The dimensional formula of a physical quantit | vedclass

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(B) The expression given is $\frac{e^2}{4 \pi \varepsilon_0 \hbar}$.We know that the Coulomb force is $F = \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^

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$[L^1 T^{-1}]$

Vedclass · Answer

(B) The expression given is $\frac{e^2}{4 \pi \varepsilon_0 \hbar}$.We know that the Coulomb force is $F = \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}$,so $\frac{e^2}{4 \pi \varepsilon_0} = F r^2$.The dimensions of $\frac{e^2}{4 \pi \varepsilon_0}$ are $[M L T^{-2}] [L^2] = [M L^3 T^{-2}]$.The dimensions of Planck's constant $h$ (or $\hbar$) are $[M L^2 T^{-1}]$.Therefore,the dimensions of the expression are $\frac{[M L^3 T^{-2}]}{[M L^2 T^{-1}]} = [L^1 T^{-1}]$.This represents the dimensions of velocity (speed of light $c$).

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