I have been searching the internet for answers to this question, but haven't found a convincing one. I would appreciate any response.
I understand why objects are opaque/black. For example when (white) light is incident on a closed book (let's say it's blue), all wavelengths of visible light except blue are absorbed and hence we see it as blue.
Now, from what I read,
[0.] it is the electrons that absorb the (non-blue) photons from the white light and jump to higher energy states.
My questions are:
1 – Do the electrons stay in the same state?
2 – If they were to jump back to there original state, shouldn't they emit a photon that is exactly of the same wavelength as the one that was absorbed by it in the first place?
3 – If they are in the same state, why does the book continue to appear blue? It can't absorb additional energy unless it returns back to its original state, In doing so won't it emit a photon of the same wavelength it absorbed in the first place?
4 – And , finally, what part of the atom is reflecting the blue light? Is it a bouncing electron?
Best Answer
Great question - partial answer.
A lot of the opacity of most objects is due to a combination of scatter and absorption. At discontinuities in refractive index, light has a probability of either refracting or reflecting. When you add for example titanium dioxide particles to paint, you create many tiny scatter points. This is what makes white paint (and paper, etc) appear "white" - photons are scattered until they come out again.
Now color is the result of different probabilities of scatter and absorption. As a photon tries to make its way out of the surface (for any non-specular reflections this usually involves multiple interactions) it has a chance of exciting an electron (either the electron that belongs to an atom, or more likely the electron that is shared between the atoms of a molecular as part of their bond) and being absorbed. Now often this excited electron has many different ways of losing its energy again - if it was excited from state A to B, it does not have to make the transition back from B to A, but it might go to some (intermediate) state C instead.
And in so doing the color of the light changes...