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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 me orbits around a proton].
Solution
Mass of a negatively charged muon, mμ = 207 me
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 rμ 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,
Ee= − 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.
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