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 :

(B) $\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]$

(A) $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$

(C) $\left[\mathrm{MLT}^{-2}\right]$

(D) $\left[\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]$

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Modern Physics

The de-Broglie wavelength associated with a particle of mass

m

and energy

E

h / 2 m E

. The dimensional formula for Planck's constant is :

(A) $[ML^{- 1} T^{- 2}]$
(B) $[ML^{2} T^{- 1}]$ ✓ Correct
(C) $[MLT^{- 2}]$
(D) $[M^{2} L^{2} T^{- 2}]$

Answer: (B)

Solution

From

λ = \frac{h}{2 m E}

h = λ 2 m E

[λ] = [L]

[2 m E] = [M] [M L^{2} T^{- 2}] = [M^{2} L^{2} T^{- 2}]

, so

[2 m E] = [M L T^{- 1}]

.

Therefore

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

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option

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