Atoms individually have no colors, but when there is a large collection of atoms we see objects colorful, which leads to a question: at least how many atoms are required for us to see the color?

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    $\begingroup$ Are you asking about the perception of colour by the eye when the light comes directly from a source? $\endgroup$ – Farcher Jan 23 '17 at 8:35
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    $\begingroup$ Interesting question, not so clear, but the answers below show that it is worth asking. You should ask yourself what you mean by "color". Color is not an intrinsic property of anything, it is a visual property, something that happens, an event. In the case you are pointing at, you cannot ignore the light factor, "how many atoms" is not a sufficient condition to ask what you are asking. Keep in mind that "atoms" (or molecules) can emit and/or reflect light. $\endgroup$ – adrienlucca.wordpress.com Jan 23 '17 at 22:09
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    $\begingroup$ Why do you say atoms have no colours? $\endgroup$ – immibis Jan 24 '17 at 1:13
  • $\begingroup$ I don't think that is what the question is going for, actually. I think it's about the fact that most materials show colors that depend on their structure at scales above the size of an atom. You can look at the spectra of individual atoms, but unless they're metal atoms that's probably not going to tell you much. $\endgroup$ – hobbs Jan 24 '17 at 2:18
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    $\begingroup$ @J.Manuel: As others have already said, we should not confuse the human perception of colour with intrinsic properties of atoms. Also, a single atom undergoing photonic emission can only emit one photon at one time, and hence we can't say that the emitted photons have 'colour' corresponding to the atom's emission spectrum. For instance, the pink colour of a hydrogen gas discharge lamp does not correspond to a single wavelength, and we need multiple atoms emitting photons of different wavelengths in roughly the same place and time so that we can perceive the overall 'colour'. $\endgroup$ – user21820 Jan 24 '17 at 13:43

There are a couple of issues here.

  1. A pink (#FF00FF) object appears pink not because each atom is pink (there is no wavelength of light that is perceived to be the same pink by the ordinary human eye. What is happening is that a pink object is emitting (or reflecting) light of multiple wavelengths that enter the eye and are detected and processed to allow us to perceive its colour as pink. One single atom, therefore, would not be able to appear to us as pink under ordinary conditions because it will not emit photons of the appropriate wavelengths rapidly enough that we see no oscillation but a steady pink.

  2. Even for colours that correspond to a single wavelength of light, we would need a significant number of atoms before it emits enough photons to form a stable statistical distribution of wavelengths (called an emission spectrum), which we can then perceive and compare to the colours that we have previously experienced. How many atoms are needed would of course depend on the rate of emission, which is proportional to the power output. For reflection it would depend largely on the intensity of light incident on the object.

  3. And of course, molecules, complexes and macromolecular structures can have very different spectra compared to their individual constituent atoms, because the energy levels for electrons change drastically when bonds are formed (or broken). For example aqueous $Fe^{3+}$ is yellow while aqueous $Fe^{2+}$ is green, while solid $Fe_2O_3$ is reddish-brown.

  4. Only about 10% of the light incident on the eye actually makes it through to the retina. Even those that strike the retina may not be detected.

  5. A human eye has receptors called cones and rods. Incidentally, a rod can actually respond to a single photon that strikes an active molecule in it, ultimately triggering an electrical pulse down the optic nerve. A cone is theoretically able to respond to a single photon as well, but for the below reason a single photon is never enough for us to see its 'colour'.

  6. Each cone absorbs incident photons of different frequencies with different probabilities. This is precisely how we can see many colours using only 3 types of cones, because light of different wavelengths can be distinguished by how much they are absorbed by each type of cone.


    But since a photon can only be absorbed by a single cone, it also implies that the retina plus brain needs many photons from the same source before it can get a statistical picture of absorption by the 3 types of cones, which it then interprets as a colour. This is the main reason we need thousands of photons from a point source before we can clearly distinguish its colour from that of other objects. The lower the intensity of light, the harder it is for us to distinguish colours. And note that we perceive the combination of pure red and pure green light (namely the combination of light of two different frequencies) the same way we perceive pure yellow light (of the appropriate single frequency), because they result in the same absorption profile for the three types of cones.

  7. Rods are much denser than cones, except in the fovea where there are nearly no rods, and hence one can see better around the central spot when in the dark. In the fovea, the 'Blue'-sensitive cones (S cones) are also rarer than the other two types at about 5%, whereas the 'Red'-sensitive cones (L cones) number about 50% to 75%.

The net effect is that you need something like 100,000 photons from the same point incident on your eye before you can perceive its colour at the normal human accuracy, even more for blue light.

And finally there is Rayleigh scattering in the Earth's atmosphere, which scatters 'violet' light (400nm wavelength) about $7$ times as strongly as red light (650nm wavelength).

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    $\begingroup$ The notion of a "statistical picture" of the light absorption by the cones is very nice. Well said! $\endgroup$ – adrienlucca.wordpress.com Jan 23 '17 at 15:10
  • $\begingroup$ You miss some important point about color vision. The 19th Century theory of trichromacy is not sufficient to explain color vision in detail: the real detector is in the brain. Even without much S cones in the fovea, and after blue filtration, we can still clearly see blue in the center of the visual field. More importantly, it is possible to see the color blue in absence of blue light, as shown by experiments by Edwin Land in the 1950's (see: millenuvole.org/f/Fotografia/Per-quali-ragioni-vediamo-i-colori/… ). If we work on dot-shaped light sources in the dark... $\endgroup$ – adrienlucca.wordpress.com Jan 24 '17 at 12:59
  • $\begingroup$ ...standard colorimetric rules apply, trichromacy is fully valid. So here it works. $\endgroup$ – adrienlucca.wordpress.com Jan 24 '17 at 13:00
  • $\begingroup$ @adrienlucca.wordpress.com: Yes that's why I was careful not to say anything about the way we interpret colour, hence I said "retina plus brain" instead of trying to expand further. No matter how the retina plus brain does it, they basically have only the statistical absorption levels of the three cone types to infer everything else. I indeed knew of colour constancy, which is what you're describing; you can actually try out the images on the Wikipedia article yourself! Anyway thanks for the link! =) $\endgroup$ – user21820 Jan 24 '17 at 13:29
  • $\begingroup$ you should read the article, it will blow your mind. Land was able to reproduce all hues with two slightly different yellow monochromatic light sources at 579 & 599nm, i.e. zero stimulation of the S cones. If you have an explanation for this, let me know. $\endgroup$ – adrienlucca.wordpress.com Jan 24 '17 at 13:35

It depends what you mean by "see". In a diffraction grating, even one photon will fall in the band of the "color" its frequency/energy assigns it.

A large ensemble of photons is necessary for "seeing" light which is described by classical electrodynamics. You can get an idea for how many photons are necessary to act like classical electromagnetism, from this double slit experiment one photon at a time:

dblsit singlephoton

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

The single frame has maybe 50 photons; by 200 frames, the interference pattern starts to appear. So I would answer that, by 10000 photons, the color should be visible through the complicated optics of the retina of the eye.

Edit after down vote.

Please note that the title of the question was drastically edited after I answered. There exists an answer with perception. This covers the detection of photons.

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    $\begingroup$ I think that the question is about the number of atoms needed to produce sufficient photons to stimulating the cones in the eye to produce the sensation of colour. Difficulties in answering this question include the fact that most sources tend not produce light just in one direction and to register a non-primary colour a number of cones need to be receiving photons from the source. $\endgroup$ – Farcher Jan 23 '17 at 7:38
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    $\begingroup$ If it is 50 (approx) photons per frame, would it not be 10,000 photons in the second image (then 50,000 and then 25,000,000)? $\endgroup$ – Baldrickk Jan 23 '17 at 14:49
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    $\begingroup$ @Baldrickk You are correct, I thought that the first on the left was 200 frames :( . I will edit. thanks $\endgroup$ – anna v Jan 23 '17 at 15:44
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    $\begingroup$ Not related to color, but interesting: once the rods in our eyes are suitably adapted, they are sensitive to as few as 6 photons! $\endgroup$ – Cort Ammon Jan 24 '17 at 1:23

To see the color of a piece of matter requires (1) a light source, and (2) a piece of matter. Third case (3) the atoms themsleves can be the light source. Your question is therefore not well formulated. You wont see any color if there's no light.

It seems first to fall down to the question: how many photons are required on a certain zone of the retina to produce a color stimulus?

To observe what you are thinking of with the naked eye in a dark room, you need :

  • either a piece of matter that is sufficiently big coupled with a beam of light that is sufficiently bright

  • or a piece of matter that emits sufficient light by itself to be noticed and seen as having a color

You should know that your results will be different in a dark room and under daylight:

  • in a dark room, the light dots that you'll observe will be clearly visible above a certain threshold

  • in the daylight, you'll need more light, the surrounding light will "melt" the stimuli

If some physicist here can calculate the power in terms of photons, here's how you could test your question :

Go to Photoshop and create a black picture, now draw some 1x1 pixels dots on it and look at the picture at 100% magnification :

1 pix dots on Bk

(look closely there are red, green, blue, magenta dots)

Personally I have a 15 inch 4k screen, it is possible therefore, knowing the screen's resolution and the geometry of its subpixels, to determine the dot size. You can also calculate the number of photons emitted by a dot it you know each color's value in terms of radiometric power.

The number of photons hitting your retina will depend on the pixel size, the color of the pixel (the emission spectrum of the pixel with a given RGB color), and your distance to the screen.

What is interesting here is that the answer you are seeking depends on the pixel color:

  • it is almost impossible for me to see the Blue pixels unless I get very close from the screen

  • the same is true with red and magenta pixels, but I can still see them and their color at a greater (about 2 times more) distance.

  • green pixels are much brighter (not because they emit more photons, but more likely because my retina is more sensitive to green), I can still see the dot at +- 8 times the distance, BUT, after a certain threshold, I see only a monochromatic dot, not a green one.

Now you can see that the answer to your question is more complicated than implied in the way you asked it.

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    $\begingroup$ I think the green/blue brightness discrepancy is partly due to the fact that rods absorb more strongly near the 'green' wavelength, as shown by the response curves, and my answer also explains why after a certain point you can no longer distinguish the colour. (See the experimentally obtained curve at en.wikipedia.org/wiki/Luminosity_function.) But I'm not so sure that a 0000FF pixel emits as high intensity of light as an FF0000 pixel. I think they are calibrated to achieve the white point, which has nothing to do with actual luminosity of the primaries. $\endgroup$ – user21820 Jan 23 '17 at 13:50
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    $\begingroup$ Yes, I'm guessing the main reason is the rod's absorption profile, but unless we know the actual luminosities of the screen pixels, it's hard to tell if it has significant effect as well. $\endgroup$ – user21820 Jan 23 '17 at 14:54
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    $\begingroup$ That's a log-plot of the graph in my answer, but how does it explain the brightness thing? I think your explanation is right partly because the Wikipedia article mentions that the experimental brightness curve shifts with colour-blindness in a manner consistent with the proportion and absorption profiles of the cones. $\endgroup$ – user21820 Jan 23 '17 at 15:31
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    $\begingroup$ @user21820 I know, I put it there to show you that the sensitivity goes as far as 830 nm! $\endgroup$ – adrienlucca.wordpress.com Jan 23 '17 at 15:34
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    $\begingroup$ Oh! That's very interesting! Sorry I didn't notice it earlier. Thanks! $\endgroup$ – user21820 Jan 23 '17 at 15:39

Each type of atom has a specific absorption/transmission spectrum, as does each specific type of molecule, e.g. a molecule of water which is two hydrogen atoms and one oxygen atom. The spectrum is which frequencies/wavelengths of light get emitted or absorbed by the molecule and how much of each frequency/wavelength. We interpret different frequencies/wavelengths as different colours.

If you can stimulate a single atom to produce enough light quickly enough, you would be able to see colour (a mixture of all its emission frequencies). So the answer is only a single atom is needed, under the right conditions.

When you have more atoms, each one can emit fewer photons on average to give the same overall intensity. There is the added advantage that the atoms are spread over a wider area, meaning more of your retina can be stimulated at the same time.

The rest of your question is about biology, not physics, i.e. how sensitive is the human eye to light and how much light is needed to be able to see colour? This is the question that everyone else has answered.

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    $\begingroup$ A transmission spectrum isn't the same as a reflectance spectrum. Your answer applies to observing transmitted light. An atom may not scatter at certain angles, in which case you will never see a photon if you're standing in the wrong place. $\endgroup$ – jiggunjer Jan 24 '17 at 7:30
  • $\begingroup$ As @jiggunjer says. There is transmission, emission and absorption. Incident light that is not absorbed is transmitted. Stimulated atoms or molecular structures can emit light, and the emission and absorption spectra have the same frequencies (because of same energy levels of electron orbitals) but different intensities (en.wikipedia.org/wiki/…). $\endgroup$ – user21820 Jan 24 '17 at 8:33

Colour is a biological/mental phenomenon, not a physical one.

This is nicely illustrated by the colour pink, or, the fact that red light and blue light together make pink light:

enter image description here

The primary colours of light are red, green, and blue, not because of any physics, but because those are the wavelengths that our eyes happen to be sensitive to.

The response curves for each colour are somewhat wide, as a consequence of that: if yellow light strikes our eyes it stimulates the red receptors a bit and the green receptors a bit. If our eye is stimulated by a red light mixed with a green light, the signals produced by the cells are indistinguishable from yellow. This is how computer monitors trick us into thinking they produce a spectrum.

Pink is not a colour insofar as there is no single wavelength of light that you can meaningfully label "pink" -- it HAS to be a mixture. The reason is that if you had a wavelength in between red and blue, it wouldn't stimulate red and blue receptors in the same way that yellow light does for red and green receptors -- the wavelength in between red and blue is simply green. We perceive the red+blue "colour" not as an interpolation, but as an entirely new, non-physical hallucination of a "colour" that doesn't really "exist" in the real world.

As others have pointed out, single atoms do have a colour. These give the characteristic spectra of neon lamps, or the absorption spectra of the outer layers of stars. When different atoms are together in a mixture, we see their overall colour as a mixture (brown, perhaps). Or, if the atoms are close enough for their wavefunctions to overlap (i.e. they have a chemical bond), then the characteristic wavelength of the whole collection might change (IIRC, this is why certain metals such as gold have anomalous colours).

Basically, the wavelengths of light are all there in nature. As for colour, it's a product of our minds and our eyes.

  • $\begingroup$ Great answer; good to clear up misconceptions! Any atom or object has a spectra of wavelengths that it absorbs/emits, which interacts with our eyes in a particular way that results in us seeing some colour, not because colour is intrinsic to the object, though the spectra is. And some colours cannot be perceived without at least two different wavelengths of light. $\endgroup$ – user21820 Jan 24 '17 at 6:57
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    $\begingroup$ "Pink is not a colour" this is pure nonsense. Of course Pink is a color like Black is a color, etc. $\endgroup$ – adrienlucca.wordpress.com Jan 24 '17 at 8:13
  • $\begingroup$ @adrienlucca.wordpress.com: Oh I didn't notice that; he probably meant "pink is not a wavelength", since his very first paragraph says "the colour pink"... Spraff: Yes please amend your answer! $\endgroup$ – user21820 Jan 24 '17 at 8:25
  • $\begingroup$ That’s magenta. I consider pink to be a desaturqted red, and is a spectral color. You’re talking about “the purple line” so purple is a better name. $\endgroup$ – JDługosz Jan 25 '17 at 2:57
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    $\begingroup$ "color" doesn't have a confused meaning; it simply has multiple definitions. Do this: "color 1. (optics) wavelength 2. (colorimetry) equivalence classes of cone stimulations 3. (psychology) ..." and the confusion disappears. In your answer you start a paragraph by saying that pink(3) is not a color(1), since there's no color(1) that is pink(3). Sure. But when you explain why, you suggest yellow(3) is an average(2) of red(1) and green(1) since there is a yellow(1), but since there's no pink(1), we hallucinate(3) it. That is confusion, and has nothing to do with how color(2) works. $\endgroup$ – H Walters Jan 28 '17 at 17:49

I think the question is ill-posed. Even single atoms can "have a colour" if you define having a colour as emitting photons with a certain frequency. The question should rather be: How many of these photons per time does the human eye need to absorb in order to perceive the corresponding colour? However, this is rather a biological question than a physical one.


First of all, individual atoms have color; Color is different wavelengths released by the atom when excited. Secondly, to see color you need about 0.1 square millimeters of atoms because this how small the naked eye can see.

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    $\begingroup$ You do not need to "see" the object emitting the photons all you need is enough photons emitted by the object hitting the retina. $\endgroup$ – Farcher Jan 23 '17 at 7:29
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    $\begingroup$ @Farcher: I agree. This is why a star we could not see with naked eye might be visible as a supernova. $\endgroup$ – sampathsris Jan 23 '17 at 8:29
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    $\begingroup$ This answer lacks logic and physical reality: (1) Color is not "different wavelengths", this is nonsense. Color is a sensation. (2) 0.1 mm² has no visual meaning, can you see a 0.1 mm² object at 10 km distance? Yes, if it is VERY bright... $\endgroup$ – adrienlucca.wordpress.com Jan 23 '17 at 12:10
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    $\begingroup$ @Hammar we know, however the question itself doesn't make sense, because there are many different combinations of "many atoms" that can be visible or not, colorful or not. $\endgroup$ – adrienlucca.wordpress.com Jan 23 '17 at 15:37
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    $\begingroup$ @adrienlucca.wordpress.com I understand what you are getting at but when a single photon hits a receptor in the retina you will not be "seeing" the object. $\endgroup$ – Farcher Jan 24 '17 at 8:43

protected by Qmechanic Jan 23 '17 at 9:11

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