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Saturday, 31 October

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 λ1=496 nm\lambda_1=496\,\mathrm{nm} and λ2=620 nm\lambda_2=620\,\mathrm{nm} fall on a metal surface. Calculate the work function if the ratio of max speed of photoelectron is v1:v2=2:1v_1:v_2=\sqrt{2}:1 for corresponding λ1\lambda_1 and λ2\lambda_2 respectively. (Take hc=1240 eV−nmhc=1240\,\mathrm{eV-nm} )
  1. (1)0.5 eV
  2. (2)1.5 eVOfficial answer
  3. (3)1.8 eV
  4. (4)2.5 eV

Official answer

Option 2

NTA final key.