Chapter 2

X-Ray Diffraction and the Reciprocal Lattice

Why Reciprocal Space?

Chapter introduced the reciprocal lattice briefly, as a tool for handling planes. That was a preview. This chapter treats it properly, because the reciprocal lattice is not a computational convenience — it is the space in which the physics of a periodic solid is actually written.

The reason is Fourier analysis. Any physical property of a crystal — the electron density \(n(\vb{r})\), the electrostatic potential, the mass density — is invariant under every lattice translation:

\begin{equation} n(\vb{r} + \Rvec) = n(\vb{r}) \qquad\text{for every lattice vector }\Rvec . \end{equation}

A periodic function can be expanded in a Fourier series, but only wavevectors that share the periodicity may appear. Writing

\begin{equation} n(\vb{r}) = \sum_{\Gvec} n_{\Gvec}\, e^{i\Gvec\cdot\vb{r}} , \end{equation}

and demanding the relation above, we need \(e^{i\Gvec\cdot(\vb{r}+\Rvec)} = e^{i\Gvec\cdot\vb{r}}\), that is

\begin{equation} \boxed{\;e^{i\Gvec\cdot\Rvec} = 1 \qquad\Longleftrightarrow\qquad \Gvec\cdot\Rvec = 2\pi m, \quad m \in \mathbb{Z}. \;} \end{equation}

The one equation that defines everything

The relation above is the definition of the reciprocal lattice: the set of all wavevectors \(\Gvec\) satisfying it for every lattice vector \(\Rvec\). Every property derived in this chapter — the \(\bvec{i}\) formulas, the Laue condition, the shape of a Brillouin zone — is a consequence of this one statement.

The set of such \(\Gvec\) is itself a Bravais lattice, living in a space whose axes have dimensions of inverse length. That space goes by several names — reciprocal space, Fourier space, momentum space, or \(\kvec\)-space — and the original lattice is called the direct lattice by contrast.

Three things happen in reciprocal space that are awkward or impossible in real space:

Diffraction becomes trivial. A diffraction pattern is a photograph of the reciprocal lattice. The Laue condition \(\Delta\kvec = \Gvec\) of the section on the Laue Condition replaces the geometry of reflecting planes with vector addition.
Waves acquire a natural home. Electrons and phonons in a crystal are labelled by a wavevector \(\kvec\), and two wavevectors differing by a \(\Gvec\) describe the same state. All the independent states therefore fit inside one primitive cell of the reciprocal lattice — the Brillouin zone of the section on brillouin Zones.
Planes become vectors. A family of planes \((hkl)\), an awkward object to compute with, is replaced by a single vector \(\Gvec_{hkl}\) perpendicular to it, of length \(2\pi/\dhkl\).

Constructing the Reciprocal Lattice

The geometric construction

Before any algebra, here is the picture. Take a family of parallel planes \((hkl)\) in the direct lattice, with interplanar spacing \(\dhkl\). From a chosen origin, draw the normal \(\uv{n}\) to that family, and mark a point on it at distance \(2\pi/\dhkl\). That point is the reciprocal lattice point \(\Gvec_{hkl}\). Repeating for every possible family of planes generates the whole reciprocal lattice.

Geometric construction. The shaded face and the dotted planes behind it are the \((010)\) family, of spacing \(d_{010}\). Along the normal to that family, at distance \(2\pi/d_{010}\) from the origin, sits the reciprocal lattice point \(\Gvec_{010}\). Every family of planes contributes one such point.

This construction makes two facts obvious that the algebra will confirm: widely spaced planes give short reciprocal vectors, and the direction of a reciprocal vector is the normal to its planes.

The primitive vectors

Given primitive vectors \(\avec{1},\avec{2},\avec{3}\) of the direct lattice with primitive cell volume

\begin{equation} V = \avec{1}\cdot(\avec{2}\times\avec{3}) = \avec{2}\cdot(\avec{3}\times\avec{1}) = \avec{3}\cdot(\avec{1}\times\avec{2}), \end{equation}

define

\begin{equation} \boxed{\; \bvec{1} = \frac{2\pi}{V}\left(\avec{2}\times\avec{3}\right), \quad \bvec{2} = \frac{2\pi}{V}\left(\avec{3}\times\avec{1}\right), \quad \bvec{3} = \frac{2\pi}{V}\left(\avec{1}\times\avec{2}\right). \;} \end{equation}

Each \(\bvec{i}\) is constructed as a cross product of the two direct vectors other than \(\avec{i}\), so it is orthogonal to both of them. Combined with the normalisation, this gives the fundamental orthogonality relation

\begin{equation} \boxed{\;\avec{i}\cdot\bvec{j} = 2\pi\,\delta_{ij} = \begin{cases} 2\pi, & i = j \\ 0, & i \neq j \end{cases}\;} \end{equation}

Proof. For \(i = j = 1\): \(\avec{1}\cdot\bvec{1} = \frac{2\pi}{V}\avec{1}\cdot(\avec{2}\times\avec{3}) = \frac{2\pi}{V}\cdot V = 2\pi\). For \(i = 2\), \(j = 1\): \(\avec{2}\cdot\bvec{1} = \frac{2\pi}{V}\avec{2}\cdot(\avec{2}\times\avec{3}) = 0\), because the scalar triple product vanishes whenever two of its vectors coincide. The remaining cases follow by cyclic permutation.

The general reciprocal lattice vector is then

\begin{equation} \Gvec = h\bvec{1} + k\bvec{2} + l\bvec{3}, \qquad h,k,l \in \mathbb{Z}, \end{equation}

and the relation above immediately delivers the defining property: for \(\Rvec = n_{1}\avec{1}+n_{2}\avec{2}+n_{3}\avec{3}\),

\begin{equation} \Gvec\cdot\Rvec = 2\pi\left(hn_{1} + kn_{2} + ln_{3}\right) = 2\pi \times \text{integer}, \end{equation}

so \(e^{i\Gvec\cdot\Rvec}=1\) as required by the relation above.

The factor of \(2\pi\): two conventions, never mixed

Two conventions are in circulation and they differ by factors of \(2\pi\):

	Physics convention	Crystallography convention
Definition	\(\avec{i}\cdot\bvec{j} = 2\pi\delta_{ij}\)	\(\avec{i}\cdot\vb{a}^{*}_{j} = \delta_{ij}\)
Length	\(\|\Gvec_{hkl}\| = 2\pi/\dhkl\)	\(\|\vb{g}_{hkl}\| = 1/\dhkl\)
Ewald radius	\(2\pi/\lambda\)	\(1/\lambda\)
Laue	\(\Delta\kvec = \Gvec\)	\(\Delta\vb{s} = \vb{g}\)

Both are correct; mixing them is not. A statement such as “reciprocal lattice points lie at distance \(1/d\)” sits beside \(\avec{i}\cdot\bvec{j} = 2\pi\delta_{ij}\) in some sets of notes, and the two cannot both be true. This book uses the physics convention throughout, with the \(2\pi\) built in, because it is the one that makes \(\kvec\) a wavevector and \(\hbar\kvec\) a momentum in the chapters on electrons and phonons.

If a numerical answer disagrees with a printed key by \(2\pi\), this is almost always why.

Properties of the reciprocal lattice

Theorem 2.1. (\(\Gvec_{hkl}\) is normal to \((hkl)\)) The reciprocal lattice vector \(\Gvec_{hkl} = h\bvec{1}+k\bvec{2}+l\bvec{3}\) is perpendicular to the family of lattice planes with Miller indices \((hkl)\).

Proof. The plane of the family nearest the origin cuts the axes at \(\avec{1}/h\), \(\avec{2}/k\), \(\avec{3}/l\). Two vectors lying in that plane are

\begin{equation*} \vb{u} = \frac{\avec{1}}{h} - \frac{\avec{2}}{k}, \qquad \vb{v} = \frac{\avec{2}}{k} - \frac{\avec{3}}{l} . \end{equation*}

Using \(\avec{i}\cdot\bvec{j} = 2\pi\delta_{ij}\),

\begin{equation*} \Gvec_{hkl}\cdot\vb{u} = \frac{h\,(2\pi)}{h} - \frac{k\,(2\pi)}{k} = 0, \end{equation*}

and likewise \(\Gvec_{hkl}\cdot\vb{v} = 0\). Being orthogonal to two independent vectors in the plane, \(\Gvec_{hkl}\) is normal to the plane.

Theorem 2.2. (Length of \(\Gvec\)) The shortest reciprocal lattice vector normal to the \((hkl)\) planes has magnitude

\begin{equation} \left|\Gvec_{hkl}\right| = \frac{2\pi}{\dhkl} . \end{equation}

Proof. The interplanar spacing is the projection of any vector joining two adjacent planes onto the unit normal \(\uv{n} = \Gvec_{hkl}/|\Gvec_{hkl}|\). Take \(\avec{1}/h\), which runs from the origin to the nearest plane:

\begin{equation*} \dhkl = \frac{\avec{1}}{h}\cdot\uv{n} = \frac{\avec{1}\cdot\Gvec_{hkl}}{h\left|\Gvec_{hkl}\right|} = \frac{2\pi h}{h\left|\Gvec_{hkl}\right|} = \frac{2\pi}{\left|\Gvec_{hkl}\right|}. \end{equation*}

The relation above is the workhorse of the whole chapter. It converts every statement about interplanar spacings into a statement about vector lengths, and it reproduces the cubic spacing formula of Chapter in one line: for a simple cubic lattice \(\bvec{i} = (2\pi/a)\uv{e}_{i}\), so \(|\Gvec_{hkl}| = (2\pi/a)\sqrt{h^{2}+k^{2}+l^{2}}\) and hence \(\dhkl = a/\sqrt{h^{2}+k^{2}+l^{2}}\).

Summary of properties

The reciprocal lattice is itself a Bravais lattice.
\(\Gvec_{hkl} \perp (hkl)\), and \(|\Gvec_{hkl}| = 2\pi/\dhkl\).
The reciprocal of the reciprocal lattice is the direct lattice.
Primitive cell volumes satisfy \(V^{*} = (2\pi)^{3}/V\); in two dimensions \(A^{*} = (2\pi)^{2}/A\).
Reciprocal lattices inherit the point symmetry of the direct lattice.
sc \(\to\) sc, bcc \(\leftrightarrow\) fcc, hexagonal \(\to\) hexagonal (rotated by \(30^{\circ}\)).

Proof. Using the relation above and the vector identity \((\vb{A}\times\vb{B})\cdot\left[(\vb{C}\times\vb{D})\times(\vb{E}\times\vb{F})\right]\), or more simply by taking the determinant of the matrix relation \(\left[\bvec{1}\,\bvec{2}\,\bvec{3}\right]^{T} = 2\pi\left[\avec{1}\,\avec{2}\,\avec{3}\right]^{-1}\),

\begin{equation*} V^{*} = \det\left[\bvec{1}\,\bvec{2}\,\bvec{3}\right] = \frac{(2\pi)^{3}}{\det\left[\avec{1}\,\avec{2}\,\avec{3}\right]} = \frac{(2\pi)^{3}}{V}. \end{equation*}

Proof. Apply the relation above to the \(\bvec{i}\):

\begin{equation*} \frac{2\pi\left(\bvec{2}\times\bvec{3}\right)}{V^{*}} = \frac{2\pi}{(2\pi)^{3}/V}\cdot \frac{(2\pi)^{2}}{V^{2}}\left(\avec{3}\times\avec{1}\right) \times\left(\avec{1}\times\avec{2}\right) . \end{equation*}

The triple cross product reduces to \(V\avec{1}\), leaving \(\avec{1}\) exactly.

Reciprocal Lattices of the Common Crystals

The definition the relation above is easy to apply but the results are worth memorising, because they recur constantly.

Simple cubic

With \(\avec{1} = a\uv{x}\), \(\avec{2} = a\uv{y}\), \(\avec{3} = a\uv{z}\), the cell volume is \(V = a^{3}\) and

\begin{equation} \bvec{1} = \frac{2\pi}{a^{3}}\left(a\uv{y}\times a\uv{z}\right) = \frac{2\pi a^{2}\uv{x}}{a^{3}} = \frac{2\pi}{a}\uv{x}, \end{equation}

and cyclically \(\bvec{2} = (2\pi/a)\uv{y}\), \(\bvec{3} = (2\pi/a)\uv{z}\).

The reciprocal of a simple cubic lattice of constant \(a\) is a simple cubic lattice of constant \(2\pi/a\). Its cell volume is \((2\pi/a)^{3} = (2\pi)^{3}/V_{\text{cell}}\), as property 4 requires.

Note the inversion: a large direct lattice constant gives a small reciprocal one. This is the general rule, and it is why the reciprocal lattice of a long thin cell is short and fat.

Body-centred cubic

Using the primitive vectors of the relation in Chapter 1,

\begin{equation} \avec{1} = \frac{a}{2}\left(\uv{x}+\uv{y}-\uv{z}\right), \quad \avec{2} = \frac{a}{2}\left(-\uv{x}+\uv{y}+\uv{z}\right), \quad \avec{3} = \frac{a}{2}\left(\uv{x}-\uv{y}+\uv{z}\right), \end{equation}

whose cell volume is \(V = a^{3}/2\). Evaluating the cross products,

\begin{equation} \bvec{1} = \frac{2\pi}{a}\left(\uv{y}+\uv{z}\right), \quad \bvec{2} = \frac{2\pi}{a}\left(\uv{z}+\uv{x}\right), \quad \bvec{3} = \frac{2\pi}{a}\left(\uv{x}+\uv{y}\right). \end{equation}

Compare these with the fcc primitive vectors of the relation in Chapter 1: they have exactly the same form, with \(a/2\) replaced by \(2\pi/a\). Hence:

The reciprocal lattice of bcc with cube edge \(a\) is fcc with cube edge \(4\pi/a\).

The reciprocal cell volume is \(V^{*} = |\bvec{1}\cdot(\bvec{2}\times\bvec{3})| = 2(2\pi/a)^{3}\), which is \((2\pi)^{3}/(a^{3}/2)\), consistent as always.

Face-centred cubic

With \(\avec{1} = \frac{a}{2}(\uv{x}+\uv{y})\), \(\avec{2} = \frac{a}{2}(\uv{y}+\uv{z})\), \(\avec{3} = \frac{a}{2}(\uv{z}+\uv{x})\) and \(V = a^{3}/4\),

\begin{equation} \bvec{1} = \frac{2\pi}{a}\left(\uv{x}+\uv{y}-\uv{z}\right), \quad \bvec{2} = \frac{2\pi}{a}\left(-\uv{x}+\uv{y}+\uv{z}\right), \quad \bvec{3} = \frac{2\pi}{a}\left(\uv{x}-\uv{y}+\uv{z}\right). \end{equation}

The reciprocal lattice of fcc with cube edge \(a\) is bcc with cube edge \(4\pi/a\), and \(V^{*} = 4(2\pi/a)^{3}\).

bcc and fcc are reciprocal partners

This pairing is examined relentlessly, and it is easy to state backwards. A memory aid: the denser lattice in real space becomes the sparser one in reciprocal space. Fcc has four lattice points per conventional cell against bcc's two, so fcc \(\to\) bcc.

It also explains a diffraction fact from Chapter: the diffraction pattern of an fcc crystal shows spots arranged on a bcc lattice, which is precisely the statement that fcc reflections are the all-odd and all-even \((hkl)\).

The bcc lattice and its reciprocal. Reading the figure right to left gives the companion statement: the reciprocal of fcc is bcc.

Simple hexagonal

Take

\begin{equation} \avec{1} = \frac{\sqrt{3}a}{2}\uv{x} + \frac{a}{2}\uv{y}, \quad \avec{2} = -\frac{\sqrt{3}a}{2}\uv{x} + \frac{a}{2}\uv{y}, \quad \avec{3} = c\,\uv{z}, \end{equation}

with \(V = \frac{\sqrt{3}}{2}a^{2}c\). Then

\begin{equation} \bvec{1} = \frac{2\pi}{a}\left(\frac{\uv{x}}{\sqrt{3}}+\uv{y}\right), \quad \bvec{2} = \frac{2\pi}{a}\left(-\frac{\uv{x}}{\sqrt{3}}+\uv{y}\right), \quad \bvec{3} = \frac{2\pi}{c}\uv{z}, \end{equation}

and \(V^{*} = 2(2\pi)^{3}/(\sqrt{3}a^{2}c)\).

The reciprocal of a simple hexagonal lattice is another simple hexagonal lattice, rotated by \(30^{\circ}\) about the \(c\) axis, with basal parameter \(4\pi/(\sqrt{3}a)\) and height \(2\pi/c\).

The \(30^{\circ}\) rotation is easy to overlook and matters for graphene: it is why the corners of the hexagonal Brillouin zone (the \(K\) points, where the Dirac cones live) lie along the direction of the bonds rather than along the direct lattice vectors.

Worked example 2.1. Reciprocal vectors of a two-dimensional hexagonal lattice

A hexagonal lattice has \(\avec{1} = a\uv{x}\) and \(\avec{2} = \frac{a}{2}\left(\uv{x}+\sqrt{3}\uv{y}\right)\). Find \(\bvec{1}\) and \(\bvec{2}\).

Solution. In two dimensions, append \(\avec{3} = \uv{z}\) and use the three-dimensional formulas; the area of the direct cell is

\begin{equation*} A = \left|\avec{1}\times\avec{2}\right| = a\cdot\frac{a\sqrt{3}}{2} = \frac{\sqrt{3}}{2}a^{2}. \end{equation*}

Then

\begin{equation*} \bvec{1} = \frac{2\pi\left(\avec{2}\times\uv{z}\right)}{A} = \frac{2\pi}{a}\left(\uv{x} - \frac{\uv{y}}{\sqrt{3}}\right), \qquad \bvec{2} = \frac{2\pi\left(\uv{z}\times\avec{1}\right)}{A} = \frac{4\pi}{\sqrt{3}a}\uv{y} . \end{equation*}

Check: \(\avec{1}\cdot\bvec{1} = 2\pi\) ✓, \(\avec{1}\cdot\bvec{2} = 0\) ✓, \(\avec{2}\cdot\bvec{1} = \frac{a}{2}\cdot\frac{2\pi}{a} - \frac{a\sqrt{3}}{2}\cdot\frac{2\pi}{a\sqrt{3}} = \pi - \pi = 0\) ✓.

Always run these four dot-product checks. They catch sign errors instantly and cost ten seconds.

Worked example 2.2. Reciprocal of a honeycomb (graphene) lattice

For a honeycomb of nearest-neighbour spacing \(a\), the Bravais lattice is triangular with primitive vectors of length \(\sqrt{3}a\). Taking

\begin{equation*} \avec{1} = \frac{\sqrt{3}a}{2}\left(\sqrt{3}\uv{x}+\uv{y}\right), \qquad \avec{2} = \sqrt{3}a\,\uv{y}, \end{equation*}

the cell area is \(A = \frac{3\sqrt{3}}{2}a^{2}\), and

\begin{equation*} \bvec{1} = \frac{4\pi}{3a}\uv{x}, \qquad \bvec{2} = \frac{2\pi}{3a}\left(-\uv{x}+\sqrt{3}\uv{y}\right). \end{equation*}

Setting \(a = 1\) gives \(\bvec{1} = \left(\frac{4\pi}{3},0\right)\) and \(\bvec{2} = \left(-\frac{2\pi}{3},\frac{2\pi}{\sqrt{3}}\right)\).

Both vectors have length \(4\pi/(3a)\) and are separated by \(120^{\circ}\): the reciprocal lattice is triangular, as expected, rotated by \(30^{\circ}\) from the direct one.

Square, rectangular and oblique lattices

In two dimensions the construction simplifies. With \(\avec{3} = \uv{z}\) and direct cell area \(A\),

\begin{equation} \bvec{1} = \frac{2\pi}{A}\left(\avec{2}\times\uv{z}\right), \qquad \bvec{2} = \frac{2\pi}{A}\left(\uv{z}\times\avec{1}\right). \end{equation}

Reciprocal lattices in two dimensions.

Direct lattice	Cell area	Reciprocal lattice	Parameters
Square, side \(a\)	\(a^{2}\)	square	side \(2\pi/a\)
Rectangular, \(a,b\)	\(ab\)	rectangular	\(2\pi/a\), \(2\pi/b\)
Hexagonal, side \(a\)	\(\frac{\sqrt{3}}{2}a^{2}\)	hexagonal, rotated \(30^{\circ}\)	\(4\pi/(\sqrt{3}a)\)
Oblique, \(a,b,\gamma\)	\(ab\sin\gamma\)	oblique	\(\frac{2\pi}{a\sin\gamma}\), \(\frac{2\pi}{b\sin\gamma}\)

Note the rectangular entry: a lattice long in \(x\) and short in \(y\) has a reciprocal that is short in \(x\) and long in \(y\). In the limit of a one-dimensional chain of spacing \(a\) (that is, \(b \to \infty\)), the reciprocal lattice becomes a set of planes spaced \(2\pi/a\) apart — the origin of the streaks seen in the diffraction patterns of layered and low-dimensional materials.

Reciprocal lattices in three dimensions — summary.

Direct	Reciprocal	Reciprocal cell volume
sc, edge \(a\)	sc, edge \(2\pi/a\)	\((2\pi/a)^{3}\)
bcc, edge \(a\)	fcc, edge \(4\pi/a\)	\(2(2\pi/a)^{3}\)
fcc, edge \(a\)	bcc, edge \(4\pi/a\)	\(4(2\pi/a)^{3}\)
Hexagonal, \(a,c\)	hexagonal (rot. \(30^{\circ}\))	\(2(2\pi)^{3}/(\sqrt{3}a^{2}c)\)
Orthorhombic, \(a,b,c\)	orthorhombic, \(\frac{2\pi}{a},\frac{2\pi}{b},\frac{2\pi}{c}\)	\((2\pi)^{3}/(abc)\)

Worked example 2.3. Identifying a reciprocal lattice from primitive vectors

A lattice has primitive vectors (in Å)

\begin{equation*} \vb{a} = 2\left(\uv{y}+\uv{z}\right), \quad \vb{b} = 2\left(\uv{z}+\uv{x}\right), \quad \vb{c} = 2\left(\uv{x}+\uv{y}\right). \end{equation*}

Identify the reciprocal lattice and its cube edge.

Solution. These are fcc primitive vectors with \(a/2 = 2\), so the direct lattice is fcc with conventional cube edge \(a = \text{4 Å}\). The reciprocal is therefore bcc with cube edge

\begin{equation*} \frac{4\pi}{a} = \frac{4\pi}{4} = \pi\ \text{Å⁻¹}. \end{equation*}

Confirming directly: \(V = \vb{a}\cdot(\vb{b}\times\vb{c}) = 16\,\text{Å}\) and

\begin{equation*} \vb{a}^{*} = \frac{2\pi}{16}\left(\vb{b}\times\vb{c}\right) = \frac{\pi}{2}\left(-\uv{x}+\uv{y}+\uv{z}\right), \end{equation*}

which with its cyclic partners generates bcc of edge \(\pi\). ✓

Diffraction: The Bragg Picture

X-rays were discovered in 1895, and for seventeen years nobody knew whether they were waves or particles. In 1912 Max von Laue reasoned that if they were waves with \(\lambda \sim \text{1 Å}\), and if crystals really were periodic arrays with spacings of the same order, then a crystal should act as a three-dimensional diffraction grating. Friedrich and Knipping performed the experiment, the spots appeared, and two questions were answered at once: X-rays are waves, and crystals are periodic.

W. L. Bragg gave the simple interpretation a year later.

Bragg's law

Treat the crystal as a stack of parallel reflecting planes of spacing \(d\). A beam striking them at glancing angle \(\theta\) is partly reflected at each plane. From the figure above, the extra path travelled by the ray reflected from the second plane is

\begin{equation} \Delta = MQ + QN = 2d\sin\theta , \end{equation}

and constructive interference requires this to be a whole number of wavelengths:

\begin{equation} \boxed{\;2d\sin\theta = n\lambda\;} \qquad n = 1,2,3,\ldots \end{equation}

Bragg's construction. The ray \(PQR\) travels the extra distance \(MQ+QN = 2d\sin\theta\) relative to \(ABC\). When this equals a whole number of wavelengths the two reflected rays reinforce.

Three consequences are worth stating explicitly.

A wavelength limit.

Since \(\sin\theta \le 1\), the relation above requires \(n\lambda \le 2d\). No reflection of any order is possible unless \(\lambda \le 2d\). This is why visible light (\(\lambda \sim \text{5000 Å}\)) cannot probe crystal structure, and why increasing the wavelength progressively extinguishes the high-angle peaks.

Orders absorbed into indices.

The \(n\)th order reflection from \((hkl)\) has the same geometry as the first order from \((nh\,nk\,nl)\), since \(d_{nh\,nk\,nl} = \dhkl/n\). Convention therefore drops \(n\) and writes

\begin{equation} 2\dhkl\sin\theta_{B} = \lambda, \qquad \sin\theta_{B} = \frac{\lambda}{2a}\sqrt{h^{2}+k^{2}+l^{2}} \ \ \text{(cubic)}. \end{equation}

Angle conventions.

\(\theta_{B}\) is measured from the plane, not from the normal. The angle between incident and diffracted beams is \(2\theta_{B}\), and this — not \(\theta_{B}\) — is what a powder diffractometer plots.

\(\theta\) or \(2\theta\)?

A published pattern showing a peak “at \(\text{31°}\)” almost always means \(2\theta = \text{31°}\), hence \(\theta_{B} = \text{15.5°}\). Substituting \(\text{31°}\) into the relation above gives a \(d\)-spacing wrong by roughly a factor of two. Check the axis label before using any number from a diffractogram.

Worked example 2.4. Reading a lattice constant off a peak

X-ray diffraction from a cubic crystal gives an intensity maximum at Bragg angle \(\text{20°}\) for the \((110)\) plane, with \(\lambda = \text{0.15 nm}\). Find the lattice parameter.

Solution. From the relation above with \(n=1\),

\begin{equation*} d_{110} = \frac{\lambda}{2\sin\theta} = \frac{0.15}{2\times 0.342} = \text{0.219 nm}. \end{equation*}

For a cubic crystal \(d_{110} = a/\sqrt{2}\), so

\begin{equation*} a = \sqrt{2}\times 0.219 = \text{0.31 nm}. \end{equation*}

Worked example 2.5. When no peak can appear

Monochromatic X-rays of \(\lambda = \text{1 Å}\) fall on a simple cubic crystal of parameter \(a = \text{1 Å}\). For which of \((100)\), \((112)\), \((210)\), \((220)\) does a first-order peak appear?

Solution. From the relation above, \(\sin\theta = \frac{1}{2}\sqrt{h^{2}+k^{2}+l^{2}}\):

Plane	\(\sqrt{N}\)	\(\sin\theta\)
\((100)\)	\(1\)	\(0.50\) ✓
\((112)\)	\(\sqrt{6}\)	\(1.22\) \(\times\)
\((210)\)	\(\sqrt{5}\)	\(1.12\) \(\times\)
\((220)\)	\(\sqrt{8}\)	\(1.41\) \(\times\)

Only \((100)\) satisfies \(\sin\theta \le 1\). With \(\lambda = 2a\sin\theta\) at most \(2a\), the condition \(\sqrt{N} \le 2a/\lambda = 2\) admits only \(N = 1\) here.

The Laue Condition

Bragg's derivation is memorable but physically dishonest: atoms do not specularly reflect, they scatter in all directions. Laue's formulation makes no such assumption and is exact.

Derivation

Let a plane wave of wavevector \(\kvec\) be incident on the crystal, and consider scattering into \(\kvec'\). Two scatterers separated by \(\Rvec\) contribute amplitudes differing in phase by \((\kvec - \kvec')\cdot\Rvec\). Summing over all lattice sites, the total scattered amplitude is proportional to

\begin{equation} F \propto \sum_{\Rvec} e^{i(\kvec-\kvec')\cdot\Rvec}. \end{equation}

This sum is a sum of unit-modulus terms with essentially random phases: it cancels to zero unless every term has the same phase, which requires \((\kvec'-\kvec)\cdot\Rvec = 2\pi \times \text{integer}\) for all \(\Rvec\). By the relation above that happens exactly when the scattering vector is a reciprocal lattice vector:

\begin{equation} \boxed{\;\Delta\kvec = \kvec' - \kvec = \Gvec\;} \end{equation}

together with \(|\kvec'| = |\kvec|\) for elastic scattering.

Equivalence with Bragg's law

Theorem 2.3. The Laue condition the relation above with \(|\kvec'| = |\kvec|\) is equivalent to Bragg's law the relation above.

Proof. Write \(\kvec' = \kvec + \Gvec\) and square:

\begin{equation*} k'^{2} = k^{2} + G^{2} + 2\kvec\cdot\Gvec . \end{equation*}

Elastic scattering means \(k' = k\), so

\begin{equation} \boxed{\;2\kvec\cdot\Gvec + G^{2} = 0\;} \end{equation}

which is the alternative form of the Laue condition. Now let \(\theta\) be the angle between \(\kvec\) and the plane normal to \(\Gvec\); then the angle between \(\kvec\) and \(\Gvec\) is \(90^{\circ}+\theta\), so \(\kvec\cdot\Gvec = -kG\sin\theta\). Substituting,

\begin{equation*} 2kG\sin\theta = G^{2} \quad\Longrightarrow\quad 2\left(\frac{2\pi}{\lambda}\right)\sin\theta = G = \frac{2\pi n}{d}, \end{equation*}

where the last step uses the relation above and allows \(\Gvec\) to be \(n\) times the shortest vector of its family. Rearranging gives \(2d\sin\theta = n\lambda\).

The relation above is the form to remember: it is what examination questions quote, and it says geometrically that \(\kvec\) terminates on the perpendicular bisector plane of \(\Gvec\) — a statement that will reappear as the definition of a Brillouin zone boundary in the section on brillouin Zones.

Worked example 2.6. Diffraction angle from a reciprocal lattice vector

A neutron beam of energy 20.4 meV diffracts from a crystal. A peak is observed for a reciprocal lattice vector of magnitude \(\text{3.14 Å⁻¹}\). What angle does the incident beam make with the plane?

Continue reading — it's free

Already registered? Use the same button — just verify your number.

Chapter 2 X-Ray Diffraction and the Reciprocal Lattice