Physics · Atomic Physics

JEE Main 2024 — 8 April, Shift 1 — Question 33

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is:

(Assume $\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 \times 10^{-31} \mathrm{~kg}$ and $\mathrm{m}_{\mathrm{p}}=1836$ times $\mathrm{m}_{\mathrm{e}}$ )

Option A:
$1: 1836$
Correct
Option B:
$1: \frac{1}{1836}$
Option C:
$1: \frac{1}{\sqrt{1836}}$
Option D:
$1: \sqrt{1836}$

Answer: A

Step-by-step solution

$\lambda$ is same for both

$\mathrm{P}=\frac{\mathrm{h}}{\lambda}$ same for both $\mathrm{P}=\sqrt{2 \mathrm{mK}}$

Hence,

$\mathrm{K} \propto \frac{1}{\mathrm{~m}}$

$\Rightarrow \frac{\mathrm{KE}_{\mathrm{p}}}{\mathrm{KE}_{\mathrm{e}}}=\frac{\mathrm{m}_{\mathrm{e}}}{\mathrm{m}_{\mathrm{p}}}=\frac{1}{1836}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Physics
Chapter: Atomic Physics
Topic: Dual Nature of Matter

Answer: A

Step-by-step solution

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