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Lesson 4 · 30 min

X-ray diffraction: measuring atoms with waves

Bragg’s law, the structure factor and systematic absences: recognise simple, body-centred, face-centred and diamond lattices from powder patterns, and see why NaCl and KCl look different.

In this lesson

Derive Bragg’s law nλ = 2d sin θ from the path difference and use it to get plane spacings from diffraction angles
Recognise simple, body-centred, face-centred and diamond lattices from sin²θ ratios, and find the lattice parameter
Explain systematic absences, and the difference between the NaCl and KCl patterns, with the structure factor

Atoms are only 0.1–0.3 nm apart, while visible light has wavelengths of 400–700 nm: “seeing” an atomic arrangement with an optical microscope is like tracing an embroidery needle with a mop. In 1912 Max von Laue realised that X-rays have wavelengths close to interatomic distances, so a crystal might act as a natural three-dimensional diffraction grating. Friedrich and Knipping shone X-rays through a copper sulfate crystal, and a regular pattern of spots duly appeared on the photographic plate. Within months the young W. L. Bragg explained the spots as “reflections” of X-rays by crystal planes, and with his father W. H. Bragg went on to solve the structure of NaCl. Father and son shared the 1915 Nobel Prize in Physics.

Bragg’s law

Treat the crystal as a stack of parallel atomic planes, spacing $d$ . X-rays arrive at a glancing angle $\theta$ and each plane “reflects” a little of the beam. Rays reflected by neighbouring planes travel different distances: the ray from the lower plane goes an extra $2d\sin\theta$ . Only when that path difference is a whole number of wavelengths do the waves from all the planes add in step to give a strong diffracted beam; move slightly off that angle and the contributions of thousands of planes cancel.

n\lambda = 2d\sin\theta

Bragg’s law: diffraction appears only at these particular angles. The detector sits at

2\theta

from the incident beam.

$n$: order of the reflection, a positive integer
$\lambda$: X-ray wavelength; 154.06 pm for Cu Kα₁
$d$: spacing of the planes
$\theta$: glancing angle between the beam and the planes

Predict first

X-rays of wavelength 154 pm fall on planes 200 pm apart. As θ increases from zero, at roughly what angle does the first strong reflection appear?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

11° 22.6° 45° 67.4°

Bragg’s law · interference from parallel planes

Bragg reflection explorer

Glancing angle θ 22.65° Plane spacing d 200 pm

X-ray wavelength λ 154.06 pm

Planes that interfere, N

Path difference 2d sin θ: 154.0 pm = 1.000 λ
Phase difference between neighbours: 360.0°
Intensity from N planes: 100.0%

Constructive: Bragg condition met, a diffraction peak

Reflected intensity against θ

Raise the number of interfering planes N from 2 to 30: the peak stays put but grows ever sharper. A real crystal has millions of planes, so its reflections are extremely sharp. Then switch λ to Mo Kα and see where the same planes now reflect.

Powder diffraction: every set of planes at once

A single crystal reflects only when turned to exactly the right angle. Grind the sample to a fine powder and there are countless randomly oriented crystallites, some of which always satisfy the Bragg condition for each set of planes. Scan $2\theta$ , record the diffracted intensity, and you have a powder pattern.

For a cubic crystal, substitute $d_{hkl} = a/\sqrt{h^2+k^2+l^2}$ into Bragg’s law:

\sin^2\theta = \frac{\lambda^2}{4a^2}\,(h^2+k^2+l^2)

So the ratios of $\sin^2\theta$ between peaks are ratios of the integers $h^2+k^2+l^2$ . Which integers appear depends on the lattice.

Why do some sets of planes give no peak? Every atom in the cell scatters X-rays, and the amplitude scattered by the whole cell is the phase-weighted sum of all their contributions, the structure factor:

F(hkl) = \sum_j f_j\, e^{2\pi i\,(h x_j + k y_j + l z_j)}

Diffracted intensity is proportional to

|F|^2

. The atomic scattering factor

f

is roughly proportional to the atom’s number of electrons and falls off with angle.

$f_j$: scattering factor of atom j
$x_j, y_j, z_j$: fractional coordinates of atom j
$h, k, l$: Miller indices of the reflection

Body-centred cubic has two atoms, at (0, 0, 0) and (½, ½, ½), so $F = f\,[1 + e^{i\pi(h+k+l)}]$ : $F = 2f$ when $h+k+l$ is even and $F = 0$ when it is odd — the body-centre atoms, lying exactly halfway between the (100) planes, scatter out of phase and cancel the (100) reflection. The same reasoning gives each cubic lattice its fingerprint:

simple cubic: every $h^2+k^2+l^2$ appears: 1, 2, 3, 4, 5, 6, 8, 9 … (7 never does, because it is not a sum of three squares);
body-centred cubic: only $h+k+l$ even: 2, 4, 6, 8, 10, 12, 14, 16 …;
face-centred cubic: $h, k, l$ all odd or all even: 3, 4, 8, 11, 12, 16, 19, 20 …;
diamond: face-centred, minus the all-even peaks with $h+k+l = 4n+2$ : 3, 8, 11, 16, 19, 24 ….

Divide each peak’s $\sin^2\theta$ by the first one’s, multiply by 1, 2 and 3 in turn, and see which column turns into one of these sequences: that identifies the lattice.

Structure factor · multiplicity · Lorentz–polarisation

Powder X-ray diffractometer

Sample Compare with Label (hkl)

Anode

Computed powder pattern · λ = 154.06 pm · $\ce{Cu}$

Peak list · Copper
2θ / °	d / pm	(hkl)	h²+k²+l²	Mult.	I
43.32	208.7	111	3	8	100.0
50.45	180.7	200	4	6	45.9
74.13	127.8	220	8	12	24.2
89.94	109.0	311	11	24	27.0
95.15	104.4	222	12	8	7.9
116.94	90.4	400	16	6	4.9

Compare the patterns of Po (simple cubic), Fe (body-centred), Cu (face-centred) and Si (diamond) and the h²+k²+l² column of the peak list; then choose NaCl, compare it with KCl and watch the leftmost (111) peak.

Predict first

For a cubic metal, the first five peaks have $\sin^2\theta$ in the ratios 1 : 1.333 : 2.667 : 3.667 : 4. Which lattice is it?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Simple cubic Body-centred cubic Face-centred cubic Diamond

NaCl and KCl: one structure, two patterns

For the rock-salt structure the structure factor is $F = 4(f_+ - f_-)$ when $h, k, l$ are all odd and $4(f_+ + f_-)$ when they are all even. In NaCl, $\ce{Na+}$ has 10 electrons and $\ce{Cl-}$ 18, so the difference is sizeable and the (111) peak is weak but visible (computed at about 9 % of (200)). In KCl, $\ce{K+}$ and $\ce{Cl-}$ are isoelectronic, 18 electrons each, and scatter almost equally: the all-odd peaks (111), (311) … all but vanish, and the KCl pattern looks like that of a simple cubic crystal with lattice parameter $a/2$ . In 1913 comparisons like this helped Bragg establish the structure in which every sodium is surrounded by six chlorines.

Worked example

The lattice parameter of copper from its first peak

Copper powder measured with Cu Kα₁ (λ = 154.06 pm) shows its first peak at $2\theta = 43.30^\circ$ . Copper is face-centred cubic; find $a$ .

Concept check

The CsCl pattern shows a (100) peak. What does that tell you?

CsCl is body-centred cubic Its lattice is simple cubic: the centre and the corners hold different ions The sample is contaminated The wrong wavelength was used

Try the calculation

α-Fe (body-centred cubic, $a = 286.65$ pm) is measured with Cu Kα₁. At what 2θ does its (110) peak appear?

Key takeaways

Bragg’s law, nλ = 2d sin θ: waves from millions of planes add in step only when the path difference is a whole number of wavelengths.
For cubic crystals sin²θ is proportional to h²+k²+l²; simple, body-centred, face-centred and diamond lattices each have a characteristic integer sequence that identifies them and yields a.
The structure factor F(hkl) = Σ f e^{2πi(hx+ky+lz)} sets peak intensities and explains systematic absences.
NaCl and KCl share a structure, but K⁺ and Cl⁻ are isoelectronic, so KCl’s all-odd peaks almost vanish.

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Data and references

W. L. Bragg, “The Structure of Some Crystals as Indicated by Their Diffraction of X-rays”, Proc. R. Soc. Lond. A 89, 248 (1913).
B. D. Cullity, S. R. Stock, Elements of X-Ray Diffraction, 3rd ed. (Prentice Hall, 2001).
International Tables for Crystallography, Vol. C, Table 6.1.1.4 — Cromer–Mann scattering-factor coefficients.
G. Hölzer et al., Phys. Rev. A 56, 4554 (1997) — Cu Kα₁ and Mo Kα₁ wavelengths.
W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed. (CRC Press, 2016) — lattice parameters.