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Friday, 9 October

JEE Main 2025 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 7 April 2025, Shift 2, Physics Q43

Question 43 of 75 in this shift, Physics question 18 of 25, Section A.

A photoemissive substance is illuminated with a radiation of wavelength λi\lambda_i so that it releases electrons with de-Broglie wavelength λe\lambda_e. The longest wavelength of radiation that can emit photoelectron is λo\lambda_o. Expression for de-Broglie wavelength is given by : (mm : mass of the electron, hh : Planck's constant and cc : speed of light)
  1. (1)λe=hλi2mc\lambda_e=\sqrt{\frac{h\lambda_i}{2mc}}
  2. (2)λe=hλo2mc\lambda_e=\sqrt{\frac{h\lambda_o}{2mc}}
  3. (3)λe=h2mc(1λi−1λo)\lambda_e=\frac{h}{\sqrt{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}
  4. (4)λe=h2mc(1λi−1λo)\lambda_e=\sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}Official answer

Official answer

Option 4

NTA final key.