Obtain an expression for the frequency of radiations emitted when a hydrogen atom de-excites from level n to level (n-1). for larger n, show that the frequency equals the classical frequency of revolution of the electron in the orbit.

Explanation

The frequency v of the emitted radiation when a hydrogen atom de-excites from level n to level (n-1) is
`E=hv=E_(2)-E_(1)`
`v=(1)/(2)(mc^(2)alpha^(2))/(h)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` where
`alpha =(2 pi K e^(2))/(ch)`= fine structure constant
`v=(1)/(2) (mc^(2)alpha^(2))/(h)[(1)/((h-1)^(2))-(1)/(n^(2))]=(nc^(2)a^(2))/(2h)[(n^(2)-(n-1)^(2))/(h^(2)(n-1))]`
`=(mc^(2)alpha^(2)[(n+n-1)(n-n+1)))/(2hn^(2)(n-1)^(2))`
`v=(mc^(2)alpha^(2)(2n-1))/(2hn (n-1)^(2))`
for large `n_(1)(2n-1)=2n and (n-1)=n.`
`v=(mc^(2))/(hn^(3)).(4pi^(2)K^(2)e^(4))/(c^(2)h^(2))`
`v=(4pi^(2)m K^(2)e^(4))/(n^(3)h^(3))" "rarr(i)`
In Bohr's Atomic model, velocity of electron in `n^(th)` orbit is `v=(nh)/(2 pi mr)`
and radius of `n^(th)` orbital is `v=(n^(2)h^(2))/(4pi^(2)m K e^(2))`
Frequency of revolution of electron `v=(V)/(2pi r)=(nh)/(2pi mr)`
and radius of `n^(th)` orbital is `r=(n^(2)h^(2))/(4pi^(2)m K e^(2))`
Frequency of revolution of electron
`v=(V)/(2pi r)=(nh)/(2pi mr)[(4pi^(2) mK e^(2))/(2 pi n^(2) h^(2))]`
`v=(Ke^(2))/(nh.r)=(Ke^(2))/(nh)[(4pi ^(2) m K e^(2))/(n^(2)h^(2))]`
`v=(4pi ^(2)m K^(2) e^(4))/(n^(3) h^(3))`
which is the same as (i)
Hence for large values of `n_(1)` classical frequency of revolution of electron in `n^(th)` orbit is the same as the frequency of radiation emitted when hydrogen atom de-excites from level (n) to level (n-1)