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].
Given: Mass of negatively charged muon mμ– = 207me
The Bohr’s radius is given by the relation:
re ∝ (1/mc)
Energy of ground state electronic hydrogen atom Ee∝ m
The value of the first Bohr orbit is given as re = 0.53A
Converting into metres, re =0.53 × 10-10m
Let us consider rμ is the radius of muonic hydrogen atom
At equilibrium
mμ rμ = me re
207me × rμ = me re
rμ = (0.53 × 10-10m)/207
rμ =2.56 × 10-13m
The ratio of the energies is given as:
Ee/Eμ = me/mμ = me/207me
We get
Eμ = 207EE
Substituting the value of energy, we get
Eμ = 207 × (-13.6)
Eμ =-2.81 keV