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Moseleys law - Kalpha2

  1. Nov 13, 2014 #1
    1. The problem statement, all variables and given/known data
    I need to use Mosley law to find Kalpha2 for lead where Z=82

    What i know is that copper (z=29) and tin (z=50) have kalpha2 of 8.03 and 25 KeV respectively

    2. Relevant equations
    Mosley law

    3. The attempt at a solution
    Could someone give me a hint please
     
  2. jcsd
  3. Nov 13, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Try a google search, using the correct spelling, Moseley's Law. You had it correct in the title line.

    You'll find the formula here, and it checks out with the data you provided. http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/moseley.html

    (Disclaimer. I have no idea how realistic it is to apply that formula to values of Z as high as 82.)
     
  4. Nov 13, 2014 #3
    just use; sqrt{f} = k_1 * (Z - k_2)

    f=E/h then add inn the constant and the energy but what i dont understand is how knowing those two (for Cu and Sn) can help with finding the third?

    Also where can i acquire the constants? It just states; f8984e9f90ef34fde9f0ff38f03179d7.png and 6bba165b44a6dbc13beeebb8b4edce0e.png are constants that depend on the type of line

    thanks
     
  5. Nov 14, 2014 #4

    NascentOxygen

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    I think it is all in the article I linked to. Perhaps you didn't scroll down?

    By substituting Z for Cu and Sn you can confirm that the equation works and gives the energy in eV just as you require.
     
  6. Nov 14, 2014 #5
    okay so k1=2.47*10^15 and k2 = 1 for kalpha?
    How does knowing the energies for these 2 other elements help me figure it out? Im not getting this :P

    Also, the question is talking about Kapha2 not just Kalpha, does that matter?
     
  7. Nov 14, 2014 #6

    NascentOxygen

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    Could you please post a jpeg of the question as it's been given to you.
     
  8. Nov 14, 2014 #7
    upload_2014-11-14_13-6-40.png
     

    Attached Files:

  9. Nov 14, 2014 #8

    NascentOxygen

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    Okay. Here's my take on this .... and I could be wrong.

    I think Moseley's Law equation is all that's called for. The difference in transition energies Kα1 and Kα2 is tiny and it suffices to use Moseley's formula as for Kα. (For an explanation of the doublets, refer to the diagram here: http://en.m.wikipedia.org/wiki/K-alpha )

    Maybe the examiners didn't expect students to have memorised the constants in the formula, just the bare relationship
    E = k(Z - 1)2

    So, by providing you with a couple of its results, you can determine k on the spot without having memorised it. Finally, using the value of k you calculated or had memorised apply Moseley's equation for atomic number 82 and determine Kα.

    Sound plausible?
     
  10. Nov 14, 2014 #9

    NascentOxygen

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  11. Nov 14, 2014 #10
    E=k(Z-1)2 ó k=(Z-1)2/E
    (29-1)2/8.03=97.63

    (50-1)2/25 =96.04

    theyr pretty close but not identical so could i just the average for K to figure out the energy for lead?
     
  12. Nov 14, 2014 #11
    ye i saw that table as well, thanks
     
  13. Nov 14, 2014 #12

    NascentOxygen

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    That sounds like a good experimental approach.

    What year of the physics course is this test paper?
     
  14. Nov 14, 2014 #13
    E=k(Z-1)2 hence E=96.835(82-1)2 = 635334.435J its pretty high right? Im guessing its in joules
    when converting im getting 3.9654458121e+24eV dosnt compare with the table
     
  15. Nov 14, 2014 #14

    NascentOxygen

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    Your post #1 indicates your working is in keV!
     
  16. Nov 14, 2014 #15
    Ye but it still dosnt line up, right?
     
  17. Nov 14, 2014 #16

    NascentOxygen

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    You haven't rearranged this correctly to isolate k.
     
  18. Nov 14, 2014 #17
    aaah thats embarrassing, its E on top :p
    E=k(Z-1)2 ó k=E/(Z-1)2
    8.03 / (29-1)2=0.0102
    25 / (50-1)2 =0.0104

    Average = 0.0102+0.0104/2=0.0103
    E=k(Z-1)2 hence E=0.0103 *(82-1)2 = 64.32KeV
     
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