30
$\begingroup$

Suppose I have some colorful solid, which I cut into two halves (both are identical). Take the first and cut it into two parts, and then repeat this again and again. I know that a single atom doesn't have a color. So there will be a point at which the solid loses its property of being colorful. I'm interested in this particular point:

If we repeatedly divide a colorful solid in half, at what point will the color disappear?

$\endgroup$
12
  • 31
    $\begingroup$ “I know that a single atom doesn't have a color” How do you know that? $\endgroup$ – Dale May 7 at 20:01
  • 3
    $\begingroup$ I think "being colourful" is not just a property of the object you're looking at: The photoreceptor cells in your retina play a role. $\endgroup$ – joigus May 7 at 20:11
  • 17
    $\begingroup$ Here we see that carbon atoms are black, hydrogen white, oxygen red, and nitrogen is blue. I guess that's why the sky is blue. en.m.wikipedia.org/wiki/Molecular_model $\endgroup$ – badjohn May 8 at 8:34
  • 2
    $\begingroup$ @badjohn No! The sky is blue due to Rayleigh scattering. The concentration of nitrogen isn't anything like high enough to give colour. Also the colours there have nothing to do with what wavelengths atoms actually absorb (unless you're being sarcastic?) $\endgroup$ – Max May 8 at 14:11
  • 16
    $\begingroup$ @Max It was a joke. $\endgroup$ – badjohn May 8 at 14:57
30
$\begingroup$

What Vasily Mitch says is true (+1). But some objects are colorful because of interactions that take place over a larger region than a single atom.

Metals reflect light because electrons spread out through the metal. They can easily move, which makes them conductive. Classically, the oscillating electric field in light vibrates the electrons, and vibrating electrons radiate light. It works out that the incoming light is absorbed and the radiated light is the reflection.

Electrons can move because neighboring atoms don't have individual separate orbitals, each with the same energy. Instead, they overlap and forming bands, states with many closely spaced energy levels.

Copper is such a solid. Copper is more conductive at lower frequencies. Red and infrared light is reflected well, but blue light is absorbed. So copper is copper colored.

For this to work, there have to be enough metal atoms in a solid with a band structure. The solid must be big enough for the electrons to vibrate and radiate. The minimum size is around a wavelength of light.

Using this idea, one can make a wire grid polarizer. Fine metal wires are deposited on a glass substrate. Light polarized parallel to the wires can excite electrons to vibrate along the length of the wire, and is reflected well. Light polarized perpendicular to the wires cannot, and is not reflected. It isn't possible to make wires fine enough for visible light, but it works for infra-red wavelengths and longer.


Thin films can also reflect colored light. Thin films are spaced layers, where light reflects off each layer. If the round trip to a deeper layer adds up to an extra wavelength of light, the reflections add constructively forming a bright reflection. For a different wavelength, the round trip might add an extra half wavelength and the reflections would cancel.

Light traveling through a transparent substance interacts with the atoms it passes. The interactions slow the light down. The index of refraction describes the degree of slowing. $n = c/v$. When light passes from one medium to another, some of it is reflected, and it undergoes a phase change. These can be calculated from the two indices of refraction.

For this to work, the film must be big enough to form a solid or liquid through which light can travel. The thickness of the films must be around a wavelength of light. The area of the reflective surfaces must be at least that big.

$\endgroup$
4
  • 2
    $\begingroup$ Interesting. Some butterflies also have wings that appear colorful thanks to their structure (asknature.org/strategy/…). I guess that if you zoom long enough, you'd just see grayscale wings. $\endgroup$ – Eric Duminil May 8 at 15:23
  • $\begingroup$ So a good summary is probably that it depends on the mechanism by which the solid gets its colour, but a given colour is probably lost when the solid drops below the relevant wavelength... if for no reason other than that it becomes difficult to see a solid at those scales. $\endgroup$ – Mark Morgan Lloyd May 10 at 9:58
  • $\begingroup$ While the thin film example shows that larger structures can be colorful in ways an atom cannot, it seems like you should also highlight that it's a phenomenon that gives the solid a color that it wouldn't have if it were thicker. So in the context of the original question, a solid does not just eventually lose its color when it's subdivided below a certain scale but can also gain color by getting smaller in some dimensions. $\endgroup$ – Will May 10 at 10:39
  • 1
    $\begingroup$ @Will - True. I had not thought of that. Like an oil film on water has color that a drop of oil does not. $\endgroup$ – mmesser314 May 10 at 14:17
28
$\begingroup$

It of course depends on what you define as colour.

If it is defined as the change in the visible spectrum of a light, then a single atom can definitely absorb a photon of a preferable wavelength and thus slightly change the spectrum of the passing light. Many atoms have excitation energies falling into the visible spectrum

enter image description here

when atoms absorb the photons, they produce the colour complement to the colour of the flame. Similar to how element absorption creates black lines in the spectrum of the sun. The change in spectrum from a single atom will be barely measurable, but it will be there.

enter image description here

Source: XKCD

However, if you define colour as something that can be perceived by human as colourful, then my guess is that the dye speckle of dozen of microns should be on the boundary of perception.

$\endgroup$
8
  • 1
    $\begingroup$ For perception of color, we could also think of dilution of a dye in a solvent, say water for simplicity. You don't just get color transitioning to no color at some point, you get a dilution, e.g. a deep red gradually fading through various shades of pink until it no longer can be perceived. $\endgroup$ – jamesqf May 8 at 16:38
  • 1
    $\begingroup$ @jamesqf - In some crystals, the color comes from sites the size of atoms. See What Causes the Purple Color of Amethyst? and the RP Photonics Encyclopedia Article on Color Centers. But yours is also a good point. $\endgroup$ – mmesser314 May 8 at 16:46
  • 4
    $\begingroup$ Please credit the source of the drawing: xkcd.com/1733 $\endgroup$ – WoJ May 9 at 10:19
  • 1
    $\begingroup$ Even though the image is a joke, you can see that it is mostly correct. See Wikipedia Fraunhofer lines $\endgroup$ – mmesser314 May 9 at 17:04
  • $\begingroup$ Never forget: in the best possible conditions, with the best possible human, single photons are at the edge of detectability. nature.com/articles/ncomms12172 $\endgroup$ – Eric Towers May 9 at 21:15
7
$\begingroup$

It depends on the solid. Molecular solids derive their color, that is their optical reflection or transmission, from the molecules that they consist of. The molecules stick together by van der Waals forces, which have only little impact on their spectral properties. Some solids, like quartz (glass), diamond, ruby, derive their colors from impurities. You need an impurity to be surrounded by a few layers of matrix material to act as in the bulk case, that is a nano crystal of at least 50 to 100 Al2O3, C or SiO2 units. In metals collective electronic excitations determine the color. Again you will need a nano crystal of 50 atoms at least to approach bulk color. Colors can also be caused by interference effects such as in thin layers of oil on water or in photonic crystals, such as occur on butterfly wings. In such cases a much larger size of multiple wavelengths is needed, so 10.000 or more atoms in each direction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.