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12.4 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?

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    Answered by

    Vishal Baghel | Contributor-Level 10

    7 months ago

    12.4 Separation of two energy level of atom, E = 2.3 eV = 2.3 * 1.6 * 10 - 19 J = 3.68 * 10 - 19

    Let ? be the frequency of radiation emitted when the atom transits from upper level to lower level.

    We have the relation for energy as E = h ?  , where

    h = Planck's constant = 6.626 * 10 - 34 Js

    Then ? = E h = 3.68 * 10 - 19 6.626 * 10 - 34 Hz = 5.55 * 10 14 Hz

    Hence the frequency is 5.55 * 10 14 Hz

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