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Friday, 9 October

JEE Main 2026 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 6 April 2026, Shift 2, Physics Q47

Question 47 of 75 in this shift, Physics question 22 of 25, Section B.

The de Broglie wavelength for an electron accelerated through the potential difference of V1V_1 volt is λ1\lambda_1. When the potential difference is changed to V2V_2 volt, the associated de Broglie wavelength is increased by 50%. If (V1/V2)=(9/α)(V_1/V_2)=(9/\alpha), then the value of α\alpha is __________.

Official answer

4

NTA final key.

Chapter
Dual Nature of Radiation and Matter
Topic
Matter Waves
Idea tested
Wavelength from Accelerating Potential

Same idea in other shifts

Asked 2× in all
  1. 28 Jul 2022, Shift 1 · Q47The equation λ=1.227x\lambda=\frac{1.227}{x} nm can be used to find the de-Brogli wavelength of an electron. In this equation xx stands for :…EasySingle 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 42 of the 174 JEE Main shifts held from 2017 to 2026.