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Number density of photons

  1. Sep 15, 2013 #1
    1. The problem statement, all variables and given/known data
    In Problem 1 find the number density of photons in the sun’s core for the region of wavelengths between 599 nm and 601 nm. In problem 1 temperature is given.



    2. Relevant equations

    Planck blackbody formula and number density= constant times T^3 where T is temperature.


    3. The attempt at a solution

    Everyone is telling me to use the constant times T^3 formula but that makes no sense in this problem. The constant times T^3 is only when all frequencies are integrated. However only a slice of frequency is considered. Is this question an error or something. A lot of questions and slides had errors in this class so far.
     
  2. jcsd
  3. Sep 15, 2013 #2

    TSny

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    You are right, you can't use that formula. You need to use Planck's distribution function for the energy per unit volume for radiation that has a wavelength between λ and λ + dλ. See equation 5 here: http://www.oocities.org/vhsatheeshkumar/ModernPhysics.pdf.

    Then think about how you would use this to get the number of photons per unit volume with wavelengths in the specified range.
     
  4. Sep 15, 2013 #3
    I know I got to integrate the wavelengths to get the energy density. However I'm having a problem finding a way to get from energy density to photon density. I know the energy of each photon is hc/(wavelength).
     
  5. Sep 15, 2013 #4

    TSny

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    You won't need to integrate. Note that you are considering a range of wavelengths Δλ that is very small. So, to a very good approximation you can let dλ ≈ Δλ.

    To go from energy density to photon number density, consider the approximate energy of any one photon in the range of wavelengths that your are dealing with.
     
  6. Sep 15, 2013 #5
    So once I apply the approximation and plug in 599nm with my dλ ≈ Δλ equaling 2nm I divide by hc/599nm
     
  7. Sep 15, 2013 #6

    TSny

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    Yes, that should do it. (You could also use 600 nm as a good average wavelength in the interval you are working with, but it won't make much difference.)
     
  8. Sep 15, 2013 #7
    Thanks a lot. I think what got me was that I never had a expression with df in it before where I did not integrate. I thought a infinitesimal was so small that you can't put a number on it. I knew I had to find the energy density and then divide by something. However because I thought I was was integrating from 601 nm to 599 nm and the wavelength was changing I didn't really have a hc/v to divide by and that got me hooked up for hours today.
     
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