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.

(N/A) According to Bohr's postulate,the angular momentum of an electron in a hydrogen atom is given by:
$mvr = n \frac{h}{2\pi}$ ........$(1)$
where $n = 1, 2, 3, \dots$
According to de Broglie's equation,the wavelength $\lambda$ is given by:
$\lambda = \frac{h}{mv}$
Rearranging this gives:
$mv = \frac{h}{\lambda}$ ........$(2)$
Substituting the value of $mv$ from equation $(2)$ into equation $(1)$:
$r \left( \frac{h}{\lambda} \right) = n \frac{h}{2\pi}$
Canceling $h$ from both sides and rearranging:
$2\pi r = n\lambda$ ........$(3)$
Since $2\pi r$ represents the circumference of the Bohr orbit,equation $(3)$ proves that the circumference of the Bohr orbit is an integral multiple of the de Broglie wavelength associated with the electron.