A free electron of 2.6eV energy collides with 
ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon 
A
B
C
D
Text Solution
Verified by ExpertsThe correct answer is:
C
For every large distance P.E.= 0
& total energy = 2.6 + 0 = 2.6eV
Finally, in first excited state of H atom total energy = -3.4eV
Loss in total energy = 2.6 - (-3.4) = 6eV
It is emitted as photon
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