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Question
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?
Solution
Separation of two energy levels in an atom,
E = 2.3 eV
= 2.3 × 1.6 × 10−19
= 3.68 × 10−19 J
Let v 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 × 10−34 Js
∴ v = `"E"/"h"`
= `(3.68 xx 10^(-19))/(6.62 xx 10^(-32))`
= 5.55 × 1014 Hz
Hence, the frequency of the radiation is 5.6 × 1014 Hz.
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