I believe I have a misunderstanding of some principles, but I have not, even through quite a bit of research, been able to understand this problem.

My current understanding of transmission, reflection and absorption is as follows:

transmission occurs when the energy of an incident photon does not correspond to any electron's energy transition within the material. Therefore, the photon does not interact with the atoms / electrons and is transmitted through.

Absorption occurs when the incident photon's energy exactly equals that of an electron's energy transition. The photon is absorbed and excites an electron to a higher state.

Reflection I feel like my understanding is flawed, since I have read multiple different views. I believe that a photon is absorbed by an atom, exciting an electron. The electron, however, almost immediately transitions back into a lower energy level, emitting a photon of identical wavelength.

My question concerning reflection is:

  • Why are some wavelengths absorbed and immediately re-emitted? I presume that it is because the electron is in a type of unstable state and therefore drops back to its previous energy level?

Given a solid object that appears red to us (therefore reflects wavelengths somewhere between 625 and 740nm), how can it be possible that all other incident wavelengths are absorbed? They must be absorbed, since the only wavelength being reflected is in the "red" range, and I can clearly see there's no visible light being transmitted through the object. However, in my knowledge, the wavelengths can only be absorbed if they correspond to the energy transition of an electron, which is not the case for every wavelength in the visible spectrum.

How is it then possible that they are absorbed? Additionally, if the electron is excited to a higher level, does it just store the energy? Does it take thermal form??

I assume that perhaps I cannot simply apply these principles of absorption, that I was taught only in relation to a single atom, to a complex body consisting of billions of atoms. Could someone elaborate on this and explain my questions about absorption and reflection? Thanks very much!

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    $\begingroup$ Reflection is best understood using the wave description of light. However, as far as I understand your question you are solely interested in the particles description, right? If so, I reckon one has to describe the solid as a coherent superposition. Describing reflection as a photon interacting with a single atom will not satisfy Snell's law. $\endgroup$
    – Semoi
    Apr 28, 2020 at 18:09
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    $\begingroup$ Yes, i am interested in understanding better how these principles (reflection/absorption) work on a particle level. If anyone would be willing to link some helpful resources about my situation I would he thankful also as I have had trouble finding good sources of understandable information. $\endgroup$
    – Ultralite
    Apr 28, 2020 at 18:31
  • $\begingroup$ One thing to consider - objects absorb very few wavelengths. Visible light comprises a tiny band of wavelengths of the light that's all around us. It also neatly fits between two special "edges" - longer wavelengths are great matches for things like atoms rotating in molecules or just the movement of molecules in bulk (so it doesn't deal with electron transitions at all - they just don't have enough energy), and shorter wavelengths have enough energy to break most double bonds. Metals share electrons in a cloud, so they don't have distinct energy levels in the first place. $\endgroup$
    – Luaan
    May 1, 2020 at 18:12

4 Answers 4


Your misunderstanding is very common and quite easy to make. Basically, what students are usually introduced to first is the thermodynamics of ideal monoatomic gasses. This is good because it is simple and easy to understand, but can be problematic because features specific to the simple substance can be misunderstood as general features of all substances.

In an ideal monoatomic gas light can interact either by scattering or by absorbing an amount of energy corresponding to an atomic transition*. Note, in the latter case the photon is not absorbed by the electron but by the atom as a whole because the atom has different internal states corresponding to the absorbed energy. As a result ideal monoatomic gasses tend to be transparent except at a few narrow** frequencies.

Now, consider a molecular gas. Just like an atom has internal states that an electron does not, similarly a molecule has internal states that an atom does not. Some states correspond to electron transitions in the molecule, but others correspond to rotational or vibrational modes. The molecular electronic transitions combined with the molecular vibrational and rotational transitions gives rise to a multitude of absorption lines, often forming continuous absorption bands, so many times these are visibly not transparent.

Now, consider a solid. Just like a molecule has states that an atom does not, similarly a solid has states that a molecule does not. The rotational and vibrational modes gain additional degrees of freedom and can act over fairly large groups of molecules (e.g. phonons). These states can have energy levels that are so closely spaced they form continuous bands, and are called energy bands. Any energy in the band will be easily absorbed. This makes most solids opaque as they absorb broad bands of radiation.

Finally, when a photon is absorbed it may be re-emitted at the same wavelength to fall back to the original energy state. However, if there are other energy states available then the energy can be emitted and retained at different energy levels. For example, a UV photon could be absorbed and a visible photon could be emitted along with an increase in a rotational degree of freedom.

*Even for an ideal monoatomic gas there are other less common mechanisms like ionization and deep inelastic scattering, but for clarity these are neglected here.

**Note that even for an ideal monoatomic gas the frequency bands are not infinitely narrow but have some breadth. This is caused by two factors. First, the width of the peaks is fundamentally limited by the time-energy uncertainty relation which says that $2 \Delta T \ \Delta E \ge \hbar$ where $\Delta E$ is the width of the energy band and $\Delta T$ is the lifetime of the transition. Second, random thermal movement of the gas will cause Doppler and pressure broadening of the frequency band.

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    $\begingroup$ Thank you, I found that your progressive explanation with gases made it very understandable for me, I think it has finally clicked! $\endgroup$
    – Ultralite
    Apr 28, 2020 at 20:17
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    $\begingroup$ Dale, this is a great answer! $\endgroup$ Apr 29, 2020 at 6:21
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    $\begingroup$ Great answer. You might want to add a word on uncertainty principle impliciation that the absorption bands/states are not infinitesimally narrow, therefore if they are close enough they overlap. $\endgroup$
    – Ben
    Apr 29, 2020 at 7:50
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    $\begingroup$ @Ben thanks for the suggestion! I have added a paragraph mentioning the uncertainty principle and Doppler broadening. $\endgroup$
    – Dale
    Apr 29, 2020 at 11:17
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    $\begingroup$ @Dale: thank you for the follow-up. I took "visibly not transparent" as "does not let (visible) light through" but with your comment it becomes clear. $\endgroup$
    – WoJ
    Apr 29, 2020 at 17:21

The other answers cover almost everything, but I would like to add that at any temperature above absolute zero, there is a degree of line broadening caused by Doppler shift: some of the atoms are moving towards you, and others away, and that will mean that in your reference frame they can absorb many different frequencies of light. This is important in astronomy.

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    $\begingroup$ True, but in the Earth's atmosphere pressure broadening is much stronger than Doppler broadening. $\endgroup$
    – jkej
    Apr 29, 2020 at 17:22

Dale and Arpad already gave great answers, but I want to correct something that you said that also contributes to your confustion:

transmission occurs when the energy of an incident photon does not correspond to any electron's energy transition within the material. Therefore, the photon does not interact with the atoms / electrons and is transmitted through.

This statement is not correct. The reality is more close to the statement you gave in reflection:

I believe that a photon is absorbed by an atom, exciting an electron. The electron, however, almost immediately transitions back into a lower energy level, emitting a photon of identical wavelength.

This "momentary absorption" is what causes the refractive index of materials to emerge. The closer the frequency of the photon to the frequency of an energy transition in an atom, "the more time it spends absorbed until re-emitted", this is why the refractive index gets higher the closer you are to an absorption line.

This absorption of energy and re-emission is called Rayleigh scattering, and the re-emitted photon can be emitted in any random direction (with the probability distribution following the antenna radiation distribution). However, because this happens in multiple atoms, the waves constructively interfere only in the forward direction, and destructively interfere in any other direction. This is explained wonderfully by Boyd in his nonlinear optics book:


  • $\begingroup$ Thanks a lot for the answer, it is very helpful. I had previously read up a little on Rayleigh Scattering but had not thought of it at all in the context of this matter. Fascinating topic, thanks again. $\endgroup$
    – Ultralite
    Apr 28, 2020 at 20:28
  • $\begingroup$ @Cd01 You're welcome, happy to help! :) $\endgroup$ Apr 29, 2020 at 10:41
  • $\begingroup$ I can't relate refractive index with the time electron spends in the excited state, . Please tell me how these two are related @OfekGillon $\endgroup$ May 11, 2020 at 6:39
  • $\begingroup$ @sawankumawat The refractive index is related to the speed of light in the material, and the speed is decreased by the time the electron spends in the excited state (repeated at each atom with the new photon that is released) $\endgroup$ Jul 21, 2022 at 21:56

You have a lot of questions, I will answer as many as I can.

When a photon interacts with an atom, three things can happen:

  1. elastic scattering, that is mirror reflection, is when a photon keeps its energy level, phase and changes direction

  2. inelastic scattering, the photon gives part of its energy to the atom, and changes direction, heats up the material

  3. absorption, the photon ceases to exist, and gives all its energy to the absorbing electron/atom system

Now you are asking why a red object is red. Now it is very important to understand that reflection is elastic scattering, and mirrors (most metals) do not have their own color, they just reflect all incident light. Gold, and some other metals are exceptions.

Now why is a red wall red? It is not (just) reflection, but because most of the incident photons are as you say absorbed, and re-emitted. Most of the photons that are re-emitted are red wavelength. Whatever the incident photon's wavelength, the surface of the material is so, that the atoms absorb almost all wavelengths (reflect some), and re-emit red wavelength. Now this is by natural Sunlight, which is mostly white, containing a combination of all visible wavelengths, and most of these wavelengths get absorbed, but only red wavelength get emitted.

But how does this wall do this? The atoms on the surface of the wall have this ability to absorb all kinds of visible wavelengths, and still emit red wavelength. The atom/electron system absorbs a certain wavelength photon, gets excited. Then, the atom/electron system relaxes in a spacial way, that could be multiple photon emission, cascades, etc, but mostly emitting red wavelength.

It is very important to understand that a white wall will appear red, if you shine red light on it. What happens then, is that the atoms on the surface absorb all the red wavelengths, and re-emit the same. The white color wall is able to do that, the atoms are able to re-emit the same wavelength photons as they absorbed.

In the case of the wall, this is diffuse reflection. Mirrors, do specular reflection, which is elastic scattering.

In your case, the wall does diffuse reflection, and most of the photons get absorbed and re-emitted in random directions. Mirrors, elastically scattering, keeping the energy level, phase and relative angle of the photons, this is the only way to keep a mirror image.

The wall cannot do that. It can only re-emit the photon in random directions, and only re-emit certain wavelengths depending on the surface atoms of the wall. A white wall is able to re-emit the same wavelengths as it absorbed, while a red wall only emits red wavelength, mostly regardless of what wavelength it absorbs.

  • $\begingroup$ Ah, I see now I had another misunderstanding in that I did not consider that an incident wavelength could cause the emission of a different wavelength. Thank you very much for your answer, it was very insightful! $\endgroup$
    – Ultralite
    Apr 28, 2020 at 20:22
  • $\begingroup$ @Cd01 I am glad I helped. $\endgroup$ Apr 28, 2020 at 22:32
  • $\begingroup$ @Cd01 well actually when a photon is absorbed at one wavelength and remitted at another it is called inelastic/Raman scattering and this accounts for a minuscule amount of the emission. Elastic/Rayleigh scattering where incoming frequency equals outgoing is 1000X more in intensity $\endgroup$
    – ChemEng
    Apr 29, 2020 at 0:30
  • $\begingroup$ @Cd01 You are aware of fluorescent materials that give off visible light under UV "black" light, right? $\endgroup$
    – Kaz
    Apr 30, 2020 at 0:09
  • $\begingroup$ I'm not sure if this is a correct description. The process of converting photons of one wavelength to another longer wavelength (fluorescence) certainly exists, but AFAIK it's not the main process that makes objects have a color. What I learned is that colored materials are not 100% opaque, so light will penetrate a little bit into the interior of the object, where it gets scattered so much of the light finds its way out again. During the travel through the interior some colors will get absorbed (and mostly converted into heat), and the colored photons we see are those that don't get absorbed. $\endgroup$
    – JanKanis
    May 2, 2020 at 14:11

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