) The total energy of the $n$-th orbit is given by $E_n = -\frac{m e^4}{8 \varepsilon_0^2 h^2 n^2}$.Since $E_n \propto m$,the ratio of the energy of the $\mu^{-}$-system to the hydrogen atom is $\frac{E_{\mu}}{E_e} = \frac{m_{\mu}}{m_e} = 207$.Thus,$E_{\mu} = 207 	imes E_e = 207 	imes (-13.6 	ext{ eV}) = -13.6 	imes 207 	ext{ eV}$.The radius of the $n$-th orbit is given by $r_n = \frac{\varepsilon_0 h^2 n^2}{\pi m e^2}$.Since $r_n \propto \frac{1}{m}$,the ratio of the radius of the $\mu^{-}$-system to the hydrogen atom is $\frac{r_{\mu}}{r_H} = \frac{m_e}{m_{\mu}} = \frac{1}{207}$.Thus,$r_{\mu} = \frac{r_H}{207}$.

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