A difference of 2 . 3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

The Separation of two energy levels in an atom, E = 2 . 3 eV = 2 . 3 × 1 . 6 × 10 - 19 = 3 . 68 × 10 - 19 J The relation between energy and frequency is given by, E = h v Where, Planck’s constant h = 6 . 62 × 10 - 34 Js , and, ν is the frequency of radiation. ∴ ν = E ⁄ h = 3 . 68 × 10 - 19 6 . 62 × 10 - 34 = 5 . 55 × 10 14 Hz The radiation frequency is 5 . 6 × 10 14 Hz . Here, the symbols have their usual meaning.