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Bremsstrahlung X rays

  1. Oct 22, 2012 #1
    1. I have problems finding the truth of this statement.
    The statement is cited from "book".
    I want to draw your attention to the following words: electron is decelerated

    2. Statement: "Bremsstrahlung X rays result from Coulomb interaction between the incident electron and the nuclei of the target material. During the Coulomb interaction between the incident electron and the nuclei, the incident electron is decelerated and loses part of its kinetic energy in the form of bremsstrahlung photons(radiative loos)."

    3. My views: electron is acelerated.
  2. jcsd
  3. Oct 22, 2012 #2


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    The electron is accelerated (its velocity vector changes, without giving a specific direction), and its final velocity is lower than the initial velocity (effective deceleration).

    => both
  4. Oct 23, 2012 #3
  5. Oct 23, 2012 #4
    Wikipedia is not the best authority to quote on this. It's partly correct - but collision with a nucleus is not a good example.
    a) It's highly unlikely that an electron could encounter a 'naked' nucleus
    b) If it did, the collision would be elastic.
    (I love the vague reference to the conservation of energy as 'explanation':)

    Bremsstrahlung result when an electron ploughs into the sea of electrons inside a solid (usually a metal) The energy loss mechanisms are complex and varied, mostly involving transitions in the band structure and the atomic orbitals.
    There's also some production of characteristic x-rays from the lower atomic shells - which tends to be more pronounced in lighter elements. Al & Mg are commonly used as targets to generate narrow band x-ray probes.
  6. Oct 23, 2012 #5
    Glad you 'love' it because it's how we explain such a phenomena..... the cause of the change in velocity is unimportant.

    I think you mean the energy GAIN mechanisms are complex and varied....
  7. Oct 23, 2012 #6


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    Quote from Leighton: "When a charged particle passes through matter, it interacts electrically with the nuclei of the atoms and also with the electrons surrounding the nuclei. The second of these interactions usually results only in an electronic excitation or ionization of some of the atoms which lie near the path of the incoming particle, and the particle loses numerous small amounts of energy in these encounters, gradually diminishing in energy and eventually coming to rest. This relatively slow deceleration of the particle is not, however responsible for the X-radiation."

    "The emission of X-rays results, rather, from the less frequent, but individually more catastrophic, encounters with the atomic nuclei. In such encounters the force acting upon the incoming particle, and therefore its acceleration, is Z times greater than in the individual electronic encounters, and in addition the more massive nucleus does not recoil as readily as does a light electron. One would thus expect the radiated energy to be Z2 times greater from these nuclear encounters than from the electronic encounters."
  8. Oct 24, 2012 #7
    I'm not an expert on Bremsstrahlung itself. But I've built and installed X-ray Spectroscopy equipment for 20 years. Mostly it's designed to avoid Bremsstrahlung because we really require the more clearly defined characteristic radiation.
    The characteristic X-rays are generated by electron collision with the inner orbitals K, L and M shells, which causes excitation and subsequent decay with associated photon emission of clearly defined energy.
    In lighter elements these lines are more energetic than the Brems. radiation which is easily filtered out by window foils.
    In heavier elements, the Brems. radiation is stronger and becomes a problem. This may indeed be due to the heavier nucleus but I find it hard to believe that it's due to actual collision with the nuclear e-field. It seems much more likely to me that the nucleus acts as an anchor for the atom as a whole and that the interaction is mainly if not entirely with the electron population.

    I can think of no energy loss mechanism between a solitary nucleus and an electron that will result in photon emission. As I said, it's an elastic collision. The electron exits with [EDIT: essentially] the same energy it entered. Unless something gets excited and subsequently decays there is no way to emit a photon. [EDIT:there is obviously a small transfer of KE to the nucleus]
    Last edited: Oct 24, 2012
  9. Oct 24, 2012 #8


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    The "collision" (near miss) with a nucleus is not elastic. In a classical picture, the acceleration is so strong that the electron (accelerated charge) emits radiation. In a QM picture, the transition is allowed, and happens quite frequently.
    The electron can come very close to a localized nucleus with a large charge - something which cannot happen with other electrons.
  10. Oct 24, 2012 #9
    I'll go along with that.
    I've been pondering the question and it seems to me the classical explanations are way too simplistic. There's a LOT going on inside a solid metal when an electron beam impacts. Primary beam electrons don't get far before they are scattered and degraded. There's a massive production of secondaries and Auger electrons. It isn't at all clear to me what part the band structure of the metal plays but I'm willing to bet it's highly significant.

    Classical Brems. theory explains some observations in nebulae but I can't accept it's the be-all-end-all of interaction in dense matter. I'd much rather blame QM interactions.
  11. Oct 24, 2012 #10
    bremsstrahlung is a purely classical phenomenon but the quantum mechanical discussion of it just involves the fact that an electron moving through empty space cannot radiate .only when it interacts with the coulomb field of nucleus ,it is possible to emit photon.A standard calculation using feynman diagram will eventually show it.
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