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Question
On the basis of Bohr's theory, derive an expression for the radius of the nth orbit of an electron of the hydrogen atom.
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Solution
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.
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