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].

Mass of a negatively charged muon, mμ = 207m_e

According to Bohr’s model,

Bohr radius, re ∝ 1/ m_e

And, energy of a ground state electronic hydrogen atom, E_e ∝ m_e.

Also, the energy of a ground state muonic hydrogen atom, E_u ∝ m_u.

We have the value of the first Bohr orbit, r_e = 0.53 A = 0.53 x 10⁻¹⁰m

Let rμ,j be the radius of muonic hydrogen atom.

At equilibrium, we can write the relation as:

mμ rμ = m_e  r_e

207 m_{e x} rμ = m_e  r_e

Therefore, rμ = 0.53 x 10⁻¹⁰ /207 = 2.56 x 10⁻¹³ m

Hence, the value of the first Bohr radius of a muonic hydrogen atom is 2.56 x 10⁻¹³ m.

Hence, the value of the first Bohr radius of a muonic hydrogen atom is  2.56 x 10⁻¹³ m

We have ,

Ee = -13.6 eV

Take the ratio of these energies as:

Ee/Eμ = me/mμ = me/207me

Eμ = 207 Ee
= 207x(-13.6) = -2.8l keV
Hence, the ground state energy of a muonic hydrogen atom is -2.81 keV