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Question on atomic spectra.

  1. Mar 31, 2005 #1

    jzq

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    I did a spectroscope experiment with a hydrogen discharge tube.

    So far I got this data:

    [tex]Line 1:[/tex]

    [tex]Color=Red[/tex]

    [tex]\lambda(nm)=700[/tex]

    [tex]\nu(Hz)=4.2*10^{14}[/tex]

    [tex]n1=?[/tex]

    [tex]n2=?[/tex]

    [tex]\Delta{E}(J)=2.8*10^{-19}[/tex]

    [tex]E(J)=?[/tex]

    This is only the data for the Red line. I figured if i knew how to do this, I'll be able find the rest of the data for the other lines. I need an explanation on how to get [tex]n1[/tex]. Once I know how, I'll be able to do the rest on my own. Thanks!
     
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  3. Mar 31, 2005 #2

    dextercioby

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    Among the spectra of the H atom (Balmer,Rydberg,Pfund,Brackett,...),which has emission/absorption lines in the visible side of the spectrum...?

    Daniel.
     
  4. Mar 31, 2005 #3

    SpaceTiger

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    Remember that

    [tex]\Delta{E}=13.6 (\frac{1}{n_1^2}-\frac{1}{n_2^2})~eV[/tex]

    Since [tex]n_1[/tex] and [tex]n_2[/tex] can only be integers, there are a finite number of transitions that could give an energy near what you observed. Try a bunch of them out and see what you get.
     
  5. Mar 31, 2005 #4

    Gokul43201

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    Look up the Bohr Atomic theory for the energies of electronic levels and hence for the energies associated with the transitions between levels (you can even derive the relation from angular momentum quantization, Newton's Laws and Coulomb's Law). You might also try to Google "Rydberg".
     
  6. Mar 31, 2005 #5

    jzq

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    :confused: Sorry...I don't compute any of the responses.
     
  7. Apr 1, 2005 #6

    jtbell

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    Doesn't your textbook discuss the hydrogen spectrum, and give the n's for the energy levels involved in the visible lines?

    Or are you supposed to try to figure out what the n's are from your data, and pretend that you don't know them in advance? In that case SpaceTiger's method is the way to go. If you can tell us what you don't understand about it, someone can probably explain it in more detail. But just saying "I don't compute" doesn't give us much to go on. :confused:
     
  8. Apr 1, 2005 #7

    jzq

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    Sorry. Anyways, I figured it out...or atleast I think so. n1=2 because visible light transitions from n=2 therefore all of the n1 for the other lines will be 2(correct me if I am wrong). So I just use that and the data that I obtained and EUREKA!!! Thanks alot for your help though guys!
     
    Last edited: Apr 1, 2005
  9. Apr 1, 2005 #8

    SpaceTiger

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    Hmm, I don't think so. You're right that as n2 becomes very large, the lines become hard to distinguish, but n2=3 (the balmer alpha line) and n2=4 (the balmer beta line) are quite different.
     
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