**Question.**

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:**

Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:

$m v r=n \frac{\mathrm{h}}{2 \pi} \ldots \ldots \ldots(1)$

Where, $n=1,2,3, \ldots$

According to de Broglie’s equation:

$\lambda=\frac{\mathrm{h}}{m v}$

or $m v=\frac{h}{\lambda} \ldots \ldots \ldots(2)$

Substituting the value of ' $m v$ ' from expression (2) in expression (1):

$\frac{\mathrm{h} r}{\lambda}=n \frac{\mathrm{h}}{2 \pi}$

or $2 \pi r=n \lambda \ldots \ldots \ldots . .(3)$

Since ' $2 \pi r$ ' represents the circumference of the Bohr orbit $(r)$, it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie's wavelength associated with the electron revolving around the orbit.