Bragg's Law and the Laue Condition

Introduction

An X-ray diffraction peak appears only when waves scattered from many repeats of a crystal reinforce one another. Bragg's law expresses this as an extra path length between waves from parallel planes. The Laue condition expresses exactly the same phase-matching requirement using wavevectors and reciprocal-lattice nodes. The two pictures answer where scattering may occur. To predict whether an allowed reflection is strong or absent, one must also calculate how atoms in the basis interfere through the structure factor.

Core explanation

Consider incident radiation of wavelength λ interacting with parallel planes of spacing d. At a glancing angle θ measured between the beam and the planes, a ray scattered from the lower of two adjacent planes travels an additional distance 2d sinθ compared with one from the upper plane. Constructive interference requires the difference to be an integer number n of wavelengths: nλ = 2d sinθ . In most modern indexing, higher-order scattering can be represented as a first-order reflection from an appropriately indexed plane family, but the n form remains useful for the geometry. The diffractometer commonly reports the angle between incoming and outgoing beams, 2θ , so entering a reported 2θ directly into sinθ doubles the relevant angle and gives the wrong d.

The wavevector of a beam has magnitude k = 2π/λ. Define the scattering vector q = k out − k in. For elastic scattering the two beam magnitudes are equal and q = 4π sinθ/λ. Crystallographic translation R changes scattering phase by q · R. All translated unit cells reinforce only when exp(i q · R) = 1 for every lattice translation. Thus q = G hkl , a reciprocal-lattice vector using the physics convention G hkl = 2π/d hkl. Equating magnitudes gives 4π sinθ/λ = 2π/d hkl, or λ = 2d hkl sinθ. This is the first-order Bragg equation. The IUCr reciprocal-lattice teaching pamphlet develops the phase-matching form, while OpenStax's crystal diffraction section presents the plane-path construction.

The structure factor adds the basis. For atoms j at fractional positions (x j, y j, z j) in a cell, a common convention writes F hkl = Σ j f j exp[2πi(hx j + ky j + lz j)], where f j is the atom's scattering factor. Its squared magnitude contributes to reflection intensity, with further experimental factors also relevant. For a two-point basis at (0,0,0) and (1/2,1/2,1/2) with equal scatterers, F hkl = f[1 + exp(iπ(h+k+l))]. If h+k+l is odd the terms cancel, while if even they reinforce. This is the body-centred reflection rule in that simple model. It is not because planes with odd indices fail Bragg geometry; cancellation within the basis removes their intensity.

Several practical complications limit naive readings. Peak intensity depends on multiplicity, polarisation, Lorentz geometry, absorption, thermal motion and preferred orientation, as well as F hkl. Powder diffraction collapses a three-dimensional reciprocal lattice into a one-dimensional pattern against 2θ, and different (hkl) families can overlap. Small crystallites or strain broaden peaks. A single missing peak may be below detection rather than a strict symmetry absence; a repeated family of systematic absences is stronger evidence for lattice centring or screw/glide symmetry. X-rays scatter mainly from electron density, so light atoms beside heavy ones may contribute weakly to intensities. Neutrons and electrons interact differently and can complement X-rays.

Step-by-step reasoning

1. Convert a reported 2θ to θ and note the wavelength. 2. Use d = nλ/(2 sinθ) to find a plane spacing. 3. Index the candidate plane family using its lattice geometry. 4. In reciprocal space, check q = G hkl for phase matching. 5. Compute or reason about F hkl before predicting intensity.

Visual explanation

Draw two parallel atomic planes and two incoming rays at the same glancing angle. Mark the lower ray's extra inbound and outbound travel segments, whose sum is 2d sinθ. Beside them draw a reciprocal-space arrow G hkl normal to the planes. A circle representing elastic wavevector geometry touches that reciprocal node at the Bragg setting. A second drawing of two scatterers half a cell apart shows their wave arrows cancelling for an odd phase difference.

Real-world analogy

A row of identical claps can sound louder when their echoes return in phase. The spacing of reflecting walls selects which timing or wavelength reinforces. Yet if two people within every repeating group clap half a cycle apart, their contributions can cancel even when the groups themselves are correctly spaced. Plane spacing is the first condition; basis interference is the second.

Real-world example

Powder XRD is used to identify crystalline phases in a battery electrode. Peak positions reveal spacings and lattice changes during charging, while relative intensities help identify atomic arrangement. If a peak shifts to lower 2θ at fixed wavelength, Bragg's law implies a larger d, provided it is the same indexed reflection. A change in intensity alone does not directly establish a changed lattice parameter.

Why?

Why can Bragg and Laue pictures agree? Both count the same phase difference between waves scattered from repeated positions. Bragg groups positions into planes and follows path lengths; Laue treats translation vectors and wavevector differences algebraically. The plane normal is the corresponding reciprocal vector.

Common misconception

"Every geometrically possible reflection must be visible." Bragg or Laue conditions identify allowed directions for coherent scattering from lattice translations, but the structure factor can cancel amplitudes from atoms inside the cell. Limited instrument sensitivity can also conceal weak nonzero reflections.

Worked example

Question: Cu Kα radiation with λ = 0.154 nm gives a first-order powder peak at 2θ = 44.0°. Find d. Would a peak at the same 2θ be strong for a body-centred lattice if indexed (100)?

Reasoning: The glancing angle is θ = 22.0°, not 44.0°. Therefore d = 0.154/[2 sin(22.0°)] = 0.206 nm to three significant figures. For a body-centred arrangement, h+k+l = 1 for (100), which is odd. The contributions of the two equivalent lattice points cancel in the ideal structure factor, so that indexing is inconsistent with an ordinary strong BCC (100) reflection. Reconsider the indexing or phase assignment.

Answer: d ≈ 0.206 nm; an ideal BCC (100) reflection is systematically absent.

Quick check

1. A diffractometer reports 2θ = 60°. Which angle goes into nλ = 2d sinθ? Answer: θ = 30°. The reported scattering angle is twice the glancing angle used in Bragg's expression.

Exam focus

Show the path difference and label θ versus 2θ. State q = G separately from the intensity relation I ∝ F ². For a centred lattice, derive a simple cancellation rather than claiming the missing reflection violates Bragg's law.

Advanced insight

The measurable pattern is the Fourier transform of electron density sampled at reciprocal-lattice points. Detectors record intensities, approximately F hkl ², but usually not the complex phases of F hkl, leading to the crystallographic phase problem. Solving a structure therefore needs additional constraints or methods to infer phases before an electron-density map can be reconstructed.

Summary

Bragg's path equation and the Laue reciprocal-vector condition are equivalent statements of constructive scattering from crystal translations. They set reflection positions. Structure factors, instrumental effects and sample texture set observed intensities; symmetry can cause systematic absences even at geometrically allowed positions.

Practice questions

1. What is the path difference between rays from adjacent planes at glancing angle θ? Answer: 2d sinθ. 2. In the physics convention, what is the magnitude of a reciprocal vector for planes spaced d? Answer: 2π/d. 3. Why can a BCC (100) reflection be absent? Answer: The corner and body-centre contributions have opposite phase for h+k+l odd and cancel. 4. If a known reflection moves to lower 2θ at fixed λ, what happens to its d spacing? Answer: It increases because sinθ falls while λ = 2d sinθ remains fixed.