A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

Separation of two energy levels in an atom,

E = 2.3 eV = 2.3 x 1.6 x 10⁻¹⁹ = 3.68 x 10⁻¹⁹J

Let v be the frequency of radiation emitted when the atom transits from the upper level to the lower level. We have the relation for energy as: E = hv

Where, h = Plank’s constant = 6.62 x 10⁻³⁴ Js

Therefore , ν= E/h = (3.68 x 10⁻¹⁹)/( 6.62 x 10⁻³⁴) = 5.55 x 10¹⁴ Hz
Hence, the frequency of the radiation is 5.6 x 10¹⁴ Hz.