What Are the Exact Scherrer Shape Factors for Different Crystallite Structures?

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SUMMARY

The Scherrer equation is essential for determining crystallite sizes from X-ray diffraction (XRD) spectra, incorporating a shape factor (K) that varies based on crystallite geometry. The shape factor is 0.89 for spherical particles and 0.94 for cubic particles, with a common default of 0.9 for unknown shapes. The discussion highlights the need for a comprehensive table of shape factors for various crystallite structures, referencing the International Union of Crystallography for specific values related to cubic systems. It emphasizes the distinction between crystallite shape and unit cell structure, noting that milling can alter crystallite shapes significantly.

PREREQUISITES
  • Understanding of the Scherrer equation and its application in XRD analysis.
  • Familiarity with crystallite shapes and their influence on diffraction patterns.
  • Knowledge of the International Union of Crystallography resources.
  • Basic principles of X-ray diffraction and crystallography.
NEXT STEPS
  • Research the International Union of Crystallography for detailed shape factor tables.
  • Explore the Scherrer equation's application in various crystallite systems beyond cubic and spherical.
  • Investigate the effects of milling on crystallite shape and size determination.
  • Learn about advanced XRD analysis techniques and software, such as ExpertPlus, for crystallite size calculation.
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Researchers in materials science, crystallographers, and anyone involved in XRD analysis seeking precise crystallite size determination and shape factor references.

Salish99
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The Scherrer equation is used to determine crystallite sizes from XRD spectra.
they contain K, a shape factor, that varies from 0.89 for spherical to 0.94 for cubic particles. Usually, this is set to 0.9 for particles of unknown size.
Does anyone know where we could find a table with the exact, published values for different crystallites (e.g. monoclinic, cubic, etc.)?

Scherrer's original papers are published here
http://gdz.sub.uni-goettingen.de/dm...]=scherrer&tx_goobit3_search[squery]=scherrer
but a) in German, and b) do not contain such a table.

thanks.

The discussions in
https://www.physicsforums.com/showthread.php?t=122682&page=2
and
https://www.physicsforums.com/showthread.php?t=322033
simply assume it's 0.9. No reference is given there. I need the exact table, and a reference.
 
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I think you are mixing the crystallite shape with the unit cell of the crystal.
As you say yourself, K depends on the shape of the crystallite and not that of the unit cell.
There may be a realtion between unit cell and crystallite shape (NaCl has cubic unit cell and may crystalize in cubic crystallites) but is not unique and not well determined.

If you take cubic crystallites and grind them in a mill you may end up with spherical particles or irregularly shaped particles, depending on the milling conditions.
There are also crystals with cubic unit cell (ferrites for example) that may crystallize in elongated (needle-like) crystals.
 
Attached a table from the International Union of Crystallography, Vol 3 Section 5.2, 1986 pp.318-323. It shows some shape factors to be used for the cubic system, depending on the reflection plane, and the shape (octahedron, cube, tetrahedron) of the actual crystal.
Interesting to see is that only one value is below 1, whereas is all this forum, as well as in many published papers, 0.9 is usually used "when the shape is unknown".
I would like to have someone tell me where that information comes from, with reference please.
thanks!

(and, no, outdated student pages like http://www.d.umn.edu/~bhar0022/dpcalculator/index.php and http://www.eng.uc.edu/~gbeaucag/Classes/XRD/SathishScherrerhtml/SathishScherrerEqn.html do not have any references either - maybe people don't think about that factor usually)
 

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I have value of 2 theta 35.546 deg, d spacing 2.5235 A, fwhm (2 th) 0.148 deg, can someone use it to get size of crystallite. after getting data analysed by a program expertplus i have got this, thankyou
 

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