Physics · Atomic Physics

JEE Main 2025 — 23 January, Evening Shift — Question 53

In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2 V , what is the wavelength of the em-wave?

(Given hc $=1242 \mathrm{eVnm}$ where h is the Planck's constant and c is the speed of light in vaccum.)

(1) 400 nm

Option A:
400nm
Option B:
600nm
Option C:
200nm
Option D:
300nm
Correct

Answer: D

Step-by-step solution

$\mathrm{eV}_{\mathrm{s}}=\mathrm{E}-\phi$

$2 \mathrm{eV}=\mathrm{E}-2.14 \mathrm{eV}$

$\mathrm{E}=4.14 \mathrm{eV}$

$E=\frac{h c}{\lambda}$

$\lambda=\frac{1242}{4.14}=300 \mathrm{~nm}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Photoelectric Effect