Q 11-01-118JEE MainJEE Main 2024 (9 Apr, Shift 2)Easy
The de-Broglie wavelength associated with a particle of mass $m$ and energy $E$ is $h/\sqrt{2mE}$. The dimensional formula for Planck's constant is:
Answer: (A) $[\text{ML}^2\text{T}^{-1}]$
$h = \lambda\sqrt{2mE}$:
$$[h] = [\text{L}]\left([\text{M}][\text{ML}^2\text{T}^{-2}]\right)^{1/2} = [\text{L}][\text{ML}\text{T}^{-1}] = [\text{ML}^2\text{T}^{-1}]$$
Solution by Sreeraj P, M.Sc Physics