Q78 / 90 · JEE Main 2024 (09 Apr Shift 2) · Physics

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

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

Solution

$\begin{aligned} & \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \text { or } \mathrm{E}=\mathrm{h} \nu \\ & {\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]=\mathrm{h}\left[\mathrm{T}^{-1}\right]} \\ & \mathrm{h}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]\end{aligned}$

Hence, the correct answer is option: 'A'.

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