Question:
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?
Answer:
Separation of two energy levels in an atom
E = 2.3 eV = 2.3 x 1.6 x 10-19 = 3.68 x 10-19 J
let AZA 1/2 be the frequency of radiation emitted when the atom transits from the upper level to the lower level
So, E = hV where h = 6.62 x 10-34 Js
V = E/h = 3.68 x 10-19/ 6.62 x 10-34 = 5.55 x 1014 Hz
hence, the frequency of the radiation is 5.55 x 1014 Hz
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