QJEEM24Chc23a: JEE Main 2024 Chemistry - Class 11 JEEM24Chc23a Solution: The de-Broglie's wavelength of an electron in the orbit is…

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:

$r_4 = 4^2 a_0 = 16 a_0$

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

$\lambda_4 = \frac{2 \pi (16 a_0)}{4} = 8 \pi a_0$