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?

Question

Philoid · Accepted Answer

Given: energy = 2.3 eV

In joules, Energy E = 2.3 × 1.6 × 10^-19 = 3.68 × 10^-19J

Energy is given as:

E = hv

Where h is Planck’s constant

v is frequency of the radiation

Or

On calculating, we get

Frequency v =0.55 × 10^-15

Or Frequency v = 5.55 × 10^-14 Hz.