Suppose there are two XRD instruments with resolution/FWHM (A) 0.030 & (B) 0.30 @ two theta ~ 350 (assume these Instrument FWHM are calculated by measuring same NIST SRM powder at two instruments in range10-1400). Also Instrument resolution file (IRF) are made using FullProf software in 2 theta range 10-1400.
Now a nano-crystallite ~50nm (or strained material) powder XRD performed at both instruments A & B and contribution of respective Instrument (IRF's) are subtracted.
Then measurements with two instruments (A & B) give
(1) size or strain value comparable but not equal? if yes, how?
(2) size or strain value obtained in two cases EQUAL after removing instrument resolution? If yes, how?
(3) If size or strain value obtained in two cases differ significantly, then which one give better result compared to other (although respective instrument contribution is removed from both cases)? Why?
(4) Other options?
There is a general rule: whenever you calculate any small value as a difference between two large values, you get big error. In this case the small value is broadening, the large values are FWHMs. So with device B you will usually get the result with large uncertainty (and because of that, usually different from the result obtained with device A).
Some back of the envelope estimation:
If you use Scherrer equation at 2Theta=35, the broadening for 50 nm particles will be ~0.18 deg. If you can measure FWHM with 10% uncertainty and the broadening is simply FWHM-FWHMstand, for device A you will get FWHMstand=0.03±0.003, FWHM=0.21±0.021 and relative error in determination of broadening will be (0.021+0.003)/0.18=13%. For device B FWHMstand=0.3±0.03, FWHM=0.48±0.048 and relative error in the calculation of broadening will be (0.03+0.048)/0.18=43%.
Similar difference in the relative error will remain for the more sophisticated methods of analysis.
There is a general rule: whenever you calculate any small value as a difference between two large values, you get big error. In this case the small value is broadening, the large values are FWHMs. So with device B you will usually get the result with large uncertainty (and because of that, usually different from the result obtained with device A).
Some back of the envelope estimation:
If you use Scherrer equation at 2Theta=35, the broadening for 50 nm particles will be ~0.18 deg. If you can measure FWHM with 10% uncertainty and the broadening is simply FWHM-FWHMstand, for device A you will get FWHMstand=0.03±0.003, FWHM=0.21±0.021 and relative error in determination of broadening will be (0.021+0.003)/0.18=13%. For device B FWHMstand=0.3±0.03, FWHM=0.48±0.048 and relative error in the calculation of broadening will be (0.03+0.048)/0.18=43%.
Similar difference in the relative error will remain for the more sophisticated methods of analysis.
Now my grasp of the obvious has become even more secure Alex! Nicely articulated :-)
I guess this is the advantage of the XRD rocking curve technique using parallel beam geometry and small pixel size 2D detector. The typically large FWHM values reported using the conventional diffractogram employing a 0D detector are principally influenced by the angular width of the incident beam itself since all of it is integrated into one relative intensity value for the entire spatial sampling area in 2D. In most practical situations out of desire for higher intensity (poor detector characteristics), wavelength dispersion is Mass Martyred along with a whole lot of other 2D reciprocal space ever present XRD data and parameters like preferred orientation, CDD (coherently diffracting domain) size/distribution, 3D strain state, etc.:-)
BTW whenever I see the term "removing" instead of "deconvoluting" in XRD, I'm already suspicious :-)
You'd need to use a known standard in both instruments to ascertain the effect of the instrumental broadening in each case.
Post your two raw data sets when convenient without any corrections.
My thoughts exactly mirror those of Alexander B. Missyul. Its all to do with the
differences, small or large, between numbers, small or large.
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