JEE Main 2021PhysicsAtomic PhysicsMediumNumerical

JEE Main 2021 — Atomic Physics Question with Solution

JEE Main 2021 (27 Jul Shift 2)

Question

The Kα X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atom with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be keV. (Round off to the nearest integer ) h=4.14×10-15 eV s, c=3×108 m s-1

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Show full solutionCorrect answer: 10
Correct answer
10

Step-by-step explanation

Ekα=Ek-EL

hcλkα=Ek-EL

EL=Ek-hcλkα

=27.5 KeV-12.42×10-7eV m0.071×10-9 m

EL=(27.5-17.5) keV

=10 keV

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About this question

This is a previous-year question from JEE Main 2021, covering the Atomic Physics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.