If e is the electronic charge,c is the speed | vedclass

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(A) The Coulomb force between two charges $e$ separated by distance $r$ is given by $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$.From thi

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

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(A) The Coulomb force between two charges $e$ separated by distance $r$ is given by $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$.From this,the dimension of $\frac{e^{2}}{4 \pi \varepsilon_{0}}$ is $[F r^{2}] = [M L T^{-2} \cdot L^{2}] = [M L^{3} T^{-2}]$.The energy of a photon is $E = \frac{hc}{\lambda}$,so the dimension of $hc$ is $[E \lambda] = [M L^{2} T^{-2} \cdot L] = [M L^{3} T^{-2}]$.Dividing these two quantities,the dimensions are $\frac{[M L^{3} T^{-2}]}{[M L^{3} T^{-2}]} = [M^{0} L^{0} T^{0}]$.Thus,the quantity is dimensionless.

If $e$ is the electronic charge,$c$ is the speed of light in free space,and $h$ is Planck's constant,the quantity $\frac{1}{4 \pi \varepsilon_{0}} \frac{| e |^{2}}{h c}$ has dimensions of .......

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