Question

The kinetic energy of the $e^{-}$ in the $2^{nd}$ Bohr orbit of a hydrogen atom is:

(as in Bohr's radius)

Answer 1

Kinetic energy in $n^{th}$ orbit is,

$K_n=\frac{m{e}^4}{8{ϵ}_0^2{n}^2{h}^2}$

Now substituting values of $e,h,m,{ϵ}_0$ and $n=2$

${K.E}_2=3.40eV$