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drfrankenstein Group Title

@Jemurray3 Please help http://physicsweb.phy.uic.edu/244/Homework/Phys244Spring2013hw1.pdf

  • one year ago
  • one year ago

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  1. Jemurray3 Group Title
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    What do you need help with? I don't want to just do it and give it to you...

    • one year ago
  2. drfrankenstein Group Title
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    well, it is our first week of class and so far we have learn about black body radiation but I don't how it relate to first question , if it does at all

    • one year ago
  3. Jemurray3 Group Title
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    It does, insofar as it requires a conceptual understand of how radiation works. You should consider that the frequency of a photon is proportional to the energy it carries. A photon of a very specific energy (and frequency) is required to raise the electron from energy level 0 to energy level 2. When it falls from 2 to 1, it releases a photon whose energy is equal to the difference in E2 and E1. The same is true for the transition from 1 to 0.

    • one year ago
  4. drfrankenstein Group Title
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    so more energy equal more frequency and energy absorbed should be largest , hence f_abs>f_ph

    • one year ago
  5. drfrankenstein Group Title
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    correct?

    • one year ago
  6. Jemurray3 Group Title
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    yes, that's right.

    • one year ago
  7. Jemurray3 Group Title
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    Though if we're being picky you should say "More energy, higher frequency"

    • one year ago
  8. drfrankenstein Group Title
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    also , in second question , I am thinking two ways 1) difference between E2 to E1 is smaller than E1 to E0 so F_fl is less than F_ph 2) E2 is at higher energy level than E1 , hence f_ph is less than f_fl

    • one year ago
  9. Jemurray3 Group Title
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    (1) is correct. It doesn't matter what the energy levels are. The only thing that matters is the difference in energy between the two levels.

    • one year ago
  10. drfrankenstein Group Title
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    thank you very much for your help; I now have at least some idea going in second week of class

    • one year ago
  11. Jemurray3 Group Title
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    Glad to hear it.

    • one year ago
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