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Question
Show that the radius of the orbit in hydrogen atom varies as n2, where n is the principal quantum number of the atom.
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Solution
According to the Bohr's theory of hydrogen atom, the angular momentum of a revolving electron is given by
`mvr=(nh)/(2pi)" ............(1)"`
Here,
m = Mass of the electron
v = Velocity the electron
r = Radius of the orbit
h = Planck's constant
n = Principal quantum number of the atom
If an electron of mass m and velocity v is moving in a circular orbit of radius r, then the centripetal force required is given by `F=(mv^2)/r`
Also, if the charge on the nucleus is Ze, then the force of electrostatic attraction between the nucleus and the electron will provide the necessary centripetal force.
`F=1/(4piepsilon_0)((Ze)(e))/r^2=(kZe^2)/r^2`
Where
`k=1/(4piepsilon_0)`
`:.(mv^2)/r=(kZe^2)/r^2 " ......(ii)"`
From (i), we get
`v=(nh)/(2pimr)`
Putting this value in (ii), we get
`m/r (n^2h^2)/(4pi^2m^2r^2)=(kZe^2)/r^2`
`=>r = (n^2h^2)/(4pi^2mkZe^2)`
⇒r ∝ n2
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