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Average wavelength for blackbody radiation

  1. Aug 21, 2010 #1
    The maximum wavelength for blackboby radiation is .29/Temperature. Is the average wavelenght 1.84 times the maximum for all temperatures?
     
  2. jcsd
  3. Aug 21, 2010 #2
  4. Aug 21, 2010 #3
    As shown in both ref., the wavelength where the emission intensity is maximum is .29/T. But I want to know the avg wavelenght at that temperatue. So I think the wavelenght where the area under the intensity ghaph is equal on both sides would be the average, but I do not know how to calculate it from gragh equation. My estimate working back from other calculations was 1.84 times the max wavelenght.
     
  5. Aug 21, 2010 #4
    You can calculate the following:

    Use Planck's Law for the blackbody radiation power as a function of T and λ

    http://en.wikipedia.org/wiki/Planck's_law

    u(T,λ) = [2hc25]/[e(hc/λkT) - 1]

    The average wavelength is then given by:

    λavg = ∫λ u(T,λ) dλ / ∫ u(T,λ) dλ

    where both integrals are from 0 to ∞. The appendix of the above wiki solves the important integrals.

    Bob S
     
  6. Aug 21, 2010 #5
    Do you know if the average is 1.84 the maximum for all temperatures? I can not do the math. The following quote from the source may relate: This function peaks for hc = 4.97λkT, a factor of 1.76 shorter in wavelength (higher in frequency) than the frequency peak. Thanks jimj
     
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