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Bragg's diffraction x ray flourescence

  1. Mar 8, 2012 #1
    i understand that when we shoot x-rays at a object, it will scatter in accordance to bragg's reflection law.

    as such we position the detector as per the picture below


    my question is , for the production of characteristic x-rays, does bragg's reflection law still holds? and why does it hold?

    because say, if i shoot from a 45 degrees angle, why do i have to put my detector at 90degrees in accordance to bragg's law.

    since characteristic x-rays are due to the electron falling into a lower state, it is not the same incident x-ray, so why does it follows bragg's law, where i have to have my detector at twice the incident angle on the target?

  2. jcsd
  3. Mar 8, 2012 #2
    There are several differences between Bragg diffraction and production of characteristic radiation by fluorescence.

    • First, Bragg diffraction is photon-in photon-out. Characteristic radiation is electron-in photon-out. Try and calculate the wavelength of a 30 keV electron and compare to a 8.053 keV photon.
    • Bragg diffraction is elastic, the incident and scattered photons have the same energy.
    • Characteristic radiation is inelastic. The photon usually has a much lower energy than the electron. (for a Copper anode, you need 8.9 keV to knock out a K-shell electron, but only get 8.053 from the K-alpha radiation)


    There is an effect called X-ray holography that is due to diffraction by the photons of a characteristic line. But it is very weak and you have to work quite hard to see it.

  4. Mar 9, 2012 #3
    hmm, i would be talking about the inelastic case then.

    since it is inelastic, does it mean that if my photons come in at 45 incident angle, the characteristic x-rays will reflect at 45 degrees angle like in the first post's picture?

    but it doesn't make sense to me? because it issn't like rolling a ball at 45 angle and letting it bounce off at 45.

    for characteristic x-rays, it is a re-emission after the electron drops into a lower orbit, so why will it conserve the direction of momentum of the incident photon?

    what determines the direction which the characteristic x-rays come out?
  5. Mar 9, 2012 #4
    If you excite fluorescence with higher energy x-rays, the characteristic radiation will be emitted uniformly in all directions. It does not matter what the direction of the incident beam is. This is because the production of fluorescence is incoherent, and to the best of my knowledge it is incoherent because it is inelastic.

    This is true in more than 99% of all cases. The remaining <1% are mostly covered by the x-ray holography effect, and the intensity may be affected by the x-ray standing wave effect.
  6. Mar 11, 2012 #5
    if the characteristic x-rays come out at all directions, then why do i want my detector to be like the picture in the first post?

    that would mean i am measuring the scattered x-rays from bragg's reflection right? and not the characteristic x-rays? if this is the case, then why does my spectrum only show characteristic peaks and very little background count?

    in accordance to bragg's law, i should have a very constructive interference at that angles right? which means high background count? but why do i have to do that in this experiment since i only want the characteristic x-rays?

    also, if i were to put at any other angle not satisfying bragg's law, then how will it change?
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