The dimensions of {e^2}/(4pi varepsilon _0 hc | vedclass

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Question

(A) The given expression is $\frac{e^2}{4\pi \varepsilon _0 hc}$.We know that the Coulomb force is given by $F = \frac{1}{4\pi \varepsilon _0} \frac{e

Vedclass · Accepted Answer

$[M^0 L^0 T^0]$

Vedclass · Answer

(A) The given expression is $\frac{e^2}{4\pi \varepsilon _0 hc}$.We know that the Coulomb force is given by $F = \frac{1}{4\pi \varepsilon _0} \frac{e^2}{r^2}$,so the dimensions of $\frac{e^2}{4\pi \varepsilon _0}$ are $[F][r^2] = [MLT^{-2}][L^2] = [ML^3T^{-2}]$.Planck's constant $h$ has dimensions $[ML^2T^{-1}]$.The velocity of light $c$ has dimensions $[LT^{-1}]$.Therefore,the dimensions of the expression are $\frac{[ML^3T^{-2}]}{[ML^2T^{-1}][LT^{-1}]} = \frac{[ML^3T^{-2}]}{[ML^3T^{-2}]} = [M^0 L^0 T^0]$.This quantity is known as the fine-structure constant,which is a dimensionless quantity.

The dimensions of ${e^2}/(4\pi \varepsilon _0 hc)$,where $e, \varepsilon _0, h,$ and $c$ are electronic charge,electric permittivity,Planck's constant,and velocity of light in vacuum respectively,are:

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