Question

a hydrogen atom in the ground state absorbs a photon of light with a wavelength of 97.3nm causing the electron to jump to an unknown energy level. the electron then relaxes emitting a photon of light in the visible range, what is the wavelength of the emitted photon?

Answer 1

Energy associated with 97.3 nm wavelength of photon= (h*c)/
                                 = 6.626*10^-34 Js*3.0*10^8m/s) / 97.3*10^-9 m = 2.04*10^-18 J = 12.75 eV
Now, energy of nth energy level of hydrogen atom =
energy of 1st energy level of hydrogen atom = -13.6 eV
energy of 4th energy level of hydrogen atom = -0.85 eV
Thus, it would take E₄-E₁ = -0.85-(-13.6) eV = 12.75 eV excite the electron from the ground state to the third excited state.

When an electron drops from a higher level to a lower level it sheds the excess energy by emitting a photon. The energy of the emitted photon is given by the Rydberg Formula as below:

where n1 < n2 and E₀ = 13.6 eV.

Now, for transition emission from 4th to 1st energy level, energy associated with the photon is:

= 13.6 (1-0.0625)= 12.75 eV = 2.04*10^-18 J

Wavelength corrresponding to this energy = h*c/E_photon =(6.626*10^-34 Js*3.0*10^8m/s)/

2.04*10^-18 J = 97.3 nm

It will emit in the same wavelgth, not in visible region.