Question

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

Answer 1

According to Bohr's postulate of angular momentum of an electron,

$mvr=\frac{nh}{2\pi }\to 1$

Also, in de-Broglie equation,

$\lambda =\frac{h}{mv}$

$mv=\frac{h}{\lambda }\to 2$

Substituting $2$ in $1$

$2\pi r=\frac{nh}{mv}=n\left(\frac{h}{mv}\right)$

$2\pi r=n\lambda$

Since $2\pi r$, is the circumference of Bohr orbit (r), it is an integral multiple of de-Broglie wavelength.