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X-ray tube and photon energy question?

  1. Aug 14, 2009 #1
    Hey all, im just studying for an upcoming medical imaging exam however im having some trouble understanding a concept:

    With regards to X-ray beam production, the effective photon energy of a beam is roughly 60-70% of the maximum photon energy (Emax) at a particular Tube voltage in the range of 40-150KV. For example a 100KV tube voltage produces an effective photon energy of approx. 60keV.

    My questions are:

    1) Why does this occur, what is the physics behind the value being roughly 60% of Emax?
    2) How is the effective photon energy different to the Average photon energy (the most frequent beam) or is this the same thing?

    Any help is greatly appreciated!
     
  2. jcsd
  3. Aug 14, 2009 #2
    Most of the x-rays are coming from "bremsstrahlung" radiation. When the accelerated electron bangs around the metal target is loses energy and this is given off as radiation.

    The highest possible x-ray energy is equal to the initial kinetic energy of the incoming electron this would happen if the electron went from it's full speed to completely stopped. Most likely the electron is going to undergo several collisions, each time releasing part of it's energy.

    A curve showing a typical x-ray spectrum is shown in the below site:

    http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/xrayc.html

    Note the abrupt edge at the smallest frequency -- this is the highest possible energy. The maximum of the bremsstrahlung curve is related to the most likely deceleration and I *ASSUME* is the 60-70% that is called the "effective photon energy". This value will vary depending on the initial voltage and the metal that the target is made of.

    The "average photon energy" would be if you added up all the x-ray energies and divided by the number of x-rays. Notice that the bremsstrahlung curve is not symmetric therefore the "average photon energy" is not necessarily at the maximum of the bremsstrahlung curve.

    You should double check on the meaning of Effective Photon Energy, but to me this is what I'd guess.
     
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