touqra said:
All the while, I thought that an atom only absorbs a photon that precisely corresponds to one of the transition level (resonance), but now I read that the atom will also absorb photon with wavelengths different from the transition level in question.
So, my question is, suppose I have two energy level = E, and the photon has energy 2E, how would the atom actually absorb E amount to the transition and what happens to the other leftover E ?
There is the Kramers' dispersion theory equation which led via for instance a joint paper
from Kramers and Heisenberg to Heisenbergs discovery of his Matrix Quantum mechanics.
This plays in 1924. kramers' dispersion relation between the atom and the radiation field:
{\cal M}\ =\ E\frac{e^2}{4\pi^2 m}\left(\sum_{abs}\frac{f_i}{\nu_i^2-\nu^2} - \sum_{emis}\frac{f_j}{\nu_j^2-\nu^2} \right)
Where the f are interaction coefficients, the \nu_i,\ \nu_j are the characteristic frequencies
of the atom and \nu is the frequency of the radiation field.
You see that the the chance for interaction is highest if the incoming frequency is
equal to the characteristic frequency but is not zero if they are not equal. The break-
through in this formula was the inclusion of the second (negative) term which describes
the emission stimulated by the incoming radiation. Regards. Hans
P.S. The birth of matrix mechanics is described in van der Waerden's book: "The sources
of quantum mechanics" which includes all the relevant papers.