Dimension of Bohr Radius

Theorem

The Bohr radius has the dimension $\mathsf L$.

Proof

By definition, the Bohr radius is:

$a_0 = \dfrac {\varepsilon_0 h^2} {\pi \E^2 m_\E}$

where:

$\varepsilon_0$ denotes the vacuum permittivity

$h$ denotes Planck's constant

$\E$ denotes the elementary charge

$m_\E$ denotes the electron rest mass

We have:

\(\ds \varepsilon_0\)

\(\text {has dimension}\)

\(\ds \mathsf {M^{-1} L^{-3} T^4 I^2}\)

Definition of Vacuum Permittivity

\(\ds h\)

\(\text {has dimension}\)

\(\ds \mathsf {M L^2 T^{-1} }\)

Definition of Planck's Constant

\(\ds \E\)

\(\text {has dimension}\)

\(\ds \mathsf {I T}\)

Definition of Elementary Charge

\(\ds m_\E\)

\(\text {has dimension}\)

\(\ds \mathsf M\)

Definition of Mass of Electron

\(\ds \leadsto \ \ \)

\(\ds a_0\)

\(\text {has dimension}\)

\(\ds \dfrac {\mathsf {M^{-1} L^{-3} T^4 I^2} \cdot \paren {\mathsf {M L^2 T^{-1} } }^2} {\paren {\mathsf {I T} }^2 \cdot \mathsf M}\)

\(\ds \)

\(=\)

\(\ds \dfrac {\mathsf {M L T^2 I^2} } {\mathsf {M I^2 T^2} }\)

\(\ds \)

\(=\)

\(\ds \mathsf L\)

$\blacksquare$