Physics · Atomic Physics

JEE Main 2025 — 2 April, Evening Shift — Question 57

An electron with mass ' $m$ ' with an initial velocity $(t=0) \vec{v}=v_{0} \hat{i}\left(v_{0}>0\right)$ enters a magnetic field $\vec{B}=B_{0} \hat{j}$ . If the initial de-Broglie wavelength at $t=0$ is $\lambda_{0}$ then its value after time ' $t$ ' would be:

Option A:
$\frac{\lambda_{0}}{\sqrt{1-\frac{e^{2} B_{0}^{2} t^{2}}{m^{2}}}}$
Option B:
$\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2} B_{0}^{2} t^{2}}{m^{2}}}}$
Option C:
$\lambda_{0} \sqrt{1+\frac{e^{2} B_{0}^{2} t^{2}}{m^{2}}}$
Option D:
$\lambda_{0}$
Correct

Answer: D

Step-by-step solution

The speed of the electron will not change in presence of magnetic field.

$\therefore \lambda=\lambda_{0}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Dual Nature of Matter

Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of Li^++ ion in its ground state is given by 1/X a 0 , where…

Answer: D

Step-by-step solution

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