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Photons emitted ( Bohr's theory )

  1. Apr 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Here's the problem:
    A group of hydrogen atoms are in n=5 state. If all atoms return to ground state, how many different photon energies will emit, assuming all transitions occur? If there are 500 atoms, assuming all transitions are equally probably, what is the total number of photons to be emitted when all atoms have returned to the ground state?


    2. Relevant equations

    Equations:
    E_gamma=hu
    E_gamma=p_gamma*c
    E=-13.6(Z^2/n^2)
    N=W/E_gamma


    3. The attempt at a solution

    How many different photon energies: 4 one from 5-4 4-3 3-2 2-1each. Using E=-13.6(Z^2/n^2). I just did E(5)-E(4) for the photon energy Ereleased
    Now I need to find the work done as I already have the energies. Perhaps there is another way to find the number of photons emitted?
    I could not figure how exactly to get the KE. I had a problem similar to this and I was really shaky on the KE part of it. If I can find the KE, I can find the work.
    KE=Ereleased - Eion meaning for 2->1 it would be 10.2-3.4?
    Using this method at E(5) I get .306-.544 which is negative :/
    Anybody notice anything wrong?
     
  2. jcsd
  3. Apr 3, 2009 #2

    turin

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    You're neglecting three things: 1) transitions between states with ni-nf > 1 , 2) degeneracies, and 3) selection rules.
     
  4. Apr 4, 2009 #3
    I have not had a class on degeneracies and have had no mention of selection rules. From what I understand in the problem is that I will take the energy from each transition 5-4 4-3 3-2 2-1 and find the number of photons from that. Something else that sounds like it could be a "degenerate rule" should surely apply here or taking the number of photons each level would be about the same as just taking the number from 5-1.
     
  5. Apr 4, 2009 #4

    Redbelly98

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    Even if we don't worry about degeneracies or selection rules*, there are still transitions like 5-3, 4-1 etc. to consider.

    *Note to experts: if this is a question about the Bohr model, we are probably meant to ignore degeneracies and selection rules.
     
  6. Apr 4, 2009 #5
    Is the ground state 1? If that is so, then the energy of 5-3+3-1 is the same as 5-1
     
  7. Apr 4, 2009 #6

    Redbelly98

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    Yes, n=1 is the ground state.

    Note, 5-3 & 3-1 involves emitting 2 photons, each of a different energy than what is emitted from 5-1.

    The question asks how many different photon energies are emitted.
     
  8. Apr 4, 2009 #7
    what is the total number of photons to be emitted when all atoms have returned to the ground state?
    Assuming each transition is equally probable was what I was getting at seems just like an awkward probability problem tho
     
  9. Apr 4, 2009 #8

    Redbelly98

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    Yes, I would agree: "an awkward probability problem" is a reasonable description of what is being asked in the 2nd question.
     
  10. Apr 4, 2009 #9

    turin

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    So we are just going to ignore conservation of angular momentum? That doesn't even seem reasonable for a physics novice.
     
  11. Apr 4, 2009 #10

    Redbelly98

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    Well, for the first question we aren't really ignoring angular momentum conservation, since transitions between any two n values is allowed. The only approximation is in treating states with the same n as degenerate (eg. 2S and 2P), which seems reasonable.

    The 2nd question is not so straightforward. Yeah, given that the Bohr model does not address degeneracies, and that transitions between different states are not equally likely, it seems pretty silly to ask what they are asking.

    It's tough to know what the desired answer is without knowing at what level the OP's professor has covered the material.
     
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