A free electron of 2 . 6   eV energy collides with a H + ion. This results in the formation of a…

A free electron of 2.6 eV energy collides with a H+ 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. h=6.6×10-34 J s
  1. 9.0×1027 MHz
  2. 1.45×109 MHz
  3. 0.19×1015 MHz
  4. 1.45×1016 MHz

Solution

By Energy conservation
K.E.e=T.E.H+ +EPhoton
2.6=-13.64+hf
hf=6eV
f=6×1.6×10-196.626×10-34
f=1.45×1015 Hz=1.45×109 MHz

Asked in: JEE Main 2021 (31 Aug Shift 2)

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