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Q.
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
NTA AbhyasNTA Abhyas 2022
Solution:
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
$m = I \pi r^{2}$ $\left[\right. \because \overset{ \rightarrow }{m} = I \overset{ \rightarrow }{A} \left]\right.$
$\therefore $ $m = \frac{e \upsilon}{2 \pi r} \cdot \pi r^{2} = \frac{e \upsilon r}{2}$
