Light falls on a surface. Some wavelengths get absorbed. The other are reflected. The reflected ones are the colors that we perceive to be of the surface.

What is the property that determines, what wavelengths are reflected and what are absorbed? Is it electronic configuration of the object on which the light falls?

If yes, then if we know the electronic configuration of a surface can we make a model, which will predict the color it will show?


This question is too broad. It involves ALL the objects in the universe which have a surface, i.e., everything. I'm going to avoid giving a lecture here.

In some liquids and most gases the electronic structure of each individual atom or molecule is enough to describe their spectra.

The "property" you are looking for in the case of solids is the band structure. See this page for a good introduction, specially the section about insulators and dopings. People arrived at the point where they needed to describe the spectrum of solids by (macroscopic) parameters instead of the atomic and band transitions (on top of that, sometimes you have to consider relativistic corrections): Absorption, and scattering or fluorescence. See this page for a shorter explanation, note that this is valid not only for visible light. For normal incidence, backward scattering is "reflection", while forward scattering is "transmission". The color you see will also depend on the flatness of the surface, but this does not affect the physics of the light scattering.

One interesting fact is that you could see (actually you can't, this holds for X-rays) that light gets "reflected" to more that one spot because of Bragg's law.

Lastly, according to your OP title, you might consider reading more about gratings, in which the surface is worked to specifically manipulate colors.


Several answers here already talk in great detail about how electron orbitals affect if a photon will be absorbed or not, but this is not the whole story. The color from reflected radiation is indeed the only factor if the surface is completely flat and perfectly reflective, excluding the black-body radiation, but most surfaces are not.

Take for example all the vibrant colors in a peacocks tail. They can appear to be a wide variety of greens, blues, and other colors. Now if you look at the same feathers after you flattened the surface you will see that the pigments in the feathers are actually brown. This kind of coloration stems from how the microscopic structures interact with the radiation, and is called structural coloration. For example, a grating-like structure will produce a color pattern that goes with the gradient of the rainbow, just as it does in a proper diffraction grating experiment. This same kind of effect can also stem from differences in reflection depths, as is the case in oil-spills, thus making them reflect a rainbow-like pattern as well.

Diffraction grating

There is also black body radiation, which is associated with the temperature of the object. This effect causes hotter objects to radiate some of it's temperature energy away in such a way that hotter objects shine brighter and with a peak that is shifted towards higher wavelengths.

Black-body spectra


Interesting question. I believe it has to do the energy levels of the atoms of the surface. I assume you understand the concept of atomic orbitals - When light encounters a surface, the wavelengths absorbed are those that contain the right amount of energy to move electrons up to the next orbital.

... Upon a google search: If the energy difference between two consecutive possible orbital states is ΔE, then the photons absorbed will be of frequency ν = (2πΔE)/h, where h is Planck's Constant.

  • $\begingroup$ ...But absorbed frequencies can be also emitted when electrons return to the fundamental state. $\endgroup$ – VictorSeven May 29 '15 at 14:12
  • 5
    $\begingroup$ @V_Programmer In solids or liquids this is not that common, as there are many strongly coupled vibrational modes, which allow non-radiative decay of the excited states. $\endgroup$ – Sebastian Riese May 29 '15 at 14:38
  • 1
    $\begingroup$ @SebastianRiese bingo! This is the essential physics so many folks overlook. This is why, for example, a black body looks black despite the fact that it emits radiation with maximum efficiency. The absorbed energy converts to internal heat (phonons) and is emitted as an equilibrium (black body) spectrum. In other words, the incoming optical photons are down-converted to a room temperature spectrum by the internal interactions in the object. Since most of that room temperature spectrum is IR, we don't see it. $\endgroup$ – DanielSank Jun 21 '15 at 20:54

A material displays a color when light is reflected off a surface. When a certain wavelength reaches a surface, if the energy $E=\frac{hc}{\lambda}$ of the photon correspond to the difference between two electronic states then it has a certain probability to be absorbed. The probability of being absorbed depends on the density of electronic states of course.

If a photon isn't absorbed it'll go through the material. This material will then be said to be transparent to its wavelength $\lambda$.

If the photon is absorbed, it will be re-emitted in no particular direction. But when emitted toward the inside of the material it'll be absorbed again, therefore the direction toward the exterior of the material will be favored (because in this direction the photons won't be absorbed). Those wavelength are being reflected.

When looking at the material, you'll see a spectrum made of all the reflected wavelengths, and that will make its color.

  • $\begingroup$ What do you mean when you say that a photon which is absorbed and then re-emitted is reflected. This is a contradiction in terms unless more detailed description is given. $\endgroup$ – DanielSank Jun 21 '15 at 20:51

Paul G. Hewitt has a great non-mathy description of this in his book Conceptual Physics.

Strike a tuning fork and it vibrates at a characteristic frequency, its Natural Frequency. The tuning fork might put out other frequencies but they are dampened faster than the target frequency. Dampening sucks energy out of vibration and becomes thermal if it doesn't become sound.

You can think of many things as being made up of tiny tuning forks. So hitting a metal sounds different from hitting wood.

Thes stiffness of a spring can determine the frequency of its vibrations. Again it has a natural frequency.

Like these examples, you can think of atoms and molecules as tiny optical tuning forks that re-emit some of the light and absorb the rest. Some light passes through.

The color you see depends on various features of the atomic and molecular interactions. For example, how strongly bounded an electron is to its parent atom is roughly analogous to the stiffness of a spring.

Vibrations in general can often be approximated through the analysis of Simple Harmonic Motion.

This is why Planck was able to analyze the interaction of Black Body Radiation with matter by assuming the radiation interacting with tiny "Harmonic Oscillators" in the surface of the black body.


Great question -- it opens up a lot of physics. My favorite example is "why do some conducting metals look gold/copper/etc. rather than grey?" Turns out it's due to relativistic effects acting on the various electrons' orbitals.

Then there's the blue jay, whose blue color is entirely interferometric rather than absorptive/reflective.

and more :-)

Edited to provide info about quantum mechanisms for color in metals. Quoting from wikipedia pages,

The characteristic color of copper results from the electronic transitions between the filled 3d and half-empty 4s atomic shells – the energy difference between these shells is such that it corresponds to orange light.

(references available at the "Copper" page)

Whereas most other pure metals are gray or silvery white, gold is slightly reddish yellow. This color is determined by the density of loosely bound (valence) electrons; those electrons oscillate as a collective "plasma" medium described in terms of a quasiparticle called a plasmon. The frequency of these oscillations lies in the ultraviolet range for most metals, but it falls into the visible range for gold due to subtle relativistic effects that affect the orbitals around gold atoms. Similar effects impart a golden hue to metallic caesium.

References include this page as well as various books.

  • $\begingroup$ Some expansion on the notion that relativistic effects account for the color of metals would be of interest. A self Q&A might go over well. $\endgroup$ – DanielSank Jun 21 '15 at 20:49
  • $\begingroup$ @DanielSank Added some material for you :-) $\endgroup$ – Carl Witthoft Jun 22 '15 at 11:34

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.