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Fraction of thermally excited atoms

  1. May 8, 2015 #1
    1. The problem statement, all variables and given/known data
    There is an electronic transition between the first energy state and the ground state of a neon atom, emitting a wavelength of 746nm. The question asks for the energy of the transition and an estimate of the fraction of atoms in a sample that is thermally excited at 300K.

    2. Relevant equations
    My guess:
    E = hc/λ.
    P(E) ∝ √E⋅e-E/kT

    3. The attempt at a solution
    So surely finding the energy is easy:
    E = hc/λ = (6.63x10-34)(3x108)/(746x10-9) = 2.666x10-19 J ≈ 2.7x10-19 J

    However, I have a feeling this is too simple and might be incorrect.

    For the probability, I suppose you use the Boltzmann Distributuon:
    P(E) ∝ √E⋅e-E/kT
    ∴P(E) ∝ 6.43x10-38

    But this is probably just the probability of getting exactly the energy value above - what I really want (I think) is the probability of the energy being equal to or greater than the energy value calculated above, i.e. the probability of E ≥ 2.7x10-19 J

    Can someone point me in the right direction, and tell me where I might have gone wrong?

    Thanks.
     
    Last edited: May 8, 2015
  2. jcsd
  3. May 9, 2015 #2

    mfb

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    Where does the sqrt(E) in your formula come from? The units don't match.
    You just consider two states, there are no other energy values.
     
  4. May 9, 2015 #3
    It is proven in meticulous detail in my lecture notes, but lets just go with p(E) = Ae-E/kT as this is always correct - basically the √E was just part of the constant A.

    Ahh, because Ne has 2 electron 'shells,' correct?
    So the probability of being excited from ground state to first energy level P(E) ∝ e-2.7x10-19/(1.38x10-23)(300) ≈ 4.75x10-29
    How do I involve the second shell now? Because I don't know the energy required to get excited up to there.

    Thanks a lot for replying.
     
  5. May 9, 2015 #4

    mfb

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    A constant cannot depend on energy, that is the purpose of a constant.
    Two states: ground state and excited state.
    That's the value you calculated in the line before, with the energy you calculated in post 1.

    For much larger temperatures you would have to take into account that P(0) would deviate from 1. Then you have to calculate a prefactor (which is the same for all states, it does not depend on energy), but this is not necessary here.
     
  6. May 9, 2015 #5
    Sorry, I'm being slow, but you say to just consider two energy states, and yet if I am trying to find the probability of Ne being excited at 300K, the probability of being in the ground state is not really useful, and I am back with only one energy value that I can plug in.

    Thanks for your patience.
     
  7. May 9, 2015 #6

    mfb

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    The probability to be in the ground state is extremely close to 1. The probability to be in the excited state is the 10-29 number you calculated.
     
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