Question

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 (μ−) of mass about 207me orbits around a proton].

Answer 1

In Bohr's model, the radius of nth orbit, $r_n=\frac{n^2h^2\epsilon_0}{\pi Zme^2}$

here it is clear that radius of a orbit is directly proportional to mass of substance revolve around that orbit.

In the given muonic hydrogen atom, a negatively charged muon of mass 207$m_e$ revolves around a proton.

Therefore radius of electron and muon can be written as, $\frac{r_{muon}}{r_e}=\frac{m_e}{m_{muon}}$

= $\frac{m_e}{207m_e}$ = 1/207

so, $r_{muon}=\frac{r_e}{207}$

as we know, radius of first orbit is approximately, 0.53 A° = 5.3 × 10^-11 m

so, $r_{muon}$ = 5.3 × 10^-11/207 = 2.5 × 10^-13 m

similarly of electron in nth orbit is given by, $E=-\frac{me^4}{8\epsilon_0 n^2h^2}$

so, $\frac{E_{muon}}{E_e}=\frac{m_{muon}}{m_e}$

or, $E_{muon}=207E_e$

we know, energy of electron in 1st orbit is -13.eV

so, $E_{muon}=-207(13.6)$eV = -2.8KeV