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 (E1) 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
v1 = 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 (E2) 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
v2 = Frequency of radiation at level (n − 1)
Energy (E) released as a result of de-excitation:
E = E2 − E1
hv = E2 − E1 …............(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 nth orbit is given as:
`"v" = "e"^2/(4pi in_0 ("h"/(2pi)) "n")` .........(vi)
And, radius of the nth 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.
