Question

Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the n^th orbital state in hydrogen atom is n times the de Broglie wavelength associated with it.

Answer 1

According to Bohr’s second postulate of quantization, the electron can revolve round the nucleus only in those circular orbits in which the angular momentum of the electron is integral multiple of `h/(2pi)`where h is Planck’s constant (= 6.62 × 10⁻³⁴ Js).

So, if m is the mass of electron an v is the velocity of electron in permitted quantized orbit with radius r than

`mvr =n h/(2pi)    (1)`

Where n is the principle quantum number and can take integral values like

n = 1, 2, 3…

This is the Bohr’s quantization condition.

Now, de-Broglie wavelength is given as

`lambda = h /(mv)`

Where λ → is wavelength associated with electron.

v is the velocity of electron.

h − is the velocity of electron.

m − mass of electron

`v =h/(mlambda)     (2)`

Putting value of v from (2) in (1)

`m xx h/(mlambda) xx r = n h/(2pi)`

`(rh)/lambda = (nh)/(2pi)`

`2pir = lambdan`

Now circumference of the electron in the n^th orbital state of Hydrogen atom with radius r is 2πr.