Physics · Atomic Physics

JEE Main 2024 — 6 April, Shift 2 — Question 43

When UV light of wavelength 300 nm is incident on the metal surface having work function 2.13 eV , electron emission takes place. The stopping potential is: $($ Given hc $=1240 \mathrm{eV} \mathrm{nm})$

Option A:
4 V
Option B:
4.1 V
Option C:
2V
Correct
Option D:
1.5V

Answer: C

Step-by-step solution

$\frac{\mathrm{hc}}{\lambda}-\phi=$ e.

$\mathrm{V}_{\mathrm{s}}$ $\Rightarrow \frac{1240}{300} \mathrm{eV}-2.13 \mathrm{eV}=\mathrm{eVs}$

$\Rightarrow 4.13 \mathrm{eV}-2.13 \mathrm{eV}=\mathrm{eVs}$

$\Rightarrow \mathrm{So}, \quad \mathrm{V}_{\mathrm{s}}=2 \mathrm{volt}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Physics
Chapter: Atomic Physics
Topic: Photoelectric Effect

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Answer: C

Step-by-step solution

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