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

#### Solution

It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n − 1).

We have the relation for energy (E_{1}) of radiation at level n as:

`"E"_1 = "hv"_1 = ("hme"^4)/((4pi)^3 in_0^2 ("h"/(2pi))^3) xx (1/"n"^2) ` ...........(i)

Where

v_{1} = Frequency of radiation at level n

h = Planck's constant

m = Mass of hydrogen atom

e = Charge on an electron

`in_0` = Permittivityof free space

Now, the relation for energy (E_{2}) of radiation at level (n − 1) is given as:

`"E"_2 = "hv"_2 = ("hme"^4)/((4pi)^3 in_0^2 ("h"/(2pi))^3) xx 1/("n" - 1)^2` ...........(ii)

Where

v_{2} = Frequency of radiation at level (n − 1)

Energy (E) released as a result of de-excitation:

E = E_{2 }− E_{1}

hv = E_{2} − E_{1} …............(iii)

Where,

v = Frequency of radiation emitted

Putting values from equations (i) and (ii) in equation (iii), we get:

`"v" = ("me"^4)/((4pi)^3 in_0^2("h"/(2pi))^3) [ 1/("n" - 1)^2 - 1/"n"^2]`

= `("me"^4 (2"n" - 1))/((4pi)^3 in_0^2 ("h"/(2pi))^3 "n"^2("n" - 1)^2)`

For large n, we can write (2n − 1) ≃ 2n and (n − 1) ≃ n.

∴ v = `("me"^4)/(32pi^3 in_0^2 ("h"/(2pi))^3 "n"^3` ...............(iv)

Classical relation of frequency of revolution of an electron is given as:

`"v"_"c" = "v"/(2pi"r")` ...........(V)

Where,

Velocity of the electron in the n^{th} orbit is given as:

`"v" = "e"^2/(4pi in_0 ("h"/(2pi)) "n")` .........(vi)

And, radius of the n^{th} orbit is given as:

`"r" = (4pi in_0 ("h"/(2pi))^2)/("me"^2) "n"^2` ............(vii)

Putting the values of equations (vi) and (vii) in equation (v), we get:

`"v"_"c" = ("me"^4)/(32pi^3 in_0^2 ("h"/(2pi))^3 "n"^3)` ................(Viii)

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.