Obtain an expression for the orbital magnetic moment of an electron rotating about the nucleus in an atom and explain the gyromagnetic ratio.

(N/A) Consider an electron of charge $(-e)$ performing uniform circular motion around a stationary heavy nucleus of charge $(+Ze)$ with orbital radius $r$ and speed $v$.
The circulating electron constitutes a current $I$ given by:
$I = \frac{e}{T} \quad \dots (1)$
where $T$ is the time period of revolution.
Since $v = \frac{2 \pi r}{T}$,we have $T = \frac{2 \pi r}{v} \quad \dots (2)$
Substituting $(2)$ into $(1)$:
$I = \frac{e v}{2 \pi r} \quad \dots (3)$
The magnetic moment $\mu_l$ associated with this current loop of area $A = \pi r^2$ is:
$\mu_l = I A = \left( \frac{e v}{2 \pi r} \right) (\pi r^2) = \frac{e v r}{2} \quad \dots (4)$
Multiplying and dividing by the electron mass $m_e$:
$\mu_l = \frac{e}{2 m_e} (m_e v r)$
Since the orbital angular momentum $L = m_e v r$,we get:
$\mu_l = \frac{e}{2 m_e} L$
The ratio $\frac{\mu_l}{L} = \frac{e}{2 m_e}$ is called the gyromagnetic ratio,which is a constant for an electron.