Obtain the first Bohr 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 `207m_e` revolves around a proton).

Explanation

Radius of first orbit for normal hydrogen atom ,
`r_e=0.53xx10^(-10)` m ,
Ground state energy of normal hydrogen atom , `(E_1)_e`=13.6 eV ,
Mass of muon, `m_mu=207 m_e`
Now, according to Bohr.s atomic model , radius of `n^(th)` orbit of hydrogen atom is given by
`r=(epsilon_0n^2h^2)/(pime^2)`
Here , n=1
`therefore r prop 1/m`
Hence, `r_mu/r_e=m_e//m_mu`
and `r_mu=r_0 m_e/m_mu=(0.53xx10^(-10))/207`
`r_mu=2.56xx10^(-13)` m
Also , energy of `n^(th)` orbit of hydrogen atom is given by
`E=-(me^4)/(8epsilon_0^2 h^2n^2)`
`therefore E prop m`
Hence, `E_mu/E_e =m_mu/m_e`
`E_mu=E_e m_mu/m_e`=-13.6 x 207
=-2815.2 eV=-2.8152 keV