Obtain the first Bohr's radius and the ground state energy of a muonic hydrogen atom [i.e.,an atom in which a negatively charged muon $(\mu^-)$ of mass about $207 m_{e}$ orbits around a proton].

(N/A) The mass of a negatively charged muon is $m_{\mu} = 207 m_{e}$.
According to Bohr's model,the radius of the $n^{th}$ orbit is given by $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$. Since $r \propto \frac{1}{m}$,we have $r_{\mu} = r_{e} \times \frac{m_{e}}{m_{\mu}}$.
Given $r_{e} = 0.53 \times 10^{-10} \, m$,the radius of the muonic hydrogen atom is $r_{\mu} = \frac{0.53 \times 10^{-10}}{207} \approx 2.56 \times 10^{-13} \, m$.
The energy of the ground state is given by $E_n = -\frac{m Z^2 e^4}{8 \epsilon_0^2 n^2 h^2}$. Since $E \propto m$,we have $E_{\mu} = E_{e} \times \frac{m_{\mu}}{m_{e}}$.
Given $E_{e} = -13.6 \, eV$,the ground state energy of the muonic hydrogen atom is $E_{\mu} = -13.6 \times 207 \, eV = -2815.2 \, eV \approx -2.81 \, keV$.