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.

Explanation

According to Bohr postulate, angular momentum, `mvr=(nh)/(2pi)` [n=principle quantum number=1,2,3,4, . . ]
or, `2pir=(nh)/(mv)` [r=radius of Bohr orbit, m=mass of electron, v=velocity of electron] . . . (1)
electron, wavelength `(lamda)=(h)/(mv)`
Substituting the value of `lamda` from (2) in (1) we have,
`2pir=n lamda` (n=1,2,3,4. . .)
Thus, the circumference of Bohr orbit is an integral multiple of the de Broglie wavelength of the revolving electron.