i am getting crystallite size in the range 4 nm-20 nm corresponding to different XRD peaks of the sample using Scherrer's formula. What to say about average crystallite size?
As already mentioned above by Gerhard Martens, microstrain might contribute to the peak broadening that you observe. Williamson-Hall method is the simplest technique to separate grain size and microstrain effects. If this doesn't really work, you might need even more sophisticated methods (modified Williamson-Hall; Warren-Averbach; WPPM; etc.).
Besides, I hope that you corrected for the instrumental broadening before calculating the grain size (or better domain size). Any other contributions than grain size to peak broadening (microstrain, instrumental effects, etc.) will result in a too small calculated grain size if you apply the Scherrer equation without separating these other effects first.
Your XRD peaks seem to deliver different crystallite sizes at different 2theta values.
So please try Willliamsen Hall plot in order to eliminate strain broadening influence onto your crystallite size results.
Please have a look at the section 'size and strain broadening' of the attachments.
Good luck.
By the way: you may share what type of sample(s) do you have?
https://en.wikipedia.org/wiki/Powder_diffraction
4 nm...isn' t this a bit too small.? These are about 10 unit cells along one direction...
As already mentioned above by Gerhard Martens, microstrain might contribute to the peak broadening that you observe. Williamson-Hall method is the simplest technique to separate grain size and microstrain effects. If this doesn't really work, you might need even more sophisticated methods (modified Williamson-Hall; Warren-Averbach; WPPM; etc.).
Besides, I hope that you corrected for the instrumental broadening before calculating the grain size (or better domain size). Any other contributions than grain size to peak broadening (microstrain, instrumental effects, etc.) will result in a too small calculated grain size if you apply the Scherrer equation without separating these other effects first.
Scherrer's formula is a simplistic approximation which does not work most times. See Timo Müller's explanation.
Hello,
For each XRD peak, it's normal to found an average crystallite size.
Generally, we applicate the Sherrer formula for the intense XRD peak. Then, it will be better to make SEM or TEM images for more comprehension.
Best regards.
Hi,
Do you have SEM/TEM confirmation? You might have nano-rods or something. I am suggesting just one more possibility...
Why u still using SF. It has no direct physical interpretation unless u introduce K that depends on the how the line profile width is determined, the shape of the crystallites, and the size distribution - which is usually unknown. All the FWHM base calculation dependent on K. An independent approximation intruduced by Laue where K could be assumed as 1 when use integral breadth. But as the profile itself usually contains Gaussian and Lorentzian function than a double voight approach has to be use.
The calculated value itself cud be area weighted or volume weighted, depend on methods of analysis. And things get complicated when u compare with TEM analysis, where the number weighted is the result.
Also... planar faults can introduce a size component that varies with hkl plane
And sometimes the shape of a crystal means size broadening varies with hkl
Two orders of peaks could help, but given the limited information impossible to say...
I have seen calculations here on RG which use degree values for the 2theta FWHM value when calculating the crystallite size via Scherrer or Williamsen-Hall. One has to use FWHM in radians ( please see the link). Degree value as input in these equations will give a factor of about 60 (=180/pi) too less crystallite size !!!!. That's why you might have calculated too little crystallite sizes ( see the comments above of Alan and Gert) . However do not forget to subtract instrumental broadening.
https://en.wikipedia.org/wiki/Scherrer_equation
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