A photo emissive substance... |JEE Main 2023| Modern Physics

JEE MAIN 2025

07-04-2025 S2

Question

A photo-emissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_e$. The longest wavelength of radiation that can emit photoelectron is $\lambda_0$. Expression for de-Broglie wavelength is given by : $(\mathrm{m}$ : mass of the electron, h : Planck's constant and c. speed of light)

Select the correct option:

$\lambda_e=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mc}\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}$

$\lambda_e=\sqrt{\frac{\mathrm{h} \lambda_i}{2 \mathrm{mc}}}$

$\lambda_e=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mc}\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}$

$\lambda_e=\sqrt{\frac{\mathrm{h} \lambda_o}{2 \mathrm{mc}}}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution