Assuming the atom in the ground state, the expression for the magnetic field at a point (nucleus) in…
Assuming the atom in the ground state, the expression for the magnetic field at a point (nucleus) in hydrogen atom due to circular motion of electron is
$\left[\mu_{0}=\right.$ permeability of free space, $\epsilon_{0}=$ permittivity of free space, $\quad \mathrm{m}=$ mass of
electron, $\mathrm{e}=$ electronic charge, $\mathrm{h}=$ Planck's constant $]$
Magnetic field at the centre of a circular coil is given by
$\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \mathrm{r}}$ ....(1)
In the case of electron revolving around the nucleus, the current I can be written as
$I$ =ef where $f$ is the frequency of revolution.
$\mathrm{f}=\frac{\mathrm{V}}{2 \pi \mathrm{r}} \quad \therefore \mathrm{I}=\frac{\mathrm{eV}}{2 \pi \mathrm{r}}$
But for electron in ground state $V=\frac{e^{2}}{2 \epsilon_{0} h}$
$\therefore \mathrm{I}=\frac{\mathrm{e}^{3}}{4 \pi \epsilon_{\mathrm{o}} \mathrm{hr}}$
Putting this value in $\mathrm{Eq}(1)$ :
$\mathrm{B}=\frac{\mu_{\mathrm{o}} \mathrm{e}^{3}}{8 \pi \epsilon_{\mathrm{o}} \mathrm{hr}^{2}}$ ...(2)
But $r=\frac{h^{2} \epsilon_{0}}{\pi m e^{2}}$
Putting this value of $\mathrm{r}$ in Eq. (2)
we get : $B=\frac{\mu_{0} e^{7} \pi m^{2}}{8 \epsilon_{0}^{3} h^{5}}$