What is Scherrer equation?

The equation D = Kλ/(β cosθ) for estimating crystallite size from the broadening of XRD peaks, where β is the full width at half maximum in radians and K is about 0.9.

Japanese name: シェラーの式

Scherrer equation appears in 2 practice questions across 2 topics, including Materials characterization (SEM, TEM, XRD, XPS), Nanomaterials.

Field: Materials characterization (SEM, TEM, XRD, XPS)

Why peaks broaden

In a large crystal, scattering from many lattice planes cancels as soon as the angle deviates slightly from the Bragg angle, so the peak is sharp. When the crystallite is small and few planes diffract together, this cancellation is incomplete, and intensity remains over a wider range of angles around the Bragg angle. The Scherrer equation D = Kλ/(β cosθ) expresses the relation between this broadening and crystallite size; the broadening is inversely proportional to the size. For example, with Cu Kα radiation, a peak at 2θ = 30° with a full width at half maximum of 0.25° gives D of about 33 nm with K = 0.9.

Cautions when using the equation

β is an angle in radians, taken directly as the width measured on the 2θ scale, while θ is half of the peak position 2θ. The measured peak width also contains the broadening of the instrument itself, so the instrumental width is determined with a standard such as LaB₆ and subtracted (for Gaussian peaks, as the square root of the difference of squares). K depends on the crystallite shape and on whether β is the full width at half maximum or the integral breadth, and values around 0.9 are commonly used. As crystallites grow, the broadening is buried in the instrumental width, so the practical upper limit is roughly 100–200 nm.

Crystallites, grains and particles

What the Scherrer equation gives is the size of the region that behaves as one perfect crystal for diffraction (the crystallite), which may not match the size of grains or particles seen in a microscope. If each particle is polycrystalline and contains several crystallites, the XRD value is smaller than the particle size seen by SEM. If small particles are agglomerated, the size measured by laser diffraction is larger still. The value is a volume-weighted average size and says nothing about the width of the distribution. Comparing it with TEM images or the specific surface area from the BET method gives a more complete picture of the sample.

Separating strain broadening

Non-uniform strain in a crystal spreads the interplanar spacings and also broadens peaks. This broadening is proportional to tanθ, a different angular dependence from crystallite broadening, which is proportional to 1/cosθ. The Williamson–Hall method writes β cosθ = Kλ/D + 4ε sinθ and plots β cosθ for several peaks against 4 sinθ; the intercept gives the crystallite size and the slope the non-uniform strain ε. In samples that have been deformed or ground in a ball mill, strain broadening is large, so calculating with the Scherrer equation alone underestimates the crystallite size.

Try it: questions about Scherrer equation

In XRD, a peak at 2θ = 38.2° had a full width at half maximum (after removing instrumental broadening) of 0.40°. Using the Scherrer equation D = Kλ/(β cosθ) with K = 0.9 and λ = 0.15406 nm, which value is closest to the crystallite size?

  1. AAbout 0.37 nm
  2. BAbout 25 nm
  3. CAbout 21 nm
Show answer and explanation

Answer: C. About 21 nm

Convert the width to radians, β = 0.40 × π/180 ≈ 6.98×10⁻³ rad, and use half of 2θ, 19.1° (cosθ ≈ 0.945), for θ. D = 0.9 × 0.15406/(6.98×10⁻³ × 0.945) ≈ 21 nm. About 0.37 nm substitutes β as 0.40 in degrees, and about 25 nm uses the cosine of 2θ, cos38.2°, instead of θ. What the Scherrer equation gives is not the grain size itself but the size of the crystallites that diffract coherently. Nonuniform strain also broadens peaks, so strictly the two are separated by methods such as the Williamson–Hall method. When crystallites become large, the broadening is too small to determine the size accurately.

FAQ

Why is the Scherrer constant K about 0.9?

Because the theoretical value for nearly spherical or cubic crystallites, with β taken as the full width at half maximum, is around 0.9. It changes somewhat with shape and the definition of width, so use the same K and the same width definition when comparing samples.

Should the peak width be entered in degrees or radians?

In radians. Multiply a width measured in degrees by π/180 to convert it. Entering it in degrees makes the crystallite size come out about 1/57 of the correct value.

Why is the size from the Scherrer equation smaller than the particle size seen by SEM?

Because each particle may consist of several crystallites, or non-uniform strain may be broadening the peaks. Twins and stacking faults also shrink the regions that diffract coherently, and so reduce the value.

Scherrer equation in the explanations

Questions that mention Scherrer equation (2)

  1. In XRD, a peak at 2θ = 38.2° had a full width at half maximum (after removing instrumental broadening) of 0.40°. Using the Scherrer equation D = Kλ/(β cosθ) with K = 0.9 and λ = 0.15406 nm, which value is closest to the crystallite size?
  2. Which correctly describes the principle of the BET method for measuring the specific surface area of a powder?

Related terms

XRD (X-ray diffraction) BET method Specific surface area Bragg's law EDS (energy-dispersive X-ray spectroscopy) Extinction rules SEM (scanning electron microscope) SIMS (secondary ion mass spectrometry) XPS (X-ray photoelectron spectroscopy)

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