xraytools.

Line Broadening

Powder peaks get wider when the crystallites get smaller — and also when the lattice spacings vary. Give the positions and widths you measured and the page takes the instrument out, applies Scherrer to each peak, and then separates size from strain the only way a single peak cannot: by how the broadening grows with angle.

You supply
A wavelength, your instrument’s own width, and a list of peaks — each one a 2θ and a full width at half maximum in degrees. Three peaks are the minimum for a Williamson–Hall fit and they are worth more the more widely they are spread.
Reading it
Every answer comes in two versions, because taking the instrument’s width out of a measured width depends on the peak shape and a real peak is between the two limits: the answer is inside the bracket, not at either end. L is a coherently diffracting domain and not a particle size, K = 0.9 is a convention worth about ten per cent on its own, and the instrumental width has to be measured on a standard — the default here is a plausible diffractometer and is not yours.

Worked examples: size only, simulated at 15 nm · strain only, simulated at 0.30% · both at once, simulated at 25 nm and 0.15%

See also: Bragg Calculator · HKL Calculator · Reduced Cell and Bravais Lattice

Input

The measurement
Å
° 2θ

This wavelength is Cu Kα.

The instrumental width is measured on a standard that broadens nothing of its own — LaB6 or silicon. The default is this site’s own assumed 0.08°, which is a plausible laboratory diffractometer and is not yours.

The peaks

Anything after a # is ignored, so you can keep a header row or label your peaks. Use a dot for the decimal point — a comma separates the two numbers on a line.

These widths are simulated

Nothing was submitted, so the box holds a pattern this site generated rather than measured: the powder simulator next door, run at Cu Kα with 25 nm crystallites, 0.15% microstrain and a half-Lorentzian peak shape, on the same 0.08° instrumental width set above. So the answer is known, and the interesting thing is that neither column below gives it — the two bracket it, because the real peak shape is between the two limits each column assumes. Paste your own widths over them.

The peaks you entered, drawn

18.434.250.065.881.60.4010.4260.4570.4930.5350.5860.6482θ / °equal heights

The 7 peaks as you entered them, each drawn at its own measured width. The heights are all the same because this page was given no intensities — a Williamson–Hall analysis does not need them — so the only thing changing across the picture is the width, which is the point. Under each one, in a lighter line, is the instrumental width of 0.080° you set above: it is the same at every angle, so wherever a measured peak is barely wider than it, that peak is telling you almost nothing about the sample. Drawn as a pseudo-Voigt with η = 0.5, which is a choice about ink and not about the analysis — the two columns below refuse to choose, which is why there are two of them.

-1.23-0.620.00+0.62+1.232θ 20.00°: 0.401° wide2θ 80.00°: 0.648° widedistance from the peak centre, ° 2θ

The same peaks, moved onto one centre so the widths can be compared — at full scale above they are all thin spikes, because a peak half a degree wide is a few pixels of a sixty-degree scan, and that is what a real pattern looks like. The axis here is distance from each peak’s own centre, not 2θ: the peaks are not all at one angle, they have been stacked. Across your list the width grows by a factor of 1.62, from the low-angle peak to the high-angle one, and the dashed curve is the instrument, which does not grow at all. Everything between those two facts is the sample.

Each peak on its own

The instrument’s own width has to be taken out of each measured width, and how depends on the peak shape: widths subtract for Lorentzian peaks (β = βobs − βinst) and squares subtract for Gaussian ones (β² = βobs² − βinst²). A real laboratory peak is a mixture of the two, so both are given and the answer is between them. They differ most when a peak is only a little wider than the instrument, which is exactly when a size is least worth quoting.

Sample broadening and the size it implies, peak by peak
2θ / ° FWHM / ° Lorentzian β / ° Lorentzian L / nm Gaussian β / ° Gaussian L / nm
20.00 0.4009 0.3209 25.2 0.3928 20.6
30.00 0.4262 0.3462 23.8 0.4186 19.7
40.00 0.4566 0.3766 22.5 0.4495 18.8
50.00 0.4926 0.4126 21.3 0.4861 18.0
60.00 0.5354 0.4554 20.2 0.5294 17.3
70.00 0.5864 0.5064 19.2 0.5809 16.7
80.00 0.6477 0.5677 18.3 0.6427 16.1

A dash means the peak is not wider than the instrument under that shape rule, so it says nothing about the sample and is left out of the fit below.

Each row applies the Scherrer equation to that peak alone, which assumes all of its width beyond the instrument comes from crystallite size. Any microstrain in the sample is counted as size as well, so these are lower bounds: the real crystallites are at least this large and are usually larger. The plot below is what separates the two.

Williamson–Hall, Lorentzian peaks

The classical plot: β cos θ against 4 sin θ, with β in radians. Size broadening does not depend on the abscissa at all, so it sits in the intercept as Kλ/L; strain broadening is proportional to it, so it is the slope, and the slope is ε directly.

0.000.681.362.042.73-9.26×10⁻⁴2.26×10⁻³5.45×10⁻³8.64×10⁻³Intercept: K lambda / L, giving L = 29.6 nm2θ = 20.00°, FWHM 0.4009°, sample broadening 0.3209°, Scherrer size 25.2 nm2θ = 20.00°, FWHM 0.4009°, sample broadening 0.3209°, Scherrer size 25.2 nm2θ = 30.00°, FWHM 0.4262°, sample broadening 0.3462°, Scherrer size 23.8 nm2θ = 30.00°, FWHM 0.4262°, sample broadening 0.3462°, Scherrer size 23.8 nm2θ = 40.00°, FWHM 0.4566°, sample broadening 0.3766°, Scherrer size 22.5 nm2θ = 40.00°, FWHM 0.4566°, sample broadening 0.3766°, Scherrer size 22.5 nm2θ = 50.00°, FWHM 0.4926°, sample broadening 0.4126°, Scherrer size 21.3 nm2θ = 50.00°, FWHM 0.4926°, sample broadening 0.4126°, Scherrer size 21.3 nm2θ = 60.00°, FWHM 0.5354°, sample broadening 0.4554°, Scherrer size 20.2 nm2θ = 60.00°, FWHM 0.5354°, sample broadening 0.4554°, Scherrer size 20.2 nm2θ = 70.00°, FWHM 0.5864°, sample broadening 0.5064°, Scherrer size 19.2 nm2θ = 70.00°, FWHM 0.5864°, sample broadening 0.5064°, Scherrer size 19.2 nm2θ = 80.00°, FWHM 0.6477°, sample broadening 0.5677°, Scherrer size 18.3 nm2θ = 80.00°, FWHM 0.6477°, sample broadening 0.5677°, Scherrer size 18.3 nm4 sin θβ cos θ / rad
What the line says
Crystallite size L 29.6 nm, from the intercept Kλ/L
Microstrain ε 0.111 %, from the slope
Peaks used 7 of 7 given
R² of the fit 0.9975 how much of the points’ spread the line accounts for

Williamson–Hall, Gaussian peaks

Squares of both axes: β²cos²θ against (4 sin θ)², which is the form that follows when the two contributions add in quadrature. The intercept is (Kλ/L)² and the slope is ε², so both answers come out through a square root and a small negative value of either is not a small answer — it is no answer.

0.001.753.505.267.01-9.22×10⁻⁶2.25×10⁻⁵5.43×10⁻⁵8.60×10⁻⁵Intercept: K lambda / L, giving L = 20.7 nm2θ = 20.00°, FWHM 0.4009°, sample broadening 0.3928°, Scherrer size 20.6 nm2θ = 20.00°, FWHM 0.4009°, sample broadening 0.3928°, Scherrer size 20.6 nm2θ = 30.00°, FWHM 0.4262°, sample broadening 0.4186°, Scherrer size 19.7 nm2θ = 30.00°, FWHM 0.4262°, sample broadening 0.4186°, Scherrer size 19.7 nm2θ = 40.00°, FWHM 0.4566°, sample broadening 0.4495°, Scherrer size 18.8 nm2θ = 40.00°, FWHM 0.4566°, sample broadening 0.4495°, Scherrer size 18.8 nm2θ = 50.00°, FWHM 0.4926°, sample broadening 0.4861°, Scherrer size 18.0 nm2θ = 50.00°, FWHM 0.4926°, sample broadening 0.4861°, Scherrer size 18.0 nm2θ = 60.00°, FWHM 0.5354°, sample broadening 0.5294°, Scherrer size 17.3 nm2θ = 60.00°, FWHM 0.5354°, sample broadening 0.5294°, Scherrer size 17.3 nm2θ = 70.00°, FWHM 0.5864°, sample broadening 0.5809°, Scherrer size 16.7 nm2θ = 70.00°, FWHM 0.5864°, sample broadening 0.5809°, Scherrer size 16.7 nm2θ = 80.00°, FWHM 0.6477°, sample broadening 0.6427°, Scherrer size 16.1 nm2θ = 80.00°, FWHM 0.6477°, sample broadening 0.6427°, Scherrer size 16.1 nm(4 sin θ)²(β cos θ)² / rad²
What the line says
Crystallite size L 20.7 nm, from the intercept Kλ/L
Microstrain ε 0.213 %, from the slope
Peaks used 7 of 7 given
R² of the fit 0.9893 how much of the points’ spread the line accounts for

What the pair of them supports

Crystallite size. Between 20.7 nm (Gaussian) and 29.6 nm (Lorentzian).

Microstrain. Between 0.111% (Lorentzian) and 0.213% (Gaussian).

Both shape rules describe these peaks. Where they give a range, the answer is inside it and the width of the range is what it costs not to know the peak shape.

Four ways this goes wrong with the arithmetic right

L is not a particle size What diffraction measures is a volume-weighted mean column length along the scattering vector — the coherently diffracting domain. A particle made of several domains, or one with a fault running through it, gives an L smaller than anything an electron microscope would measure on the same powder, and the two disagreeing is the normal case rather than a discrepancy to resolve.
K is a convention, not a constant 0.9 is the usual value for the full width at half maximum of roughly spherical crystallites. 0.89, 0.94 and 1.0 are all in print, for different shapes and for the integral breadth rather than the FWHM, so a size from this page is good to about ten per cent before anything else is considered. Quote the K you used.
The instrumental width has to be measured The value here defaults to this site’s own 0.08°, which is a plausible well-aligned laboratory diffractometer and is not your instrument. It is measured on a standard with no size or strain broadening of its own — LaB6 or silicon — and it varies with angle, which this page treats as constant.
One peak cannot separate size from strain Both broadenings grow with angle, so a single width is satisfied by a whole family of (L, ε) pairs. Only the difference in how they grow — 1/cos θ against tan θ — tells them apart, and reading that difference needs peaks spread as widely in 2θ as the pattern allows. Three clustered peaks are worth less than three spread ones.

Where this comes from