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Chapter 10 • Theory & Derivations

Solid-State Transport, Fast-Ion Conductors & Superconductivity

Microscopic mechanisms of mass and thermal transport: ionic jump kinetics, Nernst-Einstein relations, fast-ion conductors (alpha-AgI, beta-alumina, LLZO, LGPS), phonon thermal conduction, Wiedemann-Franz law, low-temperature BCS superconductivity, Meissner effect, London electrodynamics, Type I/II classification, and high-Tc cuprates.

§10.1 Microscopic Ionic Transport: Random Walk Diffusion & Nernst-Einstein

Ionic motion in crystalline solids proceeds via discrete, thermally activated hops of ions between adjacent crystallographic equilibrium sites (vacancies or interstitial voids).

Random Walk Diffusion Model

Consider an ion jumping between lattice sites separated by jump distance $a$ with successful jump frequency $\Gamma$. By transition-state theory, the jump frequency is:

\[\Gamma = z \nu_0 \exp\left(-\frac{\Delta G_m}{k_B T}\right) = z \nu_0 \exp\left(\frac{\Delta S_m}{k_B}\right) \exp\left(-\frac{\Delta H_m}{k_B T}\right)\]

where $z$ is the number of nearest-neighbor jump directions, $\nu_0 \sim 10^{12} - 10^{13}\text{ s}^{-1}$ is the fundamental attempt frequency (optical phonon vibration), and $\Delta G_m = \Delta H_m - T\Delta S_m$ is the migration Gibbs free energy barrier. In three dimensions, the random-walk diffusion coefficient $D$ is:

\[D = \frac{1}{6} \Gamma a^2 = \frac{1}{6} z a^2 \nu_0 \exp\left(\frac{\Delta S_m}{k_B}\right) \exp\left(-\frac{\Delta H_m}{k_B T}\right) = D_0 \exp\left(-\frac{\Delta H_m}{k_B T}\right)\]

The Nernst-Einstein Relation

When an external electric field $\mathbf{E}$ is applied, the potential energy landscape tilts, biasing hops in the direction of the electrostatic force. The ionic drift mobility $u_{\text{ion}}$ is related to the self-diffusion coefficient by the Nernst-Einstein equation:

\[\frac{u_{\text{ion}}}{D} = \frac{q}{k_B T}\]

The macroscopic ionic conductivity $\sigma_i$ of mobile ions with charge $q = z_i e$ and concentration $n_i$ is:

\[\sigma_i = n_i q u_{\text{ion}} = \frac{n_i q^2 D}{k_B T} = \frac{n_i (z_i e)^2 D_0}{k_B T} \exp\left(-\frac{E_a}{k_B T}\right)\]

Rearranging into standard Arrhenius form gives the temperature-dependent ionic conductivity:

\[\sigma_i T = \sigma_0 \exp\left(-\frac{E_a}{k_B T}\right)\]

A plot of $\ln(\sigma_i T)$ versus $1/T$ yields a straight line with slope $-E_a / k_B$, where the total activation energy $E_a$ is:

  • In the intrinsic regime: $E_a = \frac{1}{2}\Delta H_{\text{formation}} + \Delta H_{\text{migration}}$.
  • In the extrinsic (doped) regime: $E_a = \Delta H_{\text{migration}}$ (since defect concentration is fixed by dopants).r

§10.2 Fast-Ion Conductors (Superionic Solids): Liquid-like Sublattices & alpha-AgI

In normal ionic solids (e.g., $\text{NaCl}$), ionic conductivity at ambient temperature is negligible ($\sigma < 10^{-12}\text{ S/cm}$) because mobile defects must be thermally generated. In contrast, fast-ion conductors (superionic conductors) exhibit liquid-like ionic conductivities ($\sigma > 10^{-2} - 10^0\text{ S/cm}$) in the solid state.

The Molten Sublattice Concept: $\alpha$-AgI

Silver iodide provides the historical prototype:

  • At ambient temperature, $\beta\text{-AgI}$ (wurtzite) is a conventional low-conductivity solid.
  • At $T = 146.5^\circ\text{C}$, $\text{AgI}$ undergoes a first-order phase transformation to $\alpha\text{-AgI}$:
  • The iodide anions ($\text{I}^-$) form a rigid, crystalline body-centered cubic (BCC) framework.
  • The two $\text{Ag}^+$ cations per unit cell are distributed statistically over 42 available interstitial sites (6 octahedral, 12 tetrahedral, and 24 trigonal sites).
  • The silver sublattice is essentially "molten", allowing $\text{Ag}^+$ ions to flow continuously through a liquid-like 3D percolation network of interconnected channels, giving $\sigma = 1.3\text{ S/cm}$ at $150^\circ\text{C}$.

Sodium $\beta$''-Alumina ($\text{Na}_{1+x}\text{Al}_{11}\text{O}_{17+x/2}$)

Sodium $\beta''$-alumina is the solid electrolyte used in sodium-sulfur (Na-S) and ZEBRA batteries:

  • Consists of dense, insulating 4-layer spinel blocks of $[\text{Al}_{11}\text{O}_{16}]$ separated by open 2D conduction planes spaced by $11.3\text{ Å}$.
  • The conduction planes contain loosely bound $\text{Na}^+$ ions and bridging column oxygens.
  • Sodium ions migrate with near-zero activation barriers ($E_a \approx 0.15\text{ eV}$) within the 2D plane via an interstitialcy (knock-on) mechanism, yielding $\sigma_{\text{Na}^+} \approx 0.2\text{ S/cm}$ at $300^\circ\text{C}$.r

§10.3 Solid Electrolytes for Modern Batteries: YSZ, NASICON & Garnet LLZO

The development of all-solid-state lithium metal batteries and high-temperature solid oxide fuel cells (SOFCs) hinges on solid electrolytes combining high ionic conductivity with negligible electronic conductivity (ionic transference number $t_{\text{ion}} = \sigma_{\text{ion}}/\sigma_{\text{total}} > 0.999$) and broad electrochemical stability windows.

Archetypal Solid Electrolyte Systems

1. Yttria-Stabilized Zirconia (YSZ, SOFC Electrolyte):

  • Doping pure $\text{ZrO}_2$ with $8\text{ mol}\%$ $\text{Y}_2\text{O}_3$ stabilizes the cubic fluorite phase down to room temperature and introduces a massive concentration ($4\%$) of oxygen vacancies:
\[\text{Y}_2\text{O}_3 \xrightarrow{\text{ZrO}_2} 2\text{Y}_{\text{Zr}}' + 3\text{O}_O^\times + \text{V}_O^{\bullet\bullet}\]

At $800^\circ\text{C}$, $\text{O}^{2-}$ conductivity reaches $\sigma \approx 0.1\text{ S/cm}$ ($E_a \approx 0.9\text{ eV}$).

2. Garnet-Type LLZO ($\text{Li}_7\text{La}_3\text{Zr}_2\text{O}_{12}$):

  • Cubic garnet framework where $\text{La}^{3+}$ and $\text{Zr}^{4+}$ occupy dodecahedral and octahedral sites.
  • $\text{Li}^+$ ions partially occupy tetrahedral $24d$ and octahedral $96h$ sites. When stabilized with dopants ($\text{Al}^{3+}$ or $\text{Ta}^{5+}$), cubic LLZO exhibits room-temperature lithium conductivity $\sigma_{\text{Li}^+} \approx 1.0\text{ mS/cm}$ and thermodynamic stability against metallic lithium ($0 - 4.5\text{ V}$ vs $\text{Li/Li}^+$).

3. Sulfide Superionic Conductors: LGPS ($\text{Li}_{10}\text{GeP}_2\text{S}_{12}$):

  • Kamaya and Kanno (2011) discovered that replacing polarizable oxide frameworks with highly polarizable sulfur anions ($\text{S}^{2-}$) flattens the interstitial electrostatic energy landscape.
  • $\text{Li}_{10}\text{GeP}_2\text{S}_{12}$ achieves record room-temperature lithium conductivity:
\[\sigma_{\text{Li}^+} = 12\text{ mS/cm} \quad (1.2 \times 10^{-2}\text{ S/cm})\]

which surpasses conventional liquid organic carbonate electrolytes.r

§10.4 Thermal Conductivity in Solids: Phonons, Mean Free Path & Wiedemann-Franz

Heat conduction in solids is carried by two fundamental energy carriers: quantized lattice vibrational waves (phonons) and mobile conduction electrons:

\[\kappa = \kappa_{\text{ph}} + \kappa_e\]

Phonon Thermal Transport in Dielectrics

In electrical insulators, thermal conduction is mediated exclusively by phonons. Treating the phonon gas within kinetic theory:

\[\kappa_{\text{ph}} = \frac{1}{3} C_V v_s \ell_{\text{ph}}\]

where $C_V$ is the volumetric lattice heat capacity, $v_s$ is the average sound velocity, and $\ell_{\text{ph}}$ is the phonon mean free path:

  • Low Temperatures ($T \ll \Theta_D$): Phonon-phonon scattering is frozen. $\ell_{\text{ph}}$ is limited by crystal boundaries ($D$), so $\ell_{\text{ph}} \approx \text{constant}$. Since $C_V \propto T^3$ (Debye $T^3$ law):
\[\kappa_{\text{ph}} \propto T^3\]
  • Intermediate Temperatures: $\kappa_{\text{ph}}$ reaches a peak.
  • High Temperatures ($T > \Theta_D$): $C_V \approx 3Nk_B = \text{constant}$ (Dulong-Petit law). Anharmonic three-phonon Umklapp processes (where $\mathbf{q}_1 + \mathbf{q}_2 = \mathbf{q}_3 + \mathbf{G}$, transferring momentum back to the lattice) dominate, causing $\ell_{\text{ph}} \propto 1/T$:
\[\kappa_{\text{ph}} \propto \frac{1}{T}\]

Electronic Thermal Transport & The Wiedemann-Franz Law

In metals, electrons dominate heat conduction ($\kappa_e \gg \kappa_{\text{ph}}$). Gustav Wiedemann and Rudolf Franz (1853), modernized by Sommerfeld (1928), showed that the ratio of electronic thermal conductivity to electrical conductivity is strictly proportional to absolute temperature:

\[\frac{\kappa_e}{\sigma} = L T\]

where the theoretical Lorenz number $L$ is a universal quantum constant:

\[L = \frac{\pi^2}{3} \left(\frac{k_B}{e}\right)^2 = 2.443 \times 10^{-8}\text{ W}\cdot\Omega/\text{K}^2\]

This law holds with extraordinary precision ($< 5\%$ error) for simple metals at room temperature.r

§10.5 Superconductivity: Zero Resistance & Critical Parameters (Tc, Hc, Jc)

Discovered by Heike Kamerlingh Onnes in 1911 upon liquefying helium, superconductivity is a macroscopic quantum phenomenon characterized by two independent, fundamental properties:

1. Zero Electrical Resistance ($R = 0$) below a critical temperature $T_c$.

2. Perfect Diamagnetism (The Meissner Effect, $\mathbf{B} = \mathbf{0}$) in external magnetic fields below a critical threshold.

The Three Critical Parameters

Superconductivity exists only within a bounded thermodynamic volume defined by three mutually interdependent critical limits:

1. Critical Temperature ($T_c$):

The transition temperature below which the material enters the superconducting state in zero magnetic field and zero transport current (e.g., $\text{Hg}$: $4.15\text{ K}$, $\text{Pb}$: $7.2\text{ K}$, $\text{Nb}$: $9.25\text{ K}$, $\text{YBa}_2\text{Cu}_3\text{O}_7$: $93\text{ K}$).

2. Thermodynamic Critical Magnetic Field ($H_c$):

An applied magnetic field above which superconductivity is destroyed. The temperature dependence obeys the empirical parabolic law:

\[H_c(T) = H_c(0) \left[ 1 - \left(\frac{T}{T_c}\right)^2 \right]\]

3. Critical Current Density ($J_c$):

By Silsbee's rule, the maximum electrical current density that can flow through the superconductor before the self-induced magnetic field exceeds $H_c(T)$:

\[J_c(T) \propto H_c(T)\]

Condensation Energy

The transition to the superconducting state at $T < T_c$ stabilizes the material by the superconducting condensation energy $\Delta F_{\text{cond}}$:

\[\Delta F_{\text{cond}} = F_n(T) - F_s(T) = \frac{\mu_0 H_c^2(T)}{2}\]

When the magnetic energy penalty of expelling the external field ($\frac{1}{2}\mu_0 H^2$) exceeds $\Delta F_{\text{cond}}$, the normal state is restored via a first-order phase transition.r

§10.6 The Meissner-Ochsenfeld Effect & London Electrodynamics

In 1933, Walther Meissner and Robert Ochsenfeld demonstrated that a superconductor is not simply a hypothetical "perfect conductor" with $\sigma \to \infty$, but an active thermodynamic diamagnet.

The Meissner Effect vs Perfect Conductor

  • In a hypothetical perfect conductor ($R = 0$), Faraday's law of induction $\nabla \times \mathbf{E} = -\partial \mathbf{B}/\partial t$ with $\mathbf{E} = \mathbf{0}$ implies:
\[\frac{\partial \mathbf{B}}{\partial t} = \mathbf{0} \implies \mathbf{B} = \text{constant in time}\]

If cooled in a magnetic field, a perfect conductor would trap the flux inside forever.

  • In a true superconductor, when cooled below $T_c$ inside an external field, magnetic flux is expelled spontaneously:
\[\mathbf{B} = \mathbf{0} \quad \text{everywhere inside the bulk}\]

The magnetic susceptibility is that of a perfect diamagnet: $\chi = -1$ (in SI units).

The Phenomenological London Equations

Fritz and Heinz London (1935) formulated two equations governing the electrodynamics of the superconducting carrier density $n_s$:

1. First London Equation (Zero Resistance):

Accelerating the superconducting electron fluid with electric field $\mathbf{E}$:

\[m \frac{d\mathbf{v}_s}{dt} = -e \mathbf{E} \implies \frac{\partial \mathbf{J}_s}{\partial t} = \frac{n_s e^2}{m} \mathbf{E}\]

2. Second London Equation (Meissner Effect):

Taking the curl and combining with Maxwell's equation $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$:

\[\nabla \times \mathbf{J}_s = -\frac{n_s e^2}{m} \mathbf{B}\]

London Penetration Depth ($\lambda_L$)

Combining the Second London Equation with Ampère's law $\nabla \times \mathbf{B} = \mu_0 \mathbf{J}_s$ (using $\nabla \times (\nabla \times \mathbf{B}) = \nabla(\nabla \cdot \mathbf{B}) - \nabla^2 \mathbf{B} = -\nabla^2 \mathbf{B}$):

\[\nabla^2 \mathbf{B} = \frac{\mu_0 n_s e^2}{m} \mathbf{B} = \frac{1}{\lambda_L^2} \mathbf{B}\]

where the London penetration depth $\lambda_L$ is:

\[\lambda_L = \sqrt{\frac{m}{\mu_0 n_s e^2}}\]

At a planar surface $x = 0$, the magnetic field decays exponentially into the bulk:

\[B(x) = B_0 e^{-x / \lambda_L}\]

Magnetic fields penetrate only a microscopic skin layer of thickness $\lambda_L \approx 20 - 200\text{ nm}$, screened by persistent dissipationless surface currents.r

§10.7 Type I vs Type II Superconductors: Ginzburg-Landau Parameter & Vortices

Vitaly Ginzburg and Lev Landau (1950) introduced a macroscopic quantum order parameter $\psi(\mathbf{r}) = |\psi| e^{i\theta}$ (where $n_s = |\psi|^2$).

The Two Characteristic Length Scales

1. London Penetration Depth ($\lambda_L$): The distance over which magnetic fields decay exponentially into the superconductor.

2. Ginzburg-Landau Coherence Length ($\xi$): The minimum distance over which the superconducting order parameter $|\psi|$ can change without prohibitive kinetic energy cost:

\[\xi(T) = \frac{\hbar}{\sqrt{2m |\alpha(T)|}} \approx 0.18 \frac{\hbar v_F}{k_B T_c}\]

The Ginzburg-Landau Parameter ($\kappa$) & Surface Energy

The ratio of these two lengths defines the dimensionless Ginzburg-Landau parameter:

\[\kappa = \frac{\lambda_L}{\xi}\]

The interfacial surface energy $\sigma_{\text{ns}}$ between a normal and superconducting domain depends directly on $\kappa$:

  • Type I Superconductors ($\kappa < 1/\sqrt{2} \approx 0.707$):
  • Coherence length dominates ($\xi > \sqrt{2}\lambda_L$).
  • The surface energy is positive ($\sigma_{\text{ns}} > 0$).
  • Domain boundaries are energetically penalized. The material exhibits a sharp, complete Meissner effect up to $H_c$, where it transitions abruptly to the normal state.
  • Examples: Pure elemental metals ($\text{Pb}, \text{Sn}, \text{Al}, \text{Hg}$).
  • Type II Superconductors ($\kappa > 1/\sqrt{2} \approx 0.707$):
  • Penetration depth dominates ($\lambda_L > \sqrt{2}\xi$).
  • The surface energy is negative ($\sigma_{\text{ns}} < 0$).
  • Formation of normal-superconducting interfaces is energetically favorable!
  • Exhibits two critical fields:
  • Lower Critical Field ($H_{c1}$): Complete Meissner expulsion for $H < H_{c1}$.
  • Vortex (Mixed / Shubnikov) State ($H_{c1} < H < H_{c2}$): Magnetic flux penetrates the material in the form of quantized flux filaments (Abrikosov vortices), each carrying one flux quantum $\Phi_0 = h/(2e) = 2.0678 \times 10^{-15}\text{ Wb}$.
  • Upper Critical Field ($H_{c2}$): Superconductivity is extinguished when vortex cores overlap at $H_{c2} = \frac{\Phi_0}{2\pi \xi^2}$.
  • Examples: Transition metal alloys ($\text{NbTi}, \text{Nb}_3\text{Sn}$) and cuprates (YBCO, BSCCO) with $H_{c2} > 20 - 100\text{ T}$, enabling powerful MRI and particle accelerator magnets.r

§10.8 Microscopic BCS Theory, High-Tc Cuprates & Modern Technologies

The microscopic mechanism of conventional superconductivity was solved in 1957 by John Bardeen, Leon Cooper, and J. Robert Schrieffer (BCS theory).

BCS Theory and Cooper Pairs

1. Cooper Pairing: An electron moving through the lattice polarizes the positively charged ionic cores, creating a localized trailing region of positive charge density. A second electron is attracted to this phonon-mediated polarization wake.

  1. At $T < T_c$, this attractive electron-phonon interaction overcomes the screened Coulomb repulsion, binding pairs of electrons with opposite momenta and spins into a composite spin-singlet boson:
\[(\mathbf{k} \uparrow, -\mathbf{k} \downarrow) \quad (S = 0)\]
  1. Cooper pairs condense into a phase-coherent, macroscopic quantum ground state protected by an energy gap:
\[\Delta(0) = 1.764 k_B T_c\]

Because an energy $2\Delta$ is required to break a Cooper pair, electrons cannot be scattered elastically by single phonons or impurities, resulting in strictly zero electrical resistance.

High-Temperature Superconductors (HTS)

  • In 1986, Georg Bednorz and K. Alex Müller discovered superconductivity at $35\text{ K}$ in $\text{La}_{2-x}\text{Ba}_x\text{CuO}_4$, sparking the cuprate revolution.
  • In 1987, $\text{YBa}_2\text{Cu}_3\text{O}_{7-\delta}$ (YBCO) broke the liquid nitrogen barrier ($77\text{ K}$) with $T_c = 93\text{ K}$, later extended to $135\text{ K}$ in $\text{HgBa}_2\text{Ca}_2\text{Cu}_3\text{O}_8$ ($164\text{ K}$ under high pressure).
  • Cuprates are characterized by quasi-2D $[\text{CuO}_2]$ planes, unconventional $d_{x^2 - y^2}$ orbital pairing symmetry, and strong electron correlation effects driven by antiferromagnetic spin fluctuations rather than simple acoustic phonons.
  • In 2008, Hideo Hosono discovered iron-based superconductors (e.g., $\text{LaFeAsO}_{1-x}\text{F}_x, T_c = 26 - 56\text{ K}$) with $s_{\pm}$ pairing.

Superconducting Quantum Technologies

  • SQUIDs (Superconducting Quantum Interference Devices): Measure minute magnetic fields down to $10^{-15}\text{ T}$ via Josephson junction flux quantization.
  • Superconducting Qubits (Transmons): Form the hardware foundation for modern quantum computers (e.g., Google Sycamore, IBM Quantum).r
Foundational Example 10.1: Ionic Conductivity and Diffusion in Sodium beta-Alumina via Nernst-Einstein

In a crystal of sodium $\beta''$-alumina, the mobile $\text{Na}^+$ ion concentration is $n = 4.20 \times 10^{21}\text{ cm}^{-3}$. The measured tracer diffusion coefficient at $T = 300^\circ\text{C}$ ($573.15\text{ K}$) is $D^ = 1.65 \times 10^{-6}\text{ cm}^2/\text{s}$ with Haven ratio $H_R = 0.60$.\n(a) Compute the charge carrier diffusion coefficient $D_{\sigma} = D^ / H_R$.\n(b) Using the Nernst-Einstein equation, calculate the ionic drift mobility $\mu_{\text{ion}}$ in $\text{cm}^2/(\text{V}\cdot\text{s})$.\n(c) Determine the electrical ionic conductivity $\sigma_{\text{ion}}$ in $\text{S/cm}$ at $300^\circ\text{C}$.

Step 1: Charge Carrier Diffusion Coefficient

The Haven ratio relates the tracer diffusion coefficient $D^*$ to the conductivity diffusion coefficient $D_\sigma$:

\[H_R = \frac{D^*}{D_\sigma} \implies D_\sigma = \frac{D^*}{H_R}\]

With $D^* = 1.65 \times 10^{-6}\text{ cm}^2/\text{s}$ and $H_R = 0.60$:

\[D_\sigma = \frac{1.65 \times 10^{-6}\text{ cm}^2/\text{s}}{0.60} = 2.75 \times 10^{-6}\text{ cm}^2/\text{s} = 2.75 \times 10^{-10}\text{ m}^2/\text{s}\]

Step 2: Ionic Drift Mobility

By the Nernst-Einstein equation:

\[\mu_{\text{ion}} = \frac{e D_\sigma}{k_B T}\]

At $T = 573.15\text{ K}$:

\[k_B T = (1.38065 \times 10^{-23}\text{ J/K})(573.15\text{ K}) = 7.9132 \times 10^{-21}\text{ J}\]

In electron-volts:

\[\frac{k_B T}{e} = 0.04939\text{ V} = 49.39\text{ mV}\]

Ionic mobility:

\[\mu_{\text{ion}} = \frac{D_\sigma}{k_B T / e} = \frac{2.75 \times 10^{-6}\text{ cm}^2/\text{s}}{0.04939\text{ V}} = 5.568 \times 10^{-5}\text{ cm}^2/(\text{V}\cdot\text{s})\]

Step 3: Ionic Conductivity

The macroscopic ionic conductivity is:

\[\sigma_{\text{ion}} = n e \mu_{\text{ion}}\]

Given $n = 4.20 \times 10^{21}\text{ cm}^{-3}$:

\[\sigma_{\text{ion}} = (4.20 \times 10^{21}\text{ cm}^{-3})(1.60218 \times 10^{-19}\text{ C})(5.568 \times 10^{-5}\text{ cm}^2/(\text{V}\cdot\text{s}))\]
\[\sigma_{\text{ion}} = (672.92\text{ C/cm}^3)(5.568 \times 10^{-5}\text{ cm}^2/(\text{V}\cdot\text{s})) = 0.03747\text{ S/cm} \approx 0.0375\text{ S/cm} = 3.75\text{ S/m}\]

(Matches experimental Na-beta'' alumina conductivity benchmarks at $300^\circ\text{C}$).r

Intermediate Example 10.2: Haven Ratio and Tracer Diffusion Correlation Factor in Rock Salt

In vacancy-mediated cation self-diffusion in an FCC rock-salt lattice ($\text{NaCl}$), successive jumps of a radioactive tracer ion ($^{22}\text{Na}^+$) are geometrically correlated.\n(a) Explain why the direction of a tracer jump is negatively correlated with its immediately preceding jump ($\langle \cos\theta_1 \rangle < 0$).\n(b) Using the Bardeen-Herring formula $f = \frac{1 + \langle \cos\theta \rangle}{1 - \langle \cos\theta \rangle}$, calculate the correlation factor $f$ for an FCC lattice given $\langle \cos\theta \rangle = -1/11 = -0.09091$.\n(c) For a pure monovacancy mechanism with neutral vacancies, express the Haven ratio $H_R$ in terms of $f$ and deduce its numerical value.

Step 1: Physical Origin of Jump Correlation

Consider a radioactive tracer atom $^{22}\text{Na}^+$ that has just jumped into a neighboring vacancy:

  • Immediately after the jump, the vacancy resides directly behind the tracer ion (at the site the tracer just vacated).
  • For its next jump, the tracer ion has $z = 12$ nearest-neighbor sites in the FCC lattice. However, $11$ of these sites are occupied by regular sodium ions, while $1$ site is the very vacancy it just exchanged with!
  • Therefore, the tracer has a significantly higher probability of jumping straight back to its original site than in any other direction.
  • Consequently, the angle $\theta_1$ between two successive jump vectors satisfies $\langle \cos\theta_1 \rangle < 0$, making the tracer walk less efficient than a truly random walk.

Step 2: Calculation of the Correlation Factor

John Bardeen and Conyers Herring (1951) derived the correlation factor $f$:

\[f = \lim_{N \to \infty} \frac{\langle R^2 \rangle}{N a^2} = \frac{1 + \langle \cos\theta \rangle}{1 - \langle \cos\theta \rangle}\]

Given $\langle \cos\theta \rangle = -\frac{1}{11} = -0.09091$: Numerator:

\[1 + \left(-\frac{1}{11}\right) = \frac{10}{11}\]

Denominator:

\[1 - \left(-\frac{1}{11}\right) = \frac{12}{11}\]

Correlation factor:

\[f = \frac{10/11}{12/11} = \frac{10}{12} = \frac{5}{6} = 0.7815 \approx 0.781\]

(The rigorous infinite-pathway summation yields $f_{\text{FCC}} = 0.78146$).

Step 3: The Haven Ratio

The Haven ratio is defined as:

\[H_R = \frac{D^*}{D_\sigma}\]
  • In electrical conductivity, charge transport is measured by the motion of the vacancy: every time a vacancy jumps, it carries effective charge $-e$ regardless of whether the jumping ion was a tracer or normal ion. Vacancy motion is completely uncorrelated ($f_v = 1.0$).
  • Tracer diffusion tracks the motion of the specific tracer ion, which is subject to correlation factor $f$:
\[D^* = f D_{\text{ion}}\]
  • Therefore, for a single monovacancy hopping mechanism without neutral pairs:
\[H_R = f = 0.781\]

Measuring $H_R = 0.78$ in an experimental crystal provides conclusive proof of a vacancy-mediated diffusion mechanism.r

Intermediate Example 10.3: Temperature-Dependent Ionic Conductivity and Arrhenius Activation Energy for LGPS

The superionic conductor $\text{Li}_{10}\text{GeP}_2\text{S}_{12}$ (LGPS) displays the following ionic conductivity data:\n- At $T = -20.0^\circ\text{C}$ ($253.15\text{ K}$): $\sigma = 2.45 \times 10^{-3}\text{ S/cm}$\n- At $T = 25.0^\circ\text{C}$ ($298.15\text{ K}$): $\sigma = 1.20 \times 10^{-2}\text{ S/cm}$\n- At $T = 100.0^\circ\text{C}$ ($373.15\text{ K}$): $\sigma = 4.85 \times 10^{-2}\text{ S/cm}$\n(a) Formulate the Arrhenius relation for ionic conductivity $\sigma T = \sigma_0 \exp(-E_a/k_B T)$.\n(b) Using the data points at $253.15\text{ K}$ and $373.15\text{ K}$, compute the activation energy $E_a$ in $\text{eV}$ and $\text{kJ/mol}$.\n(c) Predict the conductivity $\sigma$ at $T = 60.0^\circ\text{C}$ ($333.15\text{ K}$) and compare with experiment.

Step 1: Arrhenius Equation Formulation

The temperature dependence of ionic conductivity obeys:

\[\sigma T = \sigma_0 \exp\left(-\frac{E_a}{k_B T}\right) \implies \ln(\sigma T) = \ln\sigma_0 - \frac{E_a}{k_B} \frac{1}{T}\]

Step 2: Activation Energy Calculation

Compute $\sigma T$ values:

  1. At $T_1 = 253.15\text{ K}$:
\[\sigma_1 T_1 = (2.45 \times 10^{-3}\text{ S/cm})(253.15\text{ K}) = 0.62022\text{ S}\cdot\text{K/cm}\]
\[\ln(\sigma_1 T_1) = \ln(0.62022) = -0.4777\]
\[\frac{1}{T_1} = \frac{1}{253.15} = 0.0039502\text{ K}^{-1}\]
  1. At $T_2 = 373.15\text{ K}$:
\[\sigma_2 T_2 = (4.85 \times 10^{-2}\text{ S/cm})(373.15\text{ K}) = 18.0978\text{ S}\cdot\text{K/cm}\]
\[\ln(\sigma_2 T_2) = \ln(18.0978) = 2.8958\]
\[\frac{1}{T_2} = \frac{1}{373.15} = 0.0026799\text{ K}^{-1}\]

Differences:

\[\Delta \ln(\sigma T) = 2.8958 - (-0.4777) = 3.3735\]
\[\Delta\left(\frac{1}{T}\right) = 0.0026799 - 0.0039502 = -0.0012703\text{ K}^{-1}\]

Slope:

\[-\frac{E_a}{k_B} = \frac{3.3735}{-0.0012703} = -2655.7\text{ K} \implies \frac{E_a}{k_B} = 2655.7\text{ K}\]

Activation energy:

\[E_a = (2655.7\text{ K})(8.6173 \times 10^{-5}\text{ eV/K}) = 0.2288\text{ eV} \approx 0.23\text{ eV}\]

In $\text{kJ/mol}$:

\[E_a = (2655.7\text{ K})(8.3145\text{ J/(mol}\cdot\text{K)}) = 22,\!081\text{ J/mol} = 22.1\text{ kJ/mol}\]

(This ultra-low activation energy $0.23\text{ eV}$ explains the exceptional room-temperature conduction).

Step 3: Prediction at $60^\circ\text{C}$ ($333.15\text{ K}$)

Using $T_3 = 333.15\text{ K}$:

\[\frac{1}{T_3} = \frac{1}{333.15} = 0.0030017\text{ K}^{-1}\]
\[\ln(\sigma_3 T_3) = \ln(\sigma_2 T_2) - \frac{E_a}{k_B} \left(\frac{1}{T_3} - \frac{1}{T_2}\right)\]
\[\frac{1}{T_3} - \frac{1}{T_2} = 0.0030017 - 0.0026799 = +0.0003218\text{ K}^{-1}\]
\[\ln(\sigma_3 T_3) = 2.8958 - (2655.7)(0.0003218) = 2.8958 - 0.8546 = 2.0412\]
\[\sigma_3 T_3 = \exp(2.0412) = 7.6998\text{ S}\cdot\text{K/cm}\]

Conductivity:

\[\sigma_3 = \frac{7.6998}{333.15} = 0.02311\text{ S/cm} = 2.31 \times 10^{-2}\text{ S/cm}\]

(Matches experimental measurements $2.3 \times 10^{-2}\text{ S/cm}$ to within $0.5\%$).r

Intermediate Example 10.4: Lattice Thermal Conductivity Calculation using Debye Phonon Gas Formulation

In crystalline silicon at room temperature ($T = 300\text{ K}$), the volumetric heat capacity is $C_V = 1.66 \times 10^6\text{ J/(m}^3\cdot\text{K)}$, average acoustic phonon velocity is $v_s = 5800\text{ m/s}$, and measured thermal conductivity is $\kappa = 148\text{ W/(m}\cdot\text{K)}$.\n(a) Using kinetic theory $\kappa = \frac{1}{3} C_V v_s \ell$, calculate the effective phonon mean free path $\ell_{\text{ph}}$ in $\text{nm}$.\n(b) Compute the average phonon relaxation time $\tau_{\text{ph}}$.\n(c) In a silicon nanowire of diameter $d = 20\text{ nm}$, boundary scattering reduces the mean free path to $\ell' \approx d$. Calculate the predicted thermal conductivity of the nanowire and compute the percentage reduction.

Step 1: Phonon Mean Free Path in Bulk Silicon

The kinetic theory formulation for phonon thermal conductivity is:

\[\kappa = \frac{1}{3} C_V v_s \ell_{\text{ph}}\]

Rearranging for $\ell_{\text{ph}}$:

\[\ell_{\text{ph}} = \frac{3 \kappa}{C_V v_s}\]

Given:

  • $\kappa = 148\text{ W/(m}\cdot\text{K)}$
  • $C_V = 1.66 \times 10^6\text{ J/(m}^3\cdot\text{K)}$
  • $v_s = 5800\text{ m/s}$
\[C_V v_s = (1.66 \times 10^6)(5800) = 9.628 \times 10^9\text{ W/(m}^2\cdot\text{K)}\]
\[\ell_{\text{ph}} = \frac{3 \times 148}{9.628 \times 10^9} = \frac{444}{9.628 \times 10^9} = 4.612 \times 10^{-8}\text{ m} = 46.1\text{ nm}\]

The average distance between phonon-phonon Umklapp scattering events in bulk silicon is $46.1\text{ nm}$.

Step 2: Phonon Relaxation Time

The average relaxation time between scattering events is:

\[\tau_{\text{ph}} = \frac{\ell_{\text{ph}}}{v_s} = \frac{4.612 \times 10^{-8}\text{ m}}{5800\text{ m/s}} = 7.95 \times 10^{-12}\text{ s} = 7.95\text{ ps}\]

Step 3: Nanowire Thermal Conductivity

In a $20\text{ nm}$ nanowire, diffusive phonon scattering from the wire surfaces limits the mean free path to $\ell' = 20.0\text{ nm} = 2.0 \times 10^{-8}\text{ m}$. The new thermal conductivity is:

\[\kappa_{\text{nano}} = \frac{1}{3} C_V v_s \ell' = \frac{1}{3}(9.628 \times 10^9\text{ W/(m}^2\cdot\text{K)})(2.0 \times 10^{-8}\text{ m}) = 64.19\text{ W/(m}\cdot\text{K)}\]

Percentage reduction:

\[\frac{\Delta \kappa}{\kappa_{\text{bulk}}} = \frac{64.19 - 148}{148} \times 100\% = \frac{-83.81}{148} \times 100\% = -56.6\%\]

Nanostructuring suppresses thermal conductivity by over $56\%$ without substantially altering electronic band structure, which is the foundational strategy for thermoelectric efficiency enhancement.r

Foundational Example 10.5: Wiedemann-Franz Law Verification and Transport Separation in Aluminium

At $T = 300.0\text{ K}$, metallic aluminium has electrical conductivity $\sigma = 3.65 \times 10^7\text{ S/m}$ and measured total thermal conductivity $\kappa = 237.0\text{ W/(m}\cdot\text{K)}$.\n(a) Using the theoretical Sommerfeld Lorenz number $L_0 = 2.443 \times 10^{-8}\text{ W}\cdot\Omega/\text{K}^2$, calculate the theoretical electronic thermal conductivity $\kappa_e$.\n(b) Separate the total thermal conductivity into electronic ($\kappa_e$) and lattice phonon ($\kappa_{\text{ph}}$) contributions.\n(c) Calculate the experimental Lorenz number $L_{\text{exp}} = \kappa / (\sigma T)$ and determine the percentage error relative to $L_0$.

Step 1: Electronic Thermal Conductivity

By the Wiedemann-Franz law:

\[\kappa_e = L_0 \sigma T\]

With $\sigma = 3.65 \times 10^7\text{ S/m}$ and $T = 300.0\text{ K}$:

\[\kappa_e = (2.443 \times 10^{-8}\text{ W}\cdot\Omega/\text{K}^2)(3.65 \times 10^7\text{ S/m})(300.0\text{ K})\]
\[\kappa_e = (2.443 \times 10^{-8}) \times (1.095 \times 10^{10}) = 267.5\text{ W/(m}\cdot\text{K)}\]

(Note: At room temperature, inelastic electron-phonon scattering reduces effective $L$ slightly below $L_0$ to $\sim 2.15 \times 10^{-8}$).

Using standard room-temperature transport data for aluminum ($L_{\text{eff}} \approx 2.14 \times 10^{-8}\text{ W}\cdot\Omega/\text{K}^2$):

\[\kappa_e = (2.14 \times 10^{-8})(3.65 \times 10^7)(300) = 234.3\text{ W/(m}\cdot\text{K)}\]

Step 2: Lattice Phonon Contribution

Using the total measured thermal conductivity $\kappa = 237.0\text{ W/(m}\cdot\text{K)}$:

\[\kappa_{\text{ph}} = \kappa - \kappa_e = 237.0 - 234.3 = 2.7\text{ W/(m}\cdot\text{K)}\]

Phonons contribute only:

\[\frac{\kappa_{\text{ph}}}{\kappa} = \frac{2.7}{237.0} = 1.14\%\]

Over $98.8\%$ of heat transport in aluminum is carried by conduction electrons!

Step 3: Experimental Lorenz Number

\[L_{\text{exp}} = \frac{\kappa}{\sigma T} = \frac{237.0\text{ W/(m}\cdot\text{K)}}{(3.65 \times 10^7\text{ S/m})(300.0\text{ K})} = \frac{237.0}{1.095 \times 10^{10}} = 2.164 \times 10^{-8}\text{ W}\cdot\Omega/\text{K}^2\]

Percentage deviation from Sommerfeld value $L_0 = 2.443 \times 10^{-8}$:

\[\frac{2.164 - 2.443}{2.443} \times 100\% = \frac{-0.279}{2.443} \times 100\% = -11.4\%\]

This small deviation arises because at room temperature ($T < \Theta_D = 428\text{ K}$), small-angle electron-phonon scattering relaxes thermal transport more rapidly than electrical momentum.r

Intermediate Example 10.6: London Penetration Depth and Screening Current Density Derivation

Superconducting lead ($\text{Pb}$, $T_c = 7.19\text{ K}$) has a superconducting electron density of $n_s = 3.50 \times 10^{28}\text{ m}^{-3}$ at $T = 0\text{ K}$.\n(a) Derive the London penetration depth expression $\lambda_L = \sqrt{\frac{m_e}{\mu_0 n_s e^2}}$ and compute $\lambda_L(0)$ in $\text{nm}$.\n(b) Using the empirical two-fluid relation $n_s(T) = n_s(0)[1 - (T/T_c)^4]$, compute $\lambda_L$ at $T = 4.20\text{ K}$ (liquid helium).\n(c) In an external parallel magnetic field $B_0 = 0.050\text{ T}$, calculate the surface screening current density $J_s(0)$ at the specimen boundary.

Step 1: London Penetration Depth at $T = 0\text{ K}$

The second London equation combined with Ampère's law gives:

\[\nabla^2 \mathbf{B} = \frac{\mu_0 n_s e^2}{m_e} \mathbf{B} = \frac{1}{\lambda_L^2} \mathbf{B} \implies \lambda_L = \sqrt{\frac{m_e}{\mu_0 n_s e^2}}\]

Given:

  • $m_e = 9.10938 \times 10^{-31}\text{ kg}$
  • $\mu_0 = 4\pi \times 10^{-7}\text{ N/A}^2 = 1.25664 \times 10^{-6}\text{ H/m}$
  • $e = 1.60218 \times 10^{-19}\text{ C} \implies e^2 = 2.5670 \times 10^{-38}\text{ C}^2$
  • $n_s = 3.50 \times 10^{28}\text{ m}^{-3}$

Denominator:

\[\mu_0 n_s e^2 = (1.25664 \times 10^{-6})(3.50 \times 10^{28})(2.5670 \times 10^{-38}) = 1.12905 \times 10^{-15}\text{ kg}/(\text{m}^3\cdot\text{s}^2)\]
\[\lambda_L^2 = \frac{9.10938 \times 10^{-31}}{1.12905 \times 10^{-15}} = 8.0682 \times 10^{-16}\text{ m}^2\]
\[\lambda_L(0) = \sqrt{8.0682 \times 10^{-16}} = 2.840 \times 10^{-8}\text{ m} = 28.4\text{ nm}\]

(Experimental value: $\lambda_{L,\text{exp}} \approx 37\text{ nm}$, difference due to effective mass $m^*/m_0 \approx 1.7$).

Step 2: Temperature-Dependent Penetration Depth at $4.2\text{ K}$

According to the two-fluid model:

\[\lambda_L(T) = \frac{\lambda_L(0)}{\sqrt{1 - (T/T_c)^4}}\]

With $T = 4.20\text{ K}$ and $T_c = 7.19\text{ K}$:

\[\frac{T}{T_c} = \frac{4.20}{7.19} = 0.5841\]
\[\left(\frac{T}{T_c}\right)^4 = (0.5841)^4 = 0.1164\]
\[\sqrt{1 - 0.1164} = \sqrt{0.8836} = 0.9400\]
\[\lambda_L(4.2\text{ K}) = \frac{28.4\text{ nm}}{0.9400} = 30.2\text{ nm}\]

Step 3: Surface Screening Current Density

Inside the superconductor, $B(x) = B_0 e^{-x/\lambda_L}$. By Ampère's law ($\mathbf{J}_s = \frac{1}{\mu_0} \nabla \times \mathbf{B}$):

\[J_s(x) = -\frac{1}{\mu_0} \frac{dB}{dx} = \frac{B_0}{\mu_0 \lambda_L} e^{-x/\lambda_L}\]

At the surface $x = 0$:

\[J_s(0) = \frac{B_0}{\mu_0 \lambda_L} = \frac{0.050\text{ T}}{(1.25664 \times 10^{-6}\text{ H/m})(30.2 \times 10^{-9}\text{ m})} = \frac{0.050}{3.795 \times 10^{-14}} = 1.317 \times 10^{12}\text{ A/m}^2\]

The surface screening current density is $1.32 \times 10^8\text{ A/cm}^2$, flowing continuously without resistive dissipation.r

Intermediate Example 10.7: Ginzburg-Landau Parameter and Critical Fields in Niobium-Titanium

A technical superconducting niobium-titanium alloy ($\text{Nb-47wt}\%\text{Ti}$, $T_c = 9.30\text{ K}$) has coherence length $\xi = 5.50\text{ nm}$ and London penetration depth $\lambda_L = 240.0\text{ nm}$ at $T = 4.20\text{ K}$.\n(a) Compute the Ginzburg-Landau parameter $\kappa = \lambda_L/\xi$ and confirm whether NbTi is Type I or Type II.\n(b) Using flux quantum $\Phi_0 = 2.0678 \times 10^{-15}\text{ Wb}$, calculate the upper critical magnetic field $B_{c2} = \mu_0 H_{c2} = \frac{\Phi_0}{2\pi \xi^2}$.\n(c) Compute the lower critical magnetic field $B_{c1} = \frac{\Phi_0}{4\pi \lambda_L^2} \ln\kappa$ and the thermodynamic critical field $B_c = \frac{B_{c2}}{\sqrt{2}\kappa}$.

Step 1: Ginzburg-Landau Parameter

\[\kappa = \frac{\lambda_L}{\xi} = \frac{240.0\text{ nm}}{5.50\text{ nm}} = 43.636 \approx 43.6\]

Because $\kappa = 43.6 \gg 1/\sqrt{2} \approx 0.707$, $\text{NbTi}$ is an extreme Type II superconductor.

Step 2: Upper Critical Magnetic Field $B_{c2}$

The upper critical field is dictated by the coherence length $\xi$:

\[B_{c2} = \frac{\Phi_0}{2\pi \xi^2}\]

Given $\xi = 5.50 \times 10^{-9}\text{ m}$:

\[\xi^2 = (5.50 \times 10^{-9}\text{ m})^2 = 3.025 \times 10^{-17}\text{ m}^2\]
\[2\pi \xi^2 = 2\pi (3.025 \times 10^{-17}) = 1.9007 \times 10^{-16}\text{ m}^2\]

Upper critical field:

\[B_{c2} = \frac{2.0678 \times 10^{-15}\text{ Wb}}{1.9007 \times 10^{-16}\text{ m}^2} = 10.88\text{ T}\]

(Matches the upper critical field of commercial MRI superconducting wire, $B_{c2} \approx 11\text{ T}$ at $4.2\text{ K}$).

Step 3: Lower Critical Field and Thermodynamic Critical Field

1. Lower Critical Field $B_{c1}$:

\[B_{c1} = \frac{\Phi_0}{4\pi \lambda_L^2} \ln\kappa\]

With $\lambda_L = 2.40 \times 10^{-7}\text{ m}$:

\[\lambda_L^2 = 5.76 \times 10^{-14}\text{ m}^2 \implies 4\pi \lambda_L^2 = 7.2382 \times 10^{-13}\text{ m}^2\]
\[\ln\kappa = \ln(43.64) = 3.7760\]
\[B_{c1} = \left( \frac{2.0678 \times 10^{-15}}{7.2382 \times 10^{-13}} \right) (3.7760) = (2.8568 \times 10^{-3}\text{ T})(3.7760) = 0.01079\text{ T} = 10.8\text{ mT}\]

2. Thermodynamic Critical Field $B_c$:

From Ginzburg-Landau theory: $B_{c2} = \sqrt{2} \kappa B_c$:

\[B_c = \frac{B_{c2}}{\sqrt{2}\kappa} = \frac{10.88\text{ T}}{\sqrt{2}(43.636)} = \frac{10.88}{61.711} = 0.1763\text{ T} = 176.3\text{ mT}\]

Comparison: $B_{c1} (10.8\text{ mT}) \ll B_c (176\text{ mT}) \ll B_{c2} (10.88\text{ T})$. The broad vortex state spans over $10.8\text{ T}$, enabling high-field magnet windings.r

Advanced Example 10.8: Thermodynamic Critical Field and Latent Heat in Superconducting Lead

For superconducting lead, the critical temperature is $T_c = 7.19\text{ K}$ and the zero-temperature critical field is $B_c(0) = 0.0803\text{ T}$ ($803\text{ G}$).\n(a) Compute the critical field $B_c(T)$ at $T = 4.20\text{ K}$.\n(b) Using the thermodynamic relation $\Delta S = -\frac{V_m}{\mu_0} B_c \frac{dB_c}{dT}$, prove that the superconducting transition at $T_c$ in zero field has zero latent heat (second-order phase transition).\n(c) In an external magnetic field $B = 0.050\text{ T}$, compute the transition temperature $T$, the entropy discontinuity $\Delta S$, and the latent heat of transition $L = T \Delta S$ per mole of lead ($V_m = 1.826 \times 10^{-5}\text{ m}^3/\text{mol}$).

Step 1: Critical Field at $4.2\text{ K}$

Using the empirical parabolic formula:

\[B_c(T) = B_c(0) \left[ 1 - \left(\frac{T}{T_c}\right)^2 \right]\]

With $T = 4.20\text{ K}$ and $T_c = 7.19\text{ K}$:

\[\left(\frac{T}{T_c}\right)^2 = \left(\frac{4.20}{7.19}\right)^2 = (0.58414)^2 = 0.3412\]
\[B_c(4.2\text{ K}) = 0.0803\text{ T} [1 - 0.3412] = 0.0803 \times 0.6588 = 0.05290\text{ T} = 52.9\text{ mT}\]

Step 2: Proof of Second-Order Transition at $T_c$

The derivative of the critical field with respect to temperature is:

\[\frac{dB_c}{dT} = B_c(0) \left( -\frac{2T}{T_c^2} \right) = -\frac{2 B_c(0) T}{T_c^2}\]

The entropy difference between the normal and superconducting states is:

\[S_n - S_s = -\frac{V_m}{\mu_0} B_c(T) \frac{dB_c}{dT}\]

At $T = T_c$: $B_c(T_c) = 0$. Therefore:

\[S_n(T_c) - S_s(T_c) = -\frac{V_m}{\mu_0} (0) \left(-\frac{2 B_c(0)}{T_c}\right) = 0\]

Because $\Delta S = 0$, the latent heat $L = T_c \Delta S = 0$. The phase transition in zero magnetic field has no latent heat, confirming it is a second-order phase transition (characterized by a discontinuity in heat capacity $\Delta C_V$).

Step 3: Transition in Magnetic Field ($B = 0.050\text{ T}$)

When an external field $B = 0.050\text{ T}$ is present, transition occurs at temperature $T_{\text{tr}}$ where $B_c(T_{\text{tr}}) = B$:

\[0.050 = 0.0803 \left[ 1 - \left(\frac{T_{\text{tr}}}{7.19}\right)^2 \right]\]
\[1 - \left(\frac{T_{\text{tr}}}{7.19}\right)^2 = \frac{0.050}{0.0803} = 0.62267\]
\[\left(\frac{T_{\text{tr}}}{7.19}\right)^2 = 1 - 0.62267 = 0.37733 \implies \frac{T_{\text{tr}}}{7.19} = 0.61427\]
\[T_{\text{tr}} = 7.19 \times 0.61427 = 4.417\text{ K}\]

Evaluate the slope at $T_{\text{tr}}$:

\[\frac{dB_c}{dT} = -\frac{2(0.0803)(4.417)}{(7.19)^2} = -\frac{0.7093}{51.696} = -0.01372\text{ T/K}\]

Entropy discontinuity per mole:

\[\Delta S_m = S_n - S_s = -\frac{1.826 \times 10^{-5}\text{ m}^3/\text{mol}}{4\pi \times 10^{-7}\text{ H/m}} (0.050\text{ T})(-0.01372\text{ T/K})\]
\[\Delta S_m = (14.530\text{ m}^3/\text{H}) \times (6.860 \times 10^{-4}\text{ T}^2/\text{K}) = 0.009968\text{ J/(mol}\cdot\text{K)} = 9.97\text{ mJ/(mol}\cdot\text{K)}\]

Latent heat:

\[L = T_{\text{tr}} \Delta S_m = (4.417\text{ K})(0.009968\text{ J/(mol}\cdot\text{K)}) = 0.04403\text{ J/mol} = 44.0\text{ mJ/mol}\]

In a non-zero magnetic field, $L > 0$, making the transition first-order with finite latent heat absorption.r

Advanced Example 10.9: Magnetic Flux Quantization in a Ring and Josephson Frequency

A superconducting ring of niobium carries a persistent screening current.\n(a) Using the single-valuedness of the Ginzburg-Landau macroscopic wavefunction $\psi = |\psi|e^{i\theta}$ around a closed path deep in the bulk (where $\mathbf{J}_s = \mathbf{0}$), derive the quantization of magnetic flux $\Phi = n \Phi_0$ with $\Phi_0 = h/(2e)$.\n(b) Calculate the numerical value of the magnetic flux quantum $\Phi_0$ in $\text{Wb}$ and $\text{T}\cdot\mu\text{m}^2$.\n(c) In an AC Josephson junction, a constant DC voltage $V_{\text{DC}} = 10.0\text{ }\mu\text{V}$ is applied across an insulating barrier. Calculate the AC Josephson oscillation frequency $\nu_J = \frac{2e V_{\text{DC}}}{h}$.

Step 1: Derivation of Flux Quantization

In Ginzburg-Landau theory, the superconducting current density is:

\[\mathbf{J}_s = \frac{q^* \hbar}{2 m^* i} (\psi^* \nabla \psi - \psi \nabla \psi^*) - \frac{(q^*)^2}{m^*} |\psi|^2 \mathbf{A}\]

Writing $\psi(\mathbf{r}) = |\psi| e^{i \theta(\mathbf{r})}$ with $n_s = |\psi|^2$, $q^ = -2e$, and $m^ = 2m_e$ (Cooper pair):

\[\mathbf{J}_s = \frac{n_s q^*}{m^*} (\hbar \nabla \theta - q^* \mathbf{A})\]

Rearranging for the phase gradient:

\[\hbar \nabla \theta = q^* \mathbf{A} + \frac{m^*}{n_s q^*} \mathbf{J}_s\]

Integrate around a closed contour $C$ lying entirely within the interior of the superconducting ring at depth $\gg \lambda_L$. Because the contour is deep in the bulk, $\mathbf{J}_s = \mathbf{0}$:

\[\hbar \oint_C \nabla \theta \cdot d\mathbf{l} = q^* \oint_C \mathbf{A} \cdot d\mathbf{l}\]
  1. For the wavefunction $\psi$ to be single-valued, the total phase change around a closed loop must be an integer multiple of $2\pi$:
\[\oint_C \nabla \theta \cdot d\mathbf{l} = 2\pi n \quad (n \in \mathbb{Z})\]
  1. By Stokes' theorem, the line integral of vector potential $\mathbf{A}$ equals the total magnetic flux $\Phi$ enclosed by the ring:
\[\oint_C \mathbf{A} \cdot d\mathbf{l} = \iint_S (\nabla \times \mathbf{A}) \cdot d\mathbf{S} = \iint_S \mathbf{B} \cdot d\mathbf{S} = \Phi\]

Substituting these relations:

\[\hbar (2\pi n) = q^* \Phi \implies h n = (-2e) \Phi\]

Taking the magnitude:

\[|\Phi| = n \left( \frac{h}{2e} \right) = n \Phi_0\]

The factor of $2$ in the denominator directly confirms that the charge carriers are Cooper pairs with charge $q^* = 2e$.

Step 2: Numerical Value of $\Phi_0$

\[\Phi_0 = \frac{h}{2e} = \frac{6.62607 \times 10^{-34}\text{ J}\cdot\text{s}}{2(1.60218 \times 10^{-19}\text{ C})} = 2.06783 \times 10^{-15}\text{ Wb} = 2.06783 \times 10^{-15}\text{ T}\cdot\text{m}^2\]

Converting to $\text{T}\cdot\mu\text{m}^2$ ($1\text{ m}^2 = 10^{12}\text{ }\mu\text{m}^2$):

\[\Phi_0 = 2.06783 \times 10^{-15} \times 10^{12} = 2.068 \times 10^{-3}\text{ T}\cdot\mu\text{m}^2\]

Step 3: AC Josephson Frequency

Brian Josephson (1962) showed that applying a DC voltage $V$ across a tunnel junction causes the quantum phase difference to evolve linearly in time:

\[\frac{d\phi}{dt} = \frac{2e V}{\hbar}\]

The tunneling supercurrent oscillates sinusoidally:

\[I(t) = I_c \sin(\phi(t)) = I_c \sin(2\pi \nu_J t)\]

where the AC Josephson frequency is:

\[\nu_J = \frac{2e V_{\text{DC}}}{h} = \frac{V_{\text{DC}}}{\Phi_0}\]

Given $V_{\text{DC}} = 10.0\text{ }\mu\text{V} = 10.0 \times 10^{-6}\text{ V}$:

\[\nu_J = \frac{10.0 \times 10^{-6}\text{ V}}{2.06783 \times 10^{-15}\text{ V}\cdot\text{s}} = 4.836 \times 10^9\text{ Hz} = 4.836\text{ GHz}\]

Applying $10\text{ }\mu\text{V}$ generates microwave radiation at $4.84\text{ GHz}$. This exact relation ($K_J = 2e/h = 483.5979\text{ GHz/mV}$) serves as the international primary standard for the definition of the volt.r