### Question 2.32

NCERT Class XI Chemistry

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

### Solution in Detail

(video solution below this)

According to the fourth postulate of

Bohr's model, the angular momentum

of an electron is given as

$\displaystyle m_e\:v\:r = \frac{nh}{2\pi}\text{ . . . (1)}$

where $\displaystyle n = 1, 2, 3 \text{. . . (integral)}$

$\displaystyle \text{(1)} \implies 2\pi \times m_e\:v\:r = nh$

$\displaystyle \implies 2\pi r = n \times \frac{h}{m_e\:v}$

but momentum $\displaystyle p = m_e\:v$

$\displaystyle \implies 2\pi r = n \times \frac{h}{p}$

but de-broglie wavelength $\displaystyle \lambda = \frac{h}{p}$

$\displaystyle \implies 2\pi r = n \times \lambda$

$\displaystyle \implies $ circumference = $\displaystyle n \times \lambda \:\underline{ QED}$

### Video Explanation

Please watch this youtube video for a quick explanation of the solution:

### More Posts

### Similar Posts

This Blog Post/Article "(solved)Question 2.32 of NCERT Class XI Chemistry Chapter 2" by Parveen is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.