Physics · Atomic Physics

JEE Main 2026 — 5 April, Evening Shift — Question 17

An electron is travelling with a velocity $v$ in free space and when it enters a medium, its velocity is reduced by $20\%$ . The de Broglie wavelength of electron in the medium is $\alpha \lambda_0$ , where $\lambda_0$ is its de Broglie wavelength in free space. The value of $\alpha$ is

Option A:
1.2
Option B:
1
Option C:
1.25
Correct
Option D:
0.75

Answer: C

Step-by-step solution

$\lambda = h/(mv)$ . Velocity becomes $0.8v$ , so $\lambda_{\text{medium}} = h/(m\cdot0.8v) = (1/0.8)\lambda_0 = 1.25\lambda_0$ . Thus $\alpha = 1.25$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Dual Nature of Matter