Physics · Atomic Physics

JEE Main 2024 — 29 January, Shift 1 — Question 41

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $25 \%$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

Option A:
$\frac{1}{1}$
Option B:
$\frac{1}{8}$
Correct
Option C:
$\frac{8}{1}$
Option D:
$\frac{1}{4}$

Answer: B

Step-by-step solution

For photon $\mathrm{E}_{\mathrm{P}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}} \Rightarrow \lambda_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{P}}}$

For electron $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}=\frac{h \mathrm{v}_{\mathrm{e}}}{2 \mathrm{~K}_{\mathrm{e}}}$

Given $\mathrm{v}_{\mathrm{e}}=0.25 \mathrm{c}$

$\lambda_{\mathrm{e}}=\frac{\mathrm{h} \times 0.25 \mathrm{c}}{2 \mathrm{~K}_{\mathrm{e}}}=\frac{\mathrm{hc}}{8 \mathrm{~K}_{\mathrm{e}}}$

Also $\lambda_{p}=\lambda_{e}$

$\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{p}}}=\frac{\mathrm{hc}}{8 \mathrm{~K}_{\mathrm{e}}}$

$\frac{\mathrm{K}_{\mathrm{e}}}{\mathrm{E}_{\mathrm{p}}}=\frac{1}{8}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Atomic Physics
Topic: Dual Nature of Matter

When a hydrogen atom going from n=2 to n=1 emits a photon, its recoil speed is frac X5 m / s . Where x= (Use : mass of hydrogen atom .=1.6…

Answer: B

Step-by-step solution

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