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NCERT Class XII Chapter
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Question : 17
Total: 18
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:  
In Bohr’s model, the radius of nth orbit, r =
n2h2ε0
πZme2
In the given muonic hydrogen atom, a negatively charged muon (µ–) of mass 207 me revolve around a proton.
Therefore radius of electron and muon can be written as
rµ
re
 =
me
mµ
 =
me
207me
⇒ rµ =
re
207
 =
0.53×10−10
207
 m = 2.5 × 10−12 m
Energy of electron in the orbit, E = -
me4
8ε0n2h2
 ⇒
Ee
Eµ
 =
me
mµ
 =
me
207me
Eµ = 207 Ee = 207 [- 13.6 eV] = - 2.8 keV
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