Physics · Atomic Physics

JEE Main 2024 — 8 April, Shift 2 — Question 44

A proton and an electron have the same de Broglie wavelength. If $K_{p}$ and $K_{e}$ be the kinetic energies of proton and electron respectively. Then choose the correct relation :

Option A:
$K_{p}>K_{e}$
Option B:
$\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{e}}$
Option C:
$K_{p}=K_{e}^{2}$
Option D:
$\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}$
Correct

Answer: D

Step-by-step solution

De Broglie wavelength of proton & electron $=\lambda$

$\because \lambda=\frac{\mathrm{h}}{\mathrm{p}}$

$\therefore \mathrm{p}_{\text {proton }}=\mathrm{p}_{\text {electron }}$ $\because \mathrm{KE}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$

$\therefore \mathrm{KE}_{\text {proton }}<\mathrm{KE}_{\text {electron }}$

$\left[\mathrm{K}_{\mathrm{p}}<\mathrm{K}_{\mathrm{e}}\right]$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Physics
Chapter: Atomic Physics
Topic: Dual Nature of Matter