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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?

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4 Answers 4

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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...

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If you are not familiar with the subject this link (wiki) can give you a basic knowledge of what happens in a simple atom of H, and then you can expand through related articles.

All your 4 questions are related, and I added [0-] to the statement in your premise:

Photons in all frequencies hitting an object are absorbed in different ways (absorbed, reflected, refracted, scattered, transtormed into thermal energy) by the atoms, not only [0-] by the electrons.

Electrons (1-) do not stay in the same state, they keep jumping up to different levels of excitation and down again to the (2-) original state.

(3-) on their way down they emit all sorts of photons, of different frequencies, as you can see in the link, but usually only one in the visual spectrum. So you see blue because that is the only frequency you can detect. And the light you see comes all the time from different electrons.

(4-) you are right, the light is coming mainly from outer electrons, but it is not mainly reflected. That is what is misleading and confusing you, if the light were reflected you would see all frequencies. The light is emitted and, the frequencies are, so to speak, filtered by the electrons, they absorb all frequencies and return only a few.

I hope I covered all aspects of your doubts

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It's all linked, so answering one will elucidate the rest. They don't stay in that same state, if that were the case then you have the problem you mention in point 3). After being excited, an electron loses it's energy by several mechanisms. Some of them radiate again "light" or more generally electromagnetic waves in other frequencies outside the visible spectrum, and not necesarilly through the same wavelength. Whereas others lose their energy through transfer to other excitations in the systems like phonons.

The issue with color is more like that particular wavelength is elastically scattered by the atom (system composed of the electrons and nucleus). This scattering can be understood somehow as if the atoms are working as an antenna radiating a particular wavelength, and thus the particular color.

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i will provide an answer to this question along with answers to some of my comments on this thread (plus provide a couple of quick references).

Electromagnetic (EM) Radiation / Light interaction with matter (material objects) involves various things. Concerning opaqueness/transparency of materials the following figure shows what happens (reference: http://www.uotechnology.edu.iq/appsciences/filesPDF/material/lectures/2c/3-Materials_prperties8.pdf):

enter image description here

EM waves interact with a material through its (micro-)structure, meaning mostly electron/energy zones (plus nuclear structure, size of material, thickness, angle of incidence etc..)

The (detailed) effects are calculated using quantum-mechanics. Due to interaction of an EM wave with the structure of the material, the amount of reflection/absorption/transmission depends on the wavelength/frequency of the incident EM wave (related to Einstein-Planck relation $E=hv$, $E$ energy of EM wave, $h$ Planck's constant, $v$ frequency of EM wave)

In summary the micro-interaction is as folows:

a photon with given energy can interact with an electron (with a given energy) and make the electron jump to another orbit/energy zone, thus the photon is absorbed, or may be scattered/deflected by an electron, thus the photon is reflected/attenuated or might not interact (depending on photon energy/wavelength and energy zones of material), thus the photon is transmitted.

To characterise the amount of opacity/transparency of a material object (with respect to EM radiation of given wavelength), there are various coefficients used, based on Beer-Lambert Law (and similar formulas). For more detailed approximations one can check for example https://newton.ex.ac.uk/research/emag/pubs/pdf/Hooper_OE_2008.pdf

The (perceived) color of a material object, depends on what (visible) wavelengths reflects and absorbs. So a material which absorbs all (visible) wavelengths while reflects none would appear as "black", while a material which reflects all (visible) wavelengths while absorbs none would appear as "white". A material which reflects (visible) wavelengths around the "red" wavelength, would appear as "red", while a material which absorbs only wavelengths around (visible) "red" and reflects everything else would appear as the "complementary color of red" and so on.

It is clear form the above, that opacity/transparency of a material object is primary a function of light/EM wave frequency (or wavelength) (in general different for each material object). As such there are certain (ranges of) wavelengths of EM radiation with respect to which material objects are (mostly) transparent or (mostly) opaque.

Concerning metals (as per one comment), metals are mostly opaque in the range of visible light, while mostly transparent in the range of X-rays/$\gamma$-rays (reference: http://higheredbcs.wiley.com/legacy/college/callister/0470125373/conceptcheck_ans/ch19.pdf based on http://phys.thu.edu.tw/~hlhsiao/mse-web_ch21.pdf)

A physics.se question about transparency of plactic bags in infrared radiation

Anyone interested to experiment, here is a nice handbook on light measurement

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  • $\begingroup$ Regarding "concept check 19.3" about transparency of metals to x rays - that is quite simply false. Of course x rays penetrate more than visible light but at every energy there is a probability of interaction between electrons and photons. At lower energies, photo-electric effect dominate (below say 80 keV although it depends on the atomic number). At intermediate energies it is Compton scatter; and above 1 MeV pair production starts to play. So it depends on your definition of "transparent" I suppose. See XCOM database nist.gov/pml/data/xraycoef $\endgroup$
    – Floris
    Commented Aug 12, 2014 at 15:28
  • $\begingroup$ @Floris, i think the definition of transparent is clear and given in the answer, i dont see your definition is it sth else, or maybe you mean that absolute transparency, in this respect you will agree that glass for example is not absolutely transparent (because the factor is not exactly 1), but what would you say if the factor is 0.95? $\endgroup$
    – Nikos M.
    Commented Aug 12, 2014 at 15:37
  • $\begingroup$ @Floris, i clearly fail to see where there is a disagreement. i really would like to know, no offense $\endgroup$
    – Nikos M.
    Commented Aug 12, 2014 at 15:39
  • $\begingroup$ #Floris the probability of interaction is based on photon energy (aka wavelength) and structure of material object, that's all there is to it, as such different wavelengths and different structures give quite different probabilities, would you agree with this? $\endgroup$
    – Nikos M.
    Commented Aug 12, 2014 at 15:44
  • $\begingroup$ My point was (and is) that no material - not even high quality optical fiber - is truly transparent - there is always some extinction coefficient. I think we we're just working with a different definition of transparency. $\endgroup$
    – Floris
    Commented Aug 12, 2014 at 15:45

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