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?
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.
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