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Auger-effect, photoelectron energy

  1. Oct 7, 2008 #1
    This problem is straight from Haken & Wolf - The Physics Of Atoms And Quanta, 7th edition. It's problem 18.9 on page 336. If you happen to have the book that is.

    Problem text:

    The [tex]L_{I}[/tex] absorption edge in tungsten is at 1.02 Å. Assume that a [tex]K_{\alpha}[/tex] photon is "absorbed" by one of the 2s electrons by an Auger process. Determine the velocity of the photoelectron released.

    Attempted solution:

    I've used formulas 18.3 and 18.6 in the book. I can use

    [tex]\bar{\nu}_K_\alpha = \frac{3}{4} R (Z-1)^{2}[/tex] (18.3)

    to get the wave number for the [tex]K_\alpha[/tex] line. I inserted the Rydberg constant and tungsten atomic number, R = 10973731 and Z = 74, and got the wave number [tex]4.386 * 10^{10}[/tex] inverse meters. This corresponds to the energy [tex] = 8.7124 * 10^{-15}[/tex] Joules. Next I use

    [tex]E_{kin} = hv_K_\alpha - E_L[/tex] (18.6)

    I set [tex]hv_K_\alpha = [/tex] (the energy I calculated) [tex] = 8.7124 * 10^{-15}[/tex] and [tex]E_L = [/tex] (energy of the ebsorption adge given in the problem) [tex] = 1.9475 * 10^{-15}[/tex]. Subtracting the second energy from the first gives

    [tex]E_{kin} = 8.7124 * 10^{-15} - 1.9475 * 10^{-15} = 6.765 * 10^{-15}[/tex] Joules

    The correct answer is [tex]5.57 *10^{-15}[/tex] Joules according to the book. So I'm not that far off but I can't figure out what I've missed.
     
  2. jcsd
  3. Oct 7, 2008 #2

    Gokul43201

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    Your answer looks to be correct, except that you've calculated the KE, when the question asks for the velocity.
     
  4. Oct 7, 2008 #3
    Okay, seems like my crappy first post is no longer editable... EDIT: fixed! :)

    Gokul: The book gives both the kinetic energy and velocity of the electron in the answer. I have only compared the first value since the velocity can't turn out right if the kinetic energy is wrong. So my answer isn't correct. I should have been more clear about this, sorry.
     
    Last edited: Oct 7, 2008
  5. Oct 7, 2008 #4

    Gokul43201

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    For the KE, I get almost exactly the same number that you got (6.8*10^{-15}J), and I think Moseley's law is more than sufficiently accurate for 2 sig figs.
     
  6. Oct 8, 2008 #5
    I see. Well that obviously raises the question about how they got the answer in the book. I think you can understand that I'd rather trust the authors than you ;)
     
  7. Oct 9, 2008 #6

    Gokul43201

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    I understand.

    To make sure you've understood the principle properly, and feel confident that you have, you should work through part 4 of the solved example on pg. 334 and make sure you get the correct answer. You could also try more problems from other books.
     
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