Question
The Bohr radius is given by $\text{a}_0=\frac{\in_0\text{h}^2}{\pi\text{me}^2}$ Verify that the RHS has dimensions of length.

Answer

$\text{a}_0=\frac{\in_0\text{h}^2}{\pi\text{me}^2}=\frac{\text{A}^2\text{T}^2\big(\text{ML}^2\text{T}^{-1}\big)^{2}}{\text{L}^2\text{ML}^{-2}\text{M(AT)}^2}$$\text{a}_0=\frac{\text{M}^2\text{L}^2\text{T}^{-2}}{\text{M}^{2}\text{L}^{3}\text{T}^{-2}}=\text{L}$
$\therefore a_0$ has the dimensions of length.

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