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))/(4pi^(2) m a_(0)^(2))`
B. `(h^(2))/(16pi^(2) m a_(0)^(2))`
C. `(h^(2))/(32pi^(2) m a_(0)^(2))`
D. `(h^(2))/(64 pi^(2) m a_(0)^(2))`

Answer

C

Explanation

By Bohr's postulate, `mv r= (nh)/(2pi)`
`:. v = (nh)/(2pi mr)`
`K.E. = (1)/(2) mv^(2) = (1)/(2) m (n^(2) h^(2))/(4pi^(2) m^(2) r^(2))`..(i)
But `r = (a_(0) n^(2))/(Z)`
for 2nd Bohor orbit of hydrogen, `n = 2, Z = 1`
`:. r = (a_(0) xx 4)/(1) = 4a_(0)`
Substituting in eqn (i)
`K.E. = (1)/(2) (m(2)^(2) h^(2))/(4pi^(2) m^(2) (16a_(0)^(2))) = (h^(2))/(32pi^(2) ma_(0)^(2))`