Modern Physics JEE Main 2025 Solution

JEE MAIN 2025

22-01-2025 SHIFT-2

Question

A light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of 2 eV . If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of metal is 1 eV )

Select the correct option:

2 eV

6 eV

5 eV

3 eV

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & \frac{\mathrm{hc}}{\lambda}=\phi+\mathrm{eV} \Rightarrow \frac{\mathrm{hc}}{\lambda}=1+2=3 \mathrm{eV} \\ & \frac{\mathrm{hc}}{\lambda / 2}=6=1+\mathrm{k}_{\max } \therefore \mathrm{k}_{\max }=5 \mathrm{eV}\end{aligned}$

Question Tags

JEE Main

Physics

Easy

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