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))/(4pima_(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

C

Explanation

Nuclear attraction on the electron=Centrifugal force
`(kZe^(2))/(r^(2))=(mv^2)/(r)`
`v^(2)=(kZe^(2))/(mr)` . . . .(i)
Angular momentum, `mvr=n(h)/(2pi)`
`v^(2)=(n^(2)h^(2))/(4pi^(2)m^(2)r^(2))` . . . (ii)
Comparing (i) and (ii)
`(kZe^(2))/(mr)=(n^(2)h^(2))/(4pi^(2)m^(2)r^(2))`
`r=(n^(2))/(Z)[(h^(2))/(4pimke^(2))]=(n^(2)a_(0))/(Z)`
`thereforev_(n)=(hZ)/(2pia_(0)n)`
`KE=(1)/(2)mv_(n)^(2)=(h^(2)Z^(2))/(8pi^(2)ma_(0)^(2)n^(2))`
KE (for n=2,Z=1) `=(h^2)/(32pi^(2)ma_(0)^(2))`.