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Bragg's Law Calculator (XRD)

In short: Solve nλ = 2d sin θ for d, θ, 2θ, λ or n. Get the d-spacing of any (h k l) in any crystal system, list the allowed XRD peaks for a cell, and estimate crystallite size with the Scherrer equation. Free, in your browser.

Quick answer: d = nλ / (2 sin θ). NaCl (a = 5.6402 Å): d(200) = a/2 = 2.8201 Å, so with Cu Kα1 (1.5406 Å) the peak is at θ = 15.85°, 2θ = 31.70°. Silicon (111): d = 5.431/√3 = 3.1356 Å, 2θ = 28.44°.

1. Bragg's law: nλ = 2d sin θ

Fill in three values and leave the one to find blank. Use θ or 2θ (not both).

2. d-spacing from Miller indices and allowed reflections

3. Scherrer crystallite size: τ = Kλ / (β cos θ)

How to use the Bragg's law calculator

  1. Pick the X-ray source (Cu Kα1 for most lab diffractometers) or type λ.
  2. Enter three of n, λ, d and 2θ (or θ) and leave the unknown blank. The answer updates as you type.
  3. For predicted peaks, pick the crystal system and centring, enter the cell, and press List reflections.
  4. Change h, k, l to get one d-spacing and its 2θ.
  5. For crystallite size, enter the peak 2θ and its FWHM (and the instrumental FWHM if you have it).

The formulas

Bragg: n λ = 2 d sin θ d = n λ / (2 sin θ) θ = arcsin(n λ / 2d) 2θ = peak position

A peak is possible only when nλ ≤ 2d. Higher orders (n = 2, 3…) of the plane (h k l) appear at the same place as the first order of (nh nk nl), so most XRD software keeps n = 1 and uses d/n.

d-spacing for each crystal system

System1 / d²
Cubic(h² + k² + l²) / a²
Tetragonal(h² + k²) / a² + l² / c²
Orthorhombich²/a² + k²/b² + l²/c²
Hexagonal4(h² + hk + k²) / 3a² + l²/c²
Monoclinic (b unique)[h²/a² + k² sin²β / b² + l²/c² − 2hl cos β / (ac)] / sin²β
Rhombohedral, triclinichᵀ G* h, with G* the inverse of the metric tensor Gij = ai·aj

The calculator always uses the general form 1/d² = hᵀ G* h, which reduces to the lines above. Hexagonal indices are the three-index (h k l); the Miller–Bravais i = −(h + k) does not change d.

Systematic absences from lattice centring

LatticeReflection present whenFirst allowed (cubic)
P primitivealways100, 110, 111, 200, 210…
I body-centred (BCC)h + k + l even110, 200, 211, 220, 310…
F face-centred (FCC)h, k, l all even or all odd111, 200, 220, 311, 222…
Diamond (Si, Ge)F, and h + k + l = 4n when all even111, 220, 311, 400, 331…
C / A / B base-centredh + k / k + l / h + l even–
R (hexagonal axes)−h + k + l = 3n–

Only centring is applied. Screw axes and glide planes remove more peaks (for example 00l with l odd for a 2₁ axis along c), and atoms with similar scattering power can make allowed peaks very weak (KCl looks primitive cubic). The list gives positions, not intensities.

X-ray wavelengths

AnodeKα1 (Å)Kα, weighted mean (Å)
Cu1.5405981.541838
Mo0.7093190.710730
Co1.7889961.790260
Cr2.2897602.291000
Fe1.9360421.937355
Ag0.559421–

Kα1: G. Hölzer et al., Phys. Rev. A 56, 4554 (1997). Weighted Kα means: International Tables for Crystallography, Vol. C, Table 4.2.2.1. Rounded values often quoted: Cu Kα1 1.5406, Cu Kα 1.5418, Mo Kα 0.7107, Co Kα1 1.7890, Cr Kα 2.2910, Fe Kα 1.9374 Å.

Scherrer equation

τ = K λ / (β cos θ) β = FWHM in radians = FWHM(°) × π / 180, after removing instrumental broadening b

K ≈ 0.9 is the usual choice; published values run from about 0.89 to 0.94 depending on crystallite shape and how the width is defined. Example: Cu Kα1, 2θ = 31.70°, FWHM = 0.25°, K = 0.9 → β = 0.004363 rad, τ = 0.9 × 1.5406 / (0.004363 × cos 15.85°) = 330 Å = 33 nm.

Limitations

Frequently asked questions

How do you calculate d-spacing from 2θ?

Use Bragg's law with n = 1: d = λ / (2 sin θ), where θ is half of the 2θ peak position. For a Cu Kα1 peak at 2θ = 31.70°: d = 1.5406 / (2 × sin 15.85°) = 2.820 Å, the (200) spacing of NaCl.

What is the d-spacing formula for a cubic crystal?

d = a / √(h² + k² + l²). For silicon, a = 5.431 Å, the (111) spacing is 5.431 / √3 = 3.1356 Å, which gives 2θ = 28.44° with Cu Kα1. Other systems use the general reciprocal metric tensor, which the calculator applies for you.

Which reflections are absent for FCC and BCC lattices?

In a face-centred (F) lattice h, k, l must be all even or all odd, so (100) and (110) are absent and (111), (200), (220) are present. In a body-centred (I) lattice h + k + l must be even, so (100) and (111) are absent and (110), (200), (211) are present.

Which wavelength should I use for Cu Kα?

Use Cu Kα1 = 1.5406 Å when the Kα2 line is stripped or resolved (most modern data), and the weighted mean Cu Kα = 1.5418 Å for unresolved peaks at low angle. The difference shifts 2θ by about 0.03° near 2θ = 30°.

How accurate is the Scherrer crystallite size?

It is an estimate of the mean size of coherently diffracting domains, usually good to about ±20%. It assumes all broadening comes from size: strain and instrument broadening also widen peaks, so subtract the instrumental FWHM (measured on a standard such as LaB6) and treat sizes above about 100–200 nm as unreliable.

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