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Plancks spectrial radtion law

  1. May 30, 2006 #1
    I have a problem that states
    Show that the wavelength Lamba_max=(2892 micro meters*K)/T
    hint: set the dS/DLamba=0.
    i have no idea how to do this.
     
  2. jcsd
  3. May 31, 2006 #2

    Hootenanny

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    You need to find the maximum of a function. The maximum of a function is given when f'(x) = 0. So taking Planck's radiation formula for energy density per unit wavelength;

    [tex]S_{\lambda} = \frac{8\pi hc}{\lambda^{5}} \cdot \frac{1}{e^{\frac{hc}{\lambda kT}} - 1}[/tex]

    The constant term [itex]8\pi hc[/itex] does not effect the position of the peak and can therefore be ignored. Thus the derivative becomes;

    [tex]\frac{d}{d\lambda} = \frac{1}{\lambda^5}\cdot \frac{1}{e^{\frac{hc}{\lambda kT}} - 1} \;\; d\lambda = 0[/tex]

    Once you have found the derivative all that remains is to solve the equation. As far as I know there exists no analytical solution to the equation and it must be solved numerically. However, if I have time I may venture into the maths forums and inquire as to whether an analytical solution exists.

    HINT: Solve numerically for [itex]\lambda_{max}T[/itex] first.

    ~H
     
    Last edited: May 31, 2006
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