The de-Broglie wavelength associated with a particle of mass m and energy E is h / √(2 m E) . The dimensional formula for Planck's constant is :

Physics · Units, Dimensions & Error Analysis

JEE Main 2024 — 9 April, Shift 2 — Question 39

The de-Broglie wavelength associated with a particle of mass $m$ and energy $E$ is $\mathrm{h} / \sqrt{2 m E}$ . The dimensional formula for Planck's constant is :

Option A:
$\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$
Option B:
$\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]$
Correct
Option C:
$\left[\mathrm{MLT}^{-2}\right]$
Option D:
$\left[\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]$

Answer: B

Step-by-step solution

$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ or $\mathrm{E}=\mathrm{h} v$

$\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]=\mathrm{h}\left[\mathrm{T}^{-1}\right]$

$\mathrm{h}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Units, Dimensions & Error Analysis
Topic: Units and Dimensions Analysis