Obtain an epression for the frequency of radiation emitted when a hydrogen atom de- excites from levle (n-1). For large n, show that this frequency equals to classical frequency of revolution of the electron in the orbit .

Explanation

We know that energy of an electron in hydrogen atom in in level is given by
`E_(n) = ( me^(4))/(8 in_(0)^(2) n^(2) h^(2))`
`therefore ` Frequency of radiation emiited when a hydrogen atom de - excits from level n to level (n -1) .
` v = (1)/(h) [E_(n) - E_(n-1)] = (m e^(4))/( 8 in_(0)^(2) h^(3))[(1)/((n-1)) - (1)/(n^(2))] = (m e^(4))/(8 in_(0)^(2) h^(3)). ((2n-1))/(n^(2) (n-1)^(2))`
If n is very large , then `1 lt lt n` and hence , the above relation is modified as .
`v = (m e^(4))/(4 in_(0)^(2) h^(2))`
As per classical model of atom , linear speed of electron in an orbit of radius r is given by .
` v= (e)/(sqrt(4 pi in_(0) mr))`
and frequency of radiation = frequency of orbitat motion of electron .
`v_(0) = (v)/(2pir) = sqrt((e^(2))/(16pi^(2) in_(0) mr^(3)))`
If we calculate the value of v and `v_(0)` then we find that the answer are almost the same .