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I have a question about why the color of an object is as it is. This is something I am wondering about after watching episode 5 of the tv-series Cosmos from 2014.

From what I understand a blue object abosrbs all the colors except blue?

Also from what I understand when an atom absorbs electromagnetic waves (or a photon) with a certain wavelength the electron moves to a higher energy level. But after a while the electron moves back to the lower energy level and when that happens a photon that corresponds to that energy level is released?

So why do we for example on a blue object see only the color that is reflected? Shouldn't we also see the absorbed colors that are released when the electron moves back to the lower energy level?

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  • $\begingroup$ This might help - If we repeatedly divide a colorful solid in half, at what point will the color disappear? $\endgroup$
    – mmesser314
    Commented Aug 6, 2021 at 2:22
  • $\begingroup$ @user394334 how would you explain reflection ? Do you think that a solid photon hits a solid atom and it reflects ? $\endgroup$
    – Ankit
    Commented Aug 6, 2021 at 3:48
  • $\begingroup$ @Ankit Good question, I haven't really thought about that. I think of it as a mirror where the light bounces off. The only explanation I have is the since the atom doesn't absorb the photon it bounces off. $\endgroup$
    – user394334
    Commented Aug 6, 2021 at 4:00

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From what I understand a blue object abosrbs all the colors except blue?

Yes, or at least it reflects a range of wavelengths with a strong peak centered around blue.

But after a while the electron moves back to the lower energy level and when that happens a photon that corresponds to that energy level is released?

It emits some form of EM radiation - quite possibly visible light (depends on the atom and energy levels), but the intensity of the radiation is completely drowned out by other light so you can't see it.

If you put a substance that does this in a light-tight box and a light detector (i.e. photomultiplier tube) in it, you will detect the visible light from these atomic transitions. This is called a scintillating detector

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  • $\begingroup$ Thank you, with intensity do you mean the amplitude?, so the amplitude of the light released when the electron goes to a lower energy level is smaller than when the electron got energy? $\endgroup$
    – user394334
    Commented Aug 6, 2021 at 4:57
  • $\begingroup$ uh i mean more that theres way more light from other sources around than the tiny little atomic transition. your eye isn't gonna notice the tiny one. It's like trying to hear a whisper at a rock concert $\endgroup$
    – Señor O
    Commented Aug 6, 2021 at 5:54
  • $\begingroup$ Thank you, but there is one thing I fail to understand. If there is a certain amount of light that hits the object, some is reflected and some is absorbed. Why is the light that is reflected more visible than the light that is absorbed and then released. If there is a "fixed" amount of light hitting the object, why is the intensity of the reflected light higher than the intensity of the absorbed and then released light? $\endgroup$
    – user394334
    Commented Aug 6, 2021 at 11:03
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When an electron moves from a higher energy state to a lower energy state it loses energy in the form of electromagnetic waves.

All objects emit electromagnetic radiation which depend on the body’s temperature. This is called blackbody radiation.

When the temperature of an object is increased it begins to glow ( first red and orange then white and blue as the temperature is progressively increased).

When a body is cool and when we do not see a glow at all it is still constantly emitting radiation that is mostly in the infrared region. Night vision equipment allows us to see objects in the dark as they are emitting radiation in the infrared zone.

In short the radiation emitted may consist of various components having different wavelengths. The spectrum of waves obtained might not all be in the visible range.

EDIT:

While getting excited the electron absorbs only one quantum of energy and gets excited to a higher energy level depending upon the energy of absorbed quanta.

While de excitation the electron eventually reaches the ground state it may do so series and take paths involving intermediate orbits. Taking these different paths will lead to release of energy or light that is not in the visible spectrum.

For example: An electron de exciting from n=3 or more to n=1 it may get de excited directly to n=1 or it may choose to get de excited in steps like going to n=2 and then to n=1. Here n is orbit number and n=1 is the ground state. (Check the graph of emission spectrum of hydrogen involving Lyman balmer paschen brackett and pfund series which correspond to ultraviolet, visible and infrared(x3) respectively.)

This de excitation in intermediate steps is probably the reason why we don’t see the red light which you mentioned even after the electron gets de excited to lower states as the energy must be conserved and this may happen when done in steps.

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  • $\begingroup$ Thank you very much, is then the simple explanation for example that a blue object will absorb red light, but when the electrons move to a lower energy state it may not be red light that is released? $\endgroup$
    – user394334
    Commented Aug 6, 2021 at 4:06
  • $\begingroup$ Atoms after jumping to the higher energy state spontaneously fall back to the lower energy states and again absorb energy to get excited. The emitted light depends on factors like temperature of the object as stated above.At a given temperature we do not see any red light coming from the blue ball ( it wouldn't be blue ball anymore) also because of the reason @senor o stated. $\endgroup$
    – curiosity
    Commented Aug 6, 2021 at 6:51
  • $\begingroup$ Thank you, so when a red photon gets aborbed, it may not be a red photon that gets released when the electron falls back to the lower energy state? I thought that "what goes in must come out" in a sense that if an electron moves up an energy state because of a red photon, and then falls back, the energy is the same, so then a red photon must come out, but I guess that might not be the case? $\endgroup$
    – user394334
    Commented Aug 6, 2021 at 11:09
  • $\begingroup$ The answer to this question perhaps lies somewhere between quantum biology(perception of colour by receptors that absorb certain frequencies and give us the sense of colour)and quantum physics.(I am yet to learn this stuff) A possible explanation; electron after receiving the energy may lose it in series or take different paths and jump to intermediate energy levels like the say from a higher excited state to second excited state and then maybe ground state leading to emission of different wavelengths as energy must be conserved. That’s probably the reason why we don’t observe red light above $\endgroup$
    – curiosity
    Commented Aug 6, 2021 at 18:01
  • $\begingroup$ @user394334 refer the edited answer $\endgroup$
    – curiosity
    Commented Aug 8, 2021 at 12:33

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