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

Frequency of radiation emitted by hydrogen atom when it makes a transition from state `n_2` to state `n_1` is given as follows :
`v=Rc(1/n_1^2-1/n_2^2)`
Here R is Rydberg constant. Substituting `n_2=n` and `n_1=n-1`, we get the following :
`v=Rc[1/(n-1)^2 -1/n^2]=Rc"(2n-1)"/(n^2(n-1)^2)`
For large value of n we can replace 2n-1`approx`2n , and n-1 `approx` n , to get the following :
`v=(2Rc)/n^3`....(i)
Frequency of revolution of electron can be written as follows:
`f=v/(2pir)`...(ii)
From Bohr.s atomic model we can write velocity and radius as follows :
`v=e^2/(2hepsilon_0n)`...(iii)
`r=(n^2h^2epsilon_0)/(pime^2)`...(iv)
Substituting (iii) and (iv) in equation (ii) we get the following :
`f=v/(2pir)=(e^2/(2hepsilon_0n))/(2pi((n^2h^2epsilon_0)/(pime^2)))=(me^4)/(4epsilon_0^2n^3h^3)=(me^4)/(8epsilon_0^2 ch^3)(2c)/n^3`
`(me^4)/(8epsilon_0^2ch^3)` is equal to Rydberg.s constant R and hence classical frequency of revolution can be written as follows :
`f=(2Rc)/n^3`...(v)
From equation (i) and (v) we can see both the frequencies are same.