Question

On the basis of Bohr's theory, derive an expression for the radius of the n^th orbit of an electron of the hydrogen atom.

Answer 1

Let m and e be the mass and the charge of the electron. If the electron revolves with velocity v in the circular orbit
of radius r, then according to first Bohr’s postulate.

Centripetal force = Electrostatic force

`(mv)^2/r = e^2/(4piepsilon_0r^2)` 

`:. v^2 = e^2/(4piepsilon_0mr)` .....(1)

According to second postulate

`mvr = (nh)/(2pi)` , where n = 1, 2, 3, .... and h is planck’s constant

Squaring this expression we get,

`m^2v^2r^2  = (n^2h^2)/(4pi^2) `

`:. v^2 = (n^2h^2)/(4pim^2r^2)` ....(2)

Equating the values of v² from eq. 1 & 2

`e^2/(4piepsilon_0mr) = (n^2h^2)/(4pi^2m^2r^2)`

`:. r= ((epsilon_0h^2)/(pime^2))n^2`

This expression gives us the radius of the Bohr’s orbit. The radius of the successive orbits is given by substituting.

n = 1, 2, 3, ... etc. since `epsilon_0` , h, m, e are all constant, ∴ r α n²

Thus radius of orbit is proportional to the square of the principle quantum number.