Search any question & find its solution
Question:
Answered & Verified by Expert
The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by:
Options:
Solution:
1553 Upvotes
Verified Answer
The correct answer is:
$\vec{\mu}_L=\frac{e \vec{L}}{2 m}$
As $\overrightarrow{\mathrm{M}}=\mathrm{IA}$
$\Rightarrow|\overrightarrow{\mathrm{M}}|=\frac{\mathrm{e}}{\frac{2 \pi \mathrm{R}}{\mathrm{v}}} \pi \mathrm{R}^2$ $\left[\because \mathrm{I}=\frac{\mathrm{Q}}{\mathrm{T}}=\frac{\mathrm{e}}{\frac{2 \pi \mathrm{R}}{\mathrm{V}}}\right]$
$\Rightarrow|\vec{M}|=\frac{1}{2} e v R \Rightarrow|\vec{M}|=\frac{\mathrm{mvR}}{1} \cdot \frac{\mathrm{e}}{2 \mathrm{~m}}$
$\Rightarrow|\overrightarrow{\mathrm{M}}|=\frac{\mathrm{eL}}{2 \mathrm{~m}} \Rightarrow|\overrightarrow{\mathrm{M}}|=-\frac{\mathrm{e} \overrightarrow{\mathrm{L}}}{2 \mathrm{~m}}$
[ $\because$ Here $\overrightarrow{\mathrm{M}}$ and $\overrightarrow{\mathrm{L}}$ will always be opposite]
$\Rightarrow|\overrightarrow{\mathrm{M}}|=\frac{\mathrm{e}}{\frac{2 \pi \mathrm{R}}{\mathrm{v}}} \pi \mathrm{R}^2$ $\left[\because \mathrm{I}=\frac{\mathrm{Q}}{\mathrm{T}}=\frac{\mathrm{e}}{\frac{2 \pi \mathrm{R}}{\mathrm{V}}}\right]$
$\Rightarrow|\vec{M}|=\frac{1}{2} e v R \Rightarrow|\vec{M}|=\frac{\mathrm{mvR}}{1} \cdot \frac{\mathrm{e}}{2 \mathrm{~m}}$
$\Rightarrow|\overrightarrow{\mathrm{M}}|=\frac{\mathrm{eL}}{2 \mathrm{~m}} \Rightarrow|\overrightarrow{\mathrm{M}}|=-\frac{\mathrm{e} \overrightarrow{\mathrm{L}}}{2 \mathrm{~m}}$
[ $\because$ Here $\overrightarrow{\mathrm{M}}$ and $\overrightarrow{\mathrm{L}}$ will always be opposite]
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.