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

Mass of a negatively charged muon, `m_(mu) = 207m_(e)`
According to Bohr's model,
Bohr radius, `r_(e)prop((1)/(m_(e)))`
And, energy of a ground state electronic hydrogen atom, `E_(e) prop m_(e)`.
Also, energy of a ground state muonic hydrogen atom, `E_(mu) prop m_(mu)`.
We have the value of the first Bohr orbit, `r_(e)=0.53 Å = 0.53 xx 10^(-10) m`
Let `r_(mu)` be the radius of muonic hydrogen atom.
At equilibrium, we can write the relation as:
`m_(mu)r_(mu) = m_(e)r_(e)`
`207m_(e) xx r_(mu) = m_(e)r_(e)`
`therefore r_(mu) = (0.53 xx 10^(-10))/(207)=2.56 xx 10^(-13) m`
Hence, the value of the first Bohr radius of a muonic hydrogen atom is
`2.56 xx 10^(-13)m`.
We have,
`E_(e) = - 13.6 eV`
Take the ratio of these energies as:
`(E_(e))/(E_(mu))=(m_(e))/(m_(mu))=(m_(e))/(207m_(e))`
`E_(mu) = 207 E_(e)`
`= 207 xx (-13.6) =- 2.81 keV`
Hence, the ground state energy of a muonic hydrogen atom is - 2.81 keV.