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Assuming the atom is 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, $m$ $=$ mass of electron, $\in_0=$ permittivity of free space, $\mathrm{h}=$ Planck's constant]
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Verified Answer
The correct answer is:
$\frac{\mu_0 \mathrm{e}^7 \pi \mathrm{m}^2}{8 \epsilon_0^3 \mathrm{~h}^5}$
The magnetic field at the center of a circular coil is given by
$$
\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}
$$
If $\mathrm{T}$ is the periodic time of revolving electron, then $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{T}}$
Also, $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{V}}$
$$
\therefore \mathrm{I}=\frac{\mathrm{eV}}{2 \pi \mathrm{r}}
$$
$$
\therefore \mathrm{B}=\frac{\mu_0 \mathrm{eV}}{4 \pi \mathrm{r}^2}
$$
For electron in ground state,
$$
\mathrm{V}=\frac{\mathrm{e}^2}{2 \in_0 \mathrm{~h}} \text { and } \mathrm{r}=\frac{\mathrm{h}^2 \in_0}{\pi \mathrm{m}^2}
$$
Putting these values in eq.(1) and simplifying we get
$$
\therefore \mathrm{B}=\frac{\mu_0 \mathrm{e}^7 \pi \mathrm{m}^2}{8 \epsilon_0^3 \mathrm{~h}^5}
$$
$$
\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}
$$
If $\mathrm{T}$ is the periodic time of revolving electron, then $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{T}}$
Also, $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{V}}$
$$
\therefore \mathrm{I}=\frac{\mathrm{eV}}{2 \pi \mathrm{r}}
$$
$$
\therefore \mathrm{B}=\frac{\mu_0 \mathrm{eV}}{4 \pi \mathrm{r}^2}
$$
For electron in ground state,
$$
\mathrm{V}=\frac{\mathrm{e}^2}{2 \in_0 \mathrm{~h}} \text { and } \mathrm{r}=\frac{\mathrm{h}^2 \in_0}{\pi \mathrm{m}^2}
$$
Putting these values in eq.(1) and simplifying we get
$$
\therefore \mathrm{B}=\frac{\mu_0 \mathrm{e}^7 \pi \mathrm{m}^2}{8 \epsilon_0^3 \mathrm{~h}^5}
$$
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