HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Physics

12: Atoms

Qstn# 4 Prvs-QstnNext-Qstn
  • #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?
    Ans : 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



    Hence, the frequency of the radiation is 5.6 × 1014 Hz.