Fluorite has a clear Raman spectrum, with the dominant band at 320 cm-1. The REE-rich fluorite sample has a typical tveitite Raman spectrum, and the fluorite band is completely missing. However, the amount of REE is 13.32 %, Y is only 4.4%. Ca is 37 %, and F = 45.7 %. Can someone help?
Hello Rainer,
tveitite and fluorite are not endmembers of a solid solution series. Fuorite is isometric, tveitite is trigonal. Also their compositions are little distinct (tveitite = (Y, Na)6Ca6Ca6(Ca,Na)F42). If the Raman data is different from that of fluorite, the species is either tveitite (if the composition meets the formula and structural patterns), or something else.
Take care, Hans
Hello Hans,
Your answer is not convincing. Please consider TiO2 with three minerals: anatase, brookite, and rutile. All have the same formula; however, they have very different Raman spectra.
The IMA formula of tveitite (Y,Na)6(Ca,Na,REE)12(Ca,Na)F42 is a better description. The mineralogy of high-temperature REE-rich fluorites is not well-studied. The Raman spectrum of this phase (Zinnwald) is very different from fluorite and is similar to tveitite-(Y). Inserting water (as OH or molecular water, or H) is highly possible. The Raman spectrum shows that the REE-rich Zinnwald fluorite is maybe identical to the tveitite-(Y).
There is also the possibility that the microprobe analyses are not correct. Na determination is not simple; Li cannot be determined with a microprobe. There are a lot of questions.
Note the significance of the P/O micro-Raman method is complimentary to micro-X-ray diffraction.
Best regards,
Rainer
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