Question

How much will the kinetic energy and total energy of an electron in H atom change if the atom emits a photon of wavelength $4862\mathring{A}$?

Answer 1

Energy of photon $=\frac{hc}{\lambda }=\frac{2\times {10}^{-25}}{4862\times {10}^{-10}}=.00041\times {10}^{-}15J=2.553ev$

This energy will change in the kinetic energy,

Now $T.E.=-(K.E.)=-2.553ev$