Measuring crystal structure: Miller indices and Bragg diffraction

Atoms are too small for light, so we use waves of the right size

Intuition

You cannot feel a grain of sand with a boxing glove. To resolve something you need a probe about as fine as the thing. Atoms are tiny, so the probe has to be X-rays, which have wavelengths about the size of an atom.

Atoms are about an angstrom ($10^{-10}$ m) across. Visible light is about a micrometre ($10^{-6}$ m), so optical microscopy cannot resolve them. X-rays have angstrom wavelengths but cannot be focused by a lens, so the signal is too weak to image directly. Electrons show wave-particle duality and can be focused by an electric field, which resolves individual atoms, but microscopy has limits in locating all the atoms and how they are arranged.

The answer is diffraction. Typical interatomic distances are about 1 to 3 angstrom, which matches X-rays, hence X-ray crystallography. Neutron and electron beams also diffract, because of their wave-like properties. A large number of atoms in a crystal diffract X-rays, which combine into one image conveying the average structure. Diffraction is an inherent property of waves.

The lecture’s everyday example is opal. Its colour comes from diffraction of visible light, not absorption. Opal is a close packing of uniform silica spheres of diameter 130 to 300 nm, and visible light is about 400 to 700 nm. The spacings are on the same length scale as visible light, so the lattice diffracts it.

A short history from the slides: Röntgen discovered X-rays in 1895. von Laue confirmed their wavelength by diffraction. W. H. and W. L. Bragg solved the first crystal structure. Nobel prizes in Physics were awarded in 1901, 1914 and 1915.

Miller indices name a plane by its intercepts

Intuition

To say which slice through a crystal you mean, you need a short label. Miller indices give three whole numbers that describe how the plane cuts the cell edges.

Miller indices $(h\,k\,l)$ are integers that define the plane $h\mathbf{a} + k\mathbf{b} + l\mathbf{c}$. The reciprocal $1/h$ gives the intercept on the a axis, and likewise $1/k$ and $1/l$ for the other two. Negative integers are written with an overbar, $\bar{h}$. The lecture shows planes $(110)$, $(010)$, $(111)$ and $(201)$ in cubic cells.

Exam detail

Test 1 asks about Miller indices. The slides give the intercept rule ($1/h$, $1/k$, $1/l$) and the overbar for negatives. Learning outcome: determine Miller indices and d-spacings.

Optional aside

The slides do not give a formula for the d-spacing from Miller indices, even though the learning outcomes say to determine d-spacings. The worksheet and Tutorial 2 are where it is practised. Check those for the exact method before relying on a formula from outside the course.

Bragg treated the crystal as stacked mirrors

Intuition

Shine a wave at a stack of semi-transparent sheets. Each sheet bounces back a bit. The bounced waves only reinforce each other at certain angles, where each one has travelled an exact whole number of wavelengths further than the last. At other angles they cancel.

Two slits give the one-dimensional picture: coherent waves through two slits make constructive interference where crests meet and destructive interference where crest meets trough, and the pattern encodes the slit separation. Bragg’s explanation for crystals is that X-rays act as if reflected by lattice planes. Beam 2 travels an extra $2d\sin\theta$ compared with beam 1. Constructive interference needs that path to equal a whole number of wavelengths, giving the Bragg equation $n\lambda = 2d\sin\theta$. A diffraction spot is seen when the incidence angle gives constructive interference.

The lecture’s interference simulation compares 2 planes with 10 planes, with scattered waves 5 degrees out of phase, and asks how adding more planes changes things. The slide does not state the answer, so the page does not either.

Try it: worked example

Planes with spacing d = 2.0 Å give a first-order reflection at 30 degrees. Find the wavelength. (Practice numbers.)

Show answer and steps

Answer2.0 Å.

1. Bragg equation: $n\lambda = 2d\sin\theta$.
2. Substitute $n = 1$, $d = 2.0$ Å, $\theta = 30^\circ$: $\lambda = 2 \times 2.0 \times \sin 30^\circ$.
3. $\sin 30^\circ = 0.5$, so $\lambda = 2 \times 2.0 \times 0.5 = 2.0$ Å.

Pitfall

Do not drop the 2 in $2d\sin\theta$. The lecture calls $\theta$ the incidence angle, so check which angle a question gives you before substituting.

A diffraction pattern is a map of the reciprocal lattice

An X-ray beam sees only a 2D projection of the 3D real lattice. The observed pattern is a curved 2D slice through the 3D reciprocal lattice. A complete pattern needs the crystal rotated through all unique orientations.

Single-crystal diffraction measures the intensities of thousands of reflections and indexes them to the reciprocal lattice by Miller indices $(h\,k\,l)$. A reflection $(h\,k\,l)$ corresponds to the Miller indices of the lattice planes that produce it by Bragg’s law. Spot position and spacing determine the unit cell type. The lecture’s example is a pattern projected along [001] with spots labelled 100, 010, 200 and 110, with axes $a^*$ and $b^*$.

Two rules from the grid slides (Lecture 6, slides 13 to 16) tell you how to read the spacing. Each diffraction spot runs perpendicular to its set of Bragg planes. And the spot spacing is inversely proportional to the d-spacing, so planes further apart in the crystal give spots closer together in the pattern. The same follows from the Bragg equation: for fixed $n$ and $\lambda$, a larger $d$ gives a smaller $\sin\theta$, so a smaller angle. The pattern is not a projected image of the grid, and it rotates with the grid.

For a plane such as $(201)$, the intercept rule gives $1/2$ on $a$ and $1/1 = 1$ on $c$. The index $k = 0$ gives $1/0$, which means the plane never meets $b$ and runs parallel to it. The slides draw $(201)$ but do not write out these intercepts, so this is the rule applied, not a slide statement.

X-rays scatter off electrons, so the pattern gives a 3D electron density map. That gives precise geometry and bonding, and the structures of large biomolecules such as RNA polymerase II.

Optional aside

Aperiodic structures are beyond scope: DNA encodes information in an aperiodic base-pair sequence on a periodic double helix, and quasicrystals won the 2011 Nobel Prize in Chemistry.