JEE Main 2021 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 4 August 2021, Shift 2, Physics Q3

Question 3 of 90 in this shift, Physics question 3 of 30, Section A.

Two wavelengths

\lambda_1=496\,\mathrm{nm}

and

\lambda_2=620\,\mathrm{nm}

fall on a metal surface. Calculate the work function if the ratio of max speed of photoelectron is

v_1:v_2=\sqrt{2}:1

for corresponding

\lambda_1

and

\lambda_2

respectively. (Take

hc=1240\,\mathrm{eV-nm}

)

(1)0.5 eV
(2)1.5 eVOfficial answer
(3)1.8 eV
(4)2.5 eV

Official answer

Option 2

NTA final key.

Chapter: Dual Nature of Radiation and Matter
Topic: Photoelectric Effect
Idea tested: Two-Wavelength System of Equations

Same idea in other shifts

Asked 7× in all

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25 Feb 2021, Shift 2 · Q13The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the…MediumSingle correct
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3 Sep 2020, Shift 1 · Q18When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.