Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the 2nd and 4th orbits of hydrogen atom _________.

For an electron revolving in a Bohr orbit, the circular motion constitutes a current loop. If the orbit radius is $$r_n$$ and the speed is $$v_n$$, the current is

$$I_n = \frac{e}{T} = \frac{e\,\omega_n}{2\pi} = \frac{e\,v_n}{2\pi r_n}$$

In Bohr’s model for the hydrogen atom:

• Radius of the $$n^{\text{th}}$$ orbit: $$r_n = a_0 n^2$$, where $$a_0$$ is the Bohr radius.
• Speed in the $$n^{\text{th}}$$ orbit: $$v_n = \frac{v_0}{n}$$, where $$v_0 = \frac{e^2}{2\varepsilon_0 h}$$ is a constant.

Substituting these into $$I_n$$:

$$I_n = \frac{e}{2\pi}\,\frac{v_n}{r_n} = \frac{e}{2\pi}\,\frac{v_0/n}{a_0 n^2}= \frac{e\,v_0}{2\pi a_0}\,\frac{1}{n^3}$$

Thus $$I_n \propto \frac{1}{n^3}$$.

The magnetic field at the centre of a circular current loop of radius $$r_n$$ carrying current $$I_n$$ is

$$B_n = \frac{\mu_0 I_n}{2r_n}\,,$$

so

$$B_n \propto \frac{I_n}{r_n} \propto \frac{1/n^3}{n^2}= \frac{1}{n^5}\,.$$

Therefore, for two orbits with quantum numbers $$n_1$$ and $$n_2$$,

$$\frac{B_{n_1}}{B_{n_2}} = \left(\frac{n_2}{n_1}\right)^5$$

Taking $$n_1 = 2$$ and $$n_2 = 4$$:

$$\frac{B_2}{B_4} = \left(\frac{4}{2}\right)^5 = 2^5 = 32$$

Hence, the magnetic field produced by the electron in the 2nd orbit is 32 times larger than that in the 4th orbit.

Final answer: 32