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 207 m_e orbits around a proton].

Solution

Mass of muon,

M_{μ} = 207 m_{e}

The first Bohr's radius of Hydrogen atom is given by,

r_{1} = 4 {πε}_{0} \frac{h^{2}}{4 π^{2} m_{e} e^{2}} = 5.29 \times 10^{- 11} m

If r₁' is the first Bohr's radius of muonic hydrogen atom, then

r_{1}' = 4 {πε}_{0} \frac{h^{2}}{4 π^{2} (207 m_{e}) e^{2}} = \frac{5.29 \times 10^{- 11}}{207} = 2.5 \times 10^{- 12} m

The ground state (n = 1) energy of H-atom is given by,

E_{1} = - {(\frac{1}{4 {πε}_{0}})}^{2} . \frac{2 π^{2} m_{e} e^{2}}{h^{2}} = 13.6 eV

If E₁' is ground state energy of muonic hydrogen atom, then we have

E_{1}' = - {(\frac{1}{4 {πε}_{0}})}^{2} . \frac{2 π^{2} (207 m_{e}) e^{2}}{h^{2}} = - 13.6 \times 207 = - 2815.2 eV .

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