I have doped Ti in ZnO using a chemical method. The XRD results are showing an increase in the crystallite size were measured using Scherrer's equation after the Ti doping. Can some one suggest me that what may be the possible reason for increase in the crystallite size of ZnO after the Ti doping? Thanks
Dear colleague Shaker Ali,
Have a nice time and good day.
Due the difference of atomic radius , the Ti doping may increase the tendency for the agglomeration of crystallite size to get bigger ones.
2- A ternary compound Ti-ZnO may be formed with high crystallite size.
I wish this is useful and help you.
Yous
A. .El-Denglawey
Dear colleague Shaker Ali,
Have a nice time and good day.
Due the difference of atomic radius , the Ti doping may increase the tendency for the agglomeration of crystallite size to get bigger ones.
2- A ternary compound Ti-ZnO may be formed with high crystallite size.
I wish this is useful and help you.
Yous
A. .El-Denglawey
Dear Shaker Ali,
how to you perfom your experiment, what size difference. Do you mean Ti or Ti(4+). I am confinced that both will not be able to replace Zn(2+) in the lattice. Without knowing the procedure (chemical leads to solvents) I guess that a Ostwald ripening of the ZnO occurs. Often a supperpostion of the XRD's is a meaningful comparison (better than Scherrer(. E.g. the 002 reflex is quitesensitive to a variation in the aspect ratio.
Article Synthesis and testing of ZnO nanoparticles for photo-initiat...
If you start with ZnO powder and Ti containing solution for doping, i guess the elevated temperature for the reaction might induce grain growth, Ostward ripening as Mr.Schmitt pointed. You can anneal ZnO (without doping) at the same temperature used in the chemical reaction under Ar ambient, and do XRD to confirm it.
I expect the XRD peak shifting due to Ti atoms diffused into ZnO, instead of the change in average domain size from Scherrer equation.
due to the change in growth temperature for ZnO and Ti and the difference of atomic radius between ZnO and Ti.
Pure ZnO melts at 1975°C, however it usually evaporates at such high T. But look at the phase diagram with TiO2: The intermediate compound Zn2TiO4 melts already near 1580°C, and towards its ZnO side (relevant for ZnO excess) there starts the formation of liquid phase already at the eutectic temperature of 1537°C. Usually grain size growth and sintering starts already at the Tammann temperature somewhat below the liquidus temperature (take care: temperatures in Kelvin for that!); and so the TiO2 additive decreases the required annealing temperature for grain growth by several hundreds Kelvin.
First you shouldn't say crystalline size. This case is the unit cell size
there is a possibilities of lattice increase by doping particularly when sopping occurred with larger atoms. In this case hosted atoms will replace Zn lattice and since it has bigger size then larger lattice will be expected. Of course if the structure remains the same
Thank you prof Omar for your answer, but the doped atom (Ti+4, ionic size 0.068 nm) is less than the hosted atom ( Zn+2, 0.074 nm). Theoretically, the size will decrease while experimentally increased.
In such a case two things are necessary to be known
1-the first and second values of lattice spacing
2-you should know if there is any second phase particularly from the data of your XRD
-- for more information, if doping material is less than 0.1% mostly effects on structure will be negligible but for more than this effects are expected. That's if the doping element absorbed by the host material
Thank you so much for the question and the discussions sir. It helped me.
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