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Planck's Radiation Formula

  1. Jan 20, 2008 #1
    Okay, so I know that Planck's Law states that,

    u(v) dv = 8[tex]\pi[/tex] v[tex]^{}3[/tex] / c[tex]^{}3([/tex]e[tex]^{}(hc/kT)[/tex]-1)

    to make this formula in terms of wavelength, do you just plug in c = v[tex]\lambda[/tex], or is there more to it? because what I get is not what I find on the net to be correct.
    Last edited: Jan 20, 2008
  2. jcsd
  3. Jan 20, 2008 #2
    sry for the bad looking equation, im new to the forum :P
  4. Jan 20, 2008 #3
    you cannot simply substitute because you are missing the dv, in the right hand side of the formula.

    sure, you can just put lambda in terms of v, but at the end, you want to integrate over u(lambda) d lambda, so you need to change dv to dlambda, and that introduces the extra factors you found on the internet.
  5. Jan 20, 2008 #4
    so how should i change u(v)dv to u(lambda)d(lambda)

    i found a relation


    so i just replace dv with negative d(lambda)??
    and then integrate?
  6. Jan 20, 2008 #5


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    Staff: Mentor

    [tex]\nu = \frac{c}{\lambda}[/tex]

    [tex]\frac{d \nu}{d \lambda} = - \frac{c}{\lambda^2}[/tex]


    [tex]d \nu = - \frac{c}{\lambda^2} d \lambda[/tex]
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