Crystal structures and Miller indices

Unit cells of SC, BCC, FCC and HCP metals, atoms per cell, packing factor and density, Miller indices of planes and directions, interplanar spacing and Bragg's law.

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Why it matters

Whether a metal bends or shatters, how dense it is, and which planes it slips or cleaves on are all set by how its atoms are stacked. Crystal structure explains why copper and aluminium are easy to draw into wire while magnesium and zinc crack in cold forming, and Miller indices are the language used to name the planes and directions involved in slip, X-ray diffraction and texture.

Key ideas

Lattice, basis and unit cell. A crystal is a lattice (a regular 3-D array of points) with an atom or group of atoms (the basis) placed at each point. The unit cell is the smallest repeating block; its edge lengths a, b, c and angles α, β, γ are the lattice parameters. There are 7 crystal systems and 14 Bravais lattices; for engineering metals three structures matter most.

Counting atoms per cell. A corner atom is shared by 8 cells (counts 1/8), a face atom by 2 (counts 1/2), a body-centre atom belongs wholly to the cell.

  • Simple cubic (SC): 8 × 1/8 = 1 atom; atoms touch along the edge, so a = 2r; coordination number (CN) 6; atomic packing factor (APF) 0.52. Only polonium adopts it.
  • Body-centred cubic (BCC): 8 × 1/8 + 1 = 2 atoms; atoms touch along the body diagonal, so √3·a = 4r; CN 8; APF 0.68. Examples: α-iron (ferrite), Cr, Mo, W, V.
  • Face-centred cubic (FCC): 8 × 1/8 + 6 × 1/2 = 4 atoms; atoms touch along the face diagonal, so √2·a = 4r; CN 12; APF 0.74. Examples: γ-iron (austenite), Al, Cu, Ni, Ag, Au.
  • Hexagonal close-packed (HCP): 6 atoms in the hexagonal prism (2 in the primitive cell); CN 12; APF 0.74 when c/a = 1.633. Examples: Mg, Zn, Ti (α), Co.

FCC and HCP are both close-packed: they stack the same close-packed layers, FCC as ABCABC… and HCP as ABAB…. The difference in stacking, not in density, explains why FCC has 12 slip systems on four non-parallel {111} planes while HCP has only 3 easy (basal) slip systems at room temperature.

Polymorphism (allotropy). Some elements change structure with temperature. Iron is BCC (α) up to 912 °C, FCC (γ) from 912 to 1394 °C and BCC (δ) to its melting point at 1538 °C. This change is the basis of heat treatment of steel.

Miller indices of a plane (hkl). (1) Find the intercepts on the x, y, z axes in units of a, b, c; a plane parallel to an axis has intercept ∞. If the plane passes through the origin, shift the origin first. (2) Take reciprocals (1/∞ = 0). (3) Clear fractions to the smallest integers. (4) Write in round brackets without commas; a negative index is written with a bar over it, e.g. (1 1̄ 0). A family of equivalent planes is written {hkl}, e.g. {100} = the six cube faces.

Miller indices of a direction [uvw]. Subtract tail coordinates from head coordinates, clear fractions, write in square brackets; a family of equivalent directions is ⟨uvw⟩. In cubic crystals only, the direction [hkl] is perpendicular to the plane (hkl).

Linear and planar density. Linear density = atoms whose centres lie on a direction ÷ length; planar density = atoms whose centres lie on a plane ÷ area. Slip happens on the densest planes along the densest directions: {111}⟨110⟩ in FCC, {110}⟨111⟩ in BCC.

Hexagonal (Miller–Bravais) indices (hkil). Four axes a1, a2, a3, c with i = −(h + k). The basal plane is (0001).

Formulas

n = Nc/8 + Nf/2 + Ni n = atoms per cubic cell; Nc, Nf, Ni = corner, face and interior atoms.

APF = n·(4/3)π·r³ / Vcell r = atomic radius (m); Vcell = a³ for cubic cells (m³). Dimensionless.

ρ = n·A / (NA·a³) ρ = theoretical density (kg/m³ or g/cm³); A = atomic mass (g/mol); NA = 6.022 × 10²³ mol⁻¹; a = lattice parameter (use cm to get g/cm³). Valid for a perfect crystal; real density is slightly lower because of vacancies and porosity.

a = 2r (SC); a = 4r/√3 (BCC); a = 2√2·r (FCC)

d_hkl = a / √(h² + k² + l²) d = interplanar spacing between adjacent (hkl) planes (m). Valid for cubic crystals only.

n_r·λ = 2·d·sin θ Bragg's law: n_r = order of reflection (integer), λ = X-ray wavelength (m), θ = Bragg angle (half the diffraction angle 2θ).

Worked examples

Example 1 (standard): theoretical density of copper. Given: Cu is FCC, atomic radius r = 0.128 nm, A = 63.55 g/mol.

  1. Lattice parameter: a = 2√2·r = 2 × 1.4142 × 0.128 = 0.3620 nm = 3.620 × 10⁻⁸ cm.
  2. Cell volume: a³ = (3.620 × 10⁻⁸)³ = 4.745 × 10⁻²³ cm³.
  3. ρ = n·A / (NA·a³) = 4 × 63.55 / (6.022 × 10²³ × 4.745 × 10⁻²³) = 254.2 / 28.58.
  4. ρ = 8.90 g/cm³ (handbook value 8.96 g/cm³).

Example 2 (GATE level): X-ray peak of ferrite. Given: α-iron is BCC with a = 0.2866 nm; Cu Kα radiation, λ = 0.1542 nm. Find the diffraction angle 2θ of the first-order (110) reflection, and check the density (A = 55.85 g/mol).

  1. d_110 = a / √(1² + 1² + 0²) = 0.2866 / 1.4142 = 0.2027 nm.
  2. Bragg's law with n_r = 1: sin θ = λ / (2d) = 0.1542 / (2 × 0.2027) = 0.3804.
  3. θ = 22.36°, so 2θ = 44.7°.
  4. Density: a³ = (2.866 × 10⁻⁸ cm)³ = 2.354 × 10⁻²³ cm³; ρ = 2 × 55.85 / (6.022 × 10²³ × 2.354 × 10⁻²³) = 7.88 g/cm³ (handbook 7.87 g/cm³).

Example 3 (indices): a plane cuts the axes at x = 1/2, y = 1, z = ∞. Reciprocals 2, 1, 0 → (210). In a cubic cell the direction [210] is normal to it.

Common mistakes

  • Using a = 4r/√3 for FCC or 2√2·r for BCC: atoms touch along the body diagonal in BCC and along the face diagonal in FCC.
  • Counting every corner atom as a whole atom (gives 9 for BCC instead of 2).
  • Taking reciprocals before shifting the origin when the plane passes through it, and so getting an infinite index.
  • Applying d = a/√(h²+k²+l²) to hexagonal or tetragonal crystals; it is for cubic only.
  • Mixing nm and cm in the density formula: a must be in cm to get g/cm³.
  • Saying FCC is ductile because it has "more slip systems" than BCC. BCC has as many or more potential systems; FCC is ductile because its slip planes are truly close-packed, so dislocations move at low stress at every temperature.
  • Using the diffraction angle 2θ in Bragg's law in place of θ.

For GATE PI

Expect one-mark questions on atoms per cell, coordination number, APF and which metals are FCC, BCC or HCP, and on the Miller indices of a sketched plane. Two-mark numericals ask for theoretical density, lattice parameter from radius, interplanar spacing or a Bragg angle. Practise converting units (nm to cm), recognising {111} and {110} as the slip planes in FCC and BCC, and the iron allotropes and their transition temperatures.

Quick check

  1. How many atoms does an FCC unit cell contain, and what is its APF?
  2. Write the Miller indices of a plane with intercepts x = 1, y = 2, z = ∞.
  3. For a cubic crystal with a = 0.36 nm, what is d for the (200) planes?
  4. Which form of iron is stable at 1000 °C, and what is its structure?
  5. In a cubic crystal, what is the angle between the direction [111] and the plane (111)?

Answers: 1. 4 atoms, 0.74; 2. (210); 3. 0.18 nm; 4. γ-iron (austenite), FCC; 5. The direction is normal to the plane (90° to the plane).

Interview questions

All Engineering Materials interview questions

Try answering each one aloud before you open it.

  1. 1.What is a crystal structure in the context of engineering materials?Concept

    A crystal structure refers to the orderly and repeating arrangement of atoms in a material. It defines the geometric pattern in which atoms are organized in a solid. Common crystal structures include body-centered cubic (BCC), face-centered cubic (FCC), and hexagonal close-packed (HCP). These structures influence the material's properties, such as strength, ductility, and conductivity.

  2. 2.Explain the significance of Miller indices in crystallography.Concept

    Miller indices are a notation system in crystallography used to describe the orientation of planes and directions in a crystal lattice. They are represented as a set of three integers (h, k, l) and are crucial for identifying and categorizing different planes within a crystal. This helps in understanding the material's properties, such as slip planes in metals, which affect mechanical behavior.

  3. 3.How do you determine the Miller indices for a given crystal plane?Concept

    To determine the Miller indices for a crystal plane, follow these steps: 1) Identify the intercepts of the plane with the crystallographic axes in terms of lattice constants. 2) Take the reciprocals of these intercepts. 3) Clear fractions by multiplying by the smallest common multiple to get whole numbers. 4) Enclose the resulting integers in parentheses, e.g., (hkl).

  4. 4.Why are FCC metals such as aluminium and copper so ductile?Application

    FCC metals have 12 slip systems of the {111}⟨110⟩ type, and the {111} planes are genuinely close-packed, so dislocations glide on them at a low critical resolved shear stress. Because the four {111} planes are not parallel, a grain can always find enough independent slip systems (five are needed) to change shape arbitrarily. FCC metals also show no ductile-to-brittle transition, so they stay ductile even at cryogenic temperatures.

  5. 5.What happens to the mechanical behaviour of iron when its structure changes from BCC to FCC on heating?Application

    Above 912 °C iron changes from BCC ferrite to FCC austenite, which is softer and more ductile, so steel is hot-worked in the austenite range. The improvement is not because FCC has more slip systems (BCC has as many or more) but because FCC slip planes are truly close-packed, giving a low, temperature-insensitive resistance to dislocation motion. FCC austenite also dissolves far more carbon (up to 2.14 %) than ferrite, which is what makes hardening by quenching possible.

  6. 6.Explain why hexagonal close-packed (HCP) metals are generally less ductile than FCC metals.Application

    HCP metals are generally less ductile than FCC metals because they have fewer slip systems available for deformation. The limited number of slip planes in HCP structures restricts the ability of the material to deform plastically under stress. This results in a higher likelihood of brittle fracture compared to FCC metals, which have multiple slip systems and can accommodate more deformation.

  7. 7.Calculate the linear density of atoms along the [110] direction in a face-centred cubic (FCC) crystal with lattice parameter a.Numerical

    The [110] direction runs along a face diagonal of length √2·a. Atom centres on it are the two corner atoms (each shared, counting 1/2 per length of the diagonal segment) and the face-centre atom, giving 1/2 + 1 + 1/2 = 2 atoms. Linear density = 2 / (√2·a) = √2 / a = 1/(2r), since atoms touch along this close-packed direction.

  8. 8.What is the coordination number of atoms in a body-centered cubic (BCC) structure, and why is it important?Concept

    The coordination number of atoms in a BCC structure is 8, meaning each atom is surrounded by 8 nearest neighbors. This is important because the coordination number influences the material's density and stability. A higher coordination number generally indicates a more densely packed structure, which can affect the material's mechanical and thermal properties.

  9. 9.Calculate the atomic packing factor (APF) for a simple cubic structure.Numerical

    The atomic packing factor (APF) is the fraction of volume occupied by atoms in a unit cell. For a simple cubic structure, there is 1 atom per unit cell, and the volume of the atom is (4/3)π(r^3), where r is the atomic radius. The volume of the unit cell is a^3, where a = 2r. Thus, APF = [(4/3)π(r^3)] / (2r)^3 = π/6 ≈ 0.52.

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