Physics · Atomic Physics

JEE Main 2026 — 2 April, Morning Shift — Question 12

For a certain metal, when monochromatic light of wavelength $\lambda$ is incident, the stopping potential for photoelectrons is $3\mathrm{V}_{\mathrm{o}}$ . When the same metal is illuminated by light of wavelength $2\lambda$ , then the stopping potential becomes $\mathrm{V}_{\mathrm{o}}$ . The threshold wavelength for photoelectric emission for the given metal is $\alpha \lambda$ . The value of $\alpha$ is

Option A:
1
Option B:
4
Correct
Option C:
2
Option D:
3

Answer: B

Step-by-step solution

$\frac{\mathrm{hc}}{\lambda} -\phi = 3\mathrm{eV}_{0}$ , $\frac{\mathrm{hc}}{2\lambda} -\phi = \mathrm{eV}_{0}$ → $\phi = \frac{\mathrm{hc}}{4\lambda}$ → $\lambda_0 = 4\lambda$

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Exam: JEE Main 2026
Paper: 2 April, Morning Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Photoelectric Effect

Answer: B

Step-by-step solution

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