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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)/(16 pi^(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=1/2 mv^(2)=1/2 m. (h^(2))/(16m^(2)pi^(2)a_(0)^(2))=(h^(2))/(32mpi^(2) a_(0)^(2))`
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