For hydrogen-like atoms and ions, the ratio of the magnetic dipole...

The magnetic dipole moment of an electron revolving in a circular orbit is given by $M = I A$.

The equivalent current is $I = \dfrac{e}{T} = \dfrac{e v}{2 \pi r}$ and the area of the orbit is $A = \pi r^2$.

$M = \left( \dfrac{e v}{2 \pi r} \right) (\pi r^2) = \dfrac{e v r}{2}$

The angular momentum of the electron is $L = m_e v r$.

The ratio of the magnetic dipole moment to the angular momentum is:

$\dfrac{M}{L} = \dfrac{\dfrac{e v r}{2}}{m_e v r} = \dfrac{e}{2 m_e}$

Substituting the values $e = 1.6 \times 10^{-19} \text{ C}$ and $m_e = 9.1 \times 10^{-31} \text{ kg}$:

$\dfrac{M}{L} = \dfrac{1.6 \times 10^{-19}}{2 \times 9.1 \times 10^{-31}} = \dfrac{1.6}{18.2} \times 10^{12} \approx 8.8 \times 10^{10} \text{ C kg}^{-1}$