Physics · Atomic Physics

JEE Main 2024 — 9 April, Shift 1 — Question 31

A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as:

Option A:
$\lambda_{\mathrm{e}}>\lambda_{\alpha}>\lambda_{\mathrm{p}}$
Option B:
$\lambda_{\alpha}<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}$
Correct
Option C:
$\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}<\lambda_{\alpha}$
Option D:
$\lambda_{\mathrm{p}}>\lambda_{\mathrm{e}}>\lambda_{\alpha}$

Answer: B

Step-by-step solution

$\lambda_{\mathrm{DB}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}$ $\Rightarrow \lambda_{\mathrm{DB}} \alpha \frac{1}{\sqrt{\mathrm{~m}}}$ $\Rightarrow \lambda_{\mathrm{a}}<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}$

Answer key and solution verified before publishing.

Practise Atomic Physics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Physics
Chapter: Atomic Physics
Topic: Dual Nature of Matter