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This question already has an answer here:

As an example, I know a leaf looks green because it absorbs red and blue light, but reflects green light*. My question is how the red and blue light are absorbed, not reflected. And conversely why the green light is reflected, and not absorbed. (I know that transmission of light can also occur, and if you can explain transmission as well that would be great, but not necessary for the answer.)

My first assumption was that the absorbed light is being converted to thermal energy, since objects that appear black to us absorb most (or at least relatively more) visible light, and these surfaces become the warmest in sunlight. So one potential answer to my question will explain how EMR in the range of visible light is converted to thermal energy. However, maybe this isn't the process, and maybe coincidentally the black surface also happen to absorb a lot of IR light, and that's why the surfaces warm up.

Maybe the process can be explained in terms of filtering: red and blue light are filtered out, and green light isn't, so it's reflected....? But that seems too simple, and 'filtering' is a very vague word. If the red and blue wavelengths are absorbed, conservation of energy must happen, so that energy needs to go somewhere. Where? And if this is the case, then how does the filtering occur?

I've done a lot of searching on this topic, and have found many explanations of absorbtion and emission of light with gasses such as neon, as per Niels Bohr's work. I don't believe this example applies to the leaf, which reflects light, not emits. However the example with neon gives the style of answer I am looking for, because a physical process is described, resulting in the perceived color (electrons absorb photons with specific amounts of energy, and then emit at that wavelength of light, and the combination of those wavelengths stimulate the retina and result in the perceived color).

*I'm using 'green light' (for example) as a shorthand to mean 'the wavelengths of EMR that stimulate M cones in the retina, and result in the perception of green.' I know that the light itself physically isn't green, but the 'greenness' arises from our perception of it.

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marked as duplicate by Emilio Pisanty, stafusa, Jon Custer, Qmechanic Jan 30 '18 at 18:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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When a photon strikes an atom, it is annihilated and 100% of its energy is absorbed by the atom. Nothing "reflects" light, all light is absorbed. A green leaf looks green because all light is absorbed, then the leaf emits photons at its natural frequency, i.e. green.

Specifically, an atom absorbs the energy of the photon and exists at an elevated energy state for some time, then it emits a photon and drops back to its natural energy level. Since it is emitting the photon at its natural tendency, they are emitted at a particular frequency, i.e. green.

If an atom in the elevated energy state is hit by a second photon, it immediately emits a photon and stays at the elevated energy state.

Now the questions is, if I shine a red laser on a green leaf, why do I see the bright red spot? This happens because the laser light has so many photons, that it super-saturates the leaf. The atoms of the leaf that are under the laser are constantly at the elevated energy state and constantly get hit by a second photon. This means they never emit photon at their natural frequency, they are constantly emitting photon at the frequency they are being bombarded at, i.e. red.

This is my theory and I am sticking to it!

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    $\begingroup$ This is quite inaccurate, particularly because it completely neglects the (often dominant!) role of virtual transitions to off-resonant energy levels, in which a pretty negligible amount of population reaches the excited states while still having a strong effect (as a macroscopic sample) on the field. (As to the role of the second photon, that's just plain wrong.) However, since you're sticking to your theory, there seems to be little point in trying to convince you otherwise. $\endgroup$ – Emilio Pisanty Jan 30 '18 at 12:45

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