The radius of the first permitted Bohr orbit | vedclass

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(C) The Bohr radius is given by $r = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$, which implies $r \propto \frac{1}{m}$.Given $m_{\mu} = 207 m_e$, the new r

Vedclass · Accepted Answer

$2.56 	imes 10^{-13} \; m, -2.8 \; keV$

Vedclass · Answer

(C) The Bohr radius is given by $r = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$, which implies $r \propto \frac{1}{m}$.Given $m_{\mu} = 207 m_e$, the new radius $r_{\mu} = \frac{r_e}{207} = \frac{0.53 	imes 10^{-10} \; m}{207} \approx 2.56 	imes 10^{-13} \; m$.The ground state energy is given by $E = -\frac{m e^4}{8 \epsilon_0^2 n^2 h^2}$, which implies $E \propto m$.Given $m_{\mu} = 207 m_e$, the new energy $E_{\mu} = E_e 	imes 207 = -13.6 \; eV 	imes 207 = -2815.2 \; eV \approx -2.8 \; keV$.

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