Using Bohr's theory show that when n is very large the frequency of radiation emitted by hydrogen atom due to transition of electrom from n to `(n-1)` is equal to frequency of revolution of electron in its orbit.

Explanation

The frequency v of the emitted radiation when a hydrogen atom de-excites form level n to level (n-1) is `E=hv=E_2-E_1`
`v=1/2 (mc^2alpha^2)/h[1/(n_1^2)-1/(n_2^2)] where alpha=(2piKe^2)/(ch)`=fine structure constant
`v=1/2 (mc^2alpha^2)/h [1/((n-1)^2)-1/(n^2)]=(mc^2alpha^2)/(2h) [(n^2-(n-1)^2)/(n^2(n-1)^2)]=(mc^2alpha^2[(n+n-1)(n-n+1)])/(2hn^2(n-1)^2)`
`v=(mc^2alpha^2(2n-1))/(2hn^2(n-1)^2)`.
for large `n,(2n-1)~~2n`, and `(n-1)~~n`
`v=(mc^2alpha^(2).2n)/(2hn^2.n^2)=(mc^2alpha^2)/(hn^3)`
Putting `alpha=(2piKe^2)/(ch)`, we get `v=(mc^2)/(hn^3). (4pi^2K^2e^4)/(c^2h^2)`
`v=(4pi^2mK^2e^4)/(n^3h^3).......(i)`
In Bohr's atom model,
velocity of electron in nth orbit is `v=(nh)/(2pimr)` and radius of nth orbit is `r=(n^2h^2)/(4pi^2mKe^2)`
`:.` frequency of revolution of electron `v=v/(2pir)=(nh)/(2pimr)((4pi^2mKe^2)/(2pi.n^2h^2))`
`v=(Ke^2)/(nh.r)=(Ke^2)/(nh)((4pi^2mKe^2)/(n^2h^2))=((4pi^2mK^2e^4)/(n^3h^3))`, which is the same is (i).
Hence for large values of n, classical frequency of revolution of electron in nth orbit is the same as the frequency of radiation emitted when hydrogen atom de-excites form level (n) to level (n-1).