Physics · Atomic Physics

JEE Main 2025 — 28 January, Morning Shift — Question 64

A proton of mass ' $m_{p}$ ' has same energy as that of a photon of wavelength ' $\lambda$ '. If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

Option A:
$\frac{1}{c} \sqrt{\frac{2 E}{m_{p}}}$
Option B:
$\frac{1}{c} \sqrt{\frac{E}{m_{p}}}$
Option C:
$\frac{1}{c} \sqrt{\frac{E}{2 m_{p}}}$
Correct
Option D:
$\frac{1}{2 c} \sqrt{\frac{E}{m_{p}}}$

Answer: C

Step-by-step solution

E is missing in the question but considering E as energy,

the solution will be

$\mathrm{E}_{\text {phooon }}=\frac{\mathrm{hc}}{\lambda}=\mathrm{E} ; \mathrm{E}_{\text {proton }}=\frac{1}{2} \mathrm{~m}_{\mathrm{p}} \mathrm{v}^{2}=\mathrm{E}$

$\frac{\lambda_{\text {proton }}}{\lambda_{\text {photon }}}=\frac{\mathrm{h} / \mathrm{p}}{\mathrm{hc} / \mathrm{E}}=\frac{\mathrm{h} / \sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{E}}}{\mathrm{hc} / \mathrm{E}}$

$=\frac{E}{c \sqrt{2 m_{p} E}}$

$\frac{\lambda_{\text {proton }}}{\lambda_{\text {photon }}}=\frac{1}{c} \sqrt{\frac{\mathrm{E}}{2 \mathrm{~m}_{\mathrm{p}}}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Morning Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Dual Nature of Matter