Magnetic field at the centre of the hydrogen atom due to motion of electron in $\mathrm{n}^{ ext {th }}$ orbit is proportional to

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Magnetic field at the centre of the hydrogen atom due to motion of electron in $\mathrm{n}^{	ext {th }}$ orbit is proportional to

Aicharya · Accepted Answer

The radius of the $\mathrm{n}^{	ext {th }}$ Bohr orbit is, $r_n \propto n^2$. The angular velocity of the electron, $\omega_n \propto \frac{1}{n^3}$ Also, current $I_n=\frac{q}{T_n}=\frac{q \omega_n}{2 \pi}$ $\begin{array}{ll} 	herefore \quad & \mathrm{I}_{\mathrm{n}} \propto \omega_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^3} \ & 	ext { Now, } \mathrm{B}_{\mathrm{n}}=\frac{\mu_0 \mathrm{I}_{\mathrm{n}}}{2 \mathrm{r}_{\mathrm{n}}} \ 	herefore \quad & \mathrm{B}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^3} 	imes \frac{1}{\mathrm{n}^2} \ 	herefore \quad & \mathrm{B}_{\mathrm{n}} \propto \frac{1}{\mathrm{n}^5} \end{array}$

Magnetic field at the centre of the hydrogen atom due to motion of electron in $\mathrm{n}^{\text {th }}$…

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