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Question

(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mE}}$.Rearranging for $h$,we get $h = \lambda \sqrt{2mE}$.The dimensional formula

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$[ML^2 T^{-1}]$

Vedclass · Answer

(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mE}}$.Rearranging for $h$,we get $h = \lambda \sqrt{2mE}$.The dimensional formula for wavelength $\lambda$ is $[L]$.The dimensional formula for mass $m$ is $[M]$.The dimensional formula for energy $E$ is $[ML^2 T^{-2}]$.Substituting these into the expression for $h$:$[h] = [L] \cdot \sqrt{[M] \cdot [ML^2 T^{-2}]}$$[h] = [L] \cdot \sqrt{[M^2 L^2 T^{-2}]}$$[h] = [L] \cdot [MLT^{-1}]$$[h] = [ML^2 T^{-1}]$.Alternatively,using $E = h
u$,where $
u$ is frequency $[T^{-1}]$:$[h] = [E] / [
u] = [ML^2 T^{-2}] / [T^{-1}] = [ML^2 T^{-1}]$.

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 $h$ is:

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