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K-alpha xrays, potential difference

  1. Feb 3, 2013 #1
    Just want someone to check my work. I suspect I didn't do it right.


    The question is from Krane's Modern Physics Chapter 8, prob. 9


    In an X-ray tube electrons strike a target after being accelerated through a potential difference V. Estimate the minimum value of V required to observe the [itex]K_\alpha[/itex] X rays of copper.

    Well, I know that in order to have the emission of a K-alpha x-ray an electron from 1s would need to be "knocked" from its place. And that can only be achieved by an incoming particle (in this case another electron) of the same energy level. So what I did was utilize this equation:

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

    I got a value of 11,441 eV, using Z=29, the Rydberg constant, and n=1. The back of the book says 11 kV. I know initially it says "estimate.." so I'm wondering if I got it right or if it was coincidentally close to the back-of-the-book answer. And if wrong, what would be a better line of attack.

    Thank you.
     
  2. jcsd
  3. Feb 3, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    11,441 eV, rounded to keV, is just 11 keV, and this corresponds to 11 kV.
    And your formula looks good.
     
  4. Feb 3, 2013 #3
    Awesome. Thanks!
     
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