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

We know that , when a hydrogen atom de-excites from level n to level (n-1) , the frequency of emitted radiation is given by
`v=(me^4)/(8epsilon_0^2h^3)[1/(n-1)^2-1/n^2]`
`rArr v=(me^4 (2n-1))/([8 epsilon_0^2 h^3 n^2 (n-1)^2])`
For large n , `2n - 1 approx 2n` and n-1 `approx` n
`therefore v cong(me^4 (2n))/([8epsilon_0^2 h^3 n^2(n)^2])=(me^4)/(4epsilon_0^2 h^3 n^3)`
Orbital frequency is given by
`v_c=v/(2pir)`
Now in Bohr.s model ,
Velocity of electron in `n^(th)` orbit of hydrogen atom is
`v=(nh)/(2pimr)`
Radius of `n^(th)` orbit in hydrogen atom is
`r=(n^2h^2epsilon_0)/(pim^2e^2)` , frequency `v_e=v/(2pir)`
Putting the value of r and v , we get
`v_e=(me^4)/(4epsilon_0^2 h^3 n^3)`
which is the same as v for large n.