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Blackbody Radiation - Peak wavelength

  1. Aug 6, 2016 #1
    1. The problem statement, all variables and given/known data
    The Planck blackbody spectrum is given by

    [tex]u(ω,t)=\frac{ħω^3}{π^2c^3(e^{βħω}-1)}[/tex]

    Show that the peak of the Planck spectrum for a blackbody at a temperature T occurs at the wavelength

    [tex]λ_{max}T=0.29[/tex]

    where T is in Kelvin and λmax is in cm.
    2. Relevant equations
    [tex]\frac{d(ω,T)}{dω}=0[/tex]

    3. The attempt at a solution
    So using the equation above I get down to:

    [tex](3-βħω)(e^{βħω})-3=0[/tex]

    This is a transcendental equation. I said:

    [tex]x=βħω[/tex]

    I've tried using the bisection method to solve but my answer doesn't match up with what it should as from the Wien Displacement Law it should be 0.29 for λ in cm. If anyone could help explain the solving for x, I would greatly appreciate it.
     
  2. jcsd
  3. Aug 6, 2016 #2

    Charles Link

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    This one is somewhat complicated by the fact that you first need to convert the Planck function to a wavelength spectral distribution before you find the maximum in the wavelength spectral distribution (density) function. You get a different answer if you try to work the problem (finding the maximum) with the distribution (density) function in frequency space.
     
  4. Aug 6, 2016 #3
    If its converted to the wavelength I got.

    [tex]x=\frac{hc}{λkT}[/tex]

    [tex]5(e^x-1)=xe^x[/tex]

    Which is a similar equation to solve. Where would I go from here?
     
  5. Aug 6, 2016 #4

    Charles Link

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    ## e^{-x}=1-x/5 ##. Numeric solution with x not equal to zero. x=4.95 approximately.
     
  6. Aug 7, 2016 #5
    I don't understand how you solved for x. I know normally with an exponential, a natural log would be involved.
     
  7. Aug 7, 2016 #6

    Charles Link

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    ## e^{-x} ## is going to be nearly zero. On a first iteration, ## x=5 ##. Taylor series can get you about 2 or 3 decimal places in one or two iterations. Otherwise iterate ##x=5(1-e^{-x} ) ## without any Taylor series. That's probably much quicker.
     
    Last edited: Aug 7, 2016
  8. Aug 7, 2016 #7

    Charles Link

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    Additional comment: You already did the hardest part. Using iterative method (see post #6), you can start with ##x=5 ## on the right side and I think you get ## x=4.95 ## on the left. Take this 4.95 and insert it into the right side, and I think you get close to 4 decimal place accuracy for your answer. This is the number you need to write out Wien's law with extreme precision. (## h ## ,## c ##, and ## k_b ## are already known. Plug it in and you should have the right answer.)
     
    Last edited: Aug 7, 2016
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