A given atom emits red light and green light, but never blue light. How does Bohr's plantetary model of the atom explain this phenomena? Help please

1 Answer

valentin68's profile pic

valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

In the Bohr model for hydrogen atom the energy of the light emitted by the atom when an electron jumps between energy levels `m` (initial) and `n` final is

`E = E_1*(1/n^2-1/m^2)` where `E_1 =13.6 eV` is the energy of first electron level.

For a different atom than hydrogen, having the total nucleus charge `Z` the Bohr model of atom predicts for the energy of the photon emitted by the electron between two energy levels of the value

`E = Z^2*E_1*(1/n^2 -1/m^2)`

We know that the energy of the photon is

`E = h*F = h*c/lambda`

Blue color has a higher energy (and shortest wavelength) than red and green color.

The photon having the highest energy (shortest wavelength) will be emitted when n=1 and m=almost infinity, in other words when electron jumps between the furthest possible energy level and the first energy level.

`E ~~Z^2*E_1`

For the hydrogen atom (Z=1) the shortest wavelength that can be emitted is about 91 nm which is situated far in the ultraviolet portion of the spectrum (there are colors beyond blue, like violet and/or ultraviolet that theoretically can be emitted). Because the atom does not emit the blue color, this means that there are no such pairs of integers `n` and `m` (final and initial energy levels) that the resulting wavelength of the emitted photon is between 450 and 495 nm. However because red and green photons have lower energies, and they are emitted by the atom, inner electron transitions (possible between smaller initial values of `m` and the same final value of `n` ) does exist.