The radius of the first orbit of hydrogen is | vedclass

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(A) The 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 energy of the $\mu^{-}$-ato

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The 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 energy of the $\mu^{-}$-atom is $E_{\mu} = \frac{m_{\mu}}{m_e} 	imes E_e$. Given $m_{\mu} = 207 m_e$,we have $E_{\mu} = 207 	imes (-13.6 	ext{ eV}) = -13.6 	imes 207 	ext{ eV}$.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 radius of the $\mu^{-}$-atom is $r_{\mu} = \frac{m_e}{m_{\mu}} 	imes r_{H} = \frac{m_e}{207 m_e} 	imes r_{H} = \frac{r_{H}}{207}$.Thus,the energy is $-13.6 	imes 207 	ext{ eV}$ and the radius is $\frac{r_{H}}{207}$.

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