The dimensional formula of a physical quantity represented by $\frac{\mathrm{e}^2}{4 \varepsilon_0 \mathrm{~h}}$ is [e is the charge of electron, $\varepsilon_0$ is the permittivity of free space, and $\mathrm{h}$ is the Planck's constant]

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Aicharya · Accepted Answer

$\mathrm{e}^2=[\mathrm{q}]^2=[\mathrm{AT}]^2=\left[\mathrm{A}^2 \mathrm{~T}^2ight]$ $\begin{aligned} & \mathrm{h}=\frac{\mathrm{E}}{\mathrm{v}}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]}{\left[\mathrm{T}^{-1}ight]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}ight] \ & \epsilon_0=\frac{1}{4 \pi \mathrm{F}} \frac{\mathrm{q}^2}{\mathrm{r}^2}=\frac{\left[\mathrm{A}^2 \mathrm{~T}^2ight]}{\left[\mathrm{MLT}^{-2}ight]\left[\mathrm{L}^2ight]}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2ight] \ & \frac{\mathrm{e}^2}{4 \epsilon_0 \mathrm{~h}}=\frac{\left[\mathrm{A}^2 \mathrm{~T}^2ight]}{\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2ight]\left[\mathrm{ML}^2 \mathrm{~T}^{-1}ight]}=\left[\mathrm{L} \mathrm{T}^{-1}ight]\end{aligned}$

The dimensional formula of a physical quantity represented by $\frac{\mathrm{e}^2}{4 \varepsilon_0…

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