Solid State Physics I: Crystallography, Diffraction, Bonding, Lattice Vibrations, Free Electrons and Band Theory
1. Elements of crystallography
A crystal is a lattice (a set of points with identical surroundings) plus a basis (the atoms attached to each point). In three dimensions there are 7 crystal systems and 14 Bravais lattices. Planes are named by Miller indices (hkl): take the intercepts in units of the lattice constants, invert, clear fractions. In a cubic lattice the spacing of (hkl) planes is d = a/√(h² + k² + l²) — for bcc iron (a = 2.87 Å), d₁₁₀ = 2.03 Å.
| Lattice | Atoms per cell | Coordination | Packing fraction |
|---|---|---|---|
| Simple cubic | 1 | 6 | π/6 = 0.52 |
| Body-centred (bcc) | 2 | 8 | π√3/8 = 0.68 |
| Face-centred (fcc) | 4 | 12 | π/(3√2) = 0.74 |
The reciprocal lattice is the set of vectors G with eiG·R = 1 for every lattice vector R; its primitive vectors are b₁ = 2π(a₂ × a₃)/[a₁·(a₂ × a₃)] and cyclically. Each G is normal to a family of planes, with |G| = 2π/d. The reciprocal of fcc is bcc and vice versa. The Wigner–Seitz cell of the reciprocal lattice is the first Brillouin zone, the natural home of wavevectors in the crystal.
2. Diffraction methods for structure determination
Waves of wavelength comparable to the spacing — X-rays, electrons, neutrons — reflect constructively from a family of planes when 2d sin θ = nλ (Bragg), equivalently when the scattering vector equals a reciprocal-lattice vector, k′ − k = G (Laue). Worked: Cu Kα (1.54 Å) on the (111) planes of copper (fcc, a = 3.61 Å): d = 3.61/√3 = 2.084 Å, sin θ = 1.54/4.168 = 0.3695, θ = 21.7°. X-rays scatter from electron density; neutrons from nuclei (and magnetic moments, which reveal antiferromagnetic order); electrons interact strongly and suit surfaces and thin films.
The intensity of a reflection is |S_hkl|², with the structure factor S = Σⱼ fⱼ e−2πi(hxⱼ + kyⱼ + lzⱼ) summed over the basis. For a conventional bcc cell (atoms at 000 and ½½½), S = f[1 + (−1)h+k+l], so reflections appear only for h + k + l even: (110), (200), (211), (220). For fcc (000, ½½0, ½0½, 0½½), S ≠ 0 only when h, k, l are all even or all odd: (111), (200), (220), (311). These extinctions identify the lattice from a powder pattern.
3. Bonding in solids
- Ionic (NaCl): electron transfer and Coulomb attraction, summed over the lattice by the Madelung constant (1.748 for NaCl) and balanced by short-range repulsion; hard, brittle, insulating, high melting point.
- Covalent (diamond, Si): shared electron pairs in directional sp³ bonds, giving the open diamond structure (coordination 4, packing fraction 0.34).
- Metallic: delocalised electrons in a sea round the ion cores; non-directional, hence close-packed structures, ductility and conduction.
- Van der Waals (solid Ar): fluctuating-dipole attraction ∝ −1/r⁶ with repulsion ∝ 1/r¹², the Lennard-Jones potential 4ε[(σ/r)¹² − (σ/r)⁶], minimum at r = 21/6σ; weak, low melting points. Hydrogen bonds (ice) are intermediate.
4. Lattice vibrations and thermal properties
A chain of atoms of mass m joined by springs K, spacing a, has the dispersion ω = 2√(K/m)|sin(ka/2)|: linear (sound) at small k, flat at the zone boundary k = π/a, where the group velocity is zero and the waves are standing. For K = 10 N/m and m = 10⁻²⁶ kg the maximum is 6.32 × 10¹³ rad/s. A chain with two different masses per cell has an acoustic branch (neighbours in phase, ω → 0 as k → 0) and an optical branch (neighbours out of phase, finite ω at k = 0, which an ionic crystal couples to infrared light), separated by a gap at the zone boundary. A crystal with p atoms per primitive cell has 3 acoustic and 3p − 3 optical branches. The quanta are phonons, with energy ħω.
| Model | High T | Low T |
|---|---|---|
| Classical (Dulong–Petit) | 3R = 24.9 J mol⁻¹ K⁻¹ | 3R — wrong, violates the third law |
| Einstein (one frequency ω_E) | → 3R for T ≫ Θ_E | ∝ (Θ_E/T)² e−Θ_E/T, falls too fast |
| Debye (ω ∝ k up to ω_D) | → 3R for T ≫ Θ_D | ∝ T³, as measured |
5. Free-electron theory, band theory and the Hall effect
The Drude model gives σ = ne²τ/m and a Hall coefficient R_H = −1/(ne). The Sommerfeld model fills plane-wave states to the Fermi energy E_F = (ħ²/2m)(3π²n)2/3 (3.24 eV for sodium, n = 2.65 × 10²⁸ m⁻³), with density of states g(E) ∝ √E; only electrons within k_BT of E_F respond, which explains the small electronic heat capacity and the Wiedemann–Franz law κ/(σT) = L = (π²/3)(k_B/e)² = 2.44 × 10⁻⁸ W Ω K⁻².
In a periodic potential the states are Bloch waves, ψ = eik·ru_k(r) with u periodic. In the nearly-free-electron model a weak potential with Fourier component U_G mixes the degenerate free-electron states k and k − G at the zone boundary (k = π/a in one dimension); the two standing waves formed — one piling charge on the ions, one between them — differ in energy, opening a gap of 2|U_G|. At the boundary the group velocity dE/dk vanishes, and the effective mass m* = ħ²/(d²E/dk²) is small near a band edge, negative near the top of a band (the origin of holes).
- Metals, insulators, semiconductors: a band holds 2N electrons (N cells, two spins). A partly filled band conducts (metal); completely filled and empty bands separated by a large gap do not (insulator); a small gap (about 1 eV) lets thermal excitation create carriers (semiconductor). Divalent metals conduct because bands overlap.
- Hall effect: in a slab of thickness t carrying current I in a field B, V_H = IB/(net). Copper (n = 8.5 × 10²⁸ m⁻³) has R_H = −7.34 × 10⁻¹¹ m³/C; with I = 10 A, B = 1 T and t = 0.1 mm, V_H = 7.34 μV. The sign of R_H gives the sign of the carriers.
Key takeaways
- Cubic d = a/√(h² + k² + l²); sc/bcc/fcc have 1/2/4 atoms, coordination 6/8/12 and packing 0.52/0.68/0.74; fcc and bcc are reciprocal to each other.
- 2d sin θ = nλ; bcc reflects only h + k + l even, fcc only hkl unmixed.
- ω = 2√(K/m)|sin(ka/2)|; acoustic and optical branches; Debye C ∝ T³ at low T, every model 3R at high T.
- A weak periodic potential opens a gap 2|U_G| at the zone boundary, where v_g = 0; m* = ħ²/(d²E/dk²) can be negative.
- R_H = −1/(ne) and V_H = IB/(net); Wiedemann–Franz κ/(σT) = (π²/3)(k_B/e)².
Practice questions (14)
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.
X-rays of wavelength 1.54 Å are diffracted in first order by the (111) planes of copper, which is fcc with a = 3.61 Å. The Bragg angle θ, in degrees, to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 21.7
d₁₁₁ = 3.61/√3 = 2.084 Å; sin θ = λ/(2d) = 1.54/4.168 = 0.3695, θ = 21.7°. Using a itself as d gives 12.3°, and quoting the scattering angle 2θ gives 43.4°.For a body-centred cubic crystal (conventional cubic cell), which reflections are allowed?
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Answer: A — (110); C — (200); D — (211)
S = f[1 + (−1)h+k+l] vanishes when h + k + l is odd. (110), (200) and (211) have even sums (2, 2, 4) and appear; (111) has sum 3 and is extinct. (111) is the first fcc reflection, so its absence separates bcc from fcc.The packing fraction of a body-centred cubic structure of touching hard spheres, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.68
Two atoms per cell touch along the body diagonal: 4r = √3a. Packing = 2 × (4/3)πr³/a³ = π√3/8 = 0.68. 0.74 is fcc (touching along a face diagonal) and 0.52 simple cubic.The reciprocal lattice of a face-centred cubic lattice is:
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Answer: A — Body-centred cubic
Building b₁ = 2π(a₂ × a₃)/V from the fcc primitive vectors (a/2)(0, 1, 1), etc., gives (2π/a)(−1, 1, 1) and its permutations — the primitive vectors of a bcc lattice of side 4π/a. The relation is mutual; only simple cubic is its own reciprocal type.Iron is body-centred cubic with a = 2.87 Å. The spacing of its (110) planes, in Å, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.03
d = a/√(h² + k² + l²) = 2.87/√2 = 2.03 Å. The formula depends only on the cubic cell edge; the bcc centring decides which reflections survive, not the spacing. Multiplying by √2 instead gives 4.06 Å.A monatomic chain has atoms of mass 1.0 × 10⁻²⁶ kg joined by springs of constant 10 N/m. Its maximum phonon angular frequency, in units of 10¹³ rad/s, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 6.32
ω_max = 2√(K/m) = 2√(10/10⁻²⁶) = 2 × 3.162 × 10¹³ = 6.32 × 10¹³ rad/s, at the zone boundary k = π/a. √(K/m) alone, 3.16, is the single-oscillator frequency.Well below its Debye temperature an insulator has a lattice heat capacity C at 10 K. The ratio C(20 K)/C(10 K) is ____.
Numerical answer — type the value.
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Answer: 8
The Debye T³ law gives (20/10)³ = 8. An insulator has no electronic term, so nothing linear in T spoils the ratio. Linear scaling would give 2, and the Einstein model would give an exponential ratio that is not observed.In the Einstein model of a solid, the lattice heat capacity at temperatures far below the Einstein temperature:
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Answer: A — Falls exponentially, faster than observed
With a single frequency, excitation needs k_BT comparable to ħω_E, so C ∝ (Θ_E/T)²e−Θ_E/T, which dies faster than experiment shows. Debye’s continuum of low-frequency acoustic modes gives T³; 3R is the high-temperature limit of both.Sodium has n = 2.65 × 10²⁸ conduction electrons per m³. Using E_F = (ħ²/2m)(3π²n)2/3 with h = 6.626 × 10⁻³⁴ J s, m_e = 9.109 × 10⁻³¹ kg and e = 1.602 × 10⁻¹⁹ C, the Fermi energy in eV, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 3.24
3π²n = 7.846 × 10²⁹ m⁻³, whose 2/3 power is 8.505 × 10¹⁹ m⁻²; ħ²/2m = 6.105 × 10⁻³⁹ J m²; E_F = 5.193 × 10⁻¹⁹ J = 3.24 eV. Forgetting the 2/3 power, or using h instead of ħ, gives values that are wrong by orders of magnitude.Copper has n = 8.5 × 10²⁸ free electrons per m³. With e = 1.602 × 10⁻¹⁹ C, its Hall coefficient in units of 10⁻¹¹ m³/C, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: -7.34
R_H = −1/(ne) = −1/(8.5 × 10²⁸ × 1.602 × 10⁻¹⁹) = −7.34 × 10⁻¹¹ m³/C. The sign is negative because the carriers are electrons; +7.34 is the answer for holes of the same density. Type the answer as -7.34.A copper strip 0.10 mm thick (n = 8.5 × 10²⁸ m⁻³) carries 10 A in a perpendicular field of 1.0 T. The magnitude of the Hall voltage, in μV, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 7.34
V_H = IB/(net) = 10 × 1.0/(8.5 × 10²⁸ × 1.602 × 10⁻¹⁹ × 1.0 × 10⁻⁴) = 7.34 × 10⁻⁶ V. The thickness is measured along B; using the width would give a different, wrong number. The small voltage is why Hall probes use semiconductors.In the nearly-free-electron model in one dimension, which statements are correct?
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Answer: A — An energy gap opens at k = ±π/a; B — The group velocity vanishes at the zone boundary; C — The gap equals 2|U_G|, twice the Fourier component of the potential
At k = π/a the states k and k − 2π/a are degenerate and mixed by U_G into two standing waves split by 2|U_G|; standing waves carry no current, so dE/dk = 0. Near the top of a band the curvature is negative, so m* is negative there — the reason holes are introduced.Which type of bonding gives a crystal that is hard, brittle, electrically insulating as a solid but conducting when molten?
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Answer: A — Ionic
In an ionic crystal the ions are locked in place, so it insulates; melting frees them to carry charge. Metals conduct as solids; van der Waals solids are soft; covalent networks such as diamond are hard but do not conduct when molten because they have no ions.According to the Wiedemann–Franz law, the ratio κ/(σT) for a metal is:
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Answer: A — (π²/3)(k_B/e)², the same for all metals
Both κ and σ are carried by the same electrons near E_F with the same τ, so n and τ cancel and κ/(σT) = (π²/3)(k_B/e)² = 2.44 × 10⁻⁸ W Ω K⁻². The Drude classical value has 3/2 in place of π²/3 and is wrong by a factor of about 2.