Let's say we have a blue, opaque material. If white light was incident on that material, the blue light would be absorbed by electrons and the electrons would transition to a higher energy state, and then the electrons would transition back to the original state (ignoring any other states it could drop to) re-emitting that light to make the material appear blue, but what happens to the rest of the light? Is the rest of the light absorbed by molecular bonds and converted into heat?

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    $\begingroup$ A blue material reflects blue radiation and absorbs red etc.. There is no atomic absorption involved in that reflection. In general, a solid is not well described by atomic physics. We need to understand its inter-atomic or molecular structure to describe interaction with light. $\endgroup$
    – CuriousOne
    Jul 23, 2016 at 0:10
  • $\begingroup$ Interesting, so the electrons in the atoms/molecules of a blue material interact with the photons, but they don't transition to another energy level and just essentially 'reject' the photon and re-emit it? Is this due to the photon energy not corresponding to any available energy levels for the electrons? $\endgroup$
    – J.Doe
    Jul 23, 2016 at 1:48
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    $\begingroup$ The energy levels in atoms are nice and discrete, but when two atoms with the same energy levels interact they influence each other. Each pair of degenerate, i.e. equal energy levels splits into two levels of unequal energy. The discrete spectrum now looks a lot more complicated. Three, four, five etc. atoms will split these levels further and further. By the time we have a macroscopic solid, these "forests" of energy levels have become continuous energy bands. It becomes meaningless to talk about atomic states. The solid behaves in a different and much more complex way than its atoms. $\endgroup$
    – CuriousOne
    Jul 23, 2016 at 1:58
  • $\begingroup$ @CuriousOne what you say is correct, but in the OP's comment he or she makes no mention of atoms. That comment is correct: the blue light is reflected because there are no energy levels to absorb it. $\endgroup$
    – garyp
    Jul 23, 2016 at 11:01
  • $\begingroup$ @garyp: It's in the title and the OP asked a similar question before. I am merely trying to make him think about the problem in a wider context than atomic physics. $\endgroup$
    – CuriousOne
    Jul 23, 2016 at 17:41

1 Answer 1


I'm not sure what you mean by "the rest of the blue light." Basically three things can happen when low energy light passes though matter.

  1. Absorption and reemission. The light excites a dispersion free mode, like your electronic energy state example, in that case the light will be reemitted at the same wavelength most often. You can work out the exact distribution with the Bohr-Dirac model (bohr model with relativistic corrections).

  2. Absorption and dispersion. Light can excite phonon modes, those will disperse and heat the object up.

  3. Finally, light can pass through the material and not interact with the object at all.

  • $\begingroup$ you have skipped mentioning elastic scattering, which is not re emission . $\endgroup$
    – anna v
    Aug 8, 2016 at 3:10
  • $\begingroup$ see quantummechanics.ucsd.edu/ph130a/130_notes/node43.html $\endgroup$
    – anna v
    Aug 8, 2016 at 3:15
  • $\begingroup$ Interesting, Thank you for your comment Anna, I will update the answer when I have a more clear picture. I have always been trained to believe light can not change directions without being absorbed and reemitted... a consequence of relativity. If one wants to compute the angular distribution of final state light after bombarding a material with photons, is not true that we sum diagrams which have the light being absorbed along with diagrams in which the light is not? That is to say, what are the observable and computational differences between absorption+reemission vs. elastic scattering? $\endgroup$ Aug 8, 2016 at 3:27
  • $\begingroup$ absorption and reemission presuppose existing quantized bound energy levels. Elastic scattering is off the spill over field of the whole atom/molecule/ lattice $\endgroup$
    – anna v
    Aug 8, 2016 at 3:46

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