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Franck-Hertz Experiment

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data
    I've been given 2 values for the potential, 10.2 and 12.1V, I have to find the principle quantum numbers of which have been excited.

    3. The attempt at a solution
    I know En= -13.6/n^2

    From there I'm stuck, I know it should go up in 4.9v, but 10.2 and 12.1 aren't multiples of either! If it were, I could work out how many multiples they were, and that would be n.
     
  2. jcsd
  3. Apr 4, 2009 #2

    Redbelly98

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    Perhaps the 10.2 and 12.1 eV energies are relative to the ground state energy -13.6 eV.

    EDIT:
    Just re-read your post.

    "En= -13.6/n^2" can't be right, since that is the energy for atomic hydrogen.

    What gas are you using? 4.9 V would correspond to mercury.
     
    Last edited: Apr 4, 2009
  4. Apr 4, 2009 #3
    ARGH! yes, that is mercury, and the question is to do with hydrogen! Thats what happens when you copy willynilly from your notes :-( Before I show further working, just note that the En= -E1/n^2 formula isn't given and I'm going from notes:

    so now I have:
    En= -(unknown base number here)/n^2

    En=E1/n^2

    If I assume 10.2 is ground state:
    En=10.2/n^2

    but I still have 2 variables :-s , what should I do? I need, perhaps:
    n= SQRT 10.2/En
    Thanks for replys!
     
  5. Apr 4, 2009 #4

    Redbelly98

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    I'm a little stumped here. What I remember of the Franck-Hertz experiment, only the 1st excitation energy is involved. And in a gas, hydrogen will form H2 molecules, which wouldn't obey the E=-Eground/n2 formula.

    But if the question is definitely about atomic hydrogen (H, not H2), then we know that Eground is 13.6 eV.

    So I'm not a lot of help at this point. The question is not at all clear to me. Can you reproduce the full problem statement, exactly word for word?
     
  6. Apr 5, 2009 #5
    Thanks for reply, this is the question in full word for word:
    In a Franck Hertz type experiment, atomic hydrogen is bombareded by electrons that have been accelerated through a poential V and the rate of arrival of electrons at the anode measure as a current I. It is found that the current falls sharply when the potetial V has the values 10.2 V and 12.1 V

    What are the principle quantum numbers of the states to which the hydrogen atoms have been excited?

    If you don't mind, could you also answer the rest of the question which has me stumped:

    If the light emitted in the decay of the exited hydrogen atoms were to be analised, how many spectral lines in total would be observed as the potential is increased from 0V to 12.1 V. Identify the spectral series to which the lines belong.
    I'm guessing, this relates to the first part of the question, when I find what the "quantised" volt is, I can count how many n's there are. Once I do that, I should be able to find the spectral lines.

    If the potential were increased further to 12.6 how many more spectral lines would be observed. Justify your answer.

    I believe I would do it as above, find out what the "quantised" unit is, and then count how many n's there are and work it out from there.

    Thanks.
     
  7. Apr 5, 2009 #6

    Redbelly98

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    Okay, now I understand better.

    "10.2 V and 12.1 V" would be the excitation potentials above the ground state, which has an energy of -13.6 eV.

    In other words, an atom in the ground state at -13.6 eV has its energy increased by 10.2 eV to ____ ? Fill in the blank, then figure out what n would be.

    Looks like you have a good idea what to do for the remaining question, like you said it depends on the n-values you come up with for the 1st part.

    p.s. You don't need to worry about the following, but a few things had me confused:
    • The gas consists of H, not H2. Hmm, might be possible if the density is extremely low, but not sure.
    • Multiple excitations occur. Again, this would be possible for low enough gas density. Not so in the classic Franck-Hertz expt (1st excitation potential only)
    • The electrode work function is negligible; in practice this would cause a several-V offset in the measurements
     
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