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))/(2pi^(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

A

Explanation

K.E. of electron `=1//2m upsilon^(2)=1//2mxx(n^(2)h^(2))/(4pi^(2)m^(2)r^(2))`
`K.E. =(n^(2)h^(2))/(8pi^(2)m a_(0)^(2))`
For second Bohr's orbit, n=2
K.E. = `(h^(2))/(2pi^(2)ma_(0)^(2))`