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 ma_0^2)`
B. ` h^2/( 16pi ^2 ma_0^2)`
C. ` h^2/( 32 pi ^2 ma_0^2)`
D. ` h^2/(6 4 pi ^2 ma_0^2)`

Answer

C

Explanation

For Bohr orbit .
` K.E, = 1/2 mu^2`
and `mur_n = (nh)/(2pi) ro u = (nh)/( 2pi mr_n)`
` :. K.. 1/2 m = (n^2 h^2 )/( 4pi^2 m^2 r_n^2) = (h^2 h^2)/(8 pi^2 mr_n^2)`
For (II) Bohr orbit , n=2
`:. r_n = a_0 xx n^2 =4a_0 (a_0 = "Bohr radius" )`
Now ` K.E. = ((2)^2 h^2)/(( 8 ppi^2 m 9 4 a_0)^2) = h^2/(32 pi^2 ma_0^2)`.