1.

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)`

Answer» Correct Answer - C
`mv(4a_0)=h/pi`
so, `v=h/(4mpia_0)`
so, `KE=1/2mv^2=1/2m.(h^2)/(16m^2pi^2a_0^2)=(h^2)/(32mpi^2 a_0^2)`


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