JEE Main 2021 — Structure Of Atom Question with Solution
From: JEE Main 2021 (Online) 1st September Evening Shift
Question
A 50 watt bulb emits monochromatic red light of wavelength of 795 nm. The number of photons emitted per second by the bulb is x 1020. The value of x is __________. [Given : h = 6.63 1034 Js and c = 3.0 108 ms1]
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Show full solutionCorrect answer: 2
Correct answer
2
Step-by-step explanation
Energy of photon is given as .... (i)
where, E = energy of photon (50 W),
n = number of photon
h = Planck's constant (6.63 1034 Js)
c = speed of light (3 108 m/s)
= wavelength of light (795 109 m)
E = 50W = 50 J = energy of photon
50 J =
2 1020
x = 2
where, E = energy of photon (50 W),
n = number of photon
h = Planck's constant (6.63 1034 Js)
c = speed of light (3 108 m/s)
= wavelength of light (795 109 m)
E = 50W = 50 J = energy of photon
50 J =
2 1020
x = 2
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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.