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

Options

A. `(h^2)/(4 pi^2 m a_0^2)`
B. `(h^2)/(16 pi^2 m a_0^2)`
C. `(h^2)/(32 pi^2 m a_0^2)`
D. `(h^2)/(64 pi^2 m a_0^2)`

Answer

C

Explanation

For Bohr orbit, angular momentum is
`mvr_n = (nh)/(2pi) " or " v = (nh)/(2 pi m r_n)`
Kinetic energy , KE = `1/2 mv^2`
or K.E. = `1/2 m xx (n^2h^2)/(4pi^2 m^2 r_n^2) = (n^2 h^2)/(8 pi^2 m r_n^2)`
For `n = 2, r_n = a_0 xx n_2 = 4a_0 (a_0` = Bohr radius)
K.E. = `(2^2h^2)/(8 pi^2 m(4a_0)^(2)) = (h^2)/(32 pi^2 ma_0^2)`.