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 207 m_e orbits around a proton].

Answer 1

Mass of a negatively charged muon, m_μ = 207 m_e 

According to Bohr’s model,

Bohr radius, `"r"_"e" prop (1/"m"_"e")`

And, energy of a ground state electronic hydrogen atom, `"E"_"e" prop "m"_"e"`.

Also energy of a ground state muonic hydrogen atom `"E"_μ prop "m"_μ`.

We have the value of the first Bohr orbit, `"r"_"e"` = 0.53 Å = 0.53 × 10⁻¹⁰ m

Let r_μ be the radius of muonic hydrogen atom.

At equilibrium, we can write the relation as:

`"m"_μ "r"_μ = "m"_"e""r"_"e"`

`207  "m"_"e" xx "r"_μ = "m"_"e""r"_"e"`

∴ `"r"_μ = (0.53 xx 10^(-10))/207`

= 2.56 × 10⁻¹³ m

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

We have,

E_e= − 13.6 eV

Take the ratio of these energies as:

`"E"_"e"/"E"_μ = "m"_"e"/"m"_μ = "m"_"e"/(207  "m"_"e")`

`"E"_μ = 207  "E"_"e"`

= `207 × (−13.6)`

= −2.81 keV

Hence, the ground state energy of a muonic hydrogen atom is −2.81 keV.