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X-ray process and electrons involved

  1. Nov 11, 2007 #1
    :confused:the X-Ray the process is like this:
    An electron hit the metal so there is a photon is emitted.
    My Q is what happened to the electron that cases the X-ray
    And what is K(a),K(b),K(g).
    Or we call them K(alpha),K(beta),K(gamma).
  2. jcsd
  3. Nov 11, 2007 #2


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    The target is typically grounded, so the impinging electrons are conducted away to a large sink. Before this happens, the electrons lose energy through 2 processes:
    (i) by collision with a core electron in the metal atom, which it "kicks" out - this produces characteristic x-ray lines (K-alpha, K-beta, L-alpha, etc.)
    (ii) by deceleration through the bulk of the metal - this produces a continuous distribution of x-rays, known as brehmstrallung.

    The nomenclature is explained here: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/xrayc.html#c1
  4. Nov 11, 2007 #3
    but this is not enough

    i know the information in the article that you gave but it is not enough and i need more sophisticated information like the effect of LS (marriage) and JJ(marriage)
    and the effect of the [effective charge ] and can i calculate theoretically K(alpha),K(beta),K(gamma) and find them theoretically by using the Q mechanics without using the experiment]
    and thank you very mach
    and help !!!!!!!!!:cry::cry::confused:
  5. Nov 11, 2007 #4


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    I don't see why LS- and JJ-couplings are important in calculating the x-ray spectra. If you want to try to calculate the core energy levels of a pure target metal (say, Mo), then you will have to solve the Schrodinger equation for a giant many-electron atom, with your favorite choice of approximation (this is now a problem independent of x-rays). The only method I know of that one can get very approximate energies from is using the Slater presciption for calculating effective nuclear charges for electrons in different subshells. The approximation is way coarser than any fine-structure corrections, so orbitals within a subshell are considered degenerate.

    Calculating the power spectrum of the Brehmstrallung is also a hard problem.

    For more sophisticated calculations, the person here that can help is quetzalcoatl.
    Last edited: Nov 11, 2007
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