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A free electron of 2.6 eV energy collide...

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 photon is relased .Find the frequency of the emitted photon `(h=6.6xx10^(-34))`Js

A

`9.0 xx 10^(27)` MHz

B

`1.45 xx 10^(9)`MHz

C

`0.19 xx 10^(15)` MHz

D

`1.45 xx 10^(16)`MHz

Text Solution

Verified by Experts

The correct Answer is:
B
By energy conservation
`K.E_(e) =T.E_("H-atom") +E_("photon")`
`2.6 = (-13.6)/4 +hv`
hv = 6eV
`v = (6xx1.6 xx10^(-19))/(6.626xx10^(-34))`
` v = 1.45 xx 10^(15) " Hz " = 1.45 xx 10^(9) `MHz .
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