Electronic Properties: Drude and Quantum Free-Electron Models, Bands, Semiconductors, Hall Effect and Dielectrics

This chapter answers the second paragraph of Section XE2.3 (Properties and Applications of Materials) of XE2 Materials Science, one of the optional sections of the Engineering Sciences (XE) paper: “Electronic Properties: Drude model and classical description of electrical conductivity, Drawbacks of classical theory, Quantum mechanical description including concept of Fermi energy, Fermi surface and density of states. Band Theory to explain insulators, conductors, and semiconductors via allowed and forbidden energy bands, Effective mass concept, Intrinsic and extrinsic semiconductors, temperature dependence of conductivity, Carrier concentration and mobility, drift vs. diffusion current, Hall Effect for a simple metal or semiconductor. Dielectric behavior, piezo- and ferro-electric behavior.” It teaches the Drude result σ = ne²τ/m and where it fails, the Fermi gas, band theory and effective mass, intrinsic and doped semiconductors, drift and diffusion, the Hall coefficient and its sign, and dielectric, piezoelectric and ferroelectric response. The numericals are the Fermi energy of a metal, the Drude relaxation time, conductivity from n and μ, the temperature scaling of n_i and the Hall carrier density.

1. Drude model: classical electrical conduction and its failures

Drude model. Conduction electrons form a classical gas of density n that scatters off the lattice. Between collisions an electron accelerates in the field E; a collision resets its drift velocity to zero on average after a mean time τ (the relaxation time). Steady state: m v_d/τ = eE, so v_d = (eτ/m)E. The current density is J = n e v_d, giving σ = ne²τ/m = neμ, with mobility μ = eτ/m (units m²/V·s). The resistivity is ρ = 1/σ. Mobility is drift speed per unit field; it is not the same as speed.

Worked example (copper). Take n = 8.5 × 10²⁸ m⁻³, σ = 5.9 × 10⁷ S/m, m = 9.109 × 10⁻³¹ kg, e = 1.602 × 10⁻¹⁹ C. Then τ = σm/(ne²) = (5.9 × 10⁷ × 9.109 × 10⁻³¹)/(8.5 × 10²⁸ × 2.566 × 10⁻³⁸) = 2.46 × 10⁻¹⁴ s, and μ = eτ/m = 4.3 × 10⁻³ m²/V·s. For E = 1 V/m, v_d = μE ≈ 4 mm/s: electrons drift very slowly although the signal travels near light speed.

  • Specific heat: classical equipartition gives the electrons 3/2 R per mole of electrons on top of the lattice 3R; measured metals show almost no such extra heat capacity (the electronic part is about 1 percent of the classical value at room temperature).
  • Temperature dependence: using v ∝ √T, Drude predicts ρ ∝ √T, whereas pure metals show ρ ∝ T near room temperature.
  • Mean free path: the model assumes electrons collide with ion cores, giving a path of one atomic spacing; the real path at low temperature is thousands of spacings, so the picture of collisions with ions is wrong.
  • Others: it cannot explain insulators and semiconductors, positive Hall coefficients in some metals, or the Pauli (temperature-independent) paramagnetism of metals.
⚠️ Drift speed is not Fermi speed
Electrons in copper move at about 10⁶ m/s (the Fermi speed) in random directions at all times. The net drift in an applied field is only millimetres per second. Never use v_F in J = n e v_d.

2. Quantum free-electron gas: Fermi energy, Fermi surface, density of states

Electrons are fermions: by the Pauli principle each allowed wavevector state holds two electrons (spin up and down). At T = 0 the N electrons fill states from the bottom up to the Fermi energy E_F. In a cube of free electrons the states fill a sphere in k-space, the Fermi sphere, whose surface is the Fermi surface with radius k_F = (3π²n)^(1/3). Then E_F = ħ²k_F²/2m = (ħ²/2m)(3π²n)^(2/3), and the Fermi speed is v_F = ħk_F/m. E_F depends only on n, grows as n^(2/3), and for metals is a few eV (T_F = E_F/k_B ≈ 10⁴ to 10⁵ K).

Worked example. For n = 8.5 × 10²⁸ m⁻³ (copper), k_F = (3π² × 8.5 × 10²⁸)^(1/3) = 1.36 × 10¹⁰ m⁻¹. Then E_F = (1.055 × 10⁻³⁴)² × (1.36 × 10¹⁰)²/(2 × 9.109 × 10⁻³¹) = 1.13 × 10⁻¹⁸ J = 7.06 eV (using ħ = 1.055 × 10⁻³⁴ J·s, 1 eV = 1.602 × 10⁻¹⁹ J), and v_F = 1.58 × 10⁶ m/s, so T_F ≈ 8 × 10⁴ K. Room temperature (k_BT = 0.026 eV) is a tiny perturbation: the gas is degenerate.

Density of states. In three dimensions the number of states per unit energy per unit volume is g(E) = (1/2π²)(2m/ħ²)^(3/2) √E (spin included), so g ∝ √E. Integrating up to E_F gives n = (2/3) g(E_F) E_F, i.e. g(E_F) = 3n/(2E_F). At T > 0 the occupation is the Fermi-Dirac function f(E) = 1/[exp((E − E_F)/k_BT) + 1]; only electrons within about k_BT of E_F can change state. A fraction T/T_F of the electrons takes part in heat capacity, so C_el = γT is small (this repairs the specific-heat failure of Drude), and for conduction the relevant speed is v_F and the scattering time is that of the electrons at E_F.

🧠 Quick scaling
E_F ∝ n^(2/3). Doubling the electron density raises E_F by 2^(2/3) = 1.59. Keep E_F in eV by dividing joules by 1.602 × 10⁻¹⁹ and check it falls in the range of roughly 1 to 15 eV expected for metals.

3. Band theory: allowed and forbidden bands, insulators, conductors, semiconductors, effective mass

In a crystal the electron moves in a periodic potential. Bloch waves with wavevector k near a Brillouin-zone boundary (k = nπ/a in one dimension) are Bragg reflected, which opens an energy gap there. The result is allowed bands separated by forbidden bands; each band holds 2 electrons per primitive cell per band. Whether a solid conducts depends on how far the highest occupied band is filled, because only a partly filled band has empty states a field can use.

Classification by band filling at T = 0
ClassBand pictureConductivity trend with T
MetalHighest band partly filled, or bands overlap; Fermi level lies inside a bandFalls with T (more lattice scattering, μ falls)
InsulatorValence band full, conduction band empty, gap E_g large (about 4 eV or more)Negligible; rises steeply if heated
SemiconductorSame as insulator but E_g about 0.5 to 3 eV (Si 1.12, Ge 0.67, GaAs 1.42 eV at 300 K)Rises rapidly with T as carriers are thermally excited

Effective mass. Near a band edge E(k) is parabolic and the electron responds to forces as if it had mass m\* = ħ²/(d²E/dk²). A narrow, sharply curved band has a small m\* (light, mobile carriers); a flat band has a large m\*. Near the top of the valence band the curvature is negative, so the electrons there respond with negative mass; the empty state is described as a hole, a particle of charge +e and positive mass. Holes make p-type conduction and give positive Hall coefficients. Since μ = eτ/m\*, a smaller effective mass gives a higher mobility.

⚠️ A full band carries no current
Beryllium and magnesium have two valence electrons per atom and would fill a band, yet they conduct because the bands overlap. Do not conclude that an even number of valence electrons means an insulator without checking for overlap.

4. Intrinsic and extrinsic semiconductors: carriers, doping and temperature

In an intrinsic semiconductor electrons excited across the gap leave an equal number of holes, so n = p = n_i. With band-edge effective densities N_c and N_v, the law of mass action holds, np = n_i² = N_cN_v exp(−E_g/k_BT), and since N_c, N_v ∝ T^(3/2), n_i ∝ T^(3/2) exp(−E_g/2k_BT). The exponential dominates: for silicon (E_g = 1.12 eV, k_B = 8.617 × 10⁻⁵ eV/K) between 300 K and 360 K, n_i₂/n_i₁ = (360/300)^1.5 × exp[(1.12/(2 × 8.617 × 10⁻⁵))(1/300 − 1/360)] = 1.315 × 36.97 = 48.6. A 60 K rise multiplies n_i about 49 times.

Extrinsic semiconductors are doped. A donor (group V such as P or As in Si) adds an electron just below the conduction band (n-type, n ≈ N_D, and p = n_i²/N_D, the minority carrier). An acceptor (group III such as B) creates a hole just above the valence band (p-type, p ≈ N_A, n = n_i²/N_A). Conductivity is σ = e(nμ_e + pμ_h). Example: n-type silicon with N_D = 2.0 × 10²² m⁻³ and μ_e = 0.12 m²/V·s (holes negligible) gives σ = 1.602 × 10⁻¹⁹ × 2.0 × 10²² × 0.12 = 384 S/m.

Doped semiconductor over temperature
RegionCarriersσ behaviour
Freeze-out (low T)Dopants not fully ionised; n rises with TRises with T
Extrinsic or saturationn ≈ N_D fixed; lattice scattering lowers μNearly flat, slightly falling
Intrinsic (high T)n_i exceeds the doping; n = pRises steeply, ln σ against 1/T has slope −E_g/2k_B
🎯 Mobility has its own temperature law
Lattice (phonon) scattering gives μ ∝ T^(−3/2); ionised-impurity scattering gives μ ∝ T^(3/2). In a doped crystal μ therefore peaks at some temperature. In a metal n is fixed, so σ follows μ alone and falls with T, while in a semiconductor the exponential growth of n overrides the fall of μ.

5. Drift and diffusion current, Einstein relation, Hall effect

Carriers move by drift under a field and by diffusion down a concentration gradient. For electrons, the total current density is J_n = n e μ_n E + e D_n (dn/dx), and for holes J_p = p e μ_p E − e D_p (dp/dx), where D is the diffusion coefficient. The Einstein relation ties the two: D/μ = k_BT/e, equal to 0.0259 V at 300 K. For μ_e = 0.135 m²/V·s, D_e = 0.135 × 0.0259 = 3.5 × 10⁻³ m²/s. In a uniform doped bar there is only drift; at a p-n junction diffusion and drift currents balance in equilibrium so the net current is zero.

Hall effect. A strip of thickness t (along B) carries current I along x, with B along z. The magnetic force deflects the carriers sideways, charge piles up on the faces, and a transverse Hall voltage builds up until the electric force balances the magnetic force. For one carrier type of charge q and density n: V_H = IB/(n q t) and the Hall coefficient R_H = V_H t/(IB) = 1/(nq). For electrons R_H = −1/(ne) (negative); for holes R_H = +1/(pe) (positive). The sign of R_H therefore identifies the majority carrier, and |R_H| gives the density. The Hall mobility is μ_H = |R_H| σ. For copper R_H = −1/(8.5 × 10²⁸ × 1.602 × 10⁻¹⁹) = −7.3 × 10⁻¹¹ m³/C.

⚠️ Thickness, not width
In V_H = IB/(nqt) the t is the dimension along the magnetic field, not the width across which V_H is measured. Where both carriers matter, R_H = (pμ_h² − nμ_e²)/[e(pμ_h + nμ_e)²], so R_H can vanish or change sign as temperature or doping shifts the balance of n and p.

6. Dielectric behaviour and polarisation mechanisms

A dielectric has no free carriers; in a field E its charges shift and give a polarisation (dipole moment per volume) P = ε₀χE = ε₀(ε_r − 1)E. The displacement is D = ε₀E + P = ε₀ε_rE. For a parallel-plate capacitor C = ε_rε₀A/d, so the dielectric constant ε_r is the factor by which the dielectric raises the capacitance. Example (ε₀ = 8.854 × 10⁻¹² F/m): ε_r = 5 and E = 2 × 10⁶ V/m give P = 8.854 × 10⁻¹² × 4 × 2 × 10⁶ = 7.08 × 10⁻⁵ C/m². For a dilute gas of atoms of polarisability α and density N, ε_r − 1 ≈ Nα/ε₀; the Clausius-Mossotti form (ε_r − 1)/(ε_r + 2) = Nα/3ε₀ is used for dense non-polar solids.

Polarisation mechanisms, slowest to fastest
MechanismOriginFollows field up to about
Space-charge (interfacial)Charge trapped at grain boundaries and interfacesLow frequencies, up to audio range
Orientational (dipolar)Permanent dipoles (e.g. water) rotate toward the field; falls with TUp to microwave range
IonicCations and anions displaced in an ionic crystalInfrared, about 10¹² to 10¹³ Hz
ElectronicElectron cloud shifts relative to nucleus; present in all materialsOptical and ultraviolet, about 10¹⁵ Hz

As frequency rises ε_r drops in steps, losing the slowest mechanism first; each step comes with a peak in dielectric loss (tan δ = ε″/ε′). At optical frequency only the electronic mechanism is left, so ε_r(optical) = n², the square of the refractive index.

7. Piezoelectric and ferroelectric behaviour

A piezoelectric crystal develops charge (polarisation) under applied stress, and strains under an applied field (the converse effect). It requires a crystal structure without a centre of symmetry, so that mechanical displacement moves the centres of positive and negative charge relative to each other. Direct effect: P = dσ (d in C/N); converse: strain = dE (d in m/V). Examples: quartz, ZnO and lead zirconate titanate (PZT). Uses: sensors, ultrasonic transducers, igniters and quartz oscillators.

A ferroelectric is a piezoelectric (indeed a polar) crystal with a spontaneous polarisation that an applied field can reverse. Polarisation against field traces a hysteresis loop with remanent polarisation P_r and coercive field E_c. Domains of aligned dipoles form, as in ferromagnets. Above the Curie temperature T_C the crystal becomes paraelectric and its permittivity follows the Curie-Weiss law ε_r ≈ C/(T − T_C). Barium titanate is the standard example (T_C about 120 °C, with the Ti ion displaced off-centre in the oxygen octahedron below T_C). Ferroelectrics have very high ε_r near T_C and are used in capacitors and memories.

⚠️ Nesting of the classes
Every ferroelectric is piezoelectric, but not every piezoelectric is ferroelectric: quartz is piezoelectric yet its polarisation cannot be switched. A centrosymmetric crystal can be neither. Above T_C a ferroelectric loses its spontaneous polarisation and, if the phase has a centre of symmetry, its piezoelectricity too.

Key takeaways

  • Drude: σ = ne²τ/m = neμ with μ = eτ/m; drift speeds are mm/s while Fermi speeds are about 10⁶ m/s. Drude fails on electronic specific heat, ρ ∝ T, mean free path, insulators and positive Hall signs.
  • Free-electron gas: E_F = (ħ²/2m)(3π²n)^(2/3) ∝ n^(2/3), g(E) ∝ √E, g(E_F) = 3n/2E_F; only electrons within about k_BT of E_F matter, which is why metals have small electronic heat capacity.
  • Bands come from Bragg reflection of Bloch waves; a partly filled band conducts, a full band does not (overlap can rescue divalent metals). m\* = ħ²/(d²E/dk²); holes are the empty states of the valence band.
  • Semiconductors: n_i ∝ T^(3/2) exp(−E_g/2k_BT), np = n_i², σ = e(nμ_e + pμ_h); doping fixes the majority carrier, with freeze-out, saturation and intrinsic regions as T rises.
  • Current is drift plus diffusion, with D/μ = k_BT/e. Hall: R_H = 1/(nq) = V_Ht/(IB); negative for electrons, positive for holes, and t is the thickness along B.
  • Dielectrics: P = ε₀(ε_r − 1)E; mechanisms drop out with frequency (space-charge, orientational, ionic, electronic). Piezoelectricity needs no centre of symmetry; ferroelectrics add a switchable spontaneous polarisation below T_C.

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. Which of the following observations about metals is NOT explained by the classical Drude model of conduction?

    1. Current density proportional to the applied electric field in a metal wire, J = σE, with σ independent of E
    2. Conductivity rising in proportion to the density of conduction electrons, at a fixed relaxation time
    3. The very small share of conduction electrons in the heat capacity of a metal at room temperature
    4. Joule heating of a wire carrying current, as electrons give up their drift energy to the lattice in collisions
    Show answer

    Answer: C — The very small share of conduction electrons in the heat capacity of a metal at room temperature

    Drude gives Ohm’s law, σ ∝ n and Joule heating directly from collisions. But treating the electrons as a classical gas assigns them 3/2 k_B each, which would add visibly to the lattice heat capacity; measured metals show almost none. The Pauli principle and Fermi-Dirac statistics are needed to explain it.

  2. In the free-electron model of a metal at T = 0, the electron density is raised to twice its value (for example by applying pressure, with the electrons unchanged in number). By what factor does the Fermi energy change?

    1. It doubles, a factor of 2
    2. It becomes 2^(2/3) times, about 1.59
    3. It becomes 2^(1/3) times, about 1.26
    4. It becomes four times, a factor of 4
    Show answer

    Answer: B — It becomes 2^(2/3) times, about 1.59

    E_F = (ħ²/2m)(3π²n)^(2/3), so E_F ∝ n^(2/3) and doubling n gives 2^(2/3) = 1.587. The factor 2^(1/3) is the change in k_F, not in E_F, and E_F is not linear in n.

  3. In band theory, what separates an intrinsic semiconductor from an insulator?

    1. Only the size of the band gap relative to k_BT, which sets the density of thermally excited carriers
    2. The conduction band of a semiconductor is already partly filled at 0 K, whereas that of an insulator is empty
    3. The valence band of a semiconductor is only half full at 0 K, whereas that of an insulator is completely full
    4. A semiconductor has no forbidden band at all, while an insulator has a wide one
    Show answer

    Answer: A — Only the size of the band gap relative to k_BT, which sets the density of thermally excited carriers

    Both have a full valence band and an empty conduction band at 0 K; they differ in the gap (about 1 eV for Si against several eV for an insulator), which fixes how many carriers k_BT can excite. A partly filled band describes a metal. A semiconductor does have a forbidden gap.

  4. Two conduction bands near their minima are E₁(k) = Ak² and E₂(k) = 4Ak², with the same constant A. Electrons in both bands have the same relaxation time τ. Using m* = ħ²/(d²E/dk²), what is the ratio of electron mobilities μ₂/μ₁?

    1. 1/4, because the more strongly curved band has a heavier carrier
    2. 2, because the mass scales as the square root of the curvature
    3. 16, because the mobility scales as the square of the curvature
    4. 4, because m* is four times smaller and μ = eτ/m* is four times larger
    Show answer

    Answer: D — 4, because m* is four times smaller and μ = eτ/m* is four times larger

    d²E/dk² = 2A for band 1 and 8A for band 2, so m₂ = m₁/4. With the same τ, μ = eτ/m* gives μ₂/μ₁ = 4. The tempting 1/4 reverses the rule: strong curvature means a light carrier, not a heavy one.

  5. A silicon crystal is doped with a small amount of boron. Which description is correct?

    1. Boron is a donor; it creates an n-type crystal with a negative Hall coefficient
    2. Boron is an acceptor; holes become the majority carriers and the Hall coefficient is positive
    3. Boron is an acceptor; its levels lie just below the conduction band and electrons are the majority carriers
    4. Boron is neutral in silicon and changes only the lattice constant, leaving the carrier density unchanged
    Show answer

    Answer: B — Boron is an acceptor; holes become the majority carriers and the Hall coefficient is positive

    Boron is in group III and has one valence electron fewer than silicon, so it accepts an electron from the valence band and leaves a hole: p-type with R_H > 0. Acceptor levels lie close above the valence-band edge, not below the conduction band.

  6. An n-type silicon sample at 300 K is doped with N_D = 5.0 × 10²¹ m⁻³ donors, all ionised. The intrinsic concentration is n_i = 1.5 × 10¹⁶ m⁻³. Using the law of mass action, the hole concentration is approximately

    1. 4.5 × 10¹⁰ m⁻³
    2. 1.5 × 10¹⁶ m⁻³
    3. 5.0 × 10²¹ m⁻³
    4. 2.3 × 10³² m⁻³
    Show answer

    Answer: A — 4.5 × 10¹⁰ m⁻³

    n ≈ N_D = 5.0 × 10²¹ m⁻³ and p = n_i²/n = (1.5 × 10¹⁶)²/(5.0 × 10²¹) = 2.25 × 10³²/5.0 × 10²¹ = 4.5 × 10¹⁰ m⁻³. The value 1.5 × 10¹⁶ would hold only if the crystal were intrinsic; 2.3 × 10³² is n_i² with the division forgotten.

  7. A pure (undoped) semiconductor has equal electron and hole concentrations, n = p, but μ_e = 3 μ_h. The Hall coefficient of this sample is

    1. Positive, because holes carry positive charge and are present in equal number
    2. Exactly zero, because equal numbers of electrons and holes always cancel the Hall voltage
    3. Negative, because the more mobile electrons dominate: R_H ∝ (pμ_h² − nμ_e²)
    4. Undefined, since the Hall effect exists only when a single carrier type is present
    Show answer

    Answer: C — Negative, because the more mobile electrons dominate: R_H ∝ (pμ_h² − nμ_e²)

    R_H = (pμ_h² − nμ_e²)/[e(pμ_h + nμ_e)²]. With n = p, the numerator is n(μ_h² − μ_e²) = n μ_h²(1 − 9) < 0, so R_H is negative. It would vanish only if μ_e = μ_h. The Hall effect is well defined for mixed conduction; it simply weights each carrier by μ².

  8. Holes in a p-type silicon bar have mobility μ_h = 0.045 m²/V·s at 300 K, where k_BT/e = 0.02585 V. Using the Einstein relation, the hole diffusion coefficient D_h is

    1. 1.74 m²/s
    2. 5.7 × 10⁻² m²/s
    3. 4.5 × 10⁻² m²/s
    4. 1.16 × 10⁻³ m²/s
    Show answer

    Answer: D — 1.16 × 10⁻³ m²/s

    D = μ k_BT/e = 0.045 × 0.02585 = 1.163 × 10⁻³ m²/s. Dividing instead of multiplying gives 0.045/0.02585 = 1.74, and the other options ignore the factor k_BT/e.

  9. The permittivity of a polar ceramic is measured over a wide frequency range. At the lowest frequencies ε_r is large, and it falls in steps. In the step that occurs in the microwave region the slowest dipoles can no longer follow the field. Which mechanism stops contributing at this step?

    1. Orientational polarisation of permanent dipoles, which cannot rotate fast enough
    2. Electronic polarisation, because the electron cloud is the heaviest object that moves
    3. Ionic polarisation, which always ends at the audio range
    4. All mechanisms together, since ε_r falls to zero beyond the microwave range
    Show answer

    Answer: A — Orientational polarisation of permanent dipoles, which cannot rotate fast enough

    Rotating a permanent dipole through the lattice or fluid is the slow process and is lost by the microwave range. Ionic polarisation survives to the infrared and electronic to the ultraviolet, so those two remain. The electron cloud is the lightest, not the heaviest, moving part, and ε_r never falls below 1.

  10. Which crystal is certain NOT to show the direct piezoelectric effect?

    1. A rock-salt-type cubic crystal, whose structure has a centre of inversion symmetry
    2. Quartz, whose crystal structure has no centre of symmetry at all
    3. A poled lead zirconate titanate ceramic used in transducers
    4. Zinc oxide in its hexagonal wurtzite structure
    Show answer

    Answer: A — A rock-salt-type cubic crystal, whose structure has a centre of inversion symmetry

    Under stress, a centrosymmetric crystal moves positive and negative charge centres together, so no net dipole appears: no piezoelectricity. Quartz, wurtzite ZnO and poled PZT all lack a centre of symmetry and are piezoelectric.

  11. Which of the following statements about the free-electron (Fermi gas) model of a metal are correct?

    1. The Fermi surface is a sphere of radius k_F = (3π²n)^(1/3) in k-space
    2. At T = 0 every state below E_F is occupied by two electrons of opposite spin and every state above is empty
    3. The density of states rises as √E, and at E_F equals 3n/(2E_F) per unit volume
    4. At room temperature the average conduction electron has thermal energy of order E_F, so all electrons take part in heat conduction
    Show answer

    Answer: A — The Fermi surface is a sphere of radius k_F = (3π²n)^(1/3) in k-space; B — At T = 0 every state below E_F is occupied by two electrons of opposite spin and every state above is empty; C — The density of states rises as √E, and at E_F equals 3n/(2E_F) per unit volume

    The Fermi-sphere, T = 0 occupation and density-of-states statements are standard results of the model. The claim that all electrons are thermally active is false: k_BT is about 0.026 eV against E_F of several eV, so only the small fraction near E_F (about T/T_F) can be thermally excited.

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

    1. In an intrinsic semiconductor the carrier density grows with temperature mainly through the factor exp(−E_g/2k_BT)
    2. In thermal equilibrium the product np equals n_i² even in a doped crystal, so raising n by donor doping lowers p
    3. A donor-doped crystal carries a net negative charge, because it has extra electrons
    4. The resistivity of a pure metal falls with rising temperature as more electrons are excited across a gap
    Show answer

    Answer: A — In an intrinsic semiconductor the carrier density grows with temperature mainly through the factor exp(−E_g/2k_BT); B — In thermal equilibrium the product np equals n_i² even in a doped crystal, so raising n by donor doping lowers p

    The statements on exp(−E_g/2k_BT) growth and on np = n_i² are correct. The claim that a donor-doped crystal is net negative is false: each ionised donor leaves a positive ion, so the crystal stays neutral overall. The claim that the resistivity of a pure metal falls with temperature is false: a metal has no gap to excite across, n is fixed and increased lattice scattering makes ρ rise with T.

  13. Which of the following statements about dielectric, piezoelectric and ferroelectric behaviour are correct?

    1. Piezoelectricity requires a crystal structure without a centre of symmetry
    2. Barium titanate above its Curie temperature is paraelectric, with a permittivity that follows the Curie-Weiss law
    3. Every piezoelectric crystal is also ferroelectric, because both have a spontaneous polarisation that can be reversed
    4. For a non-magnetic, weakly absorbing solid at optical frequency, ε_r is approximately the square of the refractive index
    Show answer

    Answer: A — Piezoelectricity requires a crystal structure without a centre of symmetry; B — Barium titanate above its Curie temperature is paraelectric, with a permittivity that follows the Curie-Weiss law; D — For a non-magnetic, weakly absorbing solid at optical frequency, ε_r is approximately the square of the refractive index

    The statements on the centre of symmetry, the paraelectric phase of barium titanate and ε_r ≈ n² are correct. The claim that every piezoelectric is ferroelectric is false: quartz is piezoelectric, but its polarisation is not spontaneous or switchable, so it is not ferroelectric. Ferroelectrics are a subset of piezoelectrics.

  14. A metal has a free-electron density n = 2.65 × 10²⁸ m⁻³. Use the free-electron model at T = 0 with ħ = 1.055 × 10⁻³⁴ J·s, m = 9.109 × 10⁻³¹ kg and 1 eV = 1.602 × 10⁻¹⁹ J. The Fermi energy is ____ eV (answer to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 3.24

    k_F = (3π² × 2.65 × 10²⁸)^(1/3) = (7.846 × 10²⁹)^(1/3) = 9.22 × 10⁹ m⁻¹. E_F = ħ²k_F²/2m = (1.113 × 10⁻⁶⁸ × 8.51 × 10¹⁹)/(1.8218 × 10⁻³⁰) = 5.20 × 10⁻¹⁹ J = 3.24 eV.

  15. Copper has n = 8.5 × 10²⁸ m⁻³ and conductivity σ = 5.9 × 10⁷ S/m at room temperature. Use the Drude model with m = 9.109 × 10⁻³¹ kg and e = 1.602 × 10⁻¹⁹ C. The relaxation time is ____ × 10⁻¹⁴ s (answer to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 2.46

    τ = σm/(ne²) = (5.9 × 10⁷ × 9.109 × 10⁻³¹)/(8.5 × 10²⁸ × (1.602 × 10⁻¹⁹)²) = 5.374 × 10⁻²³/(2.181 × 10⁻⁹) = 2.46 × 10⁻¹⁴ s.

  16. A thin strip carries a current of 5.0 mA with a magnetic field B = 0.50 T applied perpendicular to its face, along the thickness direction. The thickness along B is t = 0.20 mm and the measured Hall voltage is 12.5 mV. Only one carrier type is present, with e = 1.602 × 10⁻¹⁹ C. The carrier density is ____ × 10²¹ m⁻³ (answer to two decimal places).

    Numerical answer — type the value.

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    Answer: 6.24

    V_H = IB/(n e t), so n = IB/(e t V_H) = (5.0 × 10⁻³ × 0.50)/(1.602 × 10⁻¹⁹ × 2.0 × 10⁻⁴ × 12.5 × 10⁻³) = 2.5 × 10⁻³/(4.005 × 10⁻²⁵) = 6.24 × 10²¹ m⁻³.

  17. An n-type silicon sample has donor density N_D = 3.0 × 10²² m⁻³, all ionised, with electron mobility μ_e = 0.12 m²/V·s. The hole contribution is negligible. With e = 1.602 × 10⁻¹⁹ C, the conductivity is ____ S/m (answer as a whole number).

    Numerical answer — type the value.

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    Answer: 577

    σ = n e μ_e = 3.0 × 10²² × 1.602 × 10⁻¹⁹ × 0.12 = 4.806 × 10³ × 0.12 = 576.7 S/m, which is 577 as a whole number.

  18. Take the intrinsic carrier concentration of silicon as n_i ∝ T^(3/2) exp(−E_g/2k_BT) with E_g = 1.12 eV (independent of T) and k_B = 8.617 × 10⁻⁵ eV/K. The ratio n_i(360 K)/n_i(300 K) is ____ (answer to one decimal place).

    Numerical answer — type the value.

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    Answer: 48.6

    (360/300)^1.5 = 1.3145. Exponent = (1.12/(2 × 8.617 × 10⁻⁵)) × (1/300 − 1/360) = 6498.8 × 5.5556 × 10⁻⁴ = 3.6104, and exp(3.6104) = 36.98. The ratio is 1.3145 × 36.98 = 48.6.