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Subatomic question - K alpha lines

  1. Feb 3, 2007 #1
    1. The problem statement, all variables and given/known data

    Light from the K-alpha line of an unknown material X is compared with the K-alpha line of Carbon. The wavelength ratio is [tex]\frac{\lambda_x}{\lambda_c} = 0.148[/tex]. What is the matieral X?

    2. Relevant equations



    3. The attempt at a solution

    I've got no idea where to go with this one. The only time i've used K-alpha lines is in regards to changes of energy levels. I know E=hf=hc/lamba, but i don't see how that helps :confused:
     
    Last edited: Feb 3, 2007
  2. jcsd
  3. Feb 3, 2007 #2

    Dick

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    Your first question to answer is which energy levels is a K-alpha line related to and what is the relation of those levels to atomic number?
     
  4. Feb 3, 2007 #3

    Tom Mattson

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    It looks like you're supposed to consult a table. You would have to find the entry in the table for which [itex]\lambda_x=0.248\lambda_C[/itex].

    Do you have such a table in your book?
     
  5. Feb 3, 2007 #4
    ^ Definitely no table, Tom. It's an isolated question, so i'm thinking I need to approach it a different way.
     
  6. Feb 3, 2007 #5

    Dick

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    You didn't answer my questions yet? :frown:
     
  7. Feb 3, 2007 #6
    Sorry Dick, missed that post! I'm pretty sure the K-alpha line arises when an electron transfers from the L shell (2nd) to the K shell (first), not sure about the relation :confused:
     
  8. Feb 3, 2007 #7

    Dick

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    Ok! You have the levels. Now if you ignore the other electrons around, what is the relation between the energy of these levels and the nuclear charge?
     
  9. Feb 3, 2007 #8
    OK, i'm looking through some information on the Bohr Atom and I've found this relation which seems to link the two:

    [tex] E_n = -Z^{2} \frac{E_0}{n^{2}}[/tex]

    Not sure if that's the one you're thinking about ...
     
  10. Feb 3, 2007 #9

    Dick

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    That's the one! So the energy difference is proportional to Z^2. But there are other electrons around. There is one in particular to worry about. Which one? Isn't this fun!?
     
  11. Feb 3, 2007 #10
    OK, so the transition between the n=2 and n=1 shell will result in the emission of a photon, i'd imagine, which would have a wavelength associated with it [tex]\lambda = \frac{c}{f} = \frac{hc}{E_i - E_f}[/tex]
     
  12. Feb 3, 2007 #11

    Dick

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    Yes. And the energy difference is ROUGHLY proportional to Z^2. But I'm worried about the effect of another electron besides the one making the jump.
     
  13. Feb 3, 2007 #12
    ^ Not quite sure what you mean there. I'm looking through my notes on the Bohr Model and some worked examples of Lyman series transitions (n=2 -> n=1) and they only seem to be concerned with the energy of the photon and the difference in energies of the final and initial atomic state :confused:
     
  14. Feb 3, 2007 #13

    Dick

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    Fair enough. But they are talking about fully ionized atoms. In our case we should assume that the K shell probably won't be empty before the transition occurs. How many electrons might be there? What effect would this have on the problem?
     
  15. Feb 3, 2007 #14
    Just another thought, if I use Mosely's law, and turn it into a ratio of the wavelengths , that'll let me solve for Zx which comes out as 13.9999 = 14 = Si. Not sure if that's ok...
     
  16. Feb 3, 2007 #15

    Dick

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    Dang. I was one step away from having you (at least partially) derive Moseley's law. The other K electron will screen the nuclear charge - so you only get an effective charge of (Z-1) (as you see in his formula). The only thing you can't really easily compute is whether the screening is 100% effective. But it turns out to be pretty close. So you now understand the formula as well as you use it right? Si sounds right.
     
  17. Feb 3, 2007 #16
    Yeah, i think i understand Mosely's formula - from what I've just read, the 1 accounts for the shielding due to other electrons (and is always roughly 1 for innermost levels)
     
  18. Feb 3, 2007 #17

    Dick

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    Yep! That's pretty much it.
     
  19. Feb 3, 2007 #18
    OK, thanks for the help Dick!! :)
     
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