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I heard for certain materials, about 99% of the light interacts with the electrons. I heard it also depends on whether it is UV light, IR light, etc. Does the proportion increase for UV but decrease for IR? Is there a formula to derive this?

Similarly, how much of the light will interact with the nucleus? And what happens to the nucleus when light makes contact with it, preferably of lower wavelength energies?

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Chemistry Meta, or in Chemistry Chat. Comments continuing discussion may be removed. $\endgroup$
    – andselisk
    Commented Feb 26 at 9:28

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Consider that at visual frequencies,

  • The wavelength of light is, very roughly, three orders of magnitude larger than atoms.
  • The energy in visible light, very roughly, one eV, is about three orders of magnitude less than needed to disrupt the nucleus.
  • The mass of nucleons is, again, very roughly, three orders of magnitude greater than that of electrons, yet a protons has the same charge as an electron.

Imagine visible light washing over the whole atom, illuminating all at once. The heavy nucleons barely are affected, but an electron can be raised to a higher energy level. If an electron absorbs a quantum of energy, it is not split evenly among the other particles, but at that instant of observation becomes defined in space.

Ah! But what if the "light" had more energy? Brittanica.com states,

Absorption of gamma rays by nuclei can cause them to eject neutrons or alpha particles or it can even split a nucleus like a bursting bubble in what is called photodisintegration.

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When light shines on a solid surface, what % of light hits the electrons and what % hits the nucleus?

If we replace "hit" with "scatter from or become absorbed by" and restrict ourselves to visible light, then there is a better way to look at the question.

Light is an oscillating electric field and it's definitely the electrons that are the culprit when something happens to visible light.

We can split electrons in solids up conceptually into two groups; "bound" and "free". This is definitely a simplification but it's a been a very productive one.

Once we do that, we can think of materials as dielectrics, metals, and semiconductors/semimetals. They differ by the density of free electrons.

In dielectrics (think glass, salt crystals, etc.) light can travel a long way before being absorbed. The bound electrons do wiggle back and forth in the oscillating electric field, and that is what makes light move more slowly (think index of refraction) but they usually don't scatter or absorb it. Impurities, defects ("color centers" or F-centers) or other things that perturb the ordered state of bound electrons gradually scatter or absorb light as it passes through (think obsidian (dark glass) or colorful precious gems).

In metals with their very high density of free electrons, light doesn't stand a chance! Within a very thin layer at the top (a few to a few tens of nanometers, much shorter than a wavelength) the "free" electrons run back and forth in response to the electric field of light, setting up oscillating currents. These exactly cancel the forward going beam so that nothing propagates forward, and at the same time produce an identical beam of light going back out into vacuum. This is what is happening during reflection from metal surfaces. If the frequency of the light is too high though, there may not be enough electron density to do this, and a single crystal of the metal becomes transparent in the UV.

For more on that see:

The highest frequency that can be reflected for a given free electron density is called the plasma frequency and this concept applies to anything with a significant density of free electrons, including our ionosphere, whose plasma frequency varies between about 15 and 50 MHz depending on what the Sun is doing. Ionosphere; Layers of ionization. Lower frequencies are reflected back to Earth (so we can listen to a short wave radio station thousands of km away) and higher frequencies are transmitted (so we can listen to deep space spacecraft, GPS, and Starlink internet).

In semiconductors things get very interesting and we can play all kinds of games with the band gap. The energy of visible (and near infrared) light can be absorbed in a few microns in silicon because it can kick an electron up from one band to another. But the carrier density is low enough that those can freely be pulled away from each other and collected as charge before scattering off of other carriers. So we have CCD imagers, yay!

The reverse process (charge in, light out) gives us LEDs and semiconductor lasers which revolutionized several fields of science, technology and communications.

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  • $\begingroup$ semiconductors can also give solar cells where the light is effectively converted to electricity. $\endgroup$
    – matt_black
    Commented Feb 26 at 15:45
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    $\begingroup$ @uhoh I only added that because with my reputation level I can only suggedt edits with at least six character changes. I had initially thought that this was the correct way to pluralize acronyms in English, but after your comment I did some searching and learned I was wrong with that. Sorry :( $\endgroup$
    – Geeky Guy
    Commented Feb 26 at 17:20
  • $\begingroup$ @GeekyGuy no problem, and thanks for all the useful edits! I really wasn't sure either, I'm a major apostrophe offender; I always write "let's" when I should write "lets". I've been that way my whole life, it's hard-wired. $\endgroup$
    – uhoh
    Commented Feb 27 at 0:26
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Matter is energy trapped in a field or fields that give the energy certain properties that we manage to measure or at least acknowledge. Photons seem to be pure energy, however, they have properties that are associated with mass: Kinetic energy associated with length or possibly time of traverse, momentum, an electric and magnetic field, that crazy thing of polarization or entanglement; they are complicated little, or not so little, devils.

Photons encountering matter interact with all the fields that define that matter. If a specific change in an energy level is possible within the time frame of the interaction it might happen. We work with the differences in energy of the fields that define matter not the absolutes; the absolutes are the baseline. A photon does not hit the electrons or the nucleus; it interacts with all the fields defining the matter and effects changes where the energy fits. The photon does not enter the atom looking for an electron to dropkick to a higher level. It perturbs the fields defining the atom; the atom etc. then relaxes into its lowest available quantum state. If none are available, the photon proceeds on its way.

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Edit: I've replaced occurrences of the word "hit(s)" with "interact(s) with" as the former term is not to the liking of those who prefer using perfect terminology.


All the radiation interacts with the entirety of the atoms. Think of it as a spatial wave of energy washing over you.

The incoming radiation is absorbed as internal energy, and part of it maybe reflected (or refracted) by the atoms vibrating in sync with the radiation. From there, some of the absorbed radiation gets emitted, and the rest remains as heat. You can read more about what happens to absorbed light from this answer.

If the radiation is high enough, it can result in complete emission of electrons (photoelectric effect or ionizing radiation) or of nucleons (photodisintegration).

The level of excitation or emission of particles is defined by the binding energy of the particle system.

It can be energy gap between electron shells, or excitation energy (for excitation), the threshold energy (for photoelectric emission, in the solid state), the ionization energy (for emission of electrons in the gaseous state), average nuclear binding energy per nucleon (for emission of nucleons), etc.

The structure of the material may also affect the orbital excitation energy in more complex non hydrogen-like molecules due to quantum decoherence.

The answer to your question, "what % of light hits interacts with the electrons and what % hits interacts with the nucleus" is all of it.

But how much is distributed between the electrons and the nucleus is more complicated, and is described by the equations of quantum mechanics. I will quote from the answer I linked earlier:

Heating applies to the whole system (nuclei and electrons). However, due to the large difference of mass between nuclei and electrons (larger than 1800), and taking into account that the correct description of the dynamics at atomic level is provided by quantum mechanics (QM), the way a gain of total energy is distributed among different internal degrees of freedom is quite complex.


Additional Info

The general types of interactions between light and matter involve:

At Lower Energies (At the atomic level):

  • Absorption (excitation), Reflection, Refraction, Emission (of light)
  • Photoelectric Emission
  • Mie Scattering (Coherent and elastic scattering with a forward directionality, for particles sized similar to the wavelength)
  • Rayleigh Scattering (Coherent and elastic scattering due to polarization, and is more wavelength dependent than Mie Scattering)
  • Raman Scattering (Incoherent and inelastic scattering due to absorption by molecules)

At Mid Energy Levels (At the single electron level):

  • Thomson Scattering (Coherent and elastic scattering from a free electron by the electric field; Classical low energy limit of Compton Scattering, caused by accelerating the electron)
  • Compton Scattering (Incoherent and inelastic scattering from a free or weakly bound electron, caused by absorption and emission by an electron)

At Higher Energies (At the nuclear level):

  • Pair Production (Emission of a particle-antiparticle pair after absorption by a nucleus)
  • Photodisintegration (Emission of nucleons, mainly neutrons, but also protons or alpha particles)
  • Photofission (Nuclear fission)
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  • $\begingroup$ By mid-energy levels that includes IR and UV? Or is IR for lower-energy levels? $\endgroup$ Commented Feb 26 at 11:00
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    $\begingroup$ @NealConroy well in this case, lower energy would include UV. Compton Scattering was observed for X-Rays. Thomson scattering is the limit of Compton scattering below ~1.2 EHz. And Pair Production is observed beginning with Gamma Rays. The mass of an electron is 0.511 MeV, so we need twice that to form a pair, so 1.022 MeV (~247 EHz). At this point we start referring to energy in terms of eV and not Hz. X-Rays are 100eV - 100 KeV, and Gamma Rays are upwards. Ionizing Radiation begins at around 10 eV - 33 eV in short-wave UV (UV-C / Far-UV). $\endgroup$ Commented Feb 26 at 13:09
  • $\begingroup$ Ok, so lower energy levels wavelength almost has no reaction with when it hits the nucleus, only at high energies? $\endgroup$ Commented Feb 26 at 13:44
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    $\begingroup$ It does not "hit the nucleus". $\endgroup$
    – jimchmst
    Commented Feb 26 at 19:36
  • $\begingroup$ @jimchmst sure, the orbitals could carry the wave around, but what about high frequency radiation that removes nucleons? $\endgroup$ Commented Feb 27 at 2:11

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