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Atomic Components and the Absorption of Photons

  1. Sep 17, 2009 #1
    My question concerns my understand of the mechanics of the absorption of Photons by the components of atoms. I will use a common real-world example to facilitate my understanding of the concept.

    1. When a rock (for example) sits in the sunlight, the rock warms to the touch. I believe this is a result of the absorption of photons by atoms, more specifically by electrons orbiting the nucleus inside those atoms. Is this correct, and do any nuclear objects (protons / neutrons) also absorb these sun-generated photons?

    2. Electrons that orbit the nucleus jump to a higher orbit once a photon is absorbed, but only absorbs photons that are exactly energetic enough to force it into a higher orbit. Is this correct, and if so, what happens to the less-energetic photons (or more energetic photons)? Reflected?

    3. What makes the rock "feel" hot to the touch? I assume this is a result of those same electrons falling back to a lower orbit, and discharging a photon, which is the source of the heat we "feel". Is this correct?

    4. Assuming my statement in #3 is correct, what keeps those electrons from immediately discharging their photons once the sun sets (rock in darkness), in a flash of brilliance? Why does it take so long for a rock to cool down to ambient temperature? Are these electrons in a higher orbit not unstable and naturally 'want' to fall to a lower orbit?

    Lots of questions, but I believe they are fairly related. Thanks in advance.
     
  2. jcsd
  3. Sep 17, 2009 #2
    I'll answer question 3) because I feel most confident to take that one, and it relates to 4)

    What makes the rock feel hot to the touch has very little to do with the electrons. It is mainly because of the phonons ( lattice vibrations of the rock) generated in the process of optical electron excitations.

    Let me elaborate: Say, you have a simple Silicon crystal ( instead of a more complicated disordered "rock" lattice). Now you probably know that in a big crystal, we have energy bands instead of discrete levels (those are for molecules and a rock is too big to be considered a molecule).

    So whenever an electron is excited from a lower band to an upper band (typically the bands that are involved are the valence bands and the conduction bands), some energy is absorbed from an incoming photon and this is what puts the electron into the excited state. But unfortunately (for various technology purposes), most semiconductors (where this type of optical excitation can actually be put to use for solar-cells) have an indirect band gap. This very basically means that simply providing the energy to the electron is NOT enough to excite it to a higher level. Momentum must also be conserved, and in INDIRECT band-gap materials (by definition), the state the electron jumps into has a DIFFERENT Momentum than its original state. In this case, for this transition to occur, a PHONON ( lattice vibration) must accompany the excitation process, to conserve momentum.

    In short, while exciting these electrons you end up causing a lot of vibrations in the material, which are in turn resulting in the "heat" generation at the macroscopic level...

    These lattice vibrations quickly "die away" because the material is in constant contact with the environment...(For instance, you act as a heat sink when you touch the rock and feel it's warm)..
    I think this is the dominant mechanism for the warmness of the rock. Of course, one could talk about electronic heat conduction and so on, but I think in this case, they have little effect.
     
  4. Sep 17, 2009 #3

    alxm

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    Well, to being with, let's just say what 'heat' is:

    Heat, at the temperatures we're used to (rather than inside the sun or something) is motion. A measure of the amount of motion of atoms and molecules. You can (roughly) break that motion down into three categories: The vibrational motion of atoms in a molecule or crystal lattice and the rotational motion.

    These have discrete quantum energy states, just like with the motion of electrons. But here we're talking about the motion of the nuclei of the atoms. Or the atoms as a whole if you like. (The electrons move so incredibly fast compared to the motion of the nuclei that we tend not to think of them. They'll go where the nuclei go, without complaint) These vibrational states are what which sokrates calls phonons because he's a solid-state physics guy, but which I call 'vibrational energy states' because I'm not. :wink: Anyway, so atoms can absorb photons and go to a higher vibrational or rotational energy state, and re-emit them.

    Since nuclei are heavier and move slower, the energy involved is lower. They're generally in the infrared. Which is why 'thermal radiation' or 'heat radiation' is the same thing as infrared. (Which is below the visual range in energy, recall that electronic transitions are generally from the visual range and up, through UV to X-rays) If something gets hot enough, it'll get to energy levels so high that it enters the visual range (which starts at red). So that's 'red hot' for you.

    If something's at thermal equilibrium (constant temperature) There's an equilibrium between thermal radiation and the molecular motion of matter. It's giving off the same amount of radiation as its absorbing at any given moment. (otherwise it'd heat up or cool down)

    Okay, so to answer #1: It's mostly due to absorption of infrared radiation. Sokrates explained in good detail how electronic transitions can be combined with vibrational ones. So visual light does heat things. But not as much as we might think. We're accustomed to sources of light also giving off lots of infrared (the sun, fires, light bulbs, glowing hot objects). But if you think of one that doesn't - e.g. a fluorescent bulb, you may get an idea of how small that effect is. Imagine heating something with a flourescent bulb!
    (http://www.pl.euhou.net/docupload/files/Excersises/WorldAroundUs/Spectroscope/Spectra/Halophosphate_bulb_spectrum_labelled.png" [Broken])

    To answer #2 - Yes, if a photon doesn't match the energy level between two states it can't be absorbed. (it can't always be absorbed even if it does match) More or less anyway. (there are some chains of events that make a lot of these 'forbidden' transitions just 'very improbable') If it's not absorbed it continues on its merry way. It's transparent.

    #3 - Not quite. You feel heat because the rock is warmer than you are, and heat's being transferred to you. That transferral is mostly through what you usually call heat conduction, which at the molecular level is as simple as one molecule/atom bumping into another and slowing down a bit while the other speeds up. There's also convection, which is conducting but in a liquid or gas, which is a bit special since they expand and rise when heated. And the third way of transferring heat was already mentioned: Radiation. Emission of a 'thermal' photon which is then re-absorbed by something else.

    When you're feeling a rock, conduction dominates. The light from the sun, is of course all radiation. Most of the heat from a fireplace would be radiation too (conduction would be transferred by the floor or walls, convection would be the air, but most of that is going up the chimney).

    #4 - Okay, so it's not electrons, but the question of why they don't reach equilibrium very quickly is still valid. Part of the answer here is that things are very opaque to infrared. In other words, heat can't get out of a rock by radiation except at the surface. Not because it's not emitting it all the time, but because it's getting re-absorbed all the time.

    A rock will also stay hot because it's simply stored up a lot of heat; it's a dense object.
     
    Last edited by a moderator: May 4, 2017
  5. Sep 18, 2009 #4
    I see, but apparently the process does involve electrons. But they are not the direct cause of heat. Understood. :wink:

    Do you mean "band gap", or "Gap Band"?


    http://images.uulyrics.com/cover/t/the-gap-band/album-gap-band-iv.jpg [Broken]

    Sorry. Had to do that....:tongue2:

    So the atomic electrons are excited by photons, causing phonons and vibrations, which causes the feeling of heat.

    So the electrons absorb the photons (infrared radiation), jump to a higher state, and that (indirectly) causes the vibration of the nucleus (entire atom)?

    Is a "thermal photon" just a high-energy photon? High Frequency?

    I think you are saying that infrared radiation cannot escape the rock easily. A lot keeps being bounced around inside the rock, some being reabsorbed. Right?
     
    Last edited by a moderator: May 4, 2017
  6. Sep 19, 2009 #5

    alxm

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    No. Electrons do not absorb in the infrared. I said before, their energy levels are higher than that. Electrons can absorb visual light (and higher), and some of that energy can be transferred to vibrations of the atoms/infrared.

    When infrared is absorbed directly it turns directly into vibrations (and the reverse occurs as well).

    No, being in the infrared, I'd say it's a low-energy photon. At least relative visual light and UV and x-rays. It's high in energy compared to radio-waves. But those barely correspond to any atomic/molecular states at all. Which is why they pass through most stuff without problems.

    Right, it can only escape from the surface, really. See the formula used in practice, the
    http://en.wikipedia.org/wiki/Stefan–Boltzmann_law" [Broken]. (Where the amount of energy being radiated per unit of time is proportional to the surface arae)
     
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  7. Sep 19, 2009 #6
    I'm sorry. :confused: Then what absorbs the infrared radiation? Nuclei?


    That's right. I failed to consider that infrared "light" has lower frequency. I guess the thing that confuses me here is the term "infrared". For some reason in my mind I equate the color red with infrared. And I also equate heat lamps to red color light. So I then think of infrared light as hotter. Is it possible that "higher-energy" and what we call "heat" do not necessarily go together?
     
  8. Sep 19, 2009 #7

    ZapperZ

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    May I suggest that you also read one of the https://www.physicsforums.com/showpost.php?p=899393&postcount=4" [Broken]. It might tell you why the individual atoms that to not be a significant factor in many of the physical property (including optical properties) of solids.

    Zz.
     
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  9. Sep 22, 2009 #8
    ZapperZ - Very informative post. Explains a lot. Thanks! :approve:
     
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  10. Nov 16, 2009 #9
    If I may ask a related question ,
    While the infra red excites the lattice bonds and produces heat and the visible light photons interact at an electron level , what happens to the photons that are not then released as reflected light , ie a dark or black object ? Do they not provide enough energy to excite the electrons to release further photons?
     
    Last edited: Nov 16, 2009
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