An electron in the ground state of hydrogen atom is revolving in a circular orbit of radius $\mathrm{R}$.…
An electron in the ground state of hydrogen atom is revolving in a circular orbit of radius $\mathrm{R}$. The orbital magnetic moment of the electron is $(\mathrm{m}=$ mass of electron, $\mathrm{h}=$ Planck's constant, $\mathrm{e}=$ electronic charge $)$
$\frac{-\mathrm{eh}}{\pi \mathrm{m}}$
$\frac{\mathrm{eh}}{2 \pi \mathrm{m}}$
$\frac{2 \mathrm{eh}}{\pi \mathrm{m}}$
$\frac{\mathrm{eh}}{4 \pi \mathrm{m}}$
Solution
In ground state $(\mathrm{n}=1)$ according to Bohr's theory:
$\mathrm{mvR}=\frac{\mathrm{h}}{2 \pi}$ or $\mathrm{v}=\frac{\mathrm{h}}{2 \pi \mathrm{mR}}$
Now time period, $\mathrm{T}=\frac{2 \pi \mathrm{R}}{\mathrm{v}}=\frac{2 \pi \mathrm{R}}{\mathrm{h} / 2 \pi \mathrm{mR}}=\frac{4 \pi^{2} \mathrm{mR}^{2}}{\mathrm{~h}}$
Magnetic moment, $\mathrm{M}=\mathrm{iA}$
Where, $\mathrm{I}=\frac{\text { charge }}{\text { time period }}=\frac{\mathrm{e}}{\frac{4 \pi^{2} \mathrm{mR}^{2}}{\mathrm{~h}}}=\frac{\mathrm{eh}}{4 \pi^{2} \mathrm{mR}^{2}}$
and $\mathrm{A}=\pi \mathrm{R}^{2}$
$\therefore \mathrm{M}=\left(\pi \mathrm{R}^{2}\right)\left(\frac{\text { eh }}{4 \pi^{2} \mathrm{mR}^{2}}\right)$ or $\mathrm{M}=\frac{\text { eh }}{4 \pi \mathrm{m}}$
Direction of magnetic moment $\vec{M}$ is perpendicular to the plane of orbit.