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About the radiation of blackbody

  1. Sep 30, 2007 #1
    I am reading the book about the light resource, and I can not understand that why hot gas radiates E-M waves of discrete frequency, but hot solid radiates continuous waves. As you know, high intensity gas discharging lamps give out discrete spectrums, but tungsten resistance lamps give out continuous spectrums just as the radiation of blackbody.
  2. jcsd
  3. Oct 5, 2007 #2
    Is that for I haven't clarified my question enough? As we know, it is the electron transiting from high energy level to low energy level which results the photon emiting. Tungsten filament lamp can radiate continuous spectrum, but for discharge lamp, its spectrum is discretely distributed. What causes this difference?
  4. Oct 6, 2007 #3


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    No both hot gases and solids emits same quanta. But we must detect the radiation by an apparatus, that has a certain resolution. So we get the radiation from E to E + dE.

    And also it is meaningless to ask "what is the intensity of the radiation at energy E´ ?" Because there exists an infinie number of possible energies, the answer is always 0 or 1. So therefore we must ask "What is the intensity of the radiation between energy E' and E' +dE ?"

    Now in hot gas we get absorption or emission lines, depending on what angle we observe the cloud from. This is because the atoms absorbs and re-emits light at certain wave lenghts and if there is gas clouds in front and so on, we may se absorption lines or emission lines. The same is for a lamp, you have gas inside it that makes this absorption and emission of certain wave lenghts.

    All spectras have both a continous part, due to the heat radiation, and also discrete peaks or valleys due to absorption and/or emission of energy at certain wave lenghts. The main difference depends on how much they do and what temperatures we are talking about. The discharging lamps have so small continous part of the total EM spectra that we cant see it with "normal" devices, so its line spectra dominates.

    Look for example on the solar spectrum and the plack law:
    http://www.udel.edu/igert/pvcdrom/APPEND/Spectra.png [Broken]
    http://alfven.princeton.edu/projects/MCVPImages/PlanckGraph.gif [Broken]

    Same type of graph, but the real object (sun) has absorption and emission lines.

    maybe also this can be illuminating (also quite simplified, since there exist almost no real continoous soruce; the source may be a star in space or the tungsten restance thread with gas in a glas container)
    http://www-astronomy.mps.ohio-state.edu/~pogge/Ast161/Unit4/Images/kirchoff.gif [Broken]

    I hope this explains a bit more why some devices has lines and not.
    Last edited by a moderator: May 3, 2017
  5. Oct 6, 2007 #4
    thank you malawi, I've got it.
    But still need some more explaination. In our text book, I was told that, nomatter what kinds of materials the blackbody is composed with, it will radiates the same if the temperature is the same. and as I know, it is for that the electron transits between different energy levels that give out the photons, and different atoms of different material in different temperature radiates the different E-M waves.but as a whole body they become the same, how can this happen?
  6. Oct 7, 2007 #5


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    Now a perfect black body does not exists, but if you go back (or look forward in future courses) the black body "model" will be derived, and you will understand it mathematically.
    So I will not trough the derivation here, but I leave that to someone else, or if you want just try google it, or wait til you get to the Statistichal mechanics courses :)
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