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The kinetic energy of an electron in the...

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [`a_0` is Bohr radius] :

A

`h^2/(4pi^2 ma_0^2)`

B

`h^2/(16pi^2 ma_0^2)`

C

`h^2/(32pi^2 ma_0^2)`

D

`h^2/(64pi^2 ma_0^2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`mv(4a_0)=h/pi`
so `v =(h)/(4mpia_0)`
so , `KE=1/2mv^2 =1/2 m.h^2/(16m^2pi^2a_0^2)=h^2/(32mpi^2a_0^2)`
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