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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.

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