From XRD analysis how to measure Full Width Half Maxima (FWHM)?
Movie from previous post: some true remarks, many not recommendable procedures:
-background had been subtracted and counting statistics is not appropriately considered.
-it is very likely alpha1/2 data. But it is usually inadequate to do the fitting to a resulting diffraction peak with one single symmetric peak, because the profile is intrinsically asymmetric due to the asymmetric instrumental broadening, which is actually a bimodal function. Typically one should fit with two (e.g. pseudoVoigt) funcitons with identical shape and FWHM and with positions constrained due to the relative difference of the wavelengths. The thus resulting FWHM values are different than this very primitive fitting.
In fact having that movie
https://youtu.be/PwCc4ecj2w0
NEVER use such FWHM in the Scherrer equation. The guy mentioned the instrumental resolution but then ignores it....
Dear Chandrakant Jadhav
Below is the link of a video for FWHM measurement.
Calculating FWHM for XRD Peaks using ORIGIN:
https://youtu.be/PwCc4ecj2w0
Hope this helps.
All the best.
dear Chandrakant Jadhav ;
you can use the Scherrer Equation to estimate the size of the crystallite:
FWHM (2θ) =(Κ λ / L cosθ)
The crystallite size broadening measured at large angles 2Theta.
S. Rai, B.K. Choudhary, T. Jayakumar, K.B.S. Rao and B. Raj, Int. J. Pres. Ves. Pip. 76 (1999) p.275.
[A.T. Ozdemir and L. Edwards, Fatig. Fract. Eng. Mater. Struct. 20 (1997) p.1443.
Movie from previous post: some true remarks, many not recommendable procedures:
-background had been subtracted and counting statistics is not appropriately considered.
-it is very likely alpha1/2 data. But it is usually inadequate to do the fitting to a resulting diffraction peak with one single symmetric peak, because the profile is intrinsically asymmetric due to the asymmetric instrumental broadening, which is actually a bimodal function. Typically one should fit with two (e.g. pseudoVoigt) funcitons with identical shape and FWHM and with positions constrained due to the relative difference of the wavelengths. The thus resulting FWHM values are different than this very primitive fitting.
In fact having that movie
https://youtu.be/PwCc4ecj2w0
NEVER use such FWHM in the Scherrer equation. The guy mentioned the instrumental resolution but then ignores it....
Dear Andreas Leineweber
Thank you for pointing out the errors and giving very helpful explanation.
Regards,
Nagabhushan J Choudhari
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