JEE Main 2024 — Structure Of Atom Question with Solution

The de-Broglie wavelength of an electron can be expressed through its relationship with the principal quantum number $$n$$ in a Bohr orbit. According to de Broglie, the wavelength $\lambda$ of an electron moving in an orbit can be related to its momentum. However, when considering an electron in an atom, specifically within the Bohr model, we can relate the de-Broglie wavelength to the circumference of the orbit it moves in. The formula representing the de Broglie wavelength of an electron in the $n^{\text{th}}$ orbit of a hydrogen-like atom is:

$${\lambda }_n=\frac{2\pi r_n}{n}$$

where

• ${\lambda }_n$ is the de-Broglie wavelength of the electron in the $n^{\text{th}}$ orbit,
• $r_n$ is the radius of the $n^{\text{th}}$ orbit,
• $n$ is the principal quantum number (which is 4 in this case).

According to Bohr's theory, the radius of the $n^{\text{th}}$ orbit is given by:

$$r_n=n^2{a}_0$$

where $a_0$ is the Bohr radius.

For the $4^{\text{th}}$ orbit, $n=4$, so the radius is:

$$r_4=4^2{a}_0=16a_0$$

Substituting $r_4$ into the de-Broglie wavelength formula, we get:

$${\lambda }_4=\frac{2\pi (16a_0)}{4}=8\pi a_0$$