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I Converting between λ and ν for Blackbody Radiation?!

  1. Apr 18, 2016 #1
    Forgive me for this stupid question, but how do I convert between




    I tried c = νλ but that doesn't work. This is the Rayleigh Jeans Law by the way.
    Last edited: Apr 18, 2016
  2. jcsd
  3. Apr 18, 2016 #2

    pl. give the full expression of the quoted equation and where these two are being used.
  4. Apr 18, 2016 #3
  5. Apr 18, 2016 #4
    The Rayleigh–Jeans law agrees with experimental results at large wavelengths (low frequencies) but strongly disagrees at short wavelengths (high frequencies). This inconsistency between observations and the predictions of classical physics is commonly known as the ultraviolet catastrophe,

    see the full expressions
    associated Rayleigh–Jeans limits are given by


    now you can see the approximations- actually they are not exact expressions
  6. Apr 18, 2016 #5
  7. Apr 18, 2016 #6


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    You only have to remember that these are probability distributions ##\mathrm{d} N/\mathrm{d} \nu## or ##\mathrm{d} N/\mathrm{d} \lambda##. Now you have ##\nu=c/\lambda##. This implies
    $$B_{\nu}=\frac{\mathrm{d} N}{\mathrm{d} \nu}=\frac{\mathrm{d} N}{\mathrm{d} \lambda} \left|\frac{\mathrm{d} \lambda}{\mathrm{d} \nu}\right| = B_{\lambda} \frac{c}{\nu^2}.$$
    Now with
    $$B_{\lambda}=\frac{2 c k_B T}{\lambda^4}=\frac{2 k_B T \nu^4}{c^3} \; \Rightarrow\; B_{\nu}=\frac{2 k_B T \nu^2}{c^2},$$
    and this was to be shown.
  8. Apr 18, 2016 #7
    Thank you guys.




    the peak intensities occur at different wavelengths or frequencies.

    How do scientists measure the spectral radiance of blackbodies? Are there TWO types of equipment, one for [tex]B_\lambda[/tex] and the other for [tex]B_\nu[/tex], such that each device yields a peak at a different frequency?
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