If $r$ is the orbital radius and $v$ is the orbital velocity of an electron in a hydrogen atom, then its magnetic dipole moment is

The magnetic dipole moment of a revolving electron: The electron of change (–e) performs uniforms circular motion around a stationary heavy nucleus of charge +Ze. This constitutes a current I.

 $\therefore$                         $I=\frac{e}{T}$                                                    ...(i)

Where, T is the time period of revolution. Let r be the orbital radius of electron and v the orbital speed. Then

                          $T=\frac{2\pi r}{\upsilon }$                                                ...(ii)

Substituting the value of T in (i), we get

                          $I=\frac{e}{\frac{2\pi r}{\upsilon }}=\frac{e\upsilon }{2\pi r}$

The magnetic moment (m) associated with this circular current is

                        $m=I\pi r^2$                                             $\left[\because \overrightarrow{m}=I\overrightarrow{A}\right]$

 $\therefore$                     $m=\frac{e\upsilon }{2\pi r}\cdot \pi r^2=\frac{e\upsilon r}{2}$