Question

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

Answer 1

The expression for the angular momentum is $mvr=\frac{nh}{2\pi }$, or $2\pi r=\frac{nh}{mv}$ $......\left(i\right)$
The wavelength is given by the De Broglie's equation.
$\lambda =\frac{m}{mv}$ $.....\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$, we have
$2\pi r=n\lambda$
Hence, the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie's wavelength associated with the electron revolving around the orbit.