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An electron moving around the nucleus with an angular momentum $L$ has a magnetic moment ( $e=$ charge on electron, $m=$ mass of electron $)$
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The correct answer is:
$\frac{e}{2 m} L$
The magnetic moment of a current carrying coil is given by: $\mu=i A=i \pi r^2$
The moving electron in an orbit is equivalent to a coil with current $i=\frac{e}{T}$, where $T$ is the period of rotation around the nucleus.
Therefore, $\mu=\frac{e}{T} \pi r^2$
Using definition of angular momentum $L=m v r$ or $r=\frac{L}{m v}$ and the velocity of the electron is given by $v=\frac{2 \pi r}{T}$ or $\frac{1}{T}=\frac{v}{2 \pi r}$
We can substitute $r$ and $\frac{1}{\mathrm{~T}}$ into the expression for magnetic moment,
$\mu=e\left(\frac{v}{2 \pi r}\right) \pi r^2=\frac{e v}{2} r=\frac{e v}{2} \frac{L}{m v}=\left(\frac{e}{2 m}\right) L$
The moving electron in an orbit is equivalent to a coil with current $i=\frac{e}{T}$, where $T$ is the period of rotation around the nucleus.
Therefore, $\mu=\frac{e}{T} \pi r^2$
Using definition of angular momentum $L=m v r$ or $r=\frac{L}{m v}$ and the velocity of the electron is given by $v=\frac{2 \pi r}{T}$ or $\frac{1}{T}=\frac{v}{2 \pi r}$
We can substitute $r$ and $\frac{1}{\mathrm{~T}}$ into the expression for magnetic moment,
$\mu=e\left(\frac{v}{2 \pi r}\right) \pi r^2=\frac{e v}{2} r=\frac{e v}{2} \frac{L}{m v}=\left(\frac{e}{2 m}\right) L$
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