JEE Main 2021 · PhysicsMultiple choiceSingle correctMediumConceptual

JEE Main 17 March 2021, Shift 2, Physics Q14

Question 14 of 90 in this shift, Physics question 14 of 30, Section A.

Two identical photocathodes receive the light of frequencies

f_1

and

f_2

respectively. If the velocities of the photo-electrons coming out are

v_1

and

v_2

respectively, then

(1) $v_1-v_2=\left[\frac{2\mathrm{h}}{\mathrm{m}}(f_1-f_2)\right]^{\frac{1}{2}}$
(2) $v_1^2-v_2^2=\frac{2\mathrm{h}}{\mathrm{m}}[f_1-f_2]$ Official answer
(3) $v_1+v_2=\left[\frac{2\mathrm{h}}{\mathrm{m}}(f_1+f_2)\right]^{\frac{1}{2}}$
(4) $v_1^2+v_2^2=\frac{2\mathrm{h}}{\mathrm{m}}[f_1+f_2]$

Official answer

Option 2

NTA final key.

Chapter: Dual Nature of Radiation and Matter
Topic: Photoelectric Effect
Idea tested: Einstein's Photoelectric Equation

Same idea in other shifts

Asked 14× in all

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.