by doing the hydrothermal synthesis of ZnS (zinc sulphide) doped with Samarium, I had peaks by DRX of Sm (OH) 3, the intensity of which of the peaks relating to Sm (OH) 3 increases with the increase in concetration of the dopant. you think this is due to what, knowing that I use samarium oxide.
Farid Dris
I assume you mean XRD (X-ray diffraction is the English acronym, but I suspect that this is the French rendition in North Africa).
Is your material in powder, single crystal, or film form?
Intensity in XRD is a notoriously difficult parameter to pin down. Yes, concentration, and atomic number play a role, as does the orientation of the crystal planes to the x-ray beam and the relative intensity strength of these planes. I ignore things such as amorphous versus crystallinity or separate/different phases.
In your instance I'd go for:
- Sm has an atomic number (60) almost twice that of Zn
- Increasing the [C] of Sm should increase the intensity too - all other things being equal
I am sure that those more qualified than I will correct or add to this response.
Dear Farid Dris ,
Adding some comments to Dr. Alan F Rawle answer, you need to consider the Sm atom's hydrolysis in solution.
Depending on the hydrothermal system's pH, the hydrolysis will increase quite a lot when you increase the amount of Sm in the reaction. There should be some loss of Zn as well, in a much lower amount than the Sm hydrolysis.
A way to turn around the problem is to decrease the pH, so the higher [H+] amount in solution will prevent the Sm(OH)3 formation.
Best regards,
WNM
- Badges
- Science topic
- Similar topics
- Physics
- Condensed Matter Physics
- Solid State Physics
- PEAKS