Williamson-Hall Plot Parameter Calculator

Fit a Williamson-Hall plot to your XRD peak widths and separate crystallite size from lattice microstrain in one least-squares step.

📈 Williamson-Hall Plot Parameter Calculator
X-ray wavelength1.5406 Å
Å
0.52.5
Scherrer shape factor K0.9
K
0.81.2
Instrumental FWHM0.05 deg
deg
00.5
Instrumental correction method
Peak list (one per line: 2-theta in degrees, then FWHM in degrees)
Crystallite size D (nm)
—
Microstrain ε
—
Slope (= ε)
—
Intercept (rad)
—
Fit R²
—
Peaks used
—
Step-by-step working
Per-peak Williamson-Hall data

📈 What is a Williamson-Hall Plot?

A Williamson-Hall plot is a straight-line graph used in X-ray diffraction to separate the two physical causes of peak broadening: finite crystallite size and lattice microstrain. Every diffraction peak from a real powder is wider than the instrument's resolution, and that extra width carries information about the microstructure. The trouble is that both size and strain widen peaks at once, so a single peak cannot tell you how much of the width belongs to each. The Williamson-Hall method solves that by using several peaks at once.

The insight, published by G. K. Williamson and W. H. Hall in 1953, is that the two effects depend on the diffraction angle differently. Size broadening follows the Scherrer relation and scales as one over cos(theta), while strain broadening scales as tan(theta). Multiply both by cos(theta) and the size term becomes a constant while the strain term becomes proportional to sin(theta). Plotting beta cos(theta) against 4 sin(theta) therefore produces a straight line whose intercept encodes the crystallite size and whose slope is the microstrain.

Materials scientists use this analysis routinely: to characterise ball-milled powders where heavy plastic deformation creates both nanoscale grains and huge dislocation densities, to track how annealing temperature grows crystallites and relieves strain in thin films, to compare nanoparticle batches synthesised under different conditions, and to check whether a Scherrer size quoted from a single peak is trustworthy or badly contaminated by strain.

The most common misconception is that Williamson-Hall gives a particle size. It does not. X-rays measure the size of coherently diffracting domains, so a single microscope-visible particle containing several twinned or grain-bounded domains will report a smaller size here than in electron microscopy. A second misconception is that a poor straight-line fit means the data are bad. More often it means the broadening is anisotropic, with crystallites shaped like plates or needles, which calls for a modified model that weights each reflection by its elastic modulus or dislocation contrast factor.

This calculator takes an X-ray wavelength, a shape factor, an instrumental width, and a list of peak positions with their measured widths, then performs the least-squares fit and reports crystallite size, microstrain, slope, intercept, and fit quality along with a full per-peak table and plot.

📐 Formula

β cosθ  =  Kλ ÷ D  +  4ε sinθ
β = broadening of the peak in radians, after removing the instrumental contribution
θ = Bragg angle, half the measured 2θ position
K = Scherrer shape factor, 0.9 for FWHM of roughly spherical crystallites
λ = X-ray wavelength (Å), 1.5406 for Cu K-alpha radiation
D = volume-weighted crystallite size (Å), read from the intercept as D = Kλ ÷ intercept
ε = microstrain, dimensionless, read directly from the slope of the fitted line
β  =  √(βobs² − βinst²)   or   βobs − βinst
βobs = measured FWHM of the peak in degrees
βinst = instrumental FWHM measured from a line-broadening standard such as NIST SRM 660 LaB6
Use the quadrature form for Gaussian peak shapes and the direct subtraction form for Lorentzian shapes
Example: five Cu K-alpha peaks with an intercept of 5.5473 × 10-3 rad and a slope of 1.4994 × 10-3 give D = 25.00 nm and ε = 0.1499%.

📖 How to Use This Calculator

Steps

1
Enter the X-ray wavelength and shape factor. Type the wavelength in angstroms (1.5406 for Cu K-alpha) and the Scherrer shape factor, normally 0.9.
2
Set the instrumental broadening and correction method. Enter the instrumental FWHM measured from a standard, then choose quadrature subtraction for Gaussian peaks or direct subtraction for Lorentzian peaks.
3
Paste your peak list. Enter one peak per line as the 2-theta position in degrees followed by the measured FWHM in degrees, separated by a comma.
4
Read the size, strain, and fit quality. The calculator fits beta cos(theta) against 4 sin(theta) and reports the crystallite size from the intercept, the microstrain from the slope, and the R-squared of the fit.

💡 Example Calculations

Example 1 — Nanocrystalline Powder, 25 nm with Moderate Strain

Cu K-alpha, K = 0.9, instrumental FWHM 0.05 deg, five peaks from 30 to 90 degrees 2-theta

1
Correct each FWHM in quadrature, for example at 2θ = 30: √(0.4241² − 0.05²) = 0.4211 deg
2
Convert to radians and form the plot coordinates: x = 4 sin(15°) = 1.0353, y = β cos(15°) = 7.0999 × 10-3
3
Least-squares fit over all five peaks gives intercept = 5.5473 × 10-3 rad and slope = 1.4994 × 10-3
4
D = Kλ ÷ intercept = 0.9 × 1.5406 ÷ 5.5473 × 10-3 = 249.95 Å = 25.00 nm, and ε = slope = 0.1499%
Crystallite size 25.00 nm with microstrain 0.1499%, R² = 1.0000.
Try this example →

Example 2 — Well-Annealed Sample, Large Grains and Little Strain

Cu K-alpha, K = 0.9, instrumental FWHM 0.05 deg, much narrower peaks after annealing

1
The peaks are far narrower (0.2015 to 0.3430 deg), so the corrected β values are small and the intercept is low
2
Fitted intercept = 2.7732 × 10-3 rad, exactly half the value from Example 1
3
D = 0.9 × 1.5406 ÷ 2.7732 × 10-3 = 500.0 Å = 50.00 nm, and ε = 0.0500%
Annealing doubled the crystallite size to 50.00 nm and cut microstrain to 0.0500%.
Try this example →

Example 3 — Ball-Milled Powder, 10 nm Grains and Heavy Strain

Cu K-alpha, K = 0.9, instrumental FWHM 0.05 deg, very broad peaks after extended milling

1
Peak widths run from 1.0079 to 1.8117 deg, and they grow steeply with angle, which is the signature of strain
2
Fitted intercept = 1.3866 × 10-2 rad and slope = 2.9997 × 10-3
3
D = 0.9 × 1.5406 ÷ 1.3866 × 10-2 = 100.0 Å = 10.00 nm, and ε = 0.3000%
Milling produced 10.00 nm domains carrying 0.3000% microstrain, twice the strain of Example 1.
Try this example →

❓ Frequently Asked Questions

What is the Williamson-Hall method?+
The Williamson-Hall method separates the two causes of X-ray diffraction peak broadening: finite crystallite size and lattice microstrain. It plots beta cos(theta) against 4 sin(theta) for several peaks. The intercept gives K times lambda divided by the crystallite size, and the slope gives the microstrain.
What is the Williamson-Hall equation?+
Beta cos(theta) equals K lambda divided by D, plus 4 epsilon sin(theta). Beta is the corrected integral breadth or FWHM in radians, theta is the Bragg angle, K is the shape factor (usually 0.9), lambda is the X-ray wavelength, D is the crystallite size, and epsilon is the microstrain.
How is Williamson-Hall different from the Scherrer equation?+
The Scherrer equation uses one peak and assumes all broadening comes from crystallite size, so it systematically underestimates size in strained samples. Williamson-Hall uses several peaks and exploits the different angular dependence of size and strain broadening to separate the two contributions.
Why does size broadening go as 1 over cos(theta) and strain as tan(theta)?+
Size broadening follows the Scherrer relation, which is independent of angle once multiplied by cos(theta), so it appears as a constant intercept. Strain broadening scales with tan(theta), which becomes 4 epsilon sin(theta) after the same multiplication, so it appears as a slope.
What K value should I use?+
The shape factor K is 0.9 for the FWHM of roughly spherical crystallites, which is the near-universal default. It ranges from about 0.62 to 2.08 depending on crystallite shape and whether FWHM or integral breadth is used, so quote the K value alongside any size you report.
How do I measure instrumental broadening?+
Run a line-broadening standard with crystallites large enough and strain low enough that all measured width comes from the instrument, typically NIST SRM 660 lanthanum hexaboride or SRM 640 silicon. The FWHM of its peaks near your peaks of interest is your instrumental value.
Should I subtract instrumental broadening in quadrature or directly?+
Quadrature subtraction, the square root of the observed width squared minus the instrumental width squared, is correct for Gaussian profiles. Direct subtraction is correct for Lorentzian profiles. Most real peaks are Voigt profiles, so the true answer sits between the two results.
What does a negative slope in a Williamson-Hall plot mean?+
A negative slope implies compressive microstrain, which can be genuine in a film under compression or in a nanoparticle with surface stress. More often it means anisotropic broadening or poor peak fitting, so check the R-squared value and the individual peak widths before reporting it.
How many peaks do I need?+
Two peaks are the mathematical minimum for a straight-line fit, but four to six well-separated reflections spanning a wide 2-theta range are needed for a meaningful result. More peaks reduce the effect of any single mis-fitted width on the slope and intercept.
What does the R-squared value tell me?+
It measures how well a single straight line describes all the peaks. Values above about 0.95 suggest an isotropic size-plus-strain model fits. Low values usually indicate anisotropic broadening, where crystallites are plate-like or needle-like, and call for a modified Williamson-Hall model.
What is microstrain in a crystal?+
Microstrain is the root-mean-square variation in lattice spacing within a sample, caused by dislocations, vacancies, stacking faults, or compositional gradients. It is a dimensionless fraction, often quoted as a percentage or in units of 10 to the minus 3, and it broadens peaks progressively at higher angles.
Why is my Williamson-Hall size different from a TEM particle size?+
X-ray methods measure coherently diffracting domains, not whole particles. One TEM-visible particle can contain several domains separated by grain boundaries or twins, so the diffraction size is usually equal to or smaller than the microscopy particle size.

What is the Williamson-Hall method?

The Williamson-Hall method separates the two causes of X-ray diffraction peak broadening: finite crystallite size and lattice microstrain. It plots beta cos(theta) against 4 sin(theta) for several peaks. The intercept gives K times lambda divided by the crystallite size, and the slope gives the microstrain.

What is the Williamson-Hall equation?

Beta cos(theta) equals K lambda divided by D, plus 4 epsilon sin(theta). Beta is the corrected integral breadth or FWHM in radians, theta is the Bragg angle, K is the shape factor (usually 0.9), lambda is the X-ray wavelength, D is the crystallite size, and epsilon is the microstrain.

How is Williamson-Hall different from the Scherrer equation?

The Scherrer equation uses one peak and assumes all broadening comes from crystallite size, so it systematically underestimates size in strained samples. Williamson-Hall uses several peaks and exploits the different angular dependence of size and strain broadening to separate the two contributions.

Why does size broadening go as 1 over cos(theta) and strain as tan(theta)?

Size broadening follows the Scherrer relation, which is independent of angle once multiplied by cos(theta), so it appears as a constant intercept. Strain broadening scales with tan(theta), which becomes 4 epsilon sin(theta) after the same multiplication, so it appears as a slope.

What K value should I use?

The shape factor K is 0.9 for the FWHM of roughly spherical crystallites, which is the near-universal default. It ranges from about 0.62 to 2.08 depending on crystallite shape and whether FWHM or integral breadth is used, so quote the K value alongside any size you report.

How do I measure instrumental broadening?

Run a line-broadening standard with crystallites large enough and strain low enough that all measured width comes from the instrument, typically NIST SRM 660 lanthanum hexaboride or SRM 640 silicon. The FWHM of its peaks near your peaks of interest is your instrumental value.

Should I subtract instrumental broadening in quadrature or directly?

Quadrature subtraction, the square root of the observed width squared minus the instrumental width squared, is correct for Gaussian profiles. Direct subtraction is correct for Lorentzian profiles. Most real peaks are Voigt profiles, so the true answer sits between the two results.

What does a negative slope in a Williamson-Hall plot mean?

A negative slope implies compressive microstrain, which can be genuine in a film under compression or in a nanoparticle with surface stress. More often it means anisotropic broadening or poor peak fitting, so check the R-squared value and the individual peak widths before reporting it.

How many peaks do I need?

Two peaks are the mathematical minimum for a straight-line fit, but four to six well-separated reflections spanning a wide 2-theta range are needed for a meaningful result. More peaks reduce the effect of any single mis-fitted width on the slope and intercept.

What does the R-squared value tell me?

It measures how well a single straight line describes all the peaks. Values above about 0.95 suggest an isotropic size-plus-strain model fits. Low values usually indicate anisotropic broadening, where crystallites are plate-like or needle-like, and call for a modified Williamson-Hall model.

What is microstrain in a crystal?

Microstrain is the root-mean-square variation in lattice spacing within a sample, caused by dislocations, vacancies, stacking faults, or compositional gradients. It is a dimensionless fraction, often quoted as a percentage or in units of 10 to the minus 3, and it broadens peaks progressively at higher angles.

Why is my Williamson-Hall size different from a TEM particle size?

X-ray methods measure coherently diffracting domains, not whole particles. One TEM-visible particle can contain several domains separated by grain boundaries or twins, so the diffraction size is usually equal to or smaller than the microscopy particle size.