QJEEM24Pha2cf: JEE Main 2024 Physics - Class 11 JEEM24Pha2cf Solution: The de-Broglie wavelength associated with a particle of mass…

To determine the dimensional formula for Planck's constant, we will start by analyzing the given de-Broglie wavelength equation:

$\lambda = \frac{h}{\sqrt{2 m E}}$

Here, $\lambda$ is the wavelength, $h$ is the Planck's constant, $m$ is the mass of the particle, and $E$ is the energy of the particle.

First, let's derive the dimensional formula for each term involved:

1. Wavelength $\lambda$ has the dimensional formula of length $[L]$ .

2. Mass $m$ has the dimensional formula $[\text{M}]$ .

3. Energy $E$ has the dimensional formula of work, which is force times distance:

$[E] = [F][L] = [\text{MLT}^{-2}][L] = [\text{ML}^{2}\text{T}^{-2}]$ .

Now, let's rewrite the equation in terms of the dimensions:

$[L] = \frac{[h]}{\sqrt{2 [\text{M}] [\text{ML}^{2}\text{T}^{-2}]}}$

Simplifying inside the square root:

$[L] = \frac{[h]}{\sqrt{2 [\text{M}] [\text{M}] [\text{L}^{2}\text{T}^{-2}]}}$

$[L] = \frac{[h]}{\sqrt{2 [\text{M}^{2}] [\text{L}^{2}\text{T}^{-2}]}}$

Since the constants like 2 do not affect the dimensional formula, we can simplify further:

$[L] = \frac{[h]}{[\text{M L T}^{-1}]}$

Cross multiplying to solve for the dimensional formula of $h$ :

$[h] = [L] [\text{M L T}^{-1}]$

$[h] = [\text{M L}^{2} \text{T}^{-1}]$

Therefore, the dimensional formula for Planck's constant $h$ is:

$\left[\text{ML}^{2}\text{T}^{-1}\right]$

Hence, the correct option is:

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