How do I calculate crystallite size from an XRD peak with the Scherrer equation?

The equation is D = K × λ / (β × cos θ). D is the mean crystallite size in the direction perpendicular to the diffracting planes, λ the X-ray wavelength (0.15406 nm for Cu Kα1), β the peak width and θ the Bragg angle. The three classic mistakes are: using 2θ instead of θ, leaving β in degrees instead of radians, and giving the cosine function an angle in degrees when the software expects radians.

Your peak: θ = 36.25 / 2 = 18.125°, cos θ = 0.950. β = 0.50° × π / 180 = 0.00873 rad. With K = 0.9: D = 0.9 × 0.15406 / (0.00873 × 0.950) = 0.1387 / 0.00829 = about 17 nm.

K is a shape factor of order one. 0.9 is the conventional choice when β is the full width at half maximum and the crystallites are taken as roughly spherical; values from about 0.89 to 0.94 appear in the literature, and other values apply if the integral breadth is used instead of FWHM. The choice changes the answer by a few percent, which is smaller than the other uncertainties, so state which K and which width definition you used and keep them the same across samples.

Because of this, Scherrer sizes are best quoted to two significant figures at most and used to compare samples measured the same way, rather than as absolute dimensions.

The measured width includes broadening from the diffractometer itself, which must be removed. Measure a coarse, strain-free standard such as LaB6 or well-annealed silicon under the same conditions and take its width at a nearby angle as the instrumental contribution. For peaks that are close to Gaussian, β² = β_measured² - β_instrument².

With an instrumental width of 0.10°, your 0.50° peak becomes √(0.25 - 0.01) = 0.49° and the size changes only from 16.7 to 17.1 nm. For a sharper peak it matters much more: a measured 0.20° gives 42 nm uncorrected and 48 nm corrected. Once the sample width approaches the instrument width, the method stops working, which in practice limits it to crystallites below roughly 100 nm.

Size is not the only source of broadening. Lattice microstrain from defects also widens peaks, and its contribution grows with tan θ while size broadening goes with 1 / cos θ. If you apply the Scherrer equation to several peaks and the calculated size falls steadily at higher angles, strain is present.

The Williamson-Hall method separates them: plot β × cos θ against 4 × sin θ for all peaks. The intercept equals K × λ / D and the slope is the strain. Different sizes from different peaks with no clear trend in angle usually mean the crystallites are not equiaxed, for example ZnO rods that are longer along the c-axis, so the (002) peak is sharper than (100).

On the last part of the question: the result is the size of coherently diffracting domains, not the particle size. One particle seen in SEM or TEM can consist of several crystallites, and a dense agglomerate of many, so the XRD size is normally equal to or smaller than the microscope size. If the two agree, the particles are probably single crystals.

It is also a volume-weighted average, so a few large crystallites dominate, and for a broad size distribution the number will sit above the number-average counted from images. Reporting both the XRD crystallite size and the microscopy particle size, clearly labelled, avoids the usual confusion.