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Help with Blackbody

  1. Sep 27, 2008 #1

    tony873004

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    At a given temperature, [tex]\lambda_{max}[/tex] for a blackbody cavity = 6500 angstroms. What will [tex]\lambda_{max}[/tex] be if the temperature of the cavity walls is increased so that the rate of emission of spectral radiation is doubled?

    [tex]
    R_T = \sigma T^4 \,\,\,\, \Rightarrow \,\,\,\,T^4 = \frac{{R_T }}{\sigma }\,\,\,\, \Rightarrow \,\,\,\,T = \sqrt[4]{{\frac{{R_T }}{\sigma }}}\,\,\,
    [/tex]

    [tex]\lambda _{{\rm{max}}} = \frac{\alpha }{T}[/tex]

    [tex]\lambda _{{\rm{max,2}}} = \frac{\alpha }{{T_2 }} = \frac{\alpha }{{\sqrt[4]{{\frac{{R_{T,2} }}{\sigma }}}}} = \frac{\alpha }{{\sqrt[4]{{\frac{{2R_{T,1} }}{\sigma }}}}}[/tex]

    Just looking at the formula, it seems the answer should be [tex]\frac{1}{{\sqrt[4]{2}}} = 0.84\,\lambda _{{\rm{max,1}}} [/tex]

    But shouldn't the max wavelength go up if the temperature is going up?
    1. The problem statement, all variables and given/known data



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    3. The attempt at a solution
     
  2. jcsd
  3. Sep 27, 2008 #2

    tony873004

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    Re: blackbody

    It just dawned on me. Shoudn't the frequency to up, causing the wavelength to shorten, in which case, I might have done it correctly? Can someone double check that I did this right?
     
  4. Sep 28, 2008 #3

    gabbagabbahey

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    Re: blackbody

    Yes, higher energies mean higher frequencies and shorter wavelengths; it looks good to me:approve:
     
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