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Flame test and black (or gray) body

  1. Sep 1, 2011 #1

    fluidistic

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    I'm having an extremely hard time to understand what's going on on the following.
    Say I have a copper rod and I heat it up with a lighter. I know it will emits photons/EM radiation according to Stefan-Boltzmann's law, namely [itex]P=A \sigma T ^4[/itex] per unit area and that the spectrum is continuous. So far so good.
    Now say I take copper powder and put it under the lighter flame. Each dust of copper doesn't seem to behave like a black body since the emission spectrum isn't continuous, it's monochromatic! Or at least discrete. Each dust of copper still contains say around 10 ^21 copper atoms, so that I don't understand why the spectrum isn't anything like the one of the copper rod.
    I'm clearly misunderstanding/missing something, please help me to understand what's going on. Thank you! (very very much!)

    P.S.:Reformulating my question: What is the difference between heating up black body powder and a black body?
    Why isn't the spectra of the powder continuous like the one of a black body? I do know that the electrons in copper atoms change of "orbit" and they emit a photon when doing so, but I don't understand why it happens only in the powder form of copper.
     
    Last edited: Sep 1, 2011
  2. jcsd
  3. Sep 1, 2011 #2
    I imagine in the case of the rod heat gets conducted from the source and spread fairly evenly due to the lattice structure. With the powder I think there are probably more ionization events (B/c heat cannot conduct as easily) which would lead to the discrete emission structure.
     
  4. Sep 2, 2011 #3

    xts

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    The difference is that grains of copper powder heat up to much higher temperature so some of it evapourates - providing enough gaseous copper to emit its characteristic green light (not quite monochromatic, but definitely not continuous).

    BTW - don't believe copper behaves like a black (or even gray) body - better use Stefan's law as very rough approximation only.

    You've also overestimated size of dust 10^21 atoms is not a 'dust' but rather 'coarse sand'.
     
  5. Sep 2, 2011 #4

    Andy Resnick

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    Good question- one major difference is in the mechanism of emission- for the metal bar, the mechanism is thermal emission, while for flame spectroscopy, the mechanism is chemical oxidation:

    http://www.tempe.mi.cnr.it/zizak/tutorial/cairol06-flame-emission.pdf

    http://faculty.sdmiramar.edu/fgarce...nalyticalAbsorptionMethod_AAS_PerkinElmer.PDF

    Thus, in flame spectroscopy, the emission is due to individual atomic events (redox reactions)
     
  6. Sep 2, 2011 #5

    xts

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    Are you sure? Aren't those just transitions between chemically neutral different states of single atoms (electron flipping between 3s and 3p for yellow sodium line)?
     
  7. Sep 2, 2011 #6

    Andy Resnick

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    I'm not sure, honestly. I thought flame spectroscopy hinges on combustion of the material of interest, but perhaps the flame simply provides sufficient energy to populate excited states of isolated atoms.
     
  8. Sep 10, 2011 #7
    we have one copper 8mm. rod and we draw and make it .3 mm by drawing m/c. And also doing anneling. But after some days copper become black in the cable. what is the reason for copper black & how i found the problem and solve it.
     
  9. Sep 10, 2011 #8

    xts

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    Isn't it some kind of corrosion?
     
  10. Sep 12, 2011 #9
    yes its corrosion but i didn' know the reason. Please tell me the reason
     
  11. Sep 12, 2011 #10

    xts

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    I am not the chemist, so I am not sure, I may only speculate that black corrosion on copper is probably CuS, caused by sulfur in the environment (eg contact with vulcanised rubber), or by even small amount of H2S or organic compounds conatining sulfur in the air. Someone eats lots of onion and then hiccups :yuck:
     
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