JEE Main 2023PhysicsAtomic PhysicsMediumNumerical

JEE Main 2023 — Atomic Physics Question with Solution

JEE Main 2023 (13 Apr Shift 2)

Question

An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 600 nm. The net energy absorbed by the atom in this process is n×10-4eV. The value of n is [Assume the atom to be stationary during the absorption and emission process] (Take h=6.6×10-34J s and c=3×108 m s-1 ).

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Show full solutionCorrect answer: 4125
Correct answer
4125

Step-by-step explanation

The energy of a photon is given by E=hcλ.

It is given that λ1=500 nm, λ2=600 nm.

The net energy absorbed is 

ΔE=hcλ1-hcλ2=hc10-9(1500-1600)

=6.6×10-34×3×108×100500×600×10-9

=6.6×330×10-19 J

1 eV=1.6×10-19 J

=6.6×330×1.6eV

=4125×10-4 eV

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About this question

This is a previous-year question from JEE Main 2023, 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.