Solid State Physics I: Crystallography, Diffraction, Bonding, Lattice Vibrations, Free Electrons and Band Theory
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.