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:
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