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XRD crystallite size (Scherrer) and d-spacing (Bragg) calculator

Q: What is the Scherrer equation?

D = Kλ/ (βcos θ): the mean crystallite size D from the width β (FWHM, in radians) of an X-ray diffraction peak at Bragg angle θ, the wavelength λ and a shape factor K of about 0.9. Narrower peaks mean larger crystallites.

Q: What value of K should I use?

K = 0.9 is the common default. About 0.89–0.91 suits spherical or cubic crystallites (0.94 is also used for spheres) and about 1.0–1.2 suits plates and rods. Report the value you used.

Q: Does the FWHM need to be in radians?

Yes. The Scherrer equation needs β in radians: multiply the FWHM in degrees 2θ by π/180. The tool takes degrees and converts them.

Q: How do I calculate d-spacing from 2θ?

Use Bragg’s law, d = nλ/ (2 sin θ), with θ half of 2θ. For Cu Kα1 (1.54056 Å) a first-order peak at 2θ = 36.3° gives d = 2.473 Å.

Crystallite size D = Kλ / (β cos θ), where β is the peak’s full width at half maximum in radians, θ is half the 2θ peak position, λ the X-ray wavelength and K ≈ 0.9; the d-spacing comes from Bragg’s law, d = nλ / 2 sin θ. Paste your peak positions and widths to get the size and d-spacing of every peak and their mean, with the instrument’s broadening removed if you enter it.

What it does: Calculates crystallite size from XRD peak widths with the Scherrer equation and d-spacing with Bragg’s law, for one or many peaks, with instrumental broadening removed.
Formula: $\begin{aligned} &D = K\lambda \div (\beta\cos\theta), \\ &\quad \beta = \text{FWHM in radians}, \ K \approx 0.9 \\ &n\lambda = 2d\sin\theta \\ &\beta = \sqrt{\beta_\mathrm{obs}^2 - \beta_\mathrm{inst}^2} \end{aligned}$
Example: A peak at 2θ = 36.3° with FWHM 0.40° (Cu Kα1, 1.54056 Å, K = 0.9): D = 0.9 × 0.154056 nm ÷ (0.006981 × 0.9502) = 20.9 nm, d = 2.473 Å.

Last updated 8 November 2026

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At a glance

XRD Scherrer & Bragg calculator at a glance
Formulas	$D = K\lambda / (\beta\cos\theta)$ ; $n\lambda = 2d\sin\theta$
Input	$2\theta$ and FWHM in degrees, one row per peak (optional hkl label)
Wavelength	Cu Kα1 1.54056 Å, or any value you type
Shape factor K	0.9 default; 0.89–0.91 spheres or cubes, 0.94 spheres, 1.0–1.2 plates and rods
Instrument correction	$\beta^2 = \beta_\mathrm{obs}^2 - \beta_\mathrm{inst}^2$ (Gaussian peaks)
Limits	Ignores strain; reliable up to about 100–200 nm
Output	$D$ and $d$ per peak, mean and SD; CSV and chart (PNG, SVG)
Runs in	Your web browser; nothing is uploaded
Price	Free, no account needed

What it does

How to use the XRD Scherrer & Bragg calculator

Set the radiationKeep Cu Kα1 (1.54056 Å) or choose Other and type the wavelength of your source in ångström.
Choose KUse 0.9 unless you know the crystallite shape: about 0.89–0.91 (0.94 is also used) for spheres or cubes, about 1.0–1.2 for plates and rods.
Paste the peaksOne row per peak: 2θ and FWHM in degrees, optionally an hkl label, from your peak-fitting software or a spreadsheet.
Remove the instrumentEnter the FWHM of a well-crystallised standard measured on the same instrument; it is subtracted in quadrature from each peak.
Read the sizesGet $D$ and $d$ for every peak, the mean size and its spread, and download the table (CSV) or the chart.

How it works

$D = \frac{K \cdot \lambda}{\beta \cdot \cos\theta}$ $n\lambda = 2d \cdot \sin\theta$ $\beta = \sqrt{\beta_\mathrm{obs}^2 - \beta_\mathrm{inst}^2}$

Scherrer equation. $D$ is the mean crystallite size, measured perpendicular to the diffracting planes (for plate-like crystallites it is the thickness, not the lateral size); $\lambda$ is the X-ray wavelength; $\theta$ is the Bragg angle, half the $2\theta$ peak position; $\beta$ is the peak’s full width at half maximum converted to radians; $K$ is the shape factor, about 0.89–0.91 for spheres or cubes and about 1.0–1.2 for plates or rods, with 0.9 the common default (Khorsand Zak & Hashim 2025, Sci Rep). Some studies use $K = 0.94$ for spherical particles (Canchanya-Huaman et al. 2021, Nanomaterials).

Bragg’s law. Constructive interference from planes spaced $d$ apart happens when $n\lambda = 2d\sin\theta$ , with $n$ the order of diffraction; the tool reports first-order d-spacings for the peak list and lets you set $n$ in the converter.

Instrumental broadening. A measured peak is broadened by both the sample and the instrument, so the instrument’s contribution should be removed, measured as the FWHM of a well-crystallised standard (Canchanya-Huaman et al. 2021). Variances of independent broadenings add (Khorsand Zak & Hashim 2025), and for Gaussian peaks the FWHM scales with the standard deviation, so the tool subtracts in quadrature: $\beta^2 = \beta_\mathrm{obs}^2 - \beta_\mathrm{inst}^2$ . Peaks no wider than the instrument get no size.

Limits. The Scherrer equation treats all broadening as size broadening, ignoring lattice strain; Williamson–Hall, size–strain plot or Rietveld analysis separate the two. It gives an average over the crystallites that diffract, works best for roughly isotropic crystallites, and is only reliable up to about 100–200 nm, above which the size broadening is hard to separate from the instrument’s (Canchanya-Huaman et al. 2021). The wavelength is Cu Kα1, $\lambda = 1.54056\ \text{Å}$ , as used in that study.

Worked example: one peak with Cu Kα1

A peak at $2\theta = 36.3^\circ$ has a FWHM of 0.40°, measured with Cu Kα1 ( $\lambda = 1.54056\ \text{Å}$ ); take $K = 0.9$ .

$\theta = 18.15^\circ$ , $\cos\theta = 0.9502$ .
$\beta = 0.40^\circ \times \pi/180 = 0.006981\ \mathrm{rad}$ .
$D = \dfrac{0.9 \times 0.154056\ \mathrm{nm}}{0.006981 \times 0.9502} = \mathbf{20.9\ nm}$ .
$d = \dfrac{1.54056\ \text{Å}}{2 \times \sin 18.15^\circ} = \mathbf{2.473}\ \boldsymbol{\text{Å}}$ .

If a standard on the same instrument gives a FWHM of 0.08°, the sample width is $\sqrt{0.40^2 - 0.08^2} = 0.392^\circ$ and $D = 21.3\ \mathrm{nm}$ . With $K = 0.94$ instead of 0.9 the uncorrected size is 21.8 nm, so state K and the correction with the result.

Frequently asked questions

What is the Scherrer equation?

$D = K\lambda / (\beta\cos\theta)$ : the mean crystallite size $D$ from the width $\beta$ (FWHM, in radians) of an X-ray diffraction peak at Bragg angle $\theta$ , the wavelength $\lambda$ and a shape factor $K$ of about 0.9. Narrower peaks mean larger crystallites.

What value of K should I use?

$K = 0.9$ is the common default. About 0.89–0.91 suits spherical or cubic crystallites (0.94 is also used for spheres) and about 1.0–1.2 suits plates and rods. Report the value you used.

Does the FWHM need to be in radians?

Yes. The Scherrer equation needs $\beta$ in radians: multiply the FWHM in degrees $2\theta$ by $\pi/180$ . The tool takes degrees and converts them.

How do I correct for instrumental broadening?

Measure a well-crystallised standard on the same instrument and settings, take its FWHM near your peak, and remove it. This tool subtracts it in quadrature, $\beta = \sqrt{\beta_\mathrm{obs}^2 - \beta_\mathrm{inst}^2}$ , which assumes Gaussian peak shapes.

Why is my Scherrer size smaller or larger than the particle size from TEM?

XRD measures crystallite (coherent domain) size, not particle size: one particle can contain several crystallites. The Scherrer equation also ignores strain and, without a correction, instrumental broadening, so it is an estimate; check it with microscopy or Williamson–Hall or Rietveld analysis.

What is the size limit of the Scherrer equation?

It is reliable for crystallites up to about 100–200 nm. Larger crystallites give peaks so narrow that their broadening cannot be separated from the instrument’s.

How do I calculate d-spacing from 2θ?

Use Bragg’s law, $d = n\lambda / (2\sin\theta)$ , with $\theta$ half of $2\theta$ . For Cu Kα1 (1.54056 Å) a first-order peak at $2\theta = 36.3^\circ$ gives $d = 2.473\ \text{Å}$ .

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