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X-Ray lines from inner shell ionisation

  1. Dec 3, 2005 #1
    Hi Group
    I have a question about x-ray spectra.
    If we look at a spectrum from a typical x-ray tube with a wolfram-74 anode, we get a spectrum http://ie.lbl.gov/xray/w.htm with two lines resulting from K-shell ionisation.
    Question : Why is it (mostly) the electrons from the K-shell that are being "kicked" out ?
    Why is the chance of kicking those out bigger ?
    - or what determinates the chance of all 74 possible ionisations ?

    - ZeroGravity
    Last edited by a moderator: Apr 21, 2017
  2. jcsd
  3. Dec 3, 2005 #2
    There is a possibility that electrons from any shell may be excited or freed from the atom. As the a result the characteristic radiation includes not just transitions from L and M shells to the K shell, but transitions from the M shell into the L shell.

    The energy differences between the higher shells N and above, are generally so small that they don't emit X-rays, nor signficiantly spread the lines (e.g. an N-K transition has an energy very close to that of an M-K transition).

    In a practical X-ray tube, there will always be attenuation of the beam due to the inherent filtration of the anode and the tube envelope.

    The low-energy L-characteristic radiation will be heavily attenuated as it passes from it's point of generation below the surface of the anode, through the material of the anode.

    A note about the page you linked to:
    That spectrum doesn't actually show the k-characteristic lines (which occur at approx 58 keV and 67 keV) - the energies are too low. What you are seing around 9 keV are actually the L-lines.
    Last edited: Dec 3, 2005
  4. Dec 4, 2005 #3
    Thanks for your reply!
    If I understand correctly:
    In the spectrum http://ie.lbl.gov/xray/w.htm we see the N,M->L transitions of about 10 keV. Since there are 8 electrons in the L shell these transitions will happen more frequently than the M,L-> K transitions because there are only 2 electrons in the K-shell - Am I correct ?
    The O,N-> M (1-2 keV)Transitions should be more frequent because of the 18 electrons in the M-shell, but there is the damping of this radiation when passing through the anode material, and the spectrum is not recorded in the 1-5 keV region.
    Looking at another spectrum http://www.amptek.com/medical.html, we see the M->K (67.2 keV) transition and the L->K (59.3 keV) transition. But why are the N,M->L (about 10 keV) absent ?
    If the probability of hitting an electron is proportional to the number of electrons in the shell there should be lines at 10 keV or are those lines absorbed in the material in this case ?
    Maybe the spectrum I seek is given here http://www.ep.liu.se/ea/me/1998/001/me001-contents.pdf where the transitions to the L-shell is 4 times higher than those to the K-shell.
    Please comment - and thanks again
    - ZeroGravity
    Last edited by a moderator: Apr 21, 2017
  5. Dec 4, 2005 #4
    The probability of a particular transition occurring depends on more than the number of electrons in a particular shell.

    E.g. in the event of a K-shell ionisation, the return to ground state may be by:
    capture of an L shell electron (59.3 keV) + subsequent M-> L transition (approx 9-10 keV)
    capture of an M shell electron (67.2 keV) + subsequent higher order transitions (of energy irrelevant to x-ray production)
    capture of a higher shell electron (energy approaches 69.5 keV)

    It's a probabalistic process as to which occurrs.

    Similarly, which shell is ionised is also probabilistic and depends on more than the number of electrons in a shell - in particular it depends on the energy of the particle/photon doing the ionisation.

    A bombarding electron of 10 keV can never cause a K-shell ionisation - thus there can never be any emission of k-characteristic radiation

    The probability of an electron causing an ionisation is a function of the energy of the bombarding electron and the characteristics of the atom/molecule being ionised and the type of ionisation. The probability is typically described as a 'cross-section', and can be thought of as the area that a particular atom or shell appears to present to an incoming electron/photon. It is given the symbol Q - so Qk would represent the probability of a K-shell ionisation.

    See this text for some discussion: http://www.iupac.org/publications/analytical_compendium/Cha10sec212.pdf"

    In general, the electron impact cross section decreases as the energy of the bombarding electron rises above the binding energy of the bound electron, and, in this way, is related to the photoelectric absorbtion cross section.

    So, in the event of a collision with a 70 keV electron, the K-shell presents a greater cross section than the L-shell (because the K-shell binding energy is closer to the bombarding electron energy), so a K-shell ionisation is more likely.

    However, as electrons make their way through the target material, they continually lose energy - usually as very low energy collisions, and as their energy declines they may subsequently cause l-shell ionisation. I think it is this that makes the overall number of L shell transitions greater than K-shell transitions (there are more collisions at energies capable of causing an L-shell ionisations).

    In a medical tube, low energy photons are highly undesirable, so the tube will be designed to have a certain degree of filtration to remove them - usually the glass tube envelope. An actual medical x-ray machine will invariably have aluminium (or other material e.g. molybdenum or rhodium) filters to further attenuate low energy photons. So the spectrum of a medical x-ray set is not a true representation of the actual processes occurring in the anode. (Although I'm a bit surprised to see no hint of the L-lines in the spectrum you linked to).

    Additionally, the spectrum recorded in your 3rd link was actually reconstucted to compensate for the attenuation within the tube and for scattering of the low energy photons, in an attempt to get as close to the conditions at the anode as possible.
    Last edited by a moderator: Apr 21, 2017
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