Question 4
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?
Answer
Separation of two energy levels in an atom, E = 2.3 eV = 2.3 × 1.6 × 10 −19 = 3.68 × 10 −19 J
Let ν 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 = Planck’s constant = 6.62 x 10-34 Js
∴ V = E/h = 3.68 x 10-19 / 6.62 x 10-32 = 5.55 x 1014 Hz
Hence, the frequency of the radiation is 5.6 × 10 14 Hz.
