Solid State Physics II: Semiconductor Statistics, Metal–Semiconductor Contacts, Dielectrics and Ferroelectrics, Magnetism and Superconductivity
1. Electron and hole statistics, mobility and effective mass
With the Fermi level well inside the gap, n = N_c e−(E_c − E_F)/k_BT and p = N_v e−(E_F − E_v)/k_BT, where N_c and N_v are effective densities of states that grow as T3/2 and with the effective masses. Their product is independent of E_F — the law of mass action, np = nᵢ² = N_cN_v e−E_g/k_BT. In an intrinsic semiconductor n = p = nᵢ and E_F lies near mid-gap (displaced by (3/4)k_BT ln(m_h/m_e)). In an extrinsic n-type sample with donors N_D ≫ nᵢ, n ≈ N_D and p = nᵢ²/N_D; for silicon with nᵢ = 1.5 × 10¹⁶ m⁻³ and N_D = 10²² m⁻³, p = 2.25 × 10¹⁰ m⁻³. E_F moves towards the band of the majority carrier.
- Conductivity σ = e(nμₙ + pμ_p), with mobility μ = eτ/m* (drift velocity per unit field). N_D = 10²² m⁻³ with μₙ = 0.135 m²/V s gives σ = 216 S/m.
- Temperature dependence of an extrinsic sample: freeze-out at low T (donors not ionised), a saturation range where n ≈ N_D, and an intrinsic range at high T where nᵢ overtakes N_D and ln σ vs 1/T has slope −E_g/(2k_B).
- Effective mass m* = ħ²/(d²E/dk²) is read from cyclotron resonance, ω_c = eB/m*; in a semiconductor the Hall sign distinguishes n- from p-type.
2. Metal–semiconductor junctions: ohmic and rectifying contacts
When a metal of work function φ_m touches an n-type semiconductor of electron affinity χ and work function φ_s, electrons flow until the Fermi levels align. If φ_m > φ_s, electrons leave the semiconductor, a depletion layer forms and the bands bend up: a Schottky barrier of height φ_B = φ_m − χ (5.1 − 4.05 = 1.05 eV for gold-like φ_m on silicon). It rectifies, with I = I_s(eeV/k_BT − 1), and since conduction is by majority carriers there is no minority storage — Schottky diodes switch fast and drop about 0.3 V.
If φ_m < φ_s on n-type, electrons accumulate at the interface and there is no barrier to them: the contact is ohmic, linear and symmetric. The conditions reverse for p-type. In practice ohmic contacts are made by doping the semiconductor heavily under the metal, which makes any barrier so thin that electrons tunnel through it.
3. Dielectric properties, polarisability and ferroelectricity
- Electronic polarisability (electron cloud against nucleus) responds up to optical frequencies; ionic polarisability (ions displaced) up to the infrared; orientational polarisability of permanent dipoles, α = p²/(3k_BT) (Langevin–Debye), only up to microwaves, and it falls as 1/T.
- Local field. In a cubic or isotropic solid each dipole feels E_loc = E + P/(3ε₀), which gives the Clausius–Mossotti relation (ε_r − 1)/(ε_r + 2) = Nα/(3ε₀), linking the macroscopic ε_r to the molecular α.
- Ferroelectrics (BaTiO₃) have a spontaneous polarisation below a Curie temperature, reversible by a field, with a P–E hysteresis loop and domains; above T_c the permittivity follows a Curie–Weiss law ε ∝ 1/(T − T_c). All ferroelectrics are piezoelectric.
4. Dia-, para-, ferro-, antiferro- and ferrimagnetism, and domains
| Kind | Susceptibility χ | Origin |
|---|---|---|
| Diamagnetism | small, negative, T-independent (about −10⁻⁵) | Larmor precession of all electron orbits (Lenz) |
| Paramagnetism | positive, Curie χ = C/T | permanent moments aligning against thermal disorder |
| Pauli paramagnetism | small, positive, T-independent | conduction electrons near E_F |
| Ferromagnetism | spontaneous M below T_C; above it χ = C/(T − θ) | exchange favouring parallel spins |
| Antiferromagnetism | maximum at the Néel temperature T_N; above it χ = C/(T + θ) | exchange favouring antiparallel equal sublattices |
| Ferrimagnetism | spontaneous M below T_C | antiparallel unequal sublattices (ferrites, magnetite) |
The exchange interaction that orders a ferromagnet is quantum-mechanical and electrostatic, far stronger than dipole–dipole forces. A bulk ferromagnet nevertheless breaks into domains magnetised in different directions, because that reduces the magnetostatic (stray-field) energy, at the cost of domain-wall energy (a Bloch wall turns the spins gradually over many atoms, balancing exchange against anisotropy). Magnetisation proceeds by wall motion at low fields and rotation at high fields; pinning of walls gives the hysteresis loop, with its remanence and coercivity.
5. Superconductivity
Below a critical temperature T_c the resistance vanishes, and — more fundamentally — a weak field is expelled: the Meissner effect, B = 0 inside, χ = −1, a perfect diamagnet (a perfect conductor would only freeze the field in). A field above the critical field destroys superconductivity, with H_c(T) ≈ H_c(0)[1 − (T/T_c)²]; at T = T_c/2 that is 0.75H_c(0). Type-I superconductors (most pure metals) expel the field completely up to H_c. Type-II superconductors (alloys, Nb, the cuprates) expel it fully only below H_c1; between H_c1 and H_c2 the field enters as a lattice of vortices, each carrying one flux quantum, and superconductivity survives to fields far above a type-I H_c. The dividing line is the Ginzburg–Landau parameter κ = λ/ξ at 1/√2.
- London equation ∇ × J_s = −(n_se²/m)B, with Ampère’s law, gives ∇²B = B/λ_L²: the field decays into the surface over the penetration depth λ_L = √(m/(μ₀n_se²)) — about 53 nm for n_s = 10²⁸ m⁻³.
- BCS theory: a phonon-mediated attraction binds electrons of opposite momentum and spin into Cooper pairs, which condense into one coherent state separated from excitations by a gap; at T = 0, 2Δ = 3.52 k_BT_c (2.79 meV for Nb, T_c = 9.2 K). The phonon origin shows in the isotope effect, T_c ∝ M−1/2.
- Flux quantisation: the flux through a superconducting ring is an integer multiple of Φ₀ = h/(2e) = 2.07 × 10⁻¹⁵ Wb; the 2e is the charge of a Cooper pair and was the first direct evidence for pairing.
Key takeaways
- np = nᵢ² whatever the doping; n-type with N_D ≫ nᵢ has n ≈ N_D and p = nᵢ²/N_D; σ = e(nμₙ + pμ_p).
- On n-type, φ_m > φ_s makes a Schottky barrier φ_m − χ that rectifies; φ_m < φ_s, or heavy doping, makes an ohmic contact.
- (ε_r − 1)/(ε_r + 2) = Nα/(3ε₀); orientational α = p²/(3k_BT); ferroelectrics show hysteresis and Curie–Weiss above T_c.
- Dia χ < 0 and T-independent; para χ = C/T; ferro C/(T − θ) above T_C; antiferro peaks at T_N; ferri has unequal antiparallel sublattices.
- Meissner χ = −1; λ_L = √(m/(μ₀n_se²)); 2Δ = 3.52k_BT_c; T_c ∝ M−1/2; Φ₀ = h/2e.
Practice questions (15)
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.
Silicon at 300 K has nᵢ = 1.5 × 10¹⁶ m⁻³ and is doped with 10²² donors per m³, all ionised. The hole concentration, in units of 10¹⁰ m⁻³, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.25
n ≈ N_D = 10²² m⁻³, and the law of mass action gives p = nᵢ²/n = 2.25 × 10³²/10²² = 2.25 × 10¹⁰ m⁻³. Taking p = nᵢ ignores the doping, and p = nᵢ/N_D is dimensionally wrong.An n-type sample has n = 10²² m⁻³ electrons with mobility 0.135 m² V⁻¹ s⁻¹; the hole contribution is negligible. Its conductivity, in S/m, to the nearest integer, is ____.
Numerical answer — type the value.
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Answer: 216
σ = neμₙ = 10²² × 1.602 × 10⁻¹⁹ × 0.135 = 216 S/m, a resistivity of 4.6 mΩ m. Using μ in cm² V⁻¹ s⁻¹ (1350) without converting gives a value 10⁴ times too large.A metal of work function 5.10 eV makes contact with n-type silicon of electron affinity 4.05 eV. The ideal Schottky barrier height, in eV, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.05
φ_B = φ_m − χ = 5.10 − 4.05 = 1.05 eV, the barrier electrons in the metal see. The built-in potential on the semiconductor side is φ_m − φ_s, which is smaller by E_c − E_F; real barriers are also modified by surface states.For an n-type semiconductor of work function φ_s, an ideal metal contact is ohmic when:
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Answer: A — φ_m < φ_s
With φ_m < φ_s, electrons flow into the semiconductor, accumulate at the interface, and meet no barrier in either direction. φ_m > φ_s forms a depletion layer and a rectifying Schottky barrier. Heavy, not light, doping is the practical route to an ohmic contact, by making the barrier thin enough to tunnel.The Clausius–Mossotti relation connects the relative permittivity ε_r of a cubic solid with N molecules per unit volume of polarisability α as:
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Answer: A — (ε_r − 1)/(ε_r + 2) = Nα/(3ε₀)
The Lorentz local field E + P/(3ε₀) with P = Nα E_loc gives (ε_r − 1)/(ε_r + 2) = Nα/(3ε₀). ε_r − 1 = Nα/ε₀ is the dilute-gas limit that ignores the local field; it agrees only when Nα/ε₀ ≪ 1.Which contribution to the polarisability of a polar dielectric decreases as 1/T?
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Answer: A — Orientational
Aligning permanent dipoles competes with thermal agitation, giving the Langevin–Debye α_or = p²/(3k_BT). The electronic and ionic contributions are displacements against restoring forces and are nearly independent of T.Which statements about magnetic materials are correct?
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Answer: A — A diamagnet has a small negative susceptibility that is nearly independent of temperature; B — An ideal paramagnet obeys χ = C/T; C — The susceptibility of an antiferromagnet has a maximum at the Néel temperature
Diamagnetism comes from induced orbital currents (χ ≈ −10⁻⁵, T-independent); Curie’s law is χ = C/T; an antiferromagnet’s χ rises to a cusp at T_N and falls below it. In a ferrimagnet the antiparallel sublattices are unequal, so a net moment remains — that is what distinguishes it from an antiferromagnet.A bulk ferromagnet divides into domains mainly because this:
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Answer: A — Lowers the magnetostatic (stray-field) energy
A single domain drives a large external field whose energy is stored in space; closing the flux with domains removes most of it. Domain walls cost exchange and anisotropy energy, so the domain size is set by that balance — the walls raise, not lower, the exchange energy.A type-I superconductor has a critical field of 0.080 T at 0 K. Using H_c(T) = H_c(0)[1 − (T/T_c)²], its critical field at T = T_c/2, in T, is ____.
Numerical answer — type the value.
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Answer: 0.06
H_c = 0.080 × (1 − 0.25) = 0.060 T. A linear dependence would give 0.040 T; the parabolic law is the empirical form BCS reproduces closely.Niobium has T_c = 9.2 K. Using the BCS relation 2Δ = 3.52 k_BT_c with k_B = 1.381 × 10⁻²³ J/K and e = 1.602 × 10⁻¹⁹ C, the energy gap 2Δ at T = 0, in meV, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.79
k_BT_c = 1.381 × 10⁻²³ × 9.2/1.602 × 10⁻¹⁹ = 0.793 meV, so 2Δ = 3.52 × 0.793 = 2.79 meV. Δ itself is 1.40 meV; taking 3.52 k_BT_c as Δ doubles the gap to 5.58 meV.Mercury of isotopic mass 199.5 has T_c = 4.185 K. Assuming the isotope effect T_c ∝ M−1/2, T_c for isotopic mass 203.4, in K, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 4.14
T_c′ = 4.185 × √(199.5/203.4) = 4.185 × 0.99037 = 4.145 K → 4.14 (strictly 4.1447). A heavier isotope has lower phonon frequencies and a lower T_c; inverting the ratio gives 4.23 K, which moves the wrong way.A superconductor has a superconducting electron density of 1.0 × 10²⁸ m⁻³. With m = 9.109 × 10⁻³¹ kg, e = 1.602 × 10⁻¹⁹ C and μ₀ = 4π × 10⁻⁷ H/m, the London penetration depth, in nm, to the nearest integer, is ____.
Numerical answer — type the value.
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Answer: 53
λ_L = √(m/(μ₀n_se²)) = √(9.109 × 10⁻³¹/(1.2566 × 10⁻⁶ × 10²⁸ × 2.566 × 10⁻³⁸)) = √(2.825 × 10⁻¹⁵) = 5.31 × 10⁻⁸ m = 53 nm. Forgetting the square root gives a meaningless 2.8 × 10⁻¹⁵ m.The superconducting flux quantum h/(2e), with h = 6.626 × 10⁻³⁴ J s and e = 1.602 × 10⁻¹⁹ C, in units of 10⁻¹⁵ Wb, to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.07
Φ₀ = 6.626 × 10⁻³⁴/(2 × 1.602 × 10⁻¹⁹) = 2.07 × 10⁻¹⁵ Wb. The charge 2e of the Cooper pair halves the single-electron value h/e = 4.14 × 10⁻¹⁵ Wb, and measuring it was the direct proof of pairing.Which statements about type-II superconductors are correct?
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Answer: A — They have two critical fields, H_c1 and H_c2; B — Between H_c1 and H_c2 the field penetrates as vortices, each carrying one flux quantum; C — Their Ginzburg–Landau parameter κ exceeds 1/√2
Below H_c1 the Meissner state is complete; between H_c1 and H_c2 the mixed (vortex) state lets flux in quantum by quantum; κ = λ/ξ > 1/√2 makes the wall energy negative and favours this. Above H_c2 the vortex cores overlap and the bulk is normal.The Meissner effect means that a superconductor in a weak applied field has:
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Answer: A — B = 0 in its interior, susceptibility −1
A superconductor expels flux even when cooled in a field, so it is a perfect diamagnet, χ = −1. A merely perfect conductor would keep whatever flux it had when it lost resistance — which is why the Meissner effect, not zero resistance, is the defining property.