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A M-edges in TR-XAFS in Lanthanides

  1. Apr 25, 2016 #1


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    Hello, I am new to X-ray Absorption Spectroscopy, and I have a question about XAFS, more specifically, Time-Resolved XANES.

    Question: I don't understand why M5 edges are usually not used for Lanthanides.

    Here's some background to why I came up with this question (please read if you have some time, otherwise skip because it's long):

    Lanthanides have inner 4f orbitals that are shielded by outer 5d, 6s, and 6p orbitals. Thus, 4f orbitals do not involve in bonding, and 4f-4f transitions (which is parity forbidden) should not change the structure relative to the ground state. Recently, I read this article (http://pubs.acs.org/doi/abs/10.1021/ja407924m), which is the first direct observation of change in electronic structure between ground state and 4f-4f excited state using Time-Resolved XANES. They used L3-edge which is mainly 2p → 5d transition of Eu(IIII) ion. Because theoretically, 4f-4f transition involves mixing of 5d orbitals that makes 4f-4f transition partially allowed, the difference of L3-edge XANES between ground and 4f-excited state would be slightly different. They also argue that pre-edge (which is 2p → 4f electric dipole forbidden) is slightly different.

    Now, I think understand why they used L3-edge and not L2, L1, nor K1. I interpreted this as because K1 edge and L1 edge is electron dipole-forbidden for transitions to 5d orbitals, and L3 has twice the absorption strength because of degeneracy compared to L2 edge.

    That made me wonder. If you want to observe electronic structural changes in 4f orbitals, then why not use M5 edge (5p J = 5/2) or at least M4 edge? 5p → 4f should be electric dipole allowed, That should allow direct observation of 4f orbital.
  2. jcsd
  3. Apr 30, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. May 6, 2016 #3
    The reason is mostly practical. The photon energy for Lanthanide L edges is very easy and convenient to get from a Synchrotron, and the absorption is low enough to be used in transmission mode.
    The M edges are much much lower in energy. You need to work in ultra-high vacuum, and you will be sensitive to the surface only. Also, you will have to measure the absorption via the photo-electrons, as the beam will be completely absorbed by the sample (unless it is extremely thin).

    Despite these technical difficulties, M-edge XANES is sometimes used, especially for magnetic spectroscopy (XMCD)
  5. May 10, 2016 #4


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    Thank you for the explanation.

    Well in the paper I described above, they used total fluorescence yield because they were working on a liquid sample, below 1 mmol/L (I wouldn't necessarily call this very dilute). I thought that total fluorescence yield method is suitable for dilute samples, not transmission mode because it's hard to detect the difference.

    Why are M edges more easily absorbed by the sample than L edges?
  6. May 11, 2016 #5
    Total fluorescence yield works well for concentrated and dilute samples. For dilute samples, other methods do not work so well.

    Look at the photon energies for the M edges compared to L and K edges. In general, the higher the photon energy, the smaller the absorption and the higher the penetration power.
  7. May 11, 2016 #6


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    I think I have it clear now! Thank you very much!!!!!!!!!!!!
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