Question

Obtain the first Bohr’s radius and the ground state energy of amuonic hydrogen atom [i.e., an atom in which a negatively chargedmuon (μ–) of mass about 207me orbits around a proton].

Answer 1

Given: The mass of a negatively charged muon is 207  m e .

The value of first Bohr orbit is 0.53× 10 −10  m.

The radius of muonic hydrogen atom at equilibrium is given as,

r μ = m e r e m μ

Where, the radius of first Bohr orbit is r e and the mass is m e , the radius of muonic hydrogen atom is r μ and the mass of a negatively charged muon is m μ

By substituting the given values in the above formula, we get

r μ = 0.53× 10 −10 m e 207 m μ = 0.53× 10 −10 207 =2.56× 10 −13 m

We know that in hydrogen atom the ground state energy of electron is E e =−13.6 eV.

The ratio of energy between electron and muon is given as,

E e E μ = m e m μ

Where, the energy of electron is E e and the energy of muon is E μ .

By substituting the given values in the above formula, we get

( −13.6 ) E μ = m e 207 m e E μ =207( −13.6 ) =−2.81 keV

Thus, the value of the first Bohr radius of a muonic atom is 2.56× 10 −13  m and the ground state energy of a muonic hydrogen atom is −2.81 keV.