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

It is given that a hydrogen atom de-excites from an upper level (n) to a lower lever (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 of level n
h=Planck's constant
m=Mass of hydrogen atom
e=Charge on an electron
`in_(0)` =Permittivity of free space
Now, the relation for energy `(E_(2))` of radiation at level (n-1) given as:
`E_(2)=hv_(2)=(hme^(4))/((4pi)^(3) in_(0)^(2)(h/(2pi))^(3))xx1/(n-1)^(2)`
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)
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)(2n-1))/((4pi)^(3)in_(0)^(2)(h/(2pi))^(3)n^(2)(n-1)^(2))`
For large n, we can write `(2n-1)approx 2n and (n-1) ~=n`
`:. v=(me^(4))/(32pi^(3)in_(0)^(2)(h/(2pi))^(3)n^(3))`
Classical relation of frequency of revolution of an electron is given as:
`v_(e)=v/(2pir)`
Where,
Velocity of the electron in the `n^(th)` orbit is given as:
`e^(2)/(4pi in_(0)(h/(2pi))n)`
And radius of the `n^(th)` orbit is given as:
`r=(4pi in_(0)(h/(2pi))^(2))/(me^(2))n^(2)`
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))`
Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.