Question 429: Atoms & Nuclei

Question

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

Accepted Answer

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

Answer

$$\mathrm{n}^4$$

Answer

$$\mathrm{n}^{-3}$$

Answer

$$n^3$$

Answer

$$\mathrm{n}^{-5}$$

Atoms & Nuclei

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

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