Defects in Crystalline Materials: Point, Line and Surface Defects

This chapter belongs to the Engineering Sciences (XE) paper. It answers the last heading of Section XE2.1 of XE2 Materials Science, one of the nine optional sections, any two of which a candidate chooses in the exam: “Defects in Crystalline Materials: Zero-dimensional or point defects: Vacancies, interstitials, substitutional atoms, Frenkel and Schottky defects; Equilibrium concentration of point defects. One-dimensional or line defects: Dislocations: edge, screw and mixed. Burgers vector and Burgers circuit; Burgers vectors of stable dislocations in simple cubic, body-centred cubic and face-centred cubic lattices; Dislocations meeting at a node; Line energy of a dislocation. Dislocation motion: glide and climb. Two-dimensional or surface defects: Free surfaces, Grain boundaries, twin boundaries, stacking faults, phase boundary; Surface energy of a free surface in terms of a simple bond breaking model.” It teaches the three classes of defect, why point defects exist at equilibrium, how a dislocation is described and how it moves, and what makes a surface or an internal interface costly. The numericals are vacancy concentration and its temperature dependence, Burgers vector magnitude, dislocation line energy and surface energy from broken bonds.

Point defects: vacancies, interstitials, Frenkel and Schottky

A point defect is a disturbance confined to one or two atomic sites. A vacancy is an empty lattice site. A self-interstitial is a host atom squeezed into a void; it distorts the lattice so much that its formation energy is several times that of a vacancy, so self-interstitials are rare in equilibrium. An interstitial impurity is a small foreign atom in a void (C, N, H in iron), and a substitutional impurity is a foreign atom on a lattice site. Each defect strains the lattice around it, and each acts as a scattering centre for electrons, a source of diffusion, or a pinning site for dislocations.

In an ionic crystal, defects must keep the crystal electrically neutral. A Frenkel defect is a vacancy plus an interstitial of the same ion: an ion, usually the smaller cation, leaves its site and sits in a void, so the vacancy and interstitial are created as a pair (AgCl). The volume stays almost unchanged. A Schottky defect is a set of vacancies, one cation vacancy and one anion vacancy in NaCl, in the ratio that preserves the formula’s charge balance, with the displaced ions going to the surface or to other sites. It adds new lattice sites, so the density decreases. Both are stoichiometric, as no net charge is created.

⚠️ Density: Frenkel vs Schottky
A Frenkel pair only moves an ion within the crystal, so the density is practically unchanged. A Schottky defect removes ions from the interior to the surface, so the volume grows at constant mass and the density falls. Do not link either one to a change of stoichiometry.

Equilibrium concentration of vacancies

Vacancies are the one defect that a crystal contains in equilibrium, because they lower the free energy at any temperature above 0 K. Making n vacancies among N sites costs energy nQ (Q is the formation energy per vacancy) but raises the configurational entropy S = k ln[N!/(n!(N − n)!)], so G = nQ − TS has a minimum. With Stirling’s approximation, dG/dn = 0 gives n/(N − n) = exp(−Q/kT), and since n ≪ N,

n/N = exp(−Q/kT). Here k = 1.38 × 10⁻²³ J/K = 8.617 × 10⁻⁵ eV/K; for Q per mole use R = 8.314 J/(mol K) in place of k. The concentration rises exponentially with T, reaching about 10⁻⁴ near the melting point of many metals. A plot of ln(n/N) against 1/T is a straight line of slope −Q/k, the standard way to get Q. The number of sites per unit volume is N = N_A ρ/A.

Worked example. Q = 0.90 eV, T = 1000 K: Q/kT = 0.90/(8.617 × 10⁻⁵ × 1000) = 10.44, so n/N = e⁻¹⁰·⁴⁴ = 2.91 × 10⁻⁵. If the temperature rises from 900 K to 1200 K with Q = 1.0 eV, the ratio is exp[(Q/k)(1/900 − 1/1200)] = exp[11 605 × 2.778 × 10⁻⁴] = exp(3.224) = 25. A 33 % rise in T raises the vacancy concentration 25-fold.

Ionic crystals. A Schottky pair of formation energy Q_S has n_S ≈ N exp(−Q_S/2kT), and a Frenkel pair has n_F ≈ √(N N_i) exp(−Q_F/2kT), where N_i is the number of interstitial sites. The factor 2 appears because each event creates two defects. Rapid quenching can retain a vacancy concentration far above the equilibrium value of the lower temperature, which then decays by annealing.

⚠️ Kelvin, and eV versus J
Always put T in kelvin and match the units of Q and k: Q in eV with k = 8.617 × 10⁻⁵ eV/K, or Q in J/mol with R. Using 1.38 × 10⁻²³ with Q in eV, or ignoring the factor 2 for a pair (Schottky or Frenkel), are the two commonest slips.

Dislocations: edge, screw and mixed, and the Burgers vector

A dislocation is a line defect, the boundary between the part of a slip plane that has slipped and the part that has not. An edge dislocation is the edge of an extra half-plane of atoms inserted in the crystal; the lattice is compressed on the side of the half-plane and stretched on the other. A screw dislocation is a line around which the planes spiral like a helical ramp, so that circling the line once takes you up by one plane spacing. A mixed dislocation has both characters; a curved dislocation line in a slip plane is pure edge where the line is perpendicular to the slip vector, pure screw where it is parallel, and mixed in between.

The Burgers vector b measures the lattice distortion. To find it, draw a closed circuit of N steps in each lattice direction in perfect crystal; draw the same sequence of steps around the dislocation line and it fails to close. The closure failure, from finish to start (the convention is FS/RH: finish to start, with the circuit traced by the right-hand rule relative to the line direction), is b. Edge dislocation: b ⟂ line. Screw dislocation: b ∥ line. For a mixed dislocation making angle φ between the line and b, the screw component is b cos φ and the edge component is b sin φ. The slip plane contains both the line and b, so an edge dislocation has a unique slip plane, while a screw dislocation, with b and the line parallel, can lie in any plane containing it and so can cross-slip.

  • b is conserved along a line: it has the same value everywhere on a dislocation, even where the character changes.
  • A dislocation cannot end inside a crystal: it must close into a loop, reach a free surface or a boundary, or meet other dislocations at a node.
  • Density is the line length per unit volume, in m⁻²: about 10¹⁰ to 10¹² m⁻² in annealed metals and up to about 10¹⁵ to 10¹⁶ m⁻² after heavy cold work.

Burgers vectors of stable dislocations, nodes and line energy

Line energy. The elastic strain energy around a dislocation per unit length is about E ≈ αGb², with G the shear modulus and α between roughly 0.5 and 1. More exactly, a screw dislocation has E = Gb² ln(R/r₀)/4π and an edge dislocation E = Gb² ln(R/r₀)/[4π(1 − ν)], which for ν ≈ 1/3 is about 1.5 times the screw value. Because E ∝ b², a stable dislocation has the shortest lattice translation vector as its Burgers vector, and a dislocation with a long b can lower its energy by splitting: b₁ → b₂ + b₃ is favourable when |b₁|² > |b₂|² + |b₃|².

Burgers vector of the stable dislocation
Latticeb|b|Reason
Simple cubica⟨100⟩aShortest lattice vector is the cube edge
BCC(a/2)⟨111⟩√3 a/2Atoms touch along the body diagonal
FCC(a/2)⟨110⟩a/√2Atoms touch along the face diagonal

In BCC, the a⟨100⟩ vector is longer (|b|² = a² against 3a²/4), so the (a/2)⟨111⟩ dislocation has lower energy. In FCC, (a/2)[1 −1 0] can split into the Shockley partials (a/6)[2 −1 −1] + (a/6)[1 −2 1]: the vectors add correctly, and |b|² falls from a²/2 to a²/6 + a²/6 = a²/3. The partials repel and are held apart by the stacking-fault ribbon between them.

Nodes. Where dislocations meet, the Burgers vectors obey Frank’s rule: the sum of b of the lines directed into the node equals the sum of those directed out (all directed outward, Σb = 0). In FCC, (a/2)[1 0 1] can meet two lines (a/2)[0 1 1] and (a/2)[1 −1 0] at a node: (0 + 1, 1 − 1, 1 + 0) a/2 = (a/2)[1 0 1]. Worked line energy. For an FCC metal with a = 0.3615 nm, b = a/√2 = 0.2556 nm. With G = 48 GPa and E = Gb²/2: E = 0.5 × 48 × 10⁹ × (0.2556 × 10⁻⁹)² = 1.57 × 10⁻⁹ J/m.

🎯 Why dislocations are never ‘in equilibrium’
A dislocation costs energy per unit length (about 10⁻⁹ J/m, or a few eV per atomic plane crossed), far more than the entropy it can gain. So, unlike vacancies, dislocations have no equilibrium density, and annealing removes them. The line energy also acts as a line tension, which tries to shorten a curved dislocation.

Dislocation motion: glide and climb

Glide (slip) is motion in the slip plane, the plane containing the line and b. It is conservative: atoms only shift a fraction of a spacing, nothing is added or removed, and it needs only a shear stress, not diffusion. Under a shear stress an edge dislocation glides parallel to b and a screw dislocation glides perpendicular to b, yet both produce slip by b in the same sense. Slip occurs on the close-packed planes and in the close-packed directions, {111}⟨110⟩ in FCC, {110}⟨111⟩ (among others) in BCC and (0001)⟨11-20⟩ in HCP, because the energy and the stress needed are lowest there. When a crystal slips by one Burgers vector on the whole plane, a step b high appears at the surface.

Climb is motion of an edge dislocation perpendicular to its slip plane, out of the plane that holds b. It is non-conservative: the extra half-plane grows or shrinks by gaining or losing atoms. When vacancies diffuse to the end of the half-plane and are absorbed, the half-plane shortens and the dislocation climbs up; emission of vacancies (or absorption of interstitials) makes it climb down. Climb needs atomic diffusion, so it matters only at high temperature, where it controls creep and recovery. A screw dislocation has no extra half-plane and cannot climb; it escapes obstacles by cross-slip.

Glide and climb compared
FeatureGlideClimb
Plane of motionSlip plane (contains line and b)Perpendicular to the slip plane
Dislocation typeEdge, screw and mixedEdge component only
Mass transportNone (conservative)Vacancy diffusion (non-conservative)
Driving conditionShear stress on the slip planeHigh temperature and diffusion

Free surfaces and the bond-breaking model of surface energy

Atoms at a free surface have lost neighbours, so they are bonded less strongly and the surface carries excess energy, the surface energy γ (J/m²), of the order of 1 J/m² for metals. In the simple bond-breaking model only nearest-neighbour bonds count, each of energy ε, and the surface is a perfect cut with no relaxation. Cutting the crystal breaks bonds and makes two surfaces, so each surface is charged half of the energy of each broken bond: a surface atom that has lost n_b bonds has excess energy n_b ε/2, and per unit area

γ = (n_b ε/2) × (surface atoms per unit area). For an FCC crystal (12 nearest neighbours) the number of broken bonds is 3 for a (111) surface, 4 for (100) and 5 for (110). With the planar densities 4/(√3 a²) = 2.31/a² for (111), 2/a² for (100) and √2/a² = 1.41/a² for (110), γ(100)/γ(111) = (4 × 2)/(3 × 2.309) = 1.155. The close-packed plane has the fewest broken bonds and the lowest energy, which is why crystals prefer to show close-packed facets.

Worked example. A model FCC solid has a = 0.400 nm and bond energy ε = 0.30 eV. For the (111) surface, n_b = 3 and the planar density is 4/(√3 × 0.160) = 14.43 nm⁻². γ = (3 × 0.30/2) × 14.43 = 6.50 eV/nm². Since 1 eV/nm² = 0.1602 J/m², γ = 1.04 J/m².

⚠️ Half the bond, and the right plane density
Forgetting the factor ½ doubles the answer, and counting bonds of the wrong plane or using the cube-face density 1/a² for the FCC (100) surface (instead of 2/a²) are the usual errors. The model ignores relaxation and next-nearest bonds, so use it for ratios and trends, not for exact values.

Grain boundaries, twin boundaries, stacking faults and phase boundaries

A grain boundary separates two crystals of the same phase with different orientations. A low-angle boundary (misorientation θ below about 10° to 15°) is an array of dislocations: a tilt boundary is a wall of edge dislocations of Burgers vector b, spaced D = b/[2 sin(θ/2)] ≈ b/θ (θ in radians), and a twist boundary is a grid of screw dislocations. The closer the dislocations, the larger the angle and the energy, until the cores overlap and a high-angle boundary has an energy nearly independent of θ. Grain boundaries have lower energy than a free surface, but they are the sites of fast diffusion, impurity segregation and preferential etching.

A twin boundary is a plane across which the lattice is a mirror image of itself. In FCC the coherent twin lies on {111}, and the stacking changes from ABCABC to ABCABCBACBA, the layer at the mirror being common to both sides; every atom at the boundary keeps its nearest-neighbour bonds, so its energy is much lower than that of a typical grain boundary. A stacking fault is a local error in the stacking sequence of close-packed planes with the lattice otherwise intact. In FCC, removing one plane gives ABCABABCABC, an intrinsic fault; the layers A B A B around it form a thin region of HCP stacking. The fault carries an energy per unit area, the stacking-fault energy. A low fault energy produces wide ribbons between partials, makes cross-slip hard and favours twinning.

A phase boundary separates two different phases (different structure or composition). It is coherent if the lattice planes run continuously across it with a small elastic strain from the lattice mismatch (lowest energy), semi-coherent if misfit dislocations relieve the strain at intervals, and incoherent if the lattices do not match at all (highest energy). The energy therefore ranks: coherent twin < coherent phase boundary < semi-coherent < low-angle boundary < high-angle boundary ≲ incoherent phase boundary < free surface, as a broad ordering of typical values, not a law for every material.

Surface defects at a glance
DefectWhat changes across itEnergy
Free surfaceCrystal to vacuum; bonds brokenHighest
High-angle grain boundaryOrientation, same phaseHigh, nearly independent of angle
Low-angle boundarySmall tilt or twist, as dislocationsRises with angle
Coherent twinMirror image of the latticeLow
Stacking faultStacking sequence onlyLow in FCC metals, depends on the metal
Phase boundaryStructure or compositionCoherent < semi-coherent < incoherent

Key takeaways

  • Vacancies exist in equilibrium, n/N = exp(−Q/kT), with T in kelvin and k and Q in matching units; a Schottky or Frenkel pair has Q/2kT because it creates two defects. Self-interstitials are rare because their energy is high.
  • Frenkel pairs leave the density unchanged; Schottky defects lower it. Both preserve the charge balance and the formula.
  • An edge dislocation has b ⟂ line, a screw has b ∥ line, and b is the closure failure of the Burgers circuit. b is constant along a line, and at a node Σb in equals Σb out.
  • Line energy ≈ αGb² per unit length, so stable dislocations have the shortest lattice vector: a⟨100⟩ (SC), (a/2)⟨111⟩ (BCC), (a/2)⟨110⟩ (FCC); FCC splits into Shockley partials with |b| = a/√6.
  • Glide is conservative motion in the slip plane under shear stress; climb is non-conservative, edge-only, perpendicular to the slip plane and controlled by vacancy diffusion at high temperature.
  • Surface energy in the bond-breaking model is γ = (n_b ε/2) × planar density, lowest for the close-packed plane; grain boundaries, twins, stacking faults and phase boundaries are interfaces ranked by how much lattice registry they destroy.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. In an ionic crystal, which point defect leaves the density essentially unchanged while keeping the crystal electrically neutral?

    1. A Schottky defect, in which a cation vacancy and an anion vacancy form together and the ions go to the surface, leaving the lattice otherwise intact
    2. A single cation vacancy created alone, with the charge balance restored by the electrons in the conduction band
    3. A Frenkel defect, in which an ion leaves its site and occupies an interstitial void, creating a vacancy-interstitial pair
    4. A substitutional impurity of lower charge, which always shrinks the lattice and so raises the density of the crystal
    Show answer

    Answer: C — A Frenkel defect, in which an ion leaves its site and occupies an interstitial void, creating a vacancy-interstitial pair

    A Frenkel pair only relocates an ion inside the crystal, so the mass and the volume stay about the same. A Schottky defect carries ions to the surface and increases the number of lattice sites, so the density drops. An isolated cation vacancy would leave a net charge, and an impurity of lower charge needs compensating defects and does not always shrink the lattice.

  2. A crystal contains vacancies in thermodynamic equilibrium at every temperature above 0 K, but dislocations are never present in equilibrium. What is the reason?

    1. Vacancies cost little energy and gain entropy that outweighs it, whereas a dislocation costs energy along its whole length and gains too little entropy
    2. A vacancy is a zero-dimensional defect, and the second law of thermodynamics permits only zero-dimensional defects in equilibrium, forbidding every line defect in a closed system
    3. Dislocations are always generated by plastic deformation, so a crystal can never be free of them whatever its history, and annealing only rearranges them
    4. A vacancy lowers the internal energy of the crystal by relieving strain, while a dislocation always raises it by a larger amount, so vacancies are favoured at all temperatures
    Show answer

    Answer: A — Vacancies cost little energy and gain entropy that outweighs it, whereas a dislocation costs energy along its whole length and gains too little entropy

    The free energy G = nQ − TS has a minimum at a finite vacancy fraction because S grows quickly from n = 0. A dislocation costs about Gb² per unit length, which for a line through the crystal is vastly larger than the entropy it can add, so annealing removes dislocations. The distractor that vacancies lower the internal energy is wrong, since forming a vacancy costs energy Q.

  3. In an ionic solid, a Schottky pair has formation energy Q_S = 2.4 eV, and the equilibrium fraction of pairs is n_S/N = exp(−Q_S/2kT). Take k = 8.617 × 10⁻⁵ eV/K and T = 800 K. The fraction n_S/N is closest to

    1. 7.6 × 10⁻¹⁶, which is exp(−Q_S/kT), without the factor 2 for a pair
    2. 2.8 × 10⁻⁸, which is exp(−Q_S/2kT) with Q_S/2kT = 17.4
    3. 2.3 × 10⁻⁶, which results from using 800 as a Celsius temperature, that is 1073 K
    4. 5.8 × 10⁻³¹, which is exp(−2Q_S/kT), counting the energy of a pair twice
    Show answer

    Answer: B — 2.8 × 10⁻⁸, which is exp(−Q_S/2kT) with Q_S/2kT = 17.4

    Q_S/2kT = 2.4/(2 × 8.617 × 10⁻⁵ × 800) = 2.4/0.13787 = 17.41, so n_S/N = e⁻¹⁷·⁴¹ = 2.8 × 10⁻⁸. Dropping the factor 2 gives 7.6 × 10⁻¹⁶, using 1073 K gives 2.3 × 10⁻⁶ and doubling the energy gives 5.8 × 10⁻³¹, each a result of a distinct slip.

  4. A dislocation line is parallel to its Burgers vector everywhere along its length. Which statement is correct?

    1. It is an edge dislocation, which has a unique slip plane and can climb by absorbing vacancies at high temperature
    2. It is a mixed dislocation, with equal edge and screw components, and so it cannot glide on any plane containing the Burgers vector
    3. It is a pure edge dislocation whose extra half-plane lies parallel to the line direction, so it has no well-defined slip plane
    4. It is a screw dislocation, which has no unique slip plane and can cross-slip to another plane containing the line
    Show answer

    Answer: D — It is a screw dislocation, which has no unique slip plane and can cross-slip to another plane containing the line

    b ∥ line defines a screw dislocation. Since any plane containing the line also contains b, there is no unique slip plane, and the line can cross-slip. An edge dislocation has b ⟂ line, a mixed one has an intermediate angle, and only an edge dislocation (with its extra half-plane) can climb.

  5. An FCC metal has lattice parameter a = 0.404 nm. The magnitude of the Burgers vector of its stable (perfect) slip dislocation is closest to

    1. 0.202 nm, which is a/2
    2. 0.286 nm, which is a/√2
    3. 0.404 nm, which is a
    4. 0.350 nm, which is √3 a/2
    Show answer

    Answer: B — 0.286 nm, which is a/√2

    In FCC the stable dislocation has b = (a/2)⟨110⟩, the shortest lattice vector, so |b| = (a/2)√2 = a/√2 = 0.404/1.414 = 0.286 nm. The value √3 a/2 belongs to BCC, a/2 is only half a cube edge and is not a lattice vector of FCC, and a is a longer lattice vector of higher energy.

  6. In an FCC crystal a dislocation of Burgers vector (a/2)[1 0 1] runs into a node and splits into two dislocations leaving the node. One of them has Burgers vector (a/2)[0 1 1]. The Burgers vector of the other is

    1. (a/2)[1 −1 0]
    2. (a/2)[1 1 2]
    3. (a/2)[1 1 0]
    4. (a/2)[0 −1 1]
    Show answer

    Answer: A — (a/2)[1 −1 0]

    Frank’s rule, Σb in = Σb out, gives b₃ = b₁ − b₂ = (a/2)[1 − 0, 0 − 1, 1 − 1] = (a/2)[1 −1 0], which is also a stable FCC Burgers vector. Adding the two vectors gives (a/2)[1 1 2], which is the sum, not the difference, and is longer than the stable vectors; (a/2)[1 1 0] and (a/2)[0 −1 1] do not balance the vectors.

  7. In an FCC crystal, which of the following dislocation reactions is energetically favourable under the criterion that energy is proportional to |b|² and the Burgers vectors must balance?

    1. (a/2)[1 1 0] → (a/2)[1 0 1] + (a/2)[0 1 −1], since the vectors add up correctly
    2. a[1 0 0] → (a/2)[1 1 0] + (a/2)[1 −1 0], since it splits into two vectors that are each shorter than a
    3. (a/2)[1 −1 0] → (a/6)[2 −1 −1] + (a/6)[1 −2 1], with |b|² falling from a²/2 to a²/3
    4. (a/2)[1 −1 0] → (a/6)[2 −1 −1] + (a/6)[1 −2 −1], since both partials are short
    Show answer

    Answer: C — (a/2)[1 −1 0] → (a/6)[2 −1 −1] + (a/6)[1 −2 1], with |b|² falling from a²/2 to a²/3

    The Shockley reaction balances the vectors, (2 + 1, −1 − 2, −1 + 1)/6 = (3, −3, 0)/6, and lowers Σ|b|² from a²/2 to a²/6 + a²/6 = a²/3. The first reaction balances but raises Σ|b|² from a²/2 to a². The a[1 0 0] reaction balances but leaves Σ|b|² unchanged at a², so there is no energy gain. The last does not balance, because (3, −3, −2)/6 is not (1/2, −1/2, 0).

  8. Which statement about dislocation motion is correct?

    1. Glide of an edge dislocation requires vacancy diffusion, which is why slip stops at low temperature, as for creep
    2. Climb is conservative, since the extra half-plane merely shifts sideways with no atoms added or removed and no change in its length
    3. Screw dislocations climb readily at room temperature by absorbing interstitials at their extra half-plane, as edge dislocations do
    4. Climb of an edge dislocation is non-conservative and needs vacancy diffusion, so it matters mainly at high temperature
    Show answer

    Answer: D — Climb of an edge dislocation is non-conservative and needs vacancy diffusion, so it matters mainly at high temperature

    Climb moves an edge dislocation out of its slip plane by adding or removing atoms at the end of the extra half-plane, which needs vacancies to diffuse and is thermally activated. Glide is conservative and needs no diffusion, and a screw dislocation has no extra half-plane and so cannot climb at all.

  9. In the simple bond-breaking model of an FCC crystal, only nearest-neighbour bonds count. The numbers of broken bonds per surface atom are 3, 4 and 5, and the planar densities are 2.31/a², 2.00/a² and 1.41/a², for the (111), (100) and (110) surfaces respectively. Which surface has the lowest and which the highest surface energy?

    1. Lowest (111), highest (100)
    2. Lowest (111), highest (110), because it loses the most bonds per atom
    3. Lowest (100), highest (110), because the cube face is the simplest plane
    4. Lowest (110), highest (111), because the close-packed plane has the densest atoms
    Show answer

    Answer: A — Lowest (111), highest (100)

    γ ∝ n_b × planar density: (111) gives 3 × 2.31 = 6.93, (100) gives 4 × 2.00 = 8.00 and (110) gives 5 × 1.41 = 7.07, in units of ε/(2a²). So (111) is the lowest and (100) the highest. The (110) surface loses the most bonds per atom, which is the tempting error, but it has fewer atoms per unit area, which offsets the loss.

  10. A low-angle tilt boundary is a wall of edge dislocations with Burgers vector b, spaced D = b/θ for a small misorientation θ in radians. If the misorientation of such a boundary is doubled, the spacing of its dislocations becomes

    1. four times as large, because the energy of the boundary scales with θ²
    2. twice as large, because a bigger angle needs more room for the extra half-planes
    3. half as large, because D = b/θ varies inversely with θ
    4. unchanged, because the Burgers vector alone fixes the spacing
    Show answer

    Answer: C — half as large, because D = b/θ varies inversely with θ

    D = b/θ, so doubling θ halves D: the larger the misorientation, the more densely the dislocations pack. The spacing does not grow with θ, depends on θ and not only on b, and has no θ² scaling.

  11. Which of the following statements about point defects are correct?

    1. A Frenkel defect consists of a vacancy and an interstitial of the same ion, created as a pair
    2. A Schottky defect increases the density of the crystal because ions are packed into extra interstitial sites
    3. A Schottky defect in NaCl is a single isolated Na⁺ vacancy that leaves the crystal with a net negative charge
    4. In a metal at equilibrium the self-interstitial concentration is far below the vacancy concentration, because the formation energy of a self-interstitial is much higher
    Show answer

    Answer: A — A Frenkel defect consists of a vacancy and an interstitial of the same ion, created as a pair; D — In a metal at equilibrium the self-interstitial concentration is far below the vacancy concentration, because the formation energy of a self-interstitial is much higher

    Frenkel defects are vacancy-interstitial pairs, and self-interstitials cost several times the energy of vacancies, so they are rare at equilibrium. A Schottky defect removes ions to the surface and so lowers the density, and in NaCl it is a pair of cation and anion vacancies, which is neutral, not a single charged vacancy.

  12. Which of the following statements about dislocations are correct?

    1. In an edge dislocation the Burgers vector is perpendicular to the dislocation line
    2. A screw dislocation can climb out of its slip plane by absorbing vacancies, just as an edge dislocation does
    3. A dislocation that curves in its slip plane keeps the same Burgers vector everywhere along the line, although its edge and screw character change
    4. A dislocation line can terminate abruptly in the middle of a perfect crystal without meeting a surface, a boundary or another dislocation
    Show answer

    Answer: A — In an edge dislocation the Burgers vector is perpendicular to the dislocation line; C — A dislocation that curves in its slip plane keeps the same Burgers vector everywhere along the line, although its edge and screw character change

    b ⟂ line defines the edge type, and b is constant along a line while the angle between b and the line varies on a curved line. A screw dislocation has no extra half-plane and cannot climb. A dislocation cannot end inside a crystal: it closes in a loop, leaves through a surface or boundary, or meets others at a node.

  13. Which of the following statements about surface defects are correct?

    1. An intrinsic stacking fault in FCC, made by removing one close-packed layer, creates a thin region of HCP-type stacking
    2. A coherent twin boundary has a much lower energy than a typical high-angle grain boundary, because nearest-neighbour bonds are preserved
    3. An incoherent phase boundary has a lower energy than a coherent one, because the lattices need not match
    4. A low-angle tilt boundary is an array of edge dislocations
    Show answer

    Answer: A — An intrinsic stacking fault in FCC, made by removing one close-packed layer, creates a thin region of HCP-type stacking; B — A coherent twin boundary has a much lower energy than a typical high-angle grain boundary, because nearest-neighbour bonds are preserved; D — A low-angle tilt boundary is an array of edge dislocations

    Removing a layer from ABCABCABC gives ABCABABC…, with locally HCP-type stacking. A coherent twin keeps all nearest-neighbour bonds, so it costs far less than a high-angle boundary, and a tilt boundary is a wall of edge dislocations. An incoherent phase boundary has the highest energy of the phase boundaries, because the lattices do not match at all.

  14. The formation energy of a vacancy in a metal is Q = 0.90 eV. Take k = 8.617 × 10⁻⁵ eV/K and use n/N = exp(−Q/kT). Find the equilibrium vacancy fraction at T = 1000 K, expressed in units of 10⁻⁵ (for example, an answer of 3.00 means 3.00 × 10⁻⁵), to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 2.91

    Q/kT = 0.90/(8.617 × 10⁻⁵ × 1000) = 0.90/0.08617 = 10.444. n/N = e⁻¹⁰·⁴⁴⁴ = 2.911 × 10⁻⁵, so the answer is 2.91 in units of 10⁻⁵. Using k = 1.38 × 10⁻²³ with Q in eV would make the exponent absurdly large.

  15. A metal has vacancy formation energy Q = 1.0 eV, with k = 8.617 × 10⁻⁵ eV/K. By what factor does the equilibrium vacancy concentration increase when the temperature is raised from 900 K to 1200 K? Give the ratio rounded to the nearest integer.

    Numerical answer — type the value.

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    Answer: 25

    The ratio is exp[(Q/k)(1/T₁ − 1/T₂)] with Q/k = 1.0/(8.617 × 10⁻⁵) = 11 605 K and (1/900 − 1/1200) = 2.7778 × 10⁻⁴ K⁻¹. The exponent is 3.2236 and e³·²²³⁶ = 25.1, so the answer is 25.

  16. Iron at room temperature is BCC with lattice parameter a = 0.2866 nm. The stable dislocation has Burgers vector (a/2)⟨111⟩. Find the magnitude of its Burgers vector in nm, to three decimal places.

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    Answer: 0.248

    |b| = (a/2)√(1² + 1² + 1²) = (√3/2)a = 0.8660 × 0.2866 = 0.2482 nm, which is half of the body diagonal and equals the nearest-neighbour distance in BCC. Using a itself would give 0.287.

  17. An FCC metal has a = 0.3615 nm and shear modulus G = 48 GPa. Take the stable dislocation with b = a/√2 and the estimate of its line energy per unit length E = Gb²/2. Find E in units of 10⁻⁹ J/m, to two decimal places.

    Numerical answer — type the value.

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    Answer: 1.57

    b = 0.3615/√2 = 0.2556 nm = 2.556 × 10⁻¹⁰ m, so b² = 6.534 × 10⁻²⁰ m². E = 0.5 × 48 × 10⁹ × 6.534 × 10⁻²⁰ = 1.568 × 10⁻⁹ J/m, so the answer is 1.57. Using b = a would overestimate E by a factor of 2.

  18. A model FCC solid has a = 0.400 nm. In the simple bond-breaking model each nearest-neighbour bond has energy ε = 0.30 eV, and a surface atom on the (111) face has lost 3 bonds; each surface is charged half of the energy of each broken bond. Use the planar density 4/(√3 a²) of the (111) plane and 1 eV/nm² = 0.1602 J/m². Find the surface energy in J/m², to two decimal places.

    Numerical answer — type the value.

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    Answer: 1.04

    Energy per surface atom = 3 × 0.30/2 = 0.45 eV. Planar density = 4/(1.7321 × 0.1600) = 14.43 atoms/nm². γ = 0.45 × 14.43 = 6.495 eV/nm² = 6.495 × 0.1602 = 1.04 J/m². Leaving out the factor ½ would give 2.08.