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Ratio of energy densities of black body radiation

  1. Dec 31, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-12-31_14-4-30.png

    2. Relevant equations


    3. The attempt at a solution


    The energy density is given as ## u = \frac { 8 \pi {\nu }^2}{c^3}~ \frac { h \nu} {e^{ \frac { h\nu}{k_B T}} – 1}.##

    EDIT : I put the constant C.
    ## \frac { u( 2 \nu) } {u(\nu)} = C \frac { {e^{ \frac { h\nu}{k_B T}} – 1} }{ {e^{ \frac { 2h\nu}{k_B T}} – 1} } ##
    Where C is the appropriate constant.

    Is this correct?
     

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    Last edited: Dec 31, 2017
  2. jcsd
  3. Dec 31, 2017 #2

    ehild

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    Homework Helper

    The formula is correct (apart from a constant factor), but you have to simplify it and figure out which option it corresponds to. Note that ##e^{ \frac { 2h \nu}{k_B T}}=\left(e^{ \frac { h \nu}{k_B T}}\right)^2##
     
  4. Dec 31, 2017 #3
    Thanks for this insight.
    ## \frac { u( 2 \nu) } {u(\nu)} = \frac { {e^{ \frac { h\nu}{k_B T} } – 1} } { \left ({e^{ \frac { h\nu}{k_B T}} – 1}\right ) \left( {e^{ \frac { h\nu}{k_B T}} +1} \right) } ## ## = \frac1 {e^{ \frac { h\nu}{k_B T}} +1} ##

    So, the answer is option (B).
     
    Last edited: Dec 31, 2017
  5. Dec 31, 2017 #4

    ehild

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    Yes :)
     
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