Hi all,
I calculated crystallite size for Mg(OH)2 sample from results of XRD. I used the strongest peak ( (101) ) to calculate FWHM. check other peaks to calculate the crystallite size and then average or it is enough calculation of crystallite size for the most intensive peak?
You can check for one peak if you aare comparing your results with other samples. If there is only one sample then you can check all of them and calculate an average. Schrerrer Formula gives an estimate , and is more appropriate for crystallitz size smaller than 100 nm. So if you are getting a size between 30-100 nm, then its ok. This mettjod willl never give you can accurate method. We use TEM for accurate measurement.
Hello
Sadegh Yousefi, Sherrer's formula gives us the average crystallite size, not the actual particle size. So due to this reason you should take the average of all peaks but not those which are not able to differentiate b/w two peaks as here quite difficult for peaks above 2theeta= 65. Also, you can use W-H plot for this analysis which gives the result free from strain effect in the samples. Then you should compare these results with TEM. W-H will lie more closest to TEM.
It is immediately visible that different peaks have significantly different broadening. This happens because Mg(OH)2 is a layered compound and its crystallites will have strongly anisotropic shape. So you cannot average the sizes obtained for the different directions. They are really different. As a first approximation, you can assume that the particles are platelets with a cylindrical symmetry and calculate two sizes, parallel and perpendicular to the z axis. 011 peak corresponds to some canted direction and does not tell you anything.
Even XRD can also give better crystallite size but you need to take out the instrument broadening, size effect and strain effect.
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