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Planck's Law Problem

  1. Jan 20, 2008 #1
    1. The problem statement, all variables and given/known data

    Given Planck's Radiation Formula

    Find the frequency (Vmax) at which energy density is at a maximum. This requires simple calculus and numerical solution of a simple transcendental equation.
    You only need to find the answer to 3 significant digits.

    2. Relevant equations

    [​IMG]

    3. The attempt at a solution


    So i derived the equation and set it equal to 0 and i got

    V = [3kT / h][1 - 1/ (e^(hv/kT))]

    But i cant seem to finish it because of the V in the exp....
    also we know that
    e^x = 1 + x + x^2/2! + x^3/3!+ ... and that somehow has to play a role in solving this...I think...
     
  2. jcsd
  3. Jan 20, 2008 #2

    Dick

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    You have to do this numerically. Call hv/kT=X. Then your last equation is X/3=1-e^(-X). Write this as f(X)=1-X/3-e^(-X). You want to solve f(X)=0. Plot it. For values of X around 1 f(X) is positive, for values around 4 it's negative. So there must be a root in between. You could just solve it by bisection.
     
  4. Jan 20, 2008 #3
    x/3 = 1 - e^(-x)

    why is it 'x/3' ?, and not just x
     
  5. Jan 20, 2008 #4

    Dick

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    Because you have 3kT/h. Note the '3'.
     
  6. Jan 20, 2008 #5
    wont it be 3x/v ??
    since we have

    hv / kt = x

    andd in this part we have

    3h/kt...wont that equal 3x / v
     
  7. Jan 20, 2008 #6

    Dick

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    Yes. And 3kt/h=3v/x. ????
     
  8. Jan 20, 2008 #7
    hv / Kt = x
    kt/hv = 1/x
    3kt/hv = 3/x
    3kt/h = 3v/ x

    ...i still dont understand where you get "x/3" from, please clarify
     
  9. Jan 20, 2008 #8

    Dick

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    You had:

    V = [3kT / h][1 - 1/ (e^(hv/kT))]

    this is the same as:

    Vh/(3kT)=[1 - 1/ (e^(hv/kT))]

    Vh/(3kT)=x/3.
     
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