Dears,
Solid state photoluminous measurements on Dy and Ho indicate that only Dy based complex shows a measurable emission. Why
only the dysprosium shows measurable emission peaks (cyan emission). However both are D4d and isostructural?
Dear Dr. Murad A. AlDamen
please find this chapter below
http://shodhganga.inflibnet.ac.in/bitstream/10603/9257/5/05_chapter%201.pdf
Maytham
You do not give much detail on your compounds and the problem may be complicated. Are you sure you observe f-f transitions (they should be weak and sharp).
However a first, simple answer is the energy difference between the emitting level and the next lower-lying electronic level (so-called energy-gap "law"): For Dy(III) it is 7850 cm-1 and for Ho(III) it is only 3100 cm-1 so that quenching by high-energy vibrations (O-H. N-H, C-H, etc.) is much more efficient for Ho(III)
For a didactic review on lanthanide luminescence, see attached paper.
Dear Prof. Bunzil
The question comes from a reviewer who ask me to justify why dysprosium complex is luminescent and holmium is not. the figure of the complex is attached. His question is that the reason why only emission for Dy is found should be better investigated and ask to use UV data to compare matching of the excited level of the ligand and the Ln.
Dear Murad,
Thank you for the experimental details. What the spectra show is that with Dy you have partial energy transfer only since the emission from the ligands is clearly seen as a broad band 350-700 nm. What you see in the excitation spectrum is not clear-cut but is probably arising from the 6H(9/2) to 4F(9/2) absorption (around 500 nm), and it shows that direct excitation is sizable with respect to sensitized excitation, another indication that energy transfer is not compete.
Since generally Ho emission is much weaker than Dy one it is not surprising that you do not see it.
Lanthanide emission depends on two main parameter: (i) energy transfer (or excitation if this is performed directly into the f-levels), and (ii) non-radiative deactivation. For ions having a small energy gap (Nd, Ho, Er, Yb), the second phenomenon is largely predominent, as exemplified by their short lifetime. You may have extremely good energy transfer, but poor luminescence intensity due to non-radiative deactivation. So in your case, no need to worry about energy transfer: Ho is very easily deactivated. If you want Ho luminescence, better turn to an all-inorganic environment. In similar environment for Dy and Ho, quantum yields for Ho are usually 10-100 times less than for Dy.
As an example, I attach a recent article: you see that the average lifetime for lifetimes for Ho is 5 microseconds while the one for Dy is 28 microseconds, pointing to much less non-radiative deactivation.
With best regards
JC
Dear Murad,
I completly a with Prof. Bünzli.
Therefore, please, look at the article.
https://pdfs.semanticscholar.org/aef3/aa03734ce346a1cfc8d19e15607ed115f844.pdf
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