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How can infrared photons impart energy

  1. Jul 26, 2011 #1
    And still be re-emitted at the same wavelength, even after being absorbed?

    I imagine visible colors, to hit a dark colored surface and become absorbed. In that process some of the object's electrons (I believe the outer shell) become energetic or is raised higher. To achieve equilibrium, it must re-emit a photon so that the object can cool.
    And it is usually an infrared.
    Now when that infrared photon gets absorbed by a GHG, how could it be re-emitted at the same wavelength AND impart energy to its surroundings? Would not that violate the laws of physics?
    Or is it re-emitted as a lower infrared, and so on into the microwave during successive absorption and emission?

    I also want to know if there really is a difference in the amount of infrared emitted in the following: One sq meter of sunlight concentrated in a (tiny) 1000 sun CPV cell, as compared to the same amount of light on an equally efficient flat solar panel?

    I ask because if the "world was covered" with dark panels, the infrared heat would be too much (because their efficiency is low), but it seems that if the same amount was concentrated, the higher temps of the CPV cell would emit at higher, more visible rays that would reflect back out of whatever concentration system used (dish or freznel) and go back out toward the sun unimpeded by excess CO2. Thus less infrared, right?

    Thanks!
     
  2. jcsd
  3. Jul 26, 2011 #2
    It doesn't do both, it does one or the other. Either the photon is absorbed and emitted at the same frequency, or there is a difference and the state of the molecule is left changed. Infrared radiation doesn't affect electrons, it affects vibration-rotation states of whole molecules.
     
  4. Jul 26, 2011 #3

    Drakkith

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    Matter emits EM radiation based on its temperature. See here: http://en.wikipedia.org/wiki/Thermal_radiation


    When ANY photon is emitted from anything, the emitting object loses energy equal to the energy of the photon. If my wall absorbs a photon in the infrared range that photon is simply gone. It is not "re-emitted". The wall may emit a nearly identical photon to the original, and that photon will reduce the energy of the wall when it is emitted.

    Equilibrium is simply when an object is emitting as much ENERGY through radiation as it is absorbing.

    Imagine we have an object, say a piece of metal, heated to 1,300 C. This object will be emitting yellowish-white light because of its temperature. Now, this object could easily absorb a photon in the red wavelength and emit a photon right after that in the blue wavelength. In this case the energy of the metal would decrease overall because the photon emitted has more energy than the photon that was absorbed. The excess energy is from the thermal energy of the metal.

    If the CPV cell is heated to a higher temperature than the other panel, then it WILL emit more light at higher energies and frequencies. If heated high enough it could release a large amount of its energy in the visible range yes.
     
  5. Jul 26, 2011 #4
    Thanks. I think I understand how the GHG's work now. Since infrared is a longer wavelength, they can't "mess" with the electrons of an object, but instead, they vibrate the entire molecule, which is already vibrating and thus bounce off without being changed. The extra heat occurs because more infrared is bounced back to Earth because there is more GHG's. So all infrared photons, if absorbed, will cause kinetic motion (heat) and if re-emitted will simply change direction?
    Thanks!
     
  6. Jul 26, 2011 #5

    Drakkith

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    I don't believe that is correct. To my knowledge, ALL photons can excite an atom as a whole, not just an electron. In something like an antenna I think a radio wavelength photon can be absorbed by any of the free electrons in the metal lattice or one of the metal atoms themselves.

    A greenhouse gas (Which I assume is what GHG means) that absorbs a photon has energy transferred to it. If some of the energy is absorbed by one of the electrons, causing it to move to a higher orbital, it may fall back down and emit a photon of energy equal to the amount lost when the electron dropped back down. If the energy of the photon is not enough to be able to cause any of the electrons to move to a higher orbital, then the energy is simply in the rotations, vibrations, and kinetic energy of the atom/molecule. All of this increases it's temperature and is the main source of thermal energy retained by GHG's I think.

    I'm not sure if light can be reflected off of gas or not honestly.
     
  7. Aug 5, 2011 #6
    I am still confused at atomic level, i.e. viewing a photon hitting an atom, and atoms and heat.
    My starting understanding from discussion is that if photon hits an atom some energy may go into atomic shell, but some must go into atomic motion, in order to conserve momentum.
    -Is it only this added atomic motion that causes heat? Or is any component due to increase of mass of moving atom, via m=E/c2 of absorbed energy?
    -Above almost makes sense to me, when a photon is later radiated and momentum and heat reduced.

    But now I get confused when I try to analyise meaning of heat at atomic level
    -Heat is atomic motion, but motion relative to what, CG of body/liquid/gas? If photon
    hits an isolated atom in space can 'heat' be absorbed?
    -If a photon hits a bound atom, what if motions relative to CG are head on, is the objects
    heat reduced when photon absorbed?
    -And conversly, seems radiating a photon could increase atoms speed relative to CG, and thus its heat.

    Somewhere else I read radiation came from relative motions of radiation fields around the charged atoms in a hot object. Is this true? If so what is contribution of electron shell jumps?
    I am clearly missing some points, can anyone give me a simple radiation/heat transfer explanation at atomic level?
     
  8. Aug 5, 2011 #7

    Drakkith

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    I don't follow you on this one.


    I don't think the view of heat is applicable using just one atom or molecule. Or it is simply confusing in that way. Think of it in terms of energy, not heat, and it should make sense.

    I don't know what "radiation fields" are, so I can't say anything on that. To my knowledge any acceleration of a charged particle creates EM radiation in addition to the creation from electrons dropping orbitals. The atoms and molecules in a hot object simply have much more kinetic/rotational/vibrational energy than a cold object, and they emit more radiation due to the constant acceleration/deceleration or those particles and at a higher energy.
     
  9. Aug 5, 2011 #8

    sophiecentaur

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    Atoms with electrons changing energy states are not the only interactions with em radiation (this has already been mentioned but people are so much in love with the Hydrogen Atom and all that that they want it to be the only model). Infra red photon energy is too low to cause such electronic energy changes. It can, however, alter molecular arrangements and the rotation.
    This thread seems to be a mix of various ideas and people can be confusing two different messages. Start with the macroscopic: Take a body of gas and it will have a given temperature. That means that it will be radiating energy (in all directions) over a range of frequencies and absorbing whatever energy is incident on it. Give it an extra input of infra red radiation and it will absorb some and this will raise its temperature - causing it to radiate more, until equilibrium is reached again. If the ir came from the ground and the gas is in between the ground and Space then the ground will not be losing as much net heat as without the gas being there (due to re-radiation of ir back towards the ground) so the ground temperature will rise until a higher equilibrium is reached. The original sunlight, being the source of all this energy.

    What is happening inside the gas and its molecules? They are buzzing around, interacting and exchanging energy with the outside and each other. One incident photon, with a given energy, may add to the energy of one molecule. This molecule may then interact with another molecule and produce a photon of different energy from the original incident photon. Molecules will have a distribution of energies and can radiate over a wide range of frequencies. If extra ir hits the gas then this average energy will increase and a different average energy of radiation will be radiated - the spectrum changes a bit.
     
  10. Aug 5, 2011 #9

    Drakkith

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    Thanks sophie. Much clearer than my explanations.
     
  11. Aug 6, 2011 #10
    I am new to forums so do not know reply options, so some quick points. Would like ref to guide on options.

    Still do not follow at atomic level. I understand that all radiation/matter interaction involves at lowest level the interaction of a photon and an atom, and thus should be understandable from that level up.
    If a photon hits an atom it is always at relative speed c, so it cannot know if atom is free in space or bound. Thus it seems description of this interaction should be independent of atom’s environment, but if atom is bound within an object it is claimed that some radiation energy is used to increase atom’s speed and appear as heat energy of the object.
    Additionaly, I do not see why it should not equally likely slow the atom.

    Also if atoms are constantly accelerating/deaccelerating what force is doing this, that generates energy to radiate photon – as heat radiation is claimed separate to electron shell radiation? If just collisions change velocities, then situation seems same as billiard balls colliding where no energy is involved.
     
  12. Aug 6, 2011 #11

    Drakkith

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    Dres, remember that an accelerating charge emits EM radiation. All those atoms moving back and forth are full of charges!
     
  13. Aug 6, 2011 #12

    sophiecentaur

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    Dres
    Did you not read what I wrote about radiation and molecules?
    If a molecule absorbs some energy then, when it 'collides' with another, the collision will impart a different amount of energy due to the recently absorbed ir photon. Thus, a bit nor energy has entered into the internal energy of the bulk of the gas. The temperature will have gone up.
    Your query about the 'dynamics' of the situation is explained by the slightly different quantum jump of the smaller system under different conditions of bound or free states. The momentum conservation always will apply and is not a cause for confusion.
     
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