An excited hydrogen atom emits light with a frequency of 1.141x10^14Hz to reach the energy level for which n=4 In what principlal quantum level did the electron begin

Expert Answers
ishpiro eNotes educator| Certified Educator

The energy level of Hydrogen atom are determned by the formula

`E = -E_0/n^2` , where n is the principal quantum number and `E_0 = 13.6 eV` .

The energy of the photon emitted is the difference of the energy levels of the atom in different states:

`E_ph=E_0(1/n_1^2 - 1/n_2^2)`

Here `n_2 = 4` , the principal quantum number for the final state of the atom.

The frequency and the energy of the photon are related by the formula

`E_ph= hF` , where h is the Planck constant `h= 4.14*10^-15 eV*s`

So the energy of the photon with given frequency is

`E_(ph) = 4.14*10^-15*1.41*10^14 = 0.58 eV`


`1/n_1^2 - 1/n_2^2=E_(ph)/E_0 = 0.58/13.6=0.043`

Plugging in `n_2 = 4` , we get

 `1/n_1^2 = 0.043+1/n_2^2 =0.043+1/16 = 0.1055`

From here `n_1 = 3` .

This means the electron the Hydrogen atom originally had principal quantum number n = 3.