X-ray Diffraction: Bragg's Law, Ewald Sphere and Structure Factor

Where We Left Off

From One Electron to One Atom

A one-page reminder of Lectures 15–16. Today we ask why the atomic scattering factor depends on \(\theta\) and \(\lambda\) only through \(\sin\theta/\lambda\), then stack atoms into planes.

Thomson

\[I=I_0\frac{K}{r^2}\sin^2\alpha\]

One electron, driven by \(\mathbf E\), re-radiates at the same \(\lambda\). \(K = 7.94\times10^{-30}\) m².

Polarization

\[P=\frac{1+\cos^2 2\theta}{2}\]

Unpolarised beam: the \(z\)-component of \(\mathbf E\) cannot radiate along its own direction.

Coherence

\[\lambda' = \lambda\]

Only elastic (Thomson) scattering keeps a fixed phase and gives peaks; Compton only adds background.

One atom

\[f(0)=Z\]

The atomic scattering factor \(f\) falls with \(\sin\theta/\lambda\) because electrons in the cloud scatter with path differences.

Atomic Scattering Factor

Where Does \(\sin\theta/\lambda\) Come From?

Two electrons in one atom see the same incident wave. Their scattered waves differ in phase by an amount that depends only on the change of wave vector, \(\Delta\mathbf k = \mathbf k' - \mathbf k\), whose length is \(2\sin\theta/\lambda\).

Interactive • Animated

Two Electrons, One Triangle

Inside one atom · and the wave-vector triangle

2θ40°

λ (nm)0.154

sinθ/λ

–

|Δk| = 2 sinθ/λ

–

phase difference A–B

–

f of W (Z = 74)

–

1Path difference: \(\delta = \mathbf r\cdot\mathbf s - \mathbf r\cdot\mathbf s_0 = \mathbf r\cdot(\mathbf s-\mathbf s_0)\), with \(\mathbf s_0, \mathbf s\) unit vectors along the incident and scattered beams.
2Phase: \(2\pi\delta/\lambda = 2\pi\,\mathbf r\cdot\Delta\mathbf k\), with \(\Delta\mathbf k = (\mathbf s-\mathbf s_0)/\lambda = \mathbf k'-\mathbf k\).
3\(|\mathbf k| = |\mathbf k'| = 1/\lambda\) with \(2\theta\) between them: an isosceles triangle, so \(|\Delta\mathbf k| = 2\sin\theta/\lambda\).
4Summing over the cloud, \(f = \int\rho(\mathbf r)\,e^{2\pi i\,\Delta\mathbf k\cdot\mathbf r}\,dV\) depends on \(\theta\) and \(\lambda\) only through \(\sin\theta/\lambda\). At \(\theta=0\), \(\Delta\mathbf k = 0\) and \(f = Z\).
5With Bragg’s law \(\lambda = 2d\sin\theta\): \(\sin\theta/\lambda = 1/(2d)\). For a given \((hkl)\), \(f\) is the same whatever the radiation.

Worked Example

Computing \(f\) for Tungsten (W)

One W atom, Cu Kα radiation, \((222)\) planes of bcc W with \(a = 3.1653\) Å.

1Interplanar spacing

\[d_{222}=\frac{a}{\sqrt{12}}=\frac{3.1653}{\sqrt{12}}=0.9137\ \text{Å}\]

2Scattering parameter

\[s=\frac{\sin\theta}{\lambda}=\frac{1}{2d_{222}}=0.5472\ \text{Å}^{-1}\]

3Coefficients (Table 12.1)

\(a_1 = 5.709,\ b_1 = 28.782\)
\(a_2 = 4.677,\ b_2 = 5.084\)
\(a_3 = 2.019,\ b_3 = 0.572\)

4Substitute into Eq. (12.3)

\[f_W = 74 - 41.782\,(0.5472)^2\left[5.709\,e^{-28.782(0.5472)^2}+4.677\,e^{-5.084(0.5472)^2}+2.019\,e^{-0.572(0.5472)^2}\right] = 74 - 34.06 = 39.94\]

\(f_W \approx 40 \ll Z = 74\). Strong interference inside the electron cloud at this angle.

\(s = 1/(2d)\) does not depend on \(\lambda\). \(f\) for a given \((hkl)\) is the same for any radiation (away from absorption edges); only the angle \(\theta\) at which the peak appears changes.

Next: many atoms arranged on parallel planes scatter together, which leads to Bragg’s law.

Calculator

\(f_W\) for Any Reflection of bcc W

Quantity	Value

Part 2

The Geometry of Diffraction

One plane of atoms acts like a mirror. A crystal is a stack of many such mirrors. When do the reflections from all of them add up, and why do some expected reflections go missing?

We will cover

Bragg’s law \(n\lambda = 2d\sin\theta\), and what the order \(n\) really means
Systematic absences from centring, glides and screws
Bragg’s law in reciprocal space: the Ewald sphere

Many Planes

Bragg’s Law

What happens when several parallel planes are involved? Each plane reflects a little of the beam. The reflections add up only if they arrive in step: the path difference between them must be a whole number of wavelengths, or equivalently, the phase difference must be a multiple of \(2\pi\).

Interactive • Animated

The Bragg Laboratory

Ray 1 reflects from atom A on the top plane; ray 2 from atom O one plane deeper. Ray 2 travels the extra distance PO + OQ. Change the angle and watch the waves leaving on the right slide in and out of step.

Intensity reflected by N planes vs θ

θ22.0°

d (nm)0.200

λ (nm)0.154

planes N6

Path difference

–

In wavelengths

–

Phase difference

–

Reflected intensity

–

Adjust θ

Phasors of all N reflected waves

Step-by-step

Deriving Bragg’s Law

1

Ray 1 scatters from atom A on the top plane; ray 2 from atom O directly below it. Drop perpendiculars from A onto ray 2, meeting it at P (incoming side) and Q (outgoing side). Up to P and after Q, both rays have travelled equal distances. In the right triangle APO the hypotenuse is \(AO = d\) and the angle at A is \(\theta\), so \(\;PO = d\sin\theta\). By symmetry, \(OQ = d\sin\theta\) too.

2

Path difference between rays 1′ and 2′: \[\Delta = PO + OQ = d\sin\theta + d\sin\theta = 2d\sin\theta\] where \(d\) is the interplanar spacing.

3

For constructive interference the path difference must be a whole number of wavelengths: \[n\lambda = 2d\sin\theta \qquad n = 1, 2, 3,\ldots\] This is the famous Bragg’s law.

4

For \(n=1\) the spacing between neighbouring \((hkl)\) planes is \(d_{hkl}\): \[\lambda = 2d_{hkl}\sin\theta\]

5

For a general \(n\), divide both sides by \(n\): \[\lambda = 2\,\frac{d_{hkl}}{n}\,\sin\theta = 2\,d_{nh\,nk\,nl}\,\sin\theta\] Recall that \(d/n\) is exactly the spacing of the planes \((nh\;nk\;nl)\).

6

So the \(n\)-th order reflection from \((hkl)\) is the first-order reflection from \((nh\;nk\;nl)\). Both appear at the same \(\theta\), while the first-order reflections of \((hkl)\) and of \((nh\;nk\;nl)\) appear at different \(\theta\). This is why we always write simply \[\boxed{\lambda = 2d_{hkl}\sin\theta}\] and let the indices carry the order.

Interactive

Orders ↔ Planes

A simple cubic crystal (\(a = 0.40\) nm) and Cu Kα X-rays. Choose \((hkl)\) and see every order \(n\) that fits under \(\sin\theta \le 1\), with its equivalent first-order plane.

h

k

l

Important. Bragg’s law is only a geometrical construction. It is a necessary but not a sufficient condition for a diffracted beam. Even when Bragg’s law is satisfied, atoms sitting between the planes can cancel the reflection completely. That is our next topic.

Necessary, Not Sufficient

Systematic Absences

Crystal symmetries with centring translations, or with symmetry operations that carry a translation part (screw rotations and glides), make certain reflections vanish, even though Bragg’s law is satisfied.

Interactive • A simple 2D example

Primitive vs Centred Rectangular Lattice

Solid lines are the \((hk)\) planes through the corner points. Add a point at the centre of every cell and see where the new points fall. Try the \((21)\) planes first.

Waves from the two sets of points

Present

Reference table

Conditions for Systematic Absences

Adapted from F. A. Cotton, Chemical Applications of Group Theory. A reflection listed here is absent (\(n\) is any integer).

Condition	Reflections	Absent when

Interactive

Lattice-Centring Checker

h

k

l

Phasor sum over centring points

Present

Reading the table. For a body-centred (I) lattice, reflections with \(h+k+l = 2n+1\) (odd) are absent. In the 2D example above, the centred rectangular lattice loses every reflection with \(h+k\) odd. The \((21)\) planes have \(h+k = 3\), so that spot is missing.

Reciprocal Space

Bragg’s Law in Reciprocal Space: the Ewald Sphere

Bragg’s law can be redrawn with vectors. The result is a single picture that shows, at a glance, which planes of a crystal diffract.

Step-by-step • synced with the figure

Building the Construction

1

Represent the incident beam, at angle \(\theta\) to the \((hkl)\) planes, by a reciprocal-space vector \(\mathbf k\) of length \(|\mathbf k| = 1/\lambda\).

2

Represent the planes by their reciprocal vector \(\mathbf g_{hkl}\), which is normal to them and has length \(|\mathbf g_{hkl}| = 1/d_{hkl}\).

3

The diffracted beam is \(\mathbf k'\): it also makes angle \(\theta\) with the planes, and its length is also \(1/\lambda\) (elastic scattering).

4

Slide \(\mathbf k'\) parallel to itself so it starts where \(\mathbf k\) starts. The angle between \(\mathbf k\) and \(\mathbf k'\) is now \(2\theta\).

5

\(\mathbf k\) and \(\mathbf k'\) have the same length and a common start, so their tips lie on a circle of radius \(1/\lambda\), the Ewald sphere (a circle in 2D).

6

The chord joining the two tips has length \(2|\mathbf k|\sin\theta = \dfrac{2\sin\theta}{\lambda}\). Bragg’s law says this equals \(\dfrac1{d_{hkl}} = |\mathbf g_{hkl}|\). So \(\mathbf g_{hkl}\) runs from the tip of \(\mathbf k\) and ends exactly on the circle: \[\mathbf k + \mathbf g_{hkl} = \mathbf k'\]

\[\lambda = 2d\sin\theta \;\Rightarrow\; \frac{2\sin\theta}{\lambda} = \frac1d = |\mathbf g_{hkl}| \quad\Longleftrightarrow\quad \mathbf k + \mathbf g_{hkl} = \mathbf k'\]

“A diffracted beam with wave vector \(\mathbf k'\) will be present if and only if the endpoint of the vector \(\mathbf k + \mathbf g\) lies on the Ewald sphere.”
De Graef and McHenry, Structure of Materials

“A particular plane (hkl) in a crystal is said to be in Bragg orientation if the corresponding reciprocal lattice point \(g_{hkl}\) lies on the Ewald sphere.”
De Graef and McHenry, Structure of Materials

Interactive

Rotate the Crystal, Watch Points Cross the Circle

The reciprocal lattice is attached to the crystal, so rotating the crystal rotates the lattice about its origin O (the tip of \(\mathbf k\)). Whenever a point touches the circle, that family of planes is in Bragg orientation and a diffracted beam \(\mathbf k'\) flashes out.

crystal ω0.0°

λ (nm)0.154

a = b (nm)0.400

Reflections met so far (hk : 2θ)

—

Shorter \(\lambda\) means a bigger circle, so more points can reach it: more reflections become possible. Points farther than \(2/\lambda\) from O can never touch the circle, since \(\sin\theta\) cannot exceed 1.

3D Animation

The Ewald Sphere in 3D

A cubic reciprocal lattice slowly turns about its origin O. Each reciprocal lattice point that pierces the sphere lights up, and a diffracted beam leaves the sphere centre towards it.

drag to rotate the view—

Part 3

Structure Determination Using XRD

Now we run the argument backwards: given a measured pattern, what is the crystal? We index cubic patterns by hand and identify copper as face-centred cubic.

We will cover

Databases, and the general route from pattern to structure
Indexing cubic patterns with \(\sin^2\theta/s = \lambda^2/4a^2\)
The fingerprints of SC, BCC, FCC and diamond cubic
Worked example: copper is FCC

The Big Picture

From an XRD Pattern to a Structure

Working out structures from unknown patterns is a continuous, ongoing effort in science. There is still no general recipe that works for any unknown pattern, but a lot can be learned with a few simple steps.

ICDD

International Centre for Diffraction Data: the Powder Diffraction File, a reference pattern for hundreds of thousands of materials.

ICSD

Inorganic Crystal Structure Database: solved structures (cell, space group, atom positions) of inorganic compounds.

CSD

Cambridge Structural Database: structures of organic and metal–organic molecules and crystals.

These repositories are updated all the time. For most studies, comparing a measured pattern with the databases at least narrows down the structure. For a fuller discussion, see B. D. Cullity and S. R. Stock, Elements of X-ray Diffraction, 3rd ed., from which parts of this lecture are adapted.

Interactive

Positions Tell the Cell, Intensities Tell the Atoms

A cubic crystal with atom A at \((0,0,0)\) and atom B at \((u,u,u)\). Change the lattice parameter and the peaks move. Move atom B inside the cell and the peaks stay put but their heights change.

Simulated powder pattern (Cu Kα)

a (nm)0.400

B at u0.500

f_B / f_A0.60

drag

In general: the unit cell (its shape and size) is found from the line positions, i.e. where the diffraction spots or peaks are. The positions of the atoms inside the cell are found from the line intensities.

The General Process

1

Shape & size of the cell

From the angular positions of the diffraction lines (spots).

2

Index the pattern

Assume one of the seven crystal systems / 14 Bravais lattices, then assign Miller indices to every line. This is called indexing the pattern.

3

Atoms per cell

From (a) the cell volume, (b) the chemical composition and (c) the measured density.

4

Atom positions

Finally, deduced from the relative intensities of the lines.

Calculator • Step 3

How Many Atoms per Unit Cell?

Mass of one cell = \(\rho\,a^3\). Mass of one atom (or formula unit) = \(M/N_A\). Dividing gives

\[n = \frac{\rho\,a^3\,N_A}{M}\]

ρ (g/cm³) a (nm) M (g/mol)

Atoms (formula units) per cell

–

3D Animation • How a pattern is measured

The θ–2θ Diffractometer

drag to rotate2θ = 20.0°

Pattern being recorded

The X-ray tube and the detector move together so that both make angle \(\theta\) with the sample surface; the detector therefore sits at \(2\theta\) from the incident beam. A powder contains crystallites in every orientation, so whenever some \((hkl)\) satisfies Bragg’s law, a peak is recorded.

Indexing

Indexing XRD Patterns of Cubic Crystals

Combine Bragg’s law with the reciprocal-lattice formula for \(d_{hkl}\) from the last lecture, and one simple equation connects every peak to its Miller indices.

Step-by-step

The Key Equation

1

Bragg’s law: \(\lambda = 2d_{hkl}\sin\theta\), and from the last lecture, \(d_{hkl} = \dfrac{1}{|\mathbf g_{hkl}|}\) with \(\mathbf g_{hkl} = h\mathbf a_1^* + k\mathbf a_2^* + l\mathbf a_3^*\).

2

For a cubic cell, \(|\mathbf a_1| = |\mathbf a_2| = |\mathbf a_3| = a\), all at 90°. Using \(\mathbf a_1^* = \dfrac{\mathbf a_2\times\mathbf a_3}{V}\) (and cyclic) with \(V=a^3\): \[|\mathbf a_1^*| = |\mathbf a_2^*| = |\mathbf a_3^*| = \frac{a^2}{a^3} = \frac1a\] and the \(\mathbf a_i^*\) are also mutually perpendicular.

3

So \[|\mathbf g_{hkl}| = \sqrt{h^2|\mathbf a_1^*|^2 + k^2|\mathbf a_2^*|^2 + l^2|\mathbf a_3^*|^2} = \frac{\sqrt{h^2+k^2+l^2}}{a}\]

4

Therefore \[d_{hkl} = \frac1{|\mathbf g_{hkl}|} = \frac{a}{\sqrt{h^2+k^2+l^2}} = \frac{\lambda}{2\sin\theta}\]

5

Let \(s = h^2+k^2+l^2\). Squaring and rearranging: \[\boxed{\frac{\sin^2\theta}{s} = \frac{\lambda^2}{4a^2} = \text{constant}}\] For one crystal and one wavelength the right-hand side is fixed. So for every peak, \(\sin^2\theta\) divided by the correct \(s\) must give the same number.

3D Interactive

Families of Planes and Their s

drag to rotate

Interactive • Think about systematic absences!

Which Values of s Actually Show Up?

Possible \(s\): \(\{100\}\to1\), \(\{110\}\to2\), \(\{111\}\to3\), \(\{200\}\to4\), \(\{210\}\to5\), \(\{211\}\to6\), \(\{220\}\to8\), \(\{221\},\{300\}\to9\), \(\{310\}\to10\), \(\{311\}\to11\), \(\{222\}\to12\), \(\{320\}\to13\), \(\{321\}\to14\), \(\{400\}\to16\)… But can you expect a line for every one of these in every cubic lattice? Hover or tap a cell.

Simple cubic. No absences: every \(s\) that is a sum of three squares, 1, 2, 3, 4, 5, 6, 8, 9, 10, … (7 and 15 can never occur).

Body-centred. Absent when \(h+k+l\) is odd, leaving \(s\) = 2, 4, 6, 8, 10, 12, 14, 16, …

Face-centred. Absent when \(h,k,l\) are mixed odd and even (any two sum to odd), leaving \(s\) = 3, 4, 8, 11, 12, 16, …

Diamond cubic. FCC rules plus more (the conditions are more involved), leaving \(s\) = 3, 8, 11, 16, …

The indexing recipe. Revisit \(\sin^2\theta/s = \lambda^2/4a^2\). For the observed lines, cycle through the allowed \(s\)-sequences of the different cubic lattices. The sequence that makes \(\sin^2\theta/s\) the same constant for every line identifies the lattice: simple, body-centred, face-centred or diamond cubic. The constant then gives \(a = \lambda/(2\sqrt{\text{constant}})\).

Animated • Indexing

Fingerprints of the Four Cubic Lattices

The same cubic cell (Cu Kα) drawn four times. Absent reflections are marked ×. Each lattice leaves its own sequence of \(s = h^2+k^2+l^2\).

a0.360

Read the sequence, not the angles. Simple cubic: 1, 2, 3, 4, 5, 6, 8 … (7 is impossible). BCC: only even \(s\). FCC: pairs 3, 4 · 8 · 11, 12. Diamond also loses 4 and 12.

Worked Example

Example: Copper

A copper pattern shows three peaks at \(2\theta\) = 43.36°, 50.50° and 74.20°. Let’s index it.

Reference

All (hkl) Up to s = 10 for Cu

\(a = 0.36148\) nm, \(\lambda = 0.15418\) nm (Cu Kα). Rows marked × are absent in real patterns because of systematic absences. Adapted from De Graef & McHenry, Structure of Materials.

(hkl)	s	d (nm)	2θ (°)

Measured pattern

Cu, Cu Kα (simulated from the table)

Only (111), (200) and (220) appear between 30° and 80°, exactly as in real patterns of copper nanoparticles (e.g. Phul et al., Material Sci & Eng Int J. 2018; 2(4): 90–94).

Interactive

Indexing Workbench

sin²θ / s for each line

Choose a hypothesis

Practice • A new sample every time

Mystery Cubic Sample

An unknown cubic crystal, measured with Cu Kα. Use the first few peaks, try each lattice type, and find the one that makes \(\sin^2\theta/s\) constant. Then enter the lattice parameter.

Unknown sample

a (nm)

Pick a lattice type

Part 4

Why Peaks Vanish: the Structure Factor

Bragg’s law tells us where a peak may appear. Whether it actually does depends on where the atoms sit in the cell, and that is captured by one complex number per reflection.

We will cover

Why (100), (200), (300)… peaks appear, and why (100) can vanish
The phase of one atom and the structure factor \(F_{hkl}\)
Centring rules for P, C, I and F lattices; cubic survivors
Screw axes, glide planes, Friedel’s law and ITC reflection conditions
Worked examples: CsCl (ordered vs disordered), NaCl, diamond

Putting It Together

X-ray Diffraction Revisited

A careful look at one family of planes explains two puzzles at once: why a pattern shows (100), (200), (300)… peaks, and why adding atoms between planes can wipe out the (100) peak.

Guided animation • click any step

The (100) and (200) Story

1

(100) planes, spacing a. X-rays 1–4 meet the planes at \(\theta\) with \(\lambda = 2d_{100}\sin\theta\). Path difference 1′–2′ is \(\lambda\); 1′–3′ is \(2\lambda\); 1′–4′ is \(3\lambda\). All in step: a (100) peak.

2

Yet (200), (300), (400) peaks also appear, at other angles. If (100) comes from the (100) planes, (200) must come from (200) planes. Where are they? Draw imaginary planes halfway: \(d_{200} = a/2\).

3

Raise the angle to \(\theta_1\). Now the path difference between neighbouring solid planes is \(2\lambda\): \(2\lambda = 2a\sin\theta_1\), i.e. \(\lambda = 2(a/2)\sin\theta_1 = 2d_{200}\sin\theta_1\). Rays x′, y′, z′, w′ are in step: a (200) peak, the second order of (100).

4

Atoms within one plane don’t matter. Rays 1 and 5 hit different atoms, B and C, of the same plane. Before scattering, ray 1 travels an extra distance AB; after scattering, ray 5 travels an extra distance CD. Since AB = CD, their path difference is zero, at any angle.

5

Now fill the dashed planes with atoms. Back at the (100) angle, rays 5, 6, 7 from the dashed planes are exactly \(\lambda/2\) behind rays 1–4: 180° out of phase. With equal numbers of atoms on each plane, they cancel: (100) is absent.

6

Slide the dashed-plane atoms sideways. Nothing changes: only the number of atoms per plane (planar density) matters, not where they sit on it. So a centred arrangement also shows a systematic absence of (100), while (200) survives.

Waves arriving at the detector

Resulting XRD peaks

3D Interactive

The Same Story in 3D: Simple Cubic vs BCC

In a body-centred cubic crystal the body-centre atoms form a complete second set of (100) layers, exactly halfway between the corner layers. Both layers have one atom per \(a^2\), so they cancel the (100) reflection. The same happens for every \((hkl)\) with \(h+k+l\) odd.

drag to rotate

corner atoms body-centre atoms planes through corners planes through body centres

Present

Turning Pictures into Algebra

Phase of the Wave Scattered by an Atom at \(\mathbf r\)

Recap. An extra atom at \((\tfrac12, 0, 0)\) lies on the (200) plane, midway between (100) planes. The path difference 1′–3′ is \(\lambda/2\), so the waves cancel although Bragg’s law is satisfied for (100). Moving the atom within the (200) plane changes nothing: only its distance from the (100) planes matters.

That distance is the projection of \(\mathbf r\) on the plane normal \(\mathbf g_{100}\). In units of the spacing, and turned into a phase:

\[\mathbf g_{100}\cdot\mathbf r = \mathbf a^*\cdot\tfrac12\mathbf a = \tfrac12 \quad\Rightarrow\quad \phi = 2\pi\cdot\tfrac12 = \pi\]

For a general position \(\mathbf r = (x, y, z)\) and general planes \((hkl)\):

\[\phi = 2\pi\,\mathbf g_{hkl}\cdot\mathbf r = 2\pi\,(hx+ky+lz)\]

Combining the amplitude \(f\) and the phase \(\phi\) in one complex number gives the contribution of a single atom at \(\mathbf r\) to the beam diffracted by the \((hkl)\) planes:

\[f\!\left(\frac{\sin\theta}{\lambda}\right)e^{i\phi} = f\!\left(\frac{\sin\theta}{\lambda}\right)e^{2\pi i\,\mathbf g_{hkl}\cdot\mathbf r}\]

Interactive

Slide an Atom Between the Planes

Atom A sits at the origin; atom B at \((x, 0, 0)\). Both scatter with the same \(f\). Choose the order \(h\) of the \((h00)\) reflection.

x of B0.500

Add All Atoms

The Structure Factor \(F_{hkl}\)

\[F_{hkl} = \sum_{j=1}^{N} f_j\!\left(\frac{\sin\theta_{hkl}}{\lambda}\right)e^{2\pi i\,(hx_j+ky_j+lz_j)}\]

Sum over all \(N\) atoms of the unit cell: each atom’s amplitude times its phase factor (Eq. 12.6).

\[I_{hkl}\;\propto\;|F_{hkl}|^2 = F_{hkl}F_{hkl}^*\]

\(^*\) = complex conjugate. Only \(|F|^2\) is measured; the phase is lost.

Each atom is a vector (phasor) of length \(f_j\) at angle \(\phi_j\) in the complex (Argand) plane.
\(F\) is the vector sum of all phasors; \(|F|^2\) is proportional to the diffracted intensity.
If the phasors close into a polygon, \(F = 0\): no diffracted beam even though Bragg’s law is satisfied. This is an extinction.

Interactive

Structure-Factor Phasor Lab

h

k

l

Present

Argand plane

|F|²

–

atom	x y z	φ

For a quick comparison \(f\) is taken equal to \(Z\) (the forward-scattering value); real \(f\) values fall with \(\sin\theta/\lambda\) but do not change which reflections vanish.

What Does Centring Do to \(F\)?

Lattice Centring and the Structure Factor

One atom \(f\) at \(0,0,0\) plus its centring translations. The key identity is \(e^{i\pi n} = (-1)^n\) for integer \(n\).

P

Primitive (P)

no centring vectors

\[F_{hkl} = f\]

No extinctions: every \((hkl)\) diffracts, \(I = f^2\).

C

Base-centred (C)

t = (½, ½, 0)

\[F_{hkl} = f\left[1+(-1)^{h+k}\right]\]

Absent: \(h+k\) odd (mC, oC)

I

Body-centred (I)

t = (½, ½, ½)

\[F_{hkl} = f\left[1+(-1)^{h+k+l}\right]\]

Absent: \(h+k+l\) odd (oI, tI, cI)

F

Face-centred (F)

t = (½, ½, 0), (½, 0, ½), (0, ½, ½)

\[F_{hkl} = f\left[1+(-1)^{h+k}+(-1)^{h+l}+(-1)^{k+l}\right]\]

Present only if \(h, k, l\) are unmixed (all even or all odd)

Which Cubic Reflections Survive?

Cubic Lattices: Survivors and Extinctions

(hkl)	s	cP	cI	cF

\(|F|^2\) for one atom of scattering factor \(f\) per lattice point (after Table 12.2, De Graef and McHenry).

These are the same sequences we used to index cubic patterns (Cu is cF).

Can Symmetry Operations Erase Peaks Too?

Extinctions from Symmetry: Inversion and Screw Axes

−1

Inversion centre at the origin: \(\mathbf r \to -\mathbf r\)

\[F_{hkl} = 2\sum_{j=1}^{N/2} f_j\cos\!\left(2\pi\,\mathbf g_{hkl}\cdot\mathbf r_j\right)\;\in\;\mathbb R\]

The \(e^{+i\phi}\) and \(e^{-i\phi}\) terms pair into cosines, so \(F\) is real when the origin is on the inversion centre (phase 0 or \(\pi\)). An inversion centre causes no extinctions by itself.

Screw axis \(4_1\parallel c\) (through the origin)

Equivalent atoms: \((x, y, z)\), \((-y, x, z+\tfrac14)\), \((-x, -y, z+\tfrac12)\), \((y, -x, z+\tfrac34)\). For the \(00l\) reflections only the \(z\) coordinates matter:

\[F_{00l} = S\sum_{j=1}^{N/4} f_j\,e^{2\pi i l z_j},\qquad S = \sum_{n=0}^{3} e^{i\pi n l/2} = \frac{1-e^{2\pi i l}}{1-e^{i\pi l/2}}\]

The numerator is always 0; the denominator is 0 only when \(l = 4n\) (then, by l’Hôpital, \(S = 4\)). So \(00l\) reflections with \(l \ne 4n\) are absent: projected on \(c\), the atoms repeat every \(c/4\).

Screw ∥ c	τ	00l absent for
2₁, 4₂, 6₃	c/2	l = 2n + 1
3₁, 3₂, 6₂, 6₄	±c/3	l ≠ 3n
4₁, 4₃	±c/4	l ≠ 4n
6₁, 6₅	±c/6	l ≠ 6n

After Table 12.3, De Graef and McHenry. τ is the translation that the screw adds along c in one rotation step (for 6₂, 6₄ the projected repeat is c/3).

Interactive

Screw-Axis Phasors

l

Present

Glide Planes and Friedel’s Law

Glides, Friedel’s Law and ITC Reflection Conditions

n-glide \(\parallel (001)\), \(\boldsymbol\tau = (\tfrac12, \tfrac12, 0)\)

An atom at \((x, y, z)\) has an identical partner at \((x+\tfrac12, y+\tfrac12, -z)\). For the zone \(hk0\) the \(z\) coordinate drops out:

\[F_{hk0} = \sum_{j=1}^{N/2} f_j\,e^{2\pi i(hx_j+ky_j)}\left[1+e^{i\pi(h+k)}\right]\]

⇒ \(hk0\) reflections with \(h+k\) odd are absent.

Glide ∥ (001)	τ	hk0 absent for
a	a/2	h = 2n + 1
b	b/2	k = 2n + 1
n	(a + b)/2	h + k = 2n + 1
d	(a ± b)/4	present only for h + k = 4n with h, k even

After Table 12.4, De Graef and McHenry.

Friedel’s law

\[I_{hkl} = F_{hkl}F_{hkl}^* = F_{\bar h\bar k\bar l}^*F_{\bar h\bar k\bar l} = I_{\bar h\bar k\bar l}\]

We measure intensities, not phases, so \((hkl)\) and \((\bar h\bar k\bar l)\) look identical (neglecting anomalous dispersion). Every diffraction pattern appears centrosymmetric, so only the 11 Laue classes can be told apart from intensities alone.

ITC: “Reflection conditions”

The International Tables list the conditions for a reflection to be present (the opposite of extinction conditions):

\(Cmm2\), general position: \(hkl\): \(h+k = 2n\)
\(Cmm2\), site 4c: extra condition \(hkl\): \(h = 2n\)
\(Pmna\) (n-glide \(\perp b\)): \(h0l\): \(h+l = 2n\)

→ Absent reflections fingerprint the space group.

Beyond Cubic

Structure Determination of Non-cubic Crystals

Things get harder as the number of unknowns grows. Cubic has one lattice parameter, \(a\). Tetragonal has two, \(a=b\) and \(c\). Lower symmetries have more still.

For a tetragonal crystal, \(1/d^2 = (h^2+k^2)/a^2 + l^2/c^2\). Putting this into Bragg’s law gives

\[\sin^2\theta = A\,(h^2+k^2) + C\,l^2, \qquad A = \frac{\lambda^2}{4a^2},\quad C = \frac{\lambda^2}{4c^2}\]

The trick is to find the \((hk0)\) lines first. For them \(l=0\), so

\[\sin^2\theta = A\,(h^2+k^2)\]

and their \(\sin^2\theta\) values stand in the ratio of the allowed values of \(h^2+k^2\): 1, 2, 4, 5, 8, 9, 10, …. That gives \(A\). Subtracting \(A(h^2+k^2)\) from the remaining lines leaves multiples \(Cl^2\) = \(C\), \(4C\), \(9C\), …, which gives \(C\).

We will not go through the other crystal systems. The analytical methods exist, but their complexity grows quickly as the symmetry drops.

Interactive

Tetragonal Line Finder

a0.360

c0.500

Line positions (primitive tetragonal, Cu Kα)

A from (hk0) lines

–

C from (00l) lines

–

Using the International Tables

Reading Reflection Conditions in ITC

Every space-group page in the International Tables for Crystallography (ITC) lists reflection conditions: the conditions under which a reflection is present. Everything else is a systematic absence.

Interactive

Is This Reflection Present?

h

k

l

Present

All reflections with s ≤ 20

General positions (brown box). These conditions come from the symmetry of the space group itself (centring, glides, screws). They apply to every atom in the structure, so a reflection that fails them is always absent.

Special positions (green box). Atoms sitting on special positions satisfy extra conditions: they contribute only to reflections that obey the extra rule. For \(Ia\bar3\), atoms on the 8a and 8b sites contribute only to reflections with \(h,k,l\) all even.

Worked Structure Factors

CsCl, NaCl and Diamond

\[F_{hkl} = f_{\mathrm{Cs}} + f_{\mathrm{Cl}}\,(-1)^{h+k+l}\]

Fundamental

h + k + l even

\[I = (f_{\mathrm{Cs}}+f_{\mathrm{Cl}})^2\]

Superlattice

h + k + l odd

\[I = (f_{\mathrm{Cs}}-f_{\mathrm{Cl}})^2\]

Interactive • order–disorder

Superlattice Peaks Measure the Degree of Order

Schematic pattern (|F|² only)

order S1.00

Disordered. If the two kinds of atom occupied the two sites at random, both sites would scatter with \(f = (f_A+f_B)/2\): the superlattice peaks vanish and the pattern looks bcc. Real CsCl stays ordered; ordered β′-brass (CuZn) is the classic case that disorders on heating (near 460 °C).

NaCl: two fcc lattices, \(\boldsymbol\tau = (\tfrac12,\tfrac12,\tfrac12)\)

\[F_{hkl}=\left[f_{\mathrm{Na}}+f_{\mathrm{Cl}}(-1)^{h+k+l}\right]\left[1+(-1)^{h+k}+(-1)^{h+l}+(-1)^{k+l}\right]\]

\(h,k,l\) mixed parity: \(I = 0\) (fcc condition)
Unmixed, \(h+k+l\) even: \(I = 16(f_{\mathrm{Na}}+f_{\mathrm{Cl}})^2\), strong: (200), (220), (222)
Unmixed, \(h+k+l\) odd: \(I = 16(f_{\mathrm{Na}}-f_{\mathrm{Cl}})^2\), weak: (111), (311)

Diamond: two fcc (C) lattices, \(\boldsymbol\tau = (\tfrac14,\tfrac14,\tfrac14)\)

\[F_{hkl}=2f_{\mathrm C}\,e^{\frac{i\pi}{4}(h+k+l)}\cos\!\left[\tfrac{\pi}{4}(h+k+l)\right]\left[1+(-1)^{h+k}+(-1)^{h+l}+(-1)^{k+l}\right]\]

Mixed parity: \(I = 0\)
All odd: \(I = 32f_{\mathrm C}^2\): (111), (311), (331)
All even, \(h+k+l = 4n\): \(I = 64 f_{\mathrm C}^2\): (220), (400)
All even, \(h+k+l = 4n+2\): \(I = 0\): (200), (222). Extra extinctions!

Both cases can be explored phasor by phasor in the Structure-Factor Phasor Lab above.

Part 5

From \(|F|^2\) to the Measured Pattern

A detector does not record \(|F|^2\) directly. Multiplicity, geometry, absorption, thermal vibrations and grain size all shape the pattern we actually measure.

We will cover

Multiplicity, Lorentz–polarization, absorption and Debye–Waller factors
Peak shape and crystallite size: the Scherrer formula
Putting it all together: the powder pattern of tungsten
Practice: cubic BaTiO\(_3\) from its Wyckoff positions

What the Detector Records

From Diffracted to Measured Intensity

A real experiment integrates the diffracted intensity over time with a detector of finite size. For powder diffractometry, several correction factors multiply \(|F|^2\):

\[I_{hkl} = |F_{hkl}|^2\;p_{hkl}\;L_p(\theta)\;A(\theta)\;e^{-2M}\]

Intensities are rescaled so that the strongest peak = 100, so \(I_0\), \(K\), \(r\) and other common factors cancel.

|F|²

Structure factor

Which atoms, where, and the symmetry

p

Multiplicity

Number of equivalent planes diffracting at the same 2θ

L_p

Lorentz–polarization

Polarization plus the geometry of powder diffraction

A

Absorption

Attenuation in the sample; constant for a flat-sample diffractometer

e^−2M

Temperature

Debye–Waller: thermal vibrations smear the electron cloud

Temperature factor

Debye–Waller Factor

• Atoms vibrate about their sites; the amplitude grows with \(T\).
• The time-averaged electron cloud looks larger and more diffuse, so \(f\) decreases.

\[f_T = f_0\,e^{-B(T)\,s^2},\qquad M = B(T)\,s^2\]

The intensity is multiplied by \(e^{-2M}\); \(B\) is proportional to the mean-square atomic displacement.

B (Å²) at 280 K

B0.158

W (222), Cu Kα, s = 0.5472 Å⁻¹

–

\(B\) values from De Graef and McHenry. At 290 K, \(B_\mathrm W = 0.1581\) Å²: \(f_0 = 39.94\), \(e^{-Bs^2} = 0.9538\), so \(f = 38.09\).

Absorption factor

\(A(\theta)\)

• X-rays are partly absorbed along their path in the sample; \(A\) multiplies the intensity (not the amplitude).
• Flat, thick sample in a powder diffractometer: the incoming and outgoing paths both make angle \(\theta\) with the surface, and the total irradiated depth compensates exactly, giving

\[A = \frac{1}{2\mu}\quad\text{independent of }\theta\]

\(\mu\) is the linear absorption coefficient. Being the same for every peak, it cancels when the intensities are rescaled.

• Cylindrical sample (Debye–Scherrer camera): absorption is strongest at small \(2\theta\), opposite to the thermal effect, so the two nearly cancel.
• Single crystals: \(A\) depends strongly on the sample shape.

How Many Planes Add Up?

Multiplicity Factor \(p_{hkl}\)

In a powder, all planes of a family \(\{hkl\}\) with the same \(d\) diffract into the same \(2\theta\), so their intensities add.

Calculator

List the Equivalent Planes

h

k

l

Multiplicity

–

Equivalents generated by the Laue-class symmetry (\(m\bar3m\), \(4/mmm\), \(mmm\), \(6/mmm\), \(2/m\) with b unique, \(\bar1\)). For hexagonal, \(l\) is the fourth Miller–Bravais index.

After Table 12.6, De Graef and McHenry.

Example: Cu \(\{200\}\to p = 6\): (200), (020), (002), \((\bar200)\), \((0\bar20)\), \((00\bar2)\). (220) is of type \(0kk\) ⇒ \(p = 12\); (224) is of type \(hhl\) ⇒ \(p = 24\). Lower symmetry means fewer equivalent planes and a smaller \(p\): in a tetragonal crystal (100) and (001) are no longer equivalent.

Geometry Matters

Lorentz–Polarization Factor \(L_p(\theta)\)

1

Diffraction-ring circumference

A powder diffracts onto cones; the detector samples a fixed length of a ring of circumference \(2\pi R\sin 2\theta\).

\(\propto\dfrac{1}{\sin 2\theta}\)

2

Grains in Bragg orientation

The fraction of randomly oriented grains whose plane normals fall in the narrow band \(r\,\Delta\theta\) near the Bragg orientation.

\(\propto\cos\theta\)

3

Integrated intensity

Peak width \(\propto 1/\cos\theta\) and peak height \(\propto 1/\sin\theta\).

\(\propto\dfrac{1}{\sin 2\theta}\)

\[\frac{1}{\sin 2\theta}\cdot\cos\theta\cdot\frac{1}{\sin 2\theta}\cdot\frac{1+\cos^2 2\theta}{2}\;\propto\; L_p(\theta) = \frac{1+\cos^2 2\theta}{\sin^2\theta\,\cos\theta}\]

Interactive

Bragg θ45°

L_p at this angle

–

Large at low and high angles, smallest near \(\theta \approx 50^\circ\): low-angle peaks are boosted strongly.

What Sets the Width of a Peak?

Peak Shape and Crystallite Size: the Scherrer Formula

\[I^{G}(\alpha) = I_0\exp\!\left[-4\ln 2\left(\frac{\alpha-2\theta}{w}\right)^2\right]\qquad I^{L}(\alpha) = \frac{I_0}{1+4\left(\frac{\alpha-2\theta}{w}\right)^2}\]

\(w\) is the full width at half maximum (FWHM). Gaussian tails decay quickly; Lorentzian tails are long.

• A crystal still diffracts slightly off the exact Bragg angle: the phase error between planes is small, so the interference is incomplete.
• Fewer planes (smaller grains) ⇒ a wider range of \(\Delta\theta\) ⇒ broader peaks.

\[w = \frac{0.9\,\lambda}{D\cos\theta}\]

Scherrer formula
\(D\) = mean grain diameter; \(w\) in radians

Example (illustrative): Cu nanoparticles. Cu Kα (\(\lambda = 1.5406\) Å); (111) peak at \(2\theta = 43.3^\circ\) with FWHM \(w = 0.5^\circ\) (after removing instrumental broadening): \[D = \frac{0.9\times1.5406\ \text{Å}}{(0.5^\circ\times\pi/180)\cos 21.65^\circ}\approx 171\ \text{Å}\approx 17\ \text{nm}\]

⇒ Nanocrystalline grains: this is why the nanoparticle peaks in the copper example are broad.

Interactive

Grain Size ↔ Peak Width

D (nm)17 nm

peak 2θ43.3°

FWHM from Scherrer (Cu Kα, λ = 1.5406 Å)

–

Putting It All Together

Powder Pattern of Tungsten

bcc W, \(a = 0.31653\) nm, Cu Kα, \(T = 290\) K, \(B = 0.1581\) Å²; \(F = 2f\) (\(h+k+l\) even). Relative intensity \(\propto |F|^2 e^{-2M} p\,L_p\).

(hkl)	2θ (°)	f	p	L_p	I calc	I expt

Calculated live from the formulas above (\(\lambda = 1.5418\) Å). Experimental values after Tables 12.7–12.8, De Graef and McHenry (Powder Diffraction File #04-0806).

Good agreement, considering that the experiment lists peak (not integrated) intensities.

The Big Picture

Summary: What an XRD Pattern Tells Us

2θ

Peak positions

• Bragg’s law: \(\lambda = 2d\sin\theta\)
• \(d_{hkl}\) is fixed by the size and shape of the unit cell
• Used to index the pattern and find the lattice parameters

I

Peak intensities

• Which atoms: \(f(Z, \sin\theta/\lambda)\)
• Where they are: \(F_{hkl} = \sum_j f_j e^{2\pi i(hx_j+ky_j+lz_j)}\)
• Symmetry: centring, screw axes, glides ⇒ systematic absences
• Measurement: \(p\), \(L_p\), \(A\), \(e^{-2M}\); peak width ⇒ grain size

Electron

Thomson, P(θ)

Atom

f(sinθ/λ)

Unit cell

F_hkl, extinctions

Measured

p · L_p · A · e^−2M

Practice Problem

Cubic BaTiO\(_3\) from Its Wyckoff Positions

Example Problem 10 (p. 175), Structure of Materials by Allen and Thomas: make a perspective sketch of the cubic form of barium titanate. Space group \(Pm\bar3m\), \(a = 4.009\) Å; Ba\(^{2+}\) in 1a, Ti\(^{4+}\) in 1b, O\(^{2-}\) in 3c.

No. 221 Pm3̄m (extract)

Mult.	Wyck.	Site sym.	Coordinates
3	c	4/m m.m	0,½,½ ½,0,½ ½,½,0
1	b	m3̄m	½,½,½
1	a	m3̄m	0,0,0

Reflection conditions: general, none; special, no extra conditions.

1Place Ba\(^{2+}\) at 1a (0,0,0) and apply the primitive-cubic lattice translations: Ba appears at all 8 corners (8 × ⅛ = 1 per cell).
2Ti\(^{4+}\) at 1b (½,½,½): any lattice translation takes it outside the cell, so one Ti sits at the body centre.
3O\(^{2-}\) at 3c: the three face-centre positions; translations give all 6 face centres (6 × ½ = 3 per cell). Formula: BaTiO\(_3\). Ti sits in an O\(_6\) octahedron.

3D Interactive

drag to rotate

Ba (1a)Ti (1b)O (3c)

Practice Assessment

Did It Sink In? - 17-19

Test your mastery of Bragg’s law, systematic absences, the Ewald sphere, indexing, the structure factor and intensity corrections.

Take the Quiz