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Room-temperature iron is ferromagnetic: interactions among its electronic moments produce spontaneous local order. The independent-ion Curie law therefore needs an interaction and an ordered-state description. In a Weiss mean-field approximation, replace the applied field by $H_{\mathrm{eff}}=H+\Lambda M$. Above the ordering temperature, linearizing the single-ion response yieldsThis is the Curie–Weiss approximation on the paramagnetic side. Merely substituting room temperature below $T_C$ into that high-temperature formula would produce a negative response and is not a valid model of stable iron.

Below $T_C$, use the full nonlinear moment response to the effective field; it has nonzero zero-field solutions. On a chosen ordered branch the local small-field response has the form $M\simeq M_s(T)+{\chi }_{\mathrm{diff}}(T)H$, rather than $M\propto H/T$ through the origin. Macroscopic magnetic domains and hysteresis further determine the measured net moment. Exchange interactions, arising from electronic quantum statistics and Coulomb correlations, supply the alignment energy; bare dipolar fields alone do not explain ordinary iron's strong ferromagnetism. A localized-ion Weiss model describes the qualitative ordering, while iron's itinerant electronic structure limits a literal independent-fixed-ion interpretation.

In the spin-degenerate free electron model, a monovalent metal has conduction-electron number density $n$. The density of wave-vector states is $1/(2\pi {)}^3$ per unit real-space volume, with two spin states at each wave vector. Filling a sphere at zero temperature givesFor the Debye model, there are three acoustic polarizations and $3n$ modes per unit volume in total. Replacing their dispersions by $\omega =ck$ with the effective Debye sound speed, mode counting givesThe factor two in the electron count is spin, whereas the factor three in the phonon count is polarization; the two counts are not interchangeable. If branch speeds differ, the Debye effective average obeys $c^{-3}=(c_l^{-3}+2c_t^{-3})/3$ in the isotropic approximation.

Dividing gives $T_F/{\theta }_D=\lambda (3{\pi }^{2}n/2{)}^{1/3}$, where $\lambda =\hslash /(2m_{e}c)$. A monatomic FCC conventional cell contains four atoms, so $n=4/a^3$. ThereforeHere and below $c$ is the speed of sound, not the speed of light. The Debye cutoff counts acoustic modes in a spherical approximation to the Brillouin zone; it is not an electron-state cutoff.

Using $c=2700\mathrm{m}{\mathrm{s}}^{-1}$ gives the electron-acoustic recoil lengthThe temperature ratio therefore fixesIt follows that $n=4/a^3=9.48\times 10^{28}{\mathrm{m}}^{-3}$, $k_F=1.4105\times 10^{10}{\mathrm{m}}^{-1}$ and $k_D=1.7771\times 10^{10}{\mathrm{m}}^{-1}$. Substitution givesThe resulting Fermi energy is about $7.58\mathrm{eV}$ and the temperature ratio is indeed $240$. These values follow from the supplied effective sound speed and free-electron/Debye approximations; they are not independent precision measurements of copper's parameters.

An electron losing energy to an acoustic phonon has, in a normal process, ${\mathbf{k}}^{\prime }={\mathbf{k}}_F-{\mathbf{k}}_{\mathrm{ph}}$, with possible addition of a reciprocal-lattice vector for an Umklapp process. The energy transfer is at most of order $\hslash {\omega }_D=k_B{\theta }_D$, much smaller than $E_F=k_B{T}_F$. Hence the magnitudes change very little even though the momentum direction can change substantially. At $T\sim {\theta }_D\ll T_F$, thermal vacancies near the Fermi surface allow emission into some lower-energy states; exactly at zero temperature emission into an occupied Fermi sea would be blocked by the Pauli exclusion principle.

Using $k^{\prime }=(1-\delta )k_F$ and the free-electron dispersion, energy conservation givesSince $\delta \ll 1$, the bracket is $2\delta +O({\delta }^2)$. Rearranging yieldsEquivalently $\delta \simeq \hslash c{k}_{\mathrm{ph}}/(2E_F)$. For a Debye-cutoff phonon in the preceding model, $\delta \simeq {\theta }_D/(2T_F)=1/480=0.00208$, confirming the approximation. Near-elastic scattering constrains the radial distance in wave-vector space, not the scattering angle alone.

An intrinsic semiconductor has its carrier populations set primarily by thermal excitation across the band gap; in the ideal pure, neutral case its conduction-electron and hole densities are equal, $n=p=n_i$. An extrinsic semiconductor has carrier populations significantly controlled by donor or acceptor impurities, producing predominantly n-type semiconductor or p-type semiconductor behavior. Chemical purity alone does not fix conductivity independently of temperature and gap size.

The positive charge carrier mobility is the magnitude of mean drift speed per unit weak electric field. For electrons ${\mathbf{v}}_e=-{\mu }_e\mathbf{E}$, for holes ${\mathbf{v}}_h={\mu }_h\mathbf{E}$, and its SI unit is ${\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. In an isotropic relaxation-time model, $\mu =|q|\tau /m^{\ast }$.

A band effective mass characterizes how an electron wave packet accelerates in an energy band, rather than replacing its vacuum rest mass. In one dimension and its tensor generalization,These follow by differentiating $v_i={\hslash }^{-1}\partial E/\partial k_i$ and using $\hslash {\dot{k}}_j=F_j$. Effective masses can be anisotropic and electron band curvature can be negative.

A point magnetic dipole $\mathbf{\mu }$ producesFor $|\mu |\sim {\mu }_B$ and neighbour separation $r\sim 0.2\mathrm{nm}$, its characteristic magnitude isThe axial field is twice this scale; the exact orientation and crystal geometry are unspecified. The corresponding pair magnetic dipole interaction has energy scale $U_{\mathrm{dip}}\sim {\mu }_B{B}_{\mathrm{dip}}\simeq 1.08\times 10^{-24}\mathrm{J}$, givingAn axial orientation changes this to about $0.16\mathrm{K}$. A real ordering temperature depends on a lattice sum, symmetry and fluctuations, so this is a nearest-neighbour energy estimate, not a complete ordering calculation. It establishes that ordinary dipolar coupling is far too weak to explain a thousand-kelvin ferromagnetic transition.

The inverse-susceptibility curve has a Curie–Weiss form above $T_C$ and vanishes on approach to it. The material can have spontaneous magnetization for $T<T_C$, including the zero-temperature limit, whereas for $T>T_C$ its zero-field equilibrium state is paramagnetic. At a continuous transition the spontaneous moment tends to zero at $T_C$; divergence of differential susceptibility there is not a finite saturated permanent moment.

A simple Weiss molecular field model takes the effective internal field intensity to be $H_{\mathrm{eff}}=H+\Lambda M$, with positive coupling $\Lambda$. Let $C$ be the independent-moment Curie constant. Linearizing the single-ion response on the high-temperature side givesThis explains the straight upper branch and its positive temperature intercept. At $T_C$ the zero-field linearized response diverges; finite field, finite size and critical fluctuations limit its use. For a spin-only two-state mean-field model the full equation is $m=\tanh [(T_C/T)m+{\mu }_B{\mu }_{0}H/(k_{B}T)]$, with $m=M/(N{\mu }_B)$; below $T_C$ it has nonzero zero-field solutions.

An ordering scale near $1000\mathrm{K}$ corresponds to $k_B{T}_C\simeq 0.086\mathrm{eV}$, about four orders of magnitude larger than the preceding dipolar pair scale. Exchange interaction can supply such energies: the Pauli antisymmetry of overlapping electron states changes their spatial correlations and hence Coulomb and kinetic energies according to their spin configuration. An effective ferromagnetic coupling can favor parallel moments. It is not simply the classical magnetic field of a neighbouring dipole. For example, with the convention $\mathcal{H}=-J{\sum }_{⟨ij⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$ and $J>0$, the localized-spin mean-field scale is $k_B{T}_C=zJS(S+1)/3$; this illustrates why an atomic electronic interaction, rather than a bare dipolar field, can produce a high transition temperature.

A bulk ferromagnet generally separates into magnetic domains. Within each domain many moments align and the local magnetization can be large, but different domain orientations partially or almost completely cancel the vector sum measured for the entire specimen. This reduces its external stray field and magnetostatic energy. Formation of magnetic domain walls costs exchange and anisotropy energy, so the actual pattern balances these costs with the reduction in stray-field energy.

A strong applied field can move domain walls and rotate moments toward a nearly saturation magnetization. When the field is removed, pinning and the sample's magnetic history determine the remanent moment. Thus a small net moment does not show that local ferromagnetic order has disappeared, or that moments within each domain are all randomly oriented. At finite temperature, thermal spin fluctuations also reduce the local ordered moment below its zero-temperature maximum even in a single-domain specimen. These two reductions—between-domain cancellation and within-domain thermal reduction—should be distinguished.

Take one conduction electron per atom and a three-dimensional free electron gas, with spin degeneracy two and electron mass $m_e$. State counting inside the occupied Fermi sphere givesThe highest occupied energy at zero temperature is $E_F={\hslash }^2{k}_F^2/(2m_e)$, soThis relates the atomic number density to the Fermi temperature only because monovalency and the free-electron dispersion have been assumed. A real energy band can require a different density of states and band effective mass.

For the isotropic Debye model, count three acoustic polarizations up to one spherical wavevector cutoff $q_D$. The number of vibrational modes per volume isWith $\omega =cq$ and $\hslash {\omega }_D=k_B{\theta }_D$,The effective $c$ is the Debye sound speed: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical Brillouin zone boundary.

Pauli exclusion principle prevents most filled electronic states from changing occupation at $T\ll T_F$. Only a shell of width of order $k_{B}T$ around the Fermi energy participates. Since the density of states near that energy is of order $n/E_F$ per volume and energy, the fraction of active electrons is of order $k_{B}T/E_F=T/T_F$. Their typical change of energy is of order $k_{B}T$, so the thermal energy increment per volume scales asDifferentiating with respect to temperature gives the electronic heat capacityThe simple shell argument fixes the linear law, not its coefficient. A Sommerfeld expansion using the spin-summed free-electron density of states $g(E_F)=3n/(2E_F)$ gives $C_e=({\pi }^2/3)g(E_F)k_B^{2}T$, hence $A={\pi }^2/2$, consistent with the supplied value. The result excludes strong band-structure or interaction corrections and requires a degenerate electron population.

The conventional face-centered cubic lattice cell contains four atoms, soThe preceding free-electron and Debye expressions giveAt low temperature the electronic and lattice heat-capacity crossover in a monovalent metal is found fromThus the electronic term dominates for $0<T\lesssim 3.4$ K within the normal-metal model. Both low-temperature approximations are well controlled at that crossover. At exactly zero temperature both heat capacities vanish; below sufficiently small finite-size energy scales a continuum model also needs revision.

Room temperature is not $T\ll {\theta }_D$: using the cubic formula at 300 K would incorrectly predict $C_{\mathrm{ph}}\simeq 142n{k}_B$, far above the harmonic high-temperature limit $3n{k}_B$. Instead use the Dulong-Petit law as an estimate:Keeping the finite cutoff in the Debye heat capacity givesAt 300 K this is $2.8006n{k}_B$, whereas $C_e=0.017994n{k}_B$, giving $\boxed{C_e/C_{\mathrm{ph}}=6.42\times 10^{-3}}$. Either appropriate room-temperature estimate shows that the lattice contribution dominates. The electronic approximation remains excellent because $300\ll T_F$.

An intrinsic semiconductor has thermally generated conduction electrons and valence holes, with $n=p=n_i$ at equilibrium in the ideal undoped neutral material. Their abundance is controlled by its band gap, temperature and band density of states.

An extrinsic semiconductor has its mobile carrier populations substantially modified by electrically active impurities. A donor impurity can supply electrons, producing n-type material; an acceptor impurity can capture a valence electron and leave a mobile electron hole, producing p-type material. Compensation and incomplete ionization can modify these simple counts. At low temperature dopant freeze-out can occur; at sufficiently high temperature intrinsic pair production can overwhelm a fixed dopant density. Doping and the dominant transport regime are related but are not interchangeable descriptions at every temperature.

Assume nondegenerate three-dimensional parabolic band edges, a fixed gap and approximately temperature-independent density-of-states effective masses. Let $\mu$ be the chemical potential, $E_c$ and $E_v$ the band edges, and $N_c,N_v$ their effective density of states. Boltzmann occupation givesFor one spin-degenerate spherical conduction extremum, direct integration of $g_c(E)\propto \sqrt{E-E_c}$ givesand similarly for the valence band with its appropriate density-of-states mass. Valley and additional band degeneracies can be included in these prefactors. The $T^{3/2}$ factor follows from the integral ${\int }_0^{\infty }\sqrt{ϵ}{e}^{-ϵ/(k_{B}T)}dϵ\propto T^{3/2}$.

Multiplication removes the chemical potential:The semiconductor mass-action law combined with intrinsic charge neutrality, $n=p$, therefore givesThe intrinsic chemical potential is not necessarily exactly midgap: ${\mu }_i=(E_c+E_v)/2+(k_{B}T/2)\ln (N_v/N_c)$. Equal effective masses are unnecessary for the displayed carrier-density law. A changing gap, effective mass or degeneracy would make the prefactor or activation law temperature dependent.

The sign must become positive, so use p-type acceptor impurities, such as boron substituting for tetravalent silicon. Assume uncompensated, fully ionized acceptors of density $N_A$, nondegenerate carriers and equilibrium throughout the interval. Charge neutrality and mass action giveFor a nearly constant hole-dominated Hall coefficient, the dopant plateau must persist up to the highest temperature. Its necessary concentration scale is thereforeThe order-of-magnitude onset is $N_A\sim 4\times 10^{20}{\mathrm{m}}^{-3}$, but equality is not a flat plateau: at $N_A=n_i(500)$, $p/N_A=(1+\sqrt{5})/2=1.618$, changing the hole-only Hall coefficient by about 38 percent. The word approximately does not define a unique sharp minimum without a tolerance.

For $N_A\gg n_i$, $p\simeq N_A[1+(n_i/N_A{)}^2]$, so the hole-only $R_H\simeq 1/(e{N}_A)$ varies by a fraction of order $(n_i/N_A{)}^2$. An illustrative one-percent tolerance in that approximation is obtained with $N_A\simeq 10n_i(500)=4.0\times 10^{21}{\mathrm{m}}^{-3}$. This gives $R_H\simeq +1.56\times 10^{-3}{\mathrm{m}}^3{\mathrm{C}}^{-1}$ and less than one percent hole-density-induced variation. It is an explicit plateau criterion, not an exact minimum deduced from the unspecified word approximately.

At 300 K the minority electron concentration isThus the onset-scale estimate gives about $2.5\times 10^{11}{\mathrm{m}}^{-3}$, whereas the explicit ten-times-larger plateau choice gives $2.5\times 10^{10}{\mathrm{m}}^{-3}$.

The acceptor density for a stable Hall sign can be more restrictive if both carriers matter. With mobility ratio $r_{\mu }={\mu }_e/{\mu }_h>1$, positive Hall sign requires $p>r_{\mu }{n}_i$, or $N_A>n_i(r_{\mu }-r_{\mu }^{-1})$. In the strongly extrinsic limit the full two-carrier Hall correction scales as $(1+r_{\mu }{)}^2(n_i/N_A{)}^2$, not just $(n_i/N_A{)}^2$. Hence a true measured-Hall tolerance can require a larger dopant density than the hole-only illustration. Mobilities and their temperature dependence are not supplied, so they cannot be used to manufacture a unique sharper minimum.

In the free electron model, conduction electrons occupy independent plane-wave states with energy ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)$ in a uniform mean background. Crystal-periodic band structure and explicit electron–electron correlations are omitted. Electrons still obey the Pauli exclusion principle and Fermi–Dirac statistics; free does not mean distinguishable classical particles. A scattering time can be added phenomenologically to describe transport.

The Fermi energy is the highest occupied single-particle energy at zero temperature, measured here from the bottom of the parabolic band. The Fermi temperature is the corresponding scale $T_F=E_F/k_B$, not the actual thermal temperature of the sample. Metallic electron densities are large, so filling all distinct states up to the required momentum gives Fermi energies of several electronvolts, equivalent to tens of thousands of kelvin or more. At ordinary temperatures, only a narrow region around the Fermi surface has appreciably changing occupations.

A room-temperature metal and its lattice are nevertheless at room temperature, not at $T_F$. High occupied kinetic energies are present even in the zero-temperature ground state and cannot all be liberated as heat without violating the allowed occupancy structure. A metal below skin temperature often feels cold because it transfers heat rapidly out of the skin; its thermal conductivity and thermal effusivity matter, not the value of its Fermi temperature. At the same actual temperature, a poorly conducting material can therefore feel warmer while neither material is thermodynamically hotter.

For one spin-degenerate parabolic two-dimensional band, $k_F=\sqrt{2\pi n}=1.3729\times 10^8$ m${}^{-1}$. The free-electron model with $m=m_e$ givesThe given $10^3\Omega$ is a sheet resistivity, not a bulk resistivity in ohm-metres. The two-dimensional Drude sheet conductance is $n{e}^2\tau /m_e$, soThus $v_F=\hslash k_F/m_e=1.589\times 10^4$ m/s andThe sample temperature is not given. This is specifically the Fermi-surface mean free path, appropriate in the degenerate regime $T\ll 8.3$ K for the bare-mass model. At higher temperatures the characteristic thermal speed should replace $v_F$ for a thermal transport distance. If a real two-dimensional band has mass $m^{\ast }$, the stated results scale as $E_F,T_F\propto 1/m^{\ast }$, $\tau \propto m^{\ast }$ and $v_F\propto 1/m^{\ast }$; the inferred ${\ell }_F=\hslash k_F/(n{e}^2{\rho }_{\square })$ is mass independent under these same Drude assumptions. No semiconductor effective mass or valley degeneracy is silently introduced.

A $3d^9$ shell has one hole in the otherwise filled $d$ shell. Hund's rules give $S=1/2$, $L=2$, and, for a more-than-half-filled shell, $J=L+S=5/2$: the free-ion term is ${}^2{D}_{5/2}$. Its Landé factor isDistinguish the Curie effective moment from a saturated projection:The question then explicitly asks us to use orbital quenching in the crystal. Ligand fields split and mix orbital states, suppressing their freely orientable angular momentum. The spin-only approximation has $g\simeq 2$, $S=1/2$, so ${\mu }_{\mathrm{eff}}=\sqrt{3}{\mu }_B$ but the two field-axis projections are $\pm {\mu }_B$. The latter, not $\sqrt{3}{\mu }_B$, belong in the tanh argument and saturation magnetization.

The sulfate and water environment supplies mainly closed-shell, weakly magnetic surroundings, stabilizes localized Cu2+ sites and separates their moments so that exchange can be weak on the temperature scale considered. Paramagnetism here is an approximation of weakly interacting local spins, not a claim that dilution guarantees no magnetic ordering at any temperature. Elemental copper has a filled $3d$ band and itinerant conduction electrons rather than one localized Cu2+ spin per atom. Pauli spin response and orbital/core contributions then replace a local-spin Curie law; ordinary pure copper has a weak net diamagnetic response.^[ref]

There is one Cu2+ per formula unit, hence $n_s=N_A/V_m=5.5148\times 10^{27}$ m${}^{-3}$. The spin-only weak-field SI volume susceptibility isThese neglect exchange, anisotropic $g$ factors and background diamagnetism. At high field and low temperature, saturation requires the nonlinear response in a; a field-independent Curie value then describes only the zero-field slope.

At fixed chain length, each harmonic mode contributes a Bose–Einstein thermal occupation and a temperature-independent zero-point energy. The latter contributes no heat capacity. With $x=\hslash \omega /(k_{B}T)$,At low temperature only the approximately linear part of the acoustic dispersion is excited, so $g(\omega )\simeq 1/(\pi v_s)$. Using ${\int }_0^{\infty }{x}^2{e}^x/(e^x-1{)}^{2}dx={\pi }^2/3$ givesIt is linear in temperature, rather than the three-dimensional Debye $T^3$ law. At high temperature every scalar oscillator tends to $k_B$, henceThe cutoff temperature is ${\Theta }_m=\hslash {\omega }_m/k_B=305.5$ K. The curve rises smoothly from zero toward this plateau; ${\Theta }_m$ sets a crossover scale, not a sharp transition. In normalized variables its initial slope is $C_V/(N{k}_B)\simeq (2\pi /3)T/{\Theta }_m$. The low-temperature continuum limit also requires temperature above the finite-size level spacing; uniform center-of-mass translation is treated separately. Below the lowest nonzero mode energy a finite chain does not retain a continuum linear law down to arbitrary temperature.

At fixed volume and particle number, reversible heating obeys $dS=C_{V}dT/T$. Integrating the electronic heat capacity from zero temperature givesWith no extensive residual ground-state degeneracy, the third law of thermodynamics sets the electronic $S(0)=0$. ThereforeThe entropy normalization follows that of $\gamma$: total, per volume or per mole. Lattice contributions are additional and are not part of this electronic result.

Let $g_v(E)$ denote the spin-summed density of states per volume. An electron moment with magnitude the Bohr magneton ${\mu }_B$ has Zeeman energies $\mp {\mu }_{B}B$. At low temperature, the two spin populations areThe common chemical potential changes only at order $B^2$, so it does not affect this linear imbalance. The magnetization is $M={\mu }_B(n_{+}-n_{-})={\mu }_B^2{g}_v(E_F)B$. Define SI volume magnetic susceptibility by $M={\chi }_{p}H$ and take the weak applied field as $B\simeq {\mu }_{0}H$. Then Pauli paramagnetism givesThe last equality uses the three-dimensional parabolic free electron model, $g_v(E_F)=3n/(2E_F)$. This susceptibility is nearly independent of temperature for $T\ll T_F$, unlike the Curie law for localized independent moments. The displayed result excludes orbital diamagnetism and interaction enhancements. If susceptibility instead denotes $\partial M/\partial B$, omit ${\mu }_0$; confusing that convention with $\partial M/\partial H$ changes the units.