Magnetic, Thermal and Optical Properties: Magnetism, Hysteresis, Heat Capacity, Conduction, Thermoelectricity and Light

This chapter answers the third to fifth paragraphs of Section XE2.3 (Properties and Applications of Materials) of XE2 Materials Science, one of the optional sections of the Engineering Sciences (XE) paper: “Magnetic Properties: Origin of magnetism in materials, types of Magnetism: Diamagnetism, Paramagnetism, Ferromagnetism, Ferrimagnetism and Antiferromagnetism, Magnetic Domains & Hysteresis, Hard and soft magnetic materials. Thermal Properties: Specific heat, Classical Dulong–Petit Law, Wiedemann-Franz Law, Thermal conductivity of metals and insulators (role of electrons and phonons), Einstein and Debye model, heat conduction, thermal diffusivity, thermal expansion, and thermoelectricity. Optical Properties: Refractive index, absorption and transmission of electromagnetic radiation.” It teaches the five classes of magnetic response, domains and the B-H loop, the heat-capacity models, thermal conduction and expansion, the Seebeck and Peltier effects, and refraction, Beer-Lambert absorption and the band-gap edge. The numericals are the Wiedemann-Franz conductivity, Dulong-Petit heat capacity, the Debye T³ law, thermal diffusivity, absorption through a thickness and the cut-off wavelength.

1. Origin of magnetism and the five classes of magnetic response

Magnetism comes from the orbital motion and the spin of electrons. An electron’s natural unit of moment is the Bohr magneton μ_B = eħ/2m = 9.27 × 10⁻²⁴ A·m²; spin gives about 1 μ_B. Filled shells have paired spins and cancel, so a net atomic moment needs unpaired electrons, as in the 3d shell of Fe, Co and Ni. The response of a material is described by the magnetisation M, the susceptibility χ = M/H, and B = μ₀(H + M) = μ₀μ_rH with μ_r = 1 + χ.

The five classes of magnetism
ClassOrigin and moment arrangementχ and temperature behaviour
DiamagnetismInduced orbital currents oppose the field (Lenz); all materials show itNegative, about −10⁻⁵; independent of T; χ = −1 in a superconductor
ParamagnetismUnpaired, non-interacting moments partly aligned against thermal agitationSmall positive; Curie law χ = C/T
FerromagnetismExchange interaction aligns neighbouring moments parallel: large spontaneous M in domains (Fe, Co, Ni)Very large and field dependent; above T_C paramagnetic, χ = C/(T − T_C)
AntiferromagnetismNeighbouring moments equal and antiparallel: zero net M (e.g. MnO)Small positive; maximum at the Néel temperature T_N, Curie-Weiss above
FerrimagnetismAntiparallel sublattices of unequal moment leave a net M (ferrites, e.g. Fe₃O₄)Large, hysteretic; vanishes above T_C; ceramics, so insulating
⚠️ Do not read the sign alone
Antiferromagnets and paramagnets both have small positive χ; the difference is in the T dependence (χ peaks at T_N for an antiferromagnet, but falls monotonically as 1/T for a paramagnet). Ferrimagnets look like ferromagnets in a hysteresis loop but have lower M_s and are electrical insulators, which is why ferrites suit high-frequency cores.

2. Domains, hysteresis, hard and soft magnetic materials

A ferromagnet below T_C is saturated within each small region, a domain, but the domains point in different directions so as to reduce the stray-field (magnetostatic) energy and the bulk shows no net moment. Neighbouring domains are separated by Bloch walls, thin layers over which the moment direction turns gradually. A field magnetises the specimen first by wall motion (growth of favourably oriented domains, easy and partly reversible), then by rotation of the remaining domains into the field direction, until the saturation magnetisation M_s is reached.

On reducing H to zero the material keeps a remanence (M_r, or B_r) because walls are pinned by defects, inclusions and grain boundaries; a reverse field, the coercivity H_c, is needed to bring B to zero. Repeated cycling traces the hysteresis loop. The loop area ∮H dB (J/m³ per cycle) is the energy dissipated per unit volume per cycle, so the hysteresis power is area × frequency. Example: a loop area of 500 J/m³ at 50 Hz gives 25 kW/m³, which for steel of density 7650 kg/m³ is 25000/7650 = 3.27 W/kg.

Soft against hard magnetic materials
FeatureSoftHard
Coercivity H_cLowHigh
Loop shape and lossNarrow, small area, low loss per cycleWide, large area
MicrostructureFew pinning sites, easy wall motionMany pinning sites, fine precipitates, high anisotropy
Typical useTransformer and motor cores (silicon steel), soft ferrites at high frequencyPermanent magnets (NdFeB, SmCo, Alnico, hard ferrite); figure of merit (BH)max

3. Specific heat: Dulong-Petit, Einstein and Debye models

The heat capacity of a solid is stored mainly in lattice vibrations (phonons). Dulong-Petit (classical): each atom is three oscillators of energy k_BT each, so C_V = 3R = 24.9 J/mol·K (R = 8.314 J/mol·K), independent of the material and of T. It holds at high temperature, around or above room temperature for most metallic elements, but fails at low T, where C_V falls to zero. For aluminium (M = 27.0 g/mol) it gives 24.94/0.0270 = 924 J/kg·K per kilogram.

  • Einstein model: all atoms vibrate at one frequency ν_E, with Einstein temperature θ_E = hν_E/k_B. C_V = 3R (θ_E/T)² e^(θ_E/T)/(e^(θ_E/T) − 1)². It tends to 3R at high T and explains the fall at low T, but it dies off exponentially, faster than the measured behaviour.
  • Debye model: a continuum of phonon modes with frequencies up to a cut-off ν_D, θ_D = hν_D/k_B. At T ≪ θ_D, C_V = (12π⁴/5) R (T/θ_D)³ ≈ 1944 (T/θ_D)³ J/mol·K, the T³ law, which matches experiment; at T ≫ θ_D it recovers 3R. For θ_D = 400 K and T = 10 K, C_V = 1944/40³ = 0.0304 J/mol·K.
  • Metals: at low T, C = γT + βT³, the linear term from electrons near E_F and the cubic term from phonons; the electronic term wins below a few kelvin.
⚠️ Per mole of atoms
3R is per mole of atoms, not per mole of formula units. For a compound with p atoms per formula unit, Dulong-Petit gives 3pR per mole of formula units (for example 6R for NaCl-type MX). Quoting J/kg·K needs the atomic (or formula) mass, in kg/mol.

4. Thermal conduction, Wiedemann-Franz law, diffusivity and expansion

Fourier’s law gives the heat flux q = −κ dT/dx (W/m²), so through a slab of thickness L with temperature difference ΔT, steady q = κΔT/L. In the kinetic picture κ = (1/3) C_v v ℓ, with C_v the heat capacity per unit volume, v the carrier speed and ℓ its mean free path. Metals conduct heat chiefly by electrons, so good electrical conductors are good thermal conductors. The Wiedemann-Franz law states κ_e/(σT) = L, the Lorenz number L = (π²/3)(k_B/e)² = 2.44 × 10⁻⁸ W·Ω/K², roughly the same for all metals. Example: σ = 3.5 × 10⁷ S/m at 300 K gives κ_e = 2.44 × 10⁻⁸ × 3.5 × 10⁷ × 300 = 256 W/m·K.

Insulators have no free electrons; heat is carried by phonons. At high T, phonon-phonon (umklapp) scattering makes ℓ ∝ 1/T, so κ falls as 1/T; at low T, ℓ is fixed by the specimen size and κ ∝ C ∝ T³; hence κ has a maximum at an intermediate T. Defects, grain boundaries and alloying shorten ℓ and lower κ; a stiff, light, ordered lattice (diamond) conducts heat extremely well. In a metal both electrons and phonons carry heat but the electronic part dominates.

Thermal diffusivity α_th = κ/(ρ c_p) (m²/s) controls transient conduction: the time to equalise temperature over a length L scales as L²/α_th. Example: κ = 400 W/m·K, ρ = 8960 kg/m³, c_p = 385 J/kg·K give α_th = 400/(8960 × 385) = 1.16 × 10⁻⁴ m²/s. Thermal expansion comes from the asymmetry (anharmonicity) of the bond potential: as vibrational energy rises the mean separation grows, ΔL = αLΔT. A perfectly symmetric (harmonic) well would give zero expansion, and a deep narrow well (strong bond, high melting point) gives small α. If a bar is rigidly held, the thermal stress is σ = EαΔT; for E = 200 GPa, α = 12 × 10⁻⁶ K⁻¹ and ΔT = 100 K this is 240 MPa.

🎯 Why the Lorenz number can fail
Wiedemann-Franz needs the same scattering to limit heat and charge. It holds at high T and at very low T (elastic impurity scattering) but deviates at intermediate T, where inelastic electron-phonon collisions change the energy of an electron without changing its direction much. It also gives only the electronic κ_e: in an alloy or an insulator the phonon part must be added.

5. Thermoelectricity: Seebeck, Peltier and Thomson effects

Seebeck effect: a temperature difference ΔT across a conductor drives carriers from the hot end toward the cold end (higher thermal energy gives them higher mean speed), building up a voltage, V = SΔT, with Seebeck coefficient S in V/K. For a thermocouple of two materials A and B with junctions at different temperatures, the emf is V = (S_B − S_A)ΔT. In an n-type material the electrons collect at the cold end, so the hot terminal is positive; in a p-type material the holes collect at the cold end and the hot terminal is negative. The sign of S thus identifies the carrier type. Metals have small S (a few µV/K), while doped semiconductors have values of hundreds of µV/K.

Peltier effect is the reverse: a current I through a junction of A and B absorbs or releases heat at the rate Q = Π_AB I, with the Peltier coefficient Π = S T (the Kelvin relation, with S taken as the difference of the two materials). Reversing the current swaps heating and cooling, which is the basis of solid-state refrigerators. The Thomson effect is heat absorbed or released along a homogeneous conductor carrying current in a temperature gradient. Example: S = 40 µV/K at 300 K gives Π = 40 × 10⁻⁶ × 300 = 12 mV, so 2.0 A pumps 24 mW at the junction. A good thermoelectric needs large S, high σ and low κ together, summarised by the figure of merit ZT = S²σT/κ; heavily doped semiconductors give the best compromise.

6. Optical properties: refractive index, reflection and transmission

The refractive index is n = c/v, the ratio of the speed of light in vacuum to that in the solid. For a non-magnetic material, n = √ε_r at the frequency of the light, so it is set by electronic polarisation and rises with the density and polarisability of the atoms. Light striking a surface is partly reflected, partly absorbed and partly transmitted, with R + A + T = 1. For normal incidence from air, the reflectance is R = ((n − 1)/(n + 1))²: for n = 1.5 this is 0.04 (4 percent per glass surface), and for n = 3.5 (a semiconductor) it is 0.31.

Inside the solid, intensity decays by the Beer-Lambert law, I = I₀ exp(−αx), where α (m⁻¹) is the absorption coefficient and 1/α the penetration depth. For a slab of thickness x with two surfaces, neglecting multiple reflections, the transmitted fraction is T = (1 − R)² exp(−αx). Example: with α = 2.0 × 10⁴ m⁻¹ and reflection ignored, the thickness that lets 20 percent through is x = ln 5/α = 1.609/(2.0 × 10⁴) = 80.5 µm. Metals are opaque and shiny because their free electrons respond to the field and reflect visible light strongly.

7. Absorption edge, colour and transparency

A photon of energy hν lifts an electron across the gap only if hν ≥ E_g. Photons below the gap cannot be absorbed by interband transitions, so a pure crystal is transparent there, and absorption switches on sharply at the absorption edge, wavelength λ_c = hc/E_g ≈ 1240/E_g(eV) nm. Examples: GaAs, E_g = 1.42 eV, has λ_c = 873 nm, in the near infrared; silicon, 1.12 eV, 1107 nm. Visible light spans about 400 to 700 nm (3.1 to 1.8 eV), so a pure material is colourless and transparent when E_g > 3.1 eV (diamond, sapphire, silica), opaque and dark when E_g < 1.8 eV, and coloured in between (CdS appears yellow-orange as it absorbs the blue end).

  • Impurity colours: a dopant that creates levels inside the gap adds absorption at lower photon energy, which is how a trace of chromium makes alumina red (ruby).
  • Translucency: a pure oxide that is transparent as a single crystal can look white or translucent as a polycrystal with pores or second phases, because scattering at interfaces of different n diverts the light without absorbing it.
⚠️ Edge wavelength, not frequency
Longer wavelength means lower energy: a material transmits wavelengths LONGER than λ_c and absorbs those shorter. A silicon wafer is opaque to visible light yet transparent to 1.3 µm infrared. Always convert with the same unit system: hc = 1240 eV·nm.

Key takeaways

  • Magnetism comes from unpaired electron spins and orbits (μ_B = eħ/2m). Diamagnetism is universal, small and negative; paramagnetism follows Curie’s law; ferro- and ferrimagnets order below T_C, antiferromagnets below T_N.
  • Domains and pinned Bloch walls give remanence and coercivity; loop area is energy lost per cycle per volume. Soft materials (small H_c) suit cores, hard materials (large H_c and (BH)max) suit permanent magnets.
  • Dulong-Petit gives C_V = 3R at high T; Einstein fixes the low-T fall but too steeply; Debye gives C_V = 1944 (T/θ_D)³ J/mol·K at T ≪ θ_D. Metals add a γT electronic term.
  • Metals conduct heat by electrons: κ_e = LσT with L = 2.44 × 10⁻⁸ W·Ω/K². Insulators conduct by phonons, with κ rising as T³, peaking, then falling as 1/T. Thermal diffusivity is κ/(ρc_p).
  • Thermal expansion arises from anharmonicity (ΔL = αLΔT, constrained stress EαΔT). Seebeck V = SΔT, Peltier Π = ST, and ZT = S²σT/κ measures a thermoelectric.
  • Optics: n = c/v ≈ √ε_r, R = ((n − 1)/(n + 1))², I = I₀e^(−αx). Absorption starts at hν = E_g, λ_c = 1240/E_g(eV) nm, which fixes transparency and colour.

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 statement about diamagnetism is correct?

    1. It occurs only in materials with unpaired electrons and grows stronger as the temperature falls
    2. It is present in every material and gives a small negative susceptibility, nearly independent of T
    3. It gives a large positive susceptibility that vanishes above the Curie temperature
    4. It arises from the alignment of permanent atomic moments along the applied field
    Show answer

    Answer: B — It is present in every material and gives a small negative susceptibility, nearly independent of T

    Diamagnetism comes from orbital currents induced to oppose the field (Lenz’s law), so every material has it, with χ about −10⁻⁵ and no meaningful T dependence. The temperature-growing, positive and aligned-moment descriptions belong to paramagnetism and ferromagnetism.

  2. An electrically insulating oxide shows a spontaneous magnetisation and a hysteresis loop below a critical temperature. Neutron studies show two interpenetrating sublattices whose moments are antiparallel but unequal in magnitude. The material is

    1. an antiferromagnet, since the sublattice moments are antiparallel
    2. a paramagnet, since the moments are not all parallel to the field
    3. a diamagnet, since it is an electrical insulator with no free electrons
    4. a ferrimagnet, since unequal antiparallel moments leave a net magnetisation
    Show answer

    Answer: D — a ferrimagnet, since unequal antiparallel moments leave a net magnetisation

    Antiparallel sublattices of unequal moment give a net spontaneous M with hysteresis, the definition of ferrimagnetism (as in the ferrites). An antiferromagnet has equal sublattice moments and zero net M, so it would show no spontaneous magnetisation or loop.

  3. A silicon-steel core has a hysteresis loop of area 800 J/m³ per cycle (energy lost per unit volume per cycle). It is cycled at 50 Hz and the steel has a density of 7650 kg/m³. Considering hysteresis alone, the power dissipated per kilogram is about

    1. 5.23 W/kg
    2. 0.105 W/kg
    3. 10.5 W/kg
    4. 40.0 W/kg
    Show answer

    Answer: A — 5.23 W/kg

    Power per volume = area × frequency = 800 × 50 = 40000 W/m³; dividing by the density gives 40000/7650 = 5.23 W/kg. 0.105 forgets the frequency, 10.5 doubles the loss, and 40.0 is the power per m³ with the density step missed.

  4. A ferromagnet above its Curie temperature T_C = 1000 K follows the Curie-Weiss law χ = C/(T − T_C). Its susceptibility is 0.020 at 1100 K. Its susceptibility at 1200 K is

    1. 0.018, since χ ∝ 1/T
    2. 0.040, since χ doubles when T − T_C doubles
    3. 0.010, since T − T_C doubles from 100 K to 200 K and χ halves
    4. 0.017, since χ falls by the ratio of the temperatures 1100/1200 twice
    Show answer

    Answer: C — 0.010, since T − T_C doubles from 100 K to 200 K and χ halves

    χ ∝ 1/(T − T_C): at 1100 K the denominator is 100 K and at 1200 K it is 200 K, so χ halves to 0.010. Using the Curie law 1/T would give 0.020 × 1100/1200 = 0.018, the standard slip, because the Weiss shift T_C is ignored.

  5. For an insulating crystal at very low temperature (T much less than the Debye temperature), the lattice heat capacity per mole varies as

    1. a constant 3R, as in the Dulong-Petit law
    2. T³, as in the Debye model
    3. exp(−θ/T), as in the Einstein model, which is what experiment shows
    4. T, as for the electrons in a metal, so it is always linear in T
    Show answer

    Answer: B — T³, as in the Debye model

    The Debye model, with a continuum of low-frequency phonons, gives C_V ∝ T³, and experiment on insulators agrees. Dulong-Petit applies only at high T. Einstein’s exponential fall is too steep compared with the data. The linear term is the electronic contribution of a metal, absent in an insulator.

  6. A steel bar of Young’s modulus 210 GPa and linear expansion coefficient 11 × 10⁻⁶ K⁻¹ is rigidly clamped between two fixed walls at its initial temperature, then heated uniformly by 80 K. Treating the response as uniaxial and elastic, the compressive stress in the bar is

    1. 92 MPa, since only half of the free expansion is blocked
    2. 18.5 MPa, obtained by taking E in units of 10 GPa
    3. 370 MPa, since the clamped bar is stressed twice, at the two ends
    4. 185 MPa, from σ = EαΔT with the free expansion fully prevented
    Show answer

    Answer: D — 185 MPa, from σ = EαΔT with the free expansion fully prevented

    The blocked strain equals the free thermal strain αΔT = 11 × 10⁻⁶ × 80 = 8.8 × 10⁻⁴, so σ = Eα ΔT = 210 × 10⁹ × 8.8 × 10⁻⁴ = 1.85 × 10⁸ Pa = 185 MPa (184.8). The stress is uniform along a clamped bar, not additive at the ends.

  7. For a pure insulating crystal the thermal conductivity is carried by phonons. As the temperature rises from a few kelvin to well above the Debye temperature, κ is expected to

    1. rise steadily as T, because the heat capacity is constant
    2. fall steadily from the start, because more phonons always mean more scattering of heat carriers
    3. rise as T³, pass through a maximum, then fall roughly as 1/T
    4. follow the Wiedemann-Franz law with the electrical conductivity, since all solids obey it
    Show answer

    Answer: C — rise as T³, pass through a maximum, then fall roughly as 1/T

    Low T: ℓ is fixed by the specimen size and C ∝ T³, so κ ∝ T³. High T: umklapp phonon-phonon scattering shortens ℓ ∝ 1/T with C constant, so κ ∝ 1/T; hence a peak. Wiedemann-Franz relates only the electronic κ_e to σ and an insulator has no free electrons.

  8. A thermoelectric material at 300 K has Seebeck coefficient of magnitude 150 µV/K, electrical conductivity 1.0 × 10⁵ S/m and thermal conductivity 1.5 W/m·K. Its dimensionless figure of merit ZT = S²σT/κ is

    1. 1.5 × 10⁻³, since S²σ/κ has no factor of T
    2. 0.675, taking κ as 1.0 W/m·K in the denominator
    3. 0.45, from (150 × 10⁻⁶)² × 10⁵ × 300/1.5
    4. 4.5 × 10¹¹, taking S in µV/K without converting to V/K
    Show answer

    Answer: C — 0.45, from (150 × 10⁻⁶)² × 10⁵ × 300/1.5

    S = 1.5 × 10⁻⁴ V/K, so S² = 2.25 × 10⁻⁸; S²σ = 2.25 × 10⁻³; × T = 0.675; ÷ κ = 0.45. Dropping T gives 1.5 × 10⁻³, and using κ = 1.0 W/m·K gives 0.675 (the figure before dividing by 1.5). The sign of S does not matter, since it is squared.

  9. A glass plate has refractive index n = 1.5 and is placed in air with light at normal incidence. Neglecting absorption and multiple internal reflections, the fraction of the intensity transmitted through both surfaces is about

    1. 92.2 percent
    2. 96.0 percent
    3. 88.5 percent
    4. 84.9 percent
    Show answer

    Answer: A — 92.2 percent

    R = ((1.5 − 1)/(1.5 + 1))² = 0.04 per surface, so T = (1 − R)² = 0.96² = 0.9216, or 92.2 percent. The value 96 percent counts a single surface only; 88.5 and 84.9 count additional interfaces.

  10. A pure semiconductor crystal has band gap E_g = 2.0 eV (take hc = 1240 eV·nm and ignore reflection). Which statement about a thin crystal of it is correct?

    1. It absorbs wavelengths shorter than 620 nm and transmits longer ones, so it looks red or orange
    2. It transmits wavelengths shorter than 620 nm and absorbs longer wavelengths, so it looks blue
    3. It is colourless and fully transparent across the whole visible range, since 2.0 eV is above 1 eV
    4. It absorbs only wavelengths longer than 1240 nm, because the cut-off is 1240 nm for every material
    Show answer

    Answer: A — It absorbs wavelengths shorter than 620 nm and transmits longer ones, so it looks red or orange

    λ_c = 1240/2.0 = 620 nm. Photons with hν ≥ E_g (λ ≤ 620 nm: violet through green and yellow) are absorbed; longer wavelengths (orange and red) pass, so the crystal looks red or orange. The statement with the roles reversed confuses energy and wavelength.

  11. Which of the following statements about magnetic materials are correct?

    1. Diamagnetism is present in every material, but it is masked wherever a paramagnetic or ferromagnetic contribution is present
    2. Remanence and coercivity arise because domain walls are pinned by defects, inclusions and grain boundaries
    3. An antiferromagnet below its Néel temperature has equal and opposite neighbouring moments and no net spontaneous magnetisation
    4. A magnetically hard material has a low coercivity and a narrow loop, so it is ideal for transformer cores
    Show answer

    Answer: A — Diamagnetism is present in every material, but it is masked wherever a paramagnetic or ferromagnetic contribution is present; B — Remanence and coercivity arise because domain walls are pinned by defects, inclusions and grain boundaries; C — An antiferromagnet below its Néel temperature has equal and opposite neighbouring moments and no net spontaneous magnetisation

    The statements on diamagnetism, domain-wall pinning and antiferromagnets are correct. The claim about hard materials is false: low coercivity and a narrow loop describe a soft material, which suits cores. A hard material has a high coercivity and a wide loop and is used for permanent magnets.

  12. Which of the following statements about the heat capacity and heat conduction of solids are correct?

    1. The Einstein model assumes a single vibration frequency and predicts an exponential fall of C_V at low temperature, steeper than observed
    2. At temperatures far above the Debye temperature the Debye model gives C_V close to 3R per mole of atoms
    3. In a pure metal at room temperature, heat is carried mainly by conduction electrons, so κ_e/(σT) is close to the Lorenz number
    4. The Dulong-Petit law correctly predicts that C_V falls to zero as T approaches 0 K
    Show answer

    Answer: A — The Einstein model assumes a single vibration frequency and predicts an exponential fall of C_V at low temperature, steeper than observed; B — At temperatures far above the Debye temperature the Debye model gives C_V close to 3R per mole of atoms; C — In a pure metal at room temperature, heat is carried mainly by conduction electrons, so κ_e/(σT) is close to the Lorenz number

    The Einstein, Debye and Wiedemann-Franz statements are standard results. The Dulong-Petit claim is false: the classical Dulong-Petit value 3R is independent of T and is exactly what fails at low T, where quantum models (Einstein, Debye) bring C_V to zero.

  13. Which of the following statements about thermoelectric and optical behaviour are correct?

    1. The sign of the Seebeck voltage between hot and cold ends of a semiconductor bar indicates whether electrons or holes are the majority carriers
    2. Reversing the current through a Peltier junction changes heating at the junction into cooling and vice versa
    3. In the Beer-Lambert law the transmitted intensity falls linearly with depth, reaching zero at depth 1/α
    4. A pure crystal with a band gap of 5 eV absorbs visible light strongly by interband transitions
    Show answer

    Answer: A — The sign of the Seebeck voltage between hot and cold ends of a semiconductor bar indicates whether electrons or holes are the majority carriers; B — Reversing the current through a Peltier junction changes heating at the junction into cooling and vice versa

    The Seebeck-sign and Peltier statements are correct. The Beer-Lambert claim is false: Beer-Lambert decay is exponential: the intensity falls to 1/e at depth 1/α and never reaches zero. A 5 eV gap exceeds every visible photon energy (about 1.8 to 3.1 eV), so there is no interband absorption and the crystal is transparent.

  14. The electrical conductivity of a metal at 300 K is 3.5 × 10⁷ S/m. Use the Wiedemann-Franz law with Lorenz number L = 2.44 × 10⁻⁸ W·Ω/K² and take heat conduction to be entirely by electrons. The thermal conductivity is ____ W/m·K (answer as a whole number).

    Numerical answer — type the value.

    Show answer

    Answer: 256

    κ = LσT = 2.44 × 10⁻⁸ × 3.5 × 10⁷ × 300 = 0.854 × 300 = 256.2 W/m·K, which is 256 as a whole number.

  15. Apply the Dulong-Petit law with R = 8.314 J/mol·K to aluminium of molar mass 27.0 g/mol. The specific heat capacity is ____ J/kg·K (answer as a whole number).

    Numerical answer — type the value.

    Show answer

    Answer: 924

    C_V = 3R = 24.942 J/mol·K. Per kilogram: 24.942/0.0270 kg/mol = 923.8 J/kg·K, which is 924 as a whole number.

  16. At low temperature the lattice molar heat capacity of an insulator follows the Debye law C_V = (12π⁴/5) R (T/θ_D)³ with R = 8.314 J/mol·K. For θ_D = 400 K, the heat capacity at T = 10 K is ____ J/mol·K (answer to four decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 0.0304

    12π⁴/5 = 12 × 97.409/5 = 233.78; × R = 233.78 × 8.314 = 1943.7 J/mol·K. (T/θ_D)³ = (10/400)³ = 1.5625 × 10⁻⁵. C_V = 1943.7 × 1.5625 × 10⁻⁵ = 0.03037, which is 0.0304 to four decimals.

  17. Monochromatic light travels through a solid with absorption coefficient α = 350 cm⁻¹. Neglect reflection losses and scattering. The thickness through which the transmitted intensity is only 5.0 percent of the incident intensity is ____ µm (answer to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 85.6

    I/I₀ = exp(−αx) = 0.05, so x = ln(20)/α = 2.9957/350 cm⁻¹ = 8.559 × 10⁻³ cm = 85.6 µm (1 cm = 10⁴ µm).

  18. A metal has thermal conductivity κ = 237 W/m·K, density ρ = 2700 kg/m³ and specific heat capacity c_p = 900 J/kg·K. Its thermal diffusivity κ/(ρ c_p) is ____ × 10⁻⁵ m²/s (answer to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 9.75

    α_th = κ/(ρ c_p) = 237/(2700 × 900) = 237/2.43 × 10⁶ = 9.753 × 10⁻⁵ m²/s, which is 9.75 × 10⁻⁵.