Show that 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 Bohrs angular momentum
`mvr = ( nh)/( 2pi)`
`2 pi r = ( nh)/( mv ) " "...(i)`
According to De - Broglie
`lamda = ( h)/( mv) ....(ii)`
From equation (i) and (ii) we get `= 2 pi r = n lamda `
Thus, the circumference `(2pi r )` of Both orbit for hydrogen - atom is an integral multiple of de Broglie wavelength.