SELECT * FROM question_mgmt as q WHERE id=2765 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=8 AND chapterId=126 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

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?

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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