JEE Main 2021ChemistryStructure of AtomMediumNumerical

JEE Main 2021 — Structure of Atom Question with Solution

JEE Main 2021 (27 Aug Shift 2)

Question

The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x×1013. The value of x is (Nearest integer) h=6.63×10-34Js, c=3.00×108 ms-1:

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Show full solutionCorrect answer: 50
Correct answer
50

Step-by-step explanation

Energy emitted in 0.1sec

=0.1 sec×10-3 J s-1

=10-4 J

If 'n' photons of λ=1000 nm are emitted, then 10-4=nhcλ

10-4=n×6.63×10-34×3×1081000×10-9

n=5.02×1014=50.2×1013

n=50 (nearest integer)

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About this question

This is a previous-year question from JEE Main 2021, covering the Structure of Atom chapter of Chemistry. 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.