In Bohr’s model, the radius of nth orbit, r = $\frac{n^2 h^2 \varepsilon_0}{\pi Z m_e^2}$ In the given muonic hydrogen atom, a negatively charged muon ($\mu^{-}$) of mass 207 $m_e$ revolve around a proton. Therefore radius of electron and muon can be written as $\frac{r_\mu}{r_e}$ = $\frac{m_e}{m_\mu}$ = $\frac{m_e}{207 m_e}$ ⇒ $r_\mu$ = $\frac{r_e}{207}$ = $\frac{0.53 \times 10^{-10}}{207}$ m = 2.5 × $10^{-12}$ m Energy of electron in the orbit, E = - $\frac{m e^4}{8 \varepsilon_0 n^2 h^2}$ ⇒ $\frac{E_e}{E_\mu}$ = $\frac{m_e}{m_\mu}$ = $\frac{m_e}{207 m_e}$ $E_\mu$ = 207 $E_e$ = 207 [- 13.6 eV] = - 2.8 keV