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A thermodynamic potential packages the equilibrium state in variables suited to the imposed constraints. For a simple fluid, $U(S,V,N)$ has differential $dU=TdS-pdV+\mu dN$. Legendre transforms give $H=U+pV$, $F=U-TS$ and $G=U+pV-TS$, with natural variables of a thermodynamic potential $(S,p,N)$, $(T,V,N)$ and $(T,p,N)$ respectively. At fixed particle number, Helmholtz free energy is minimized at fixed $T,V$; Gibbs free energy at fixed $T,p$.

A flow vessel is an open control volume, even if its walls are enclosed. For each inlet, pressure forces carry flow work $pdV$ in addition to internal energy. The steady first-law balance isWith no heat exchange, shaft work, or significant kinetic and gravitational energy differences, the enthalpy fluxes are equal. Mass flow must also balance. This equality concerns mass-flow-weighted specific enthalpies, not an unweighted sum over pipes.

Use the liquefaction yield with separate warm-gas and liquid outlets. The wording specifies that the outgoing gas is at 14 K; it does not require the outgoing liquid to have that temperature. The two outlet streams need not be in mutual thermal equilibrium. Taking $H$ as the printed specific enthalpy, the steady-flow energy balance givesFor fixed outlet enthalpies the greatest yield occurs at the smallest inlet enthalpy. Among the tabulated pressures this is 30 atm. Approximating the atmospheric-pressure gas enthalpy by the tabulated zero-pressure limiting value givesThis uses a low-pressure approximation: $p_0=1$ atm is not literally the zero-pressure table entry. Linear interpolation to 1 atm would instead give $H(p_0)\simeq 8.651\times 10^4\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$ and $\alpha \simeq 0.193$. The sparse table does not determine an exact continuous optimum between entries.

Atmospheric-pressure liquid helium must be much colder than the 14 K gas outlet: its normal boiling point is about 4.224 K.^[ref] A counterflow heat exchanger can return cold gas through the incoming stream before that gas leaves the overall control volume, while the collected liquid leaves separately at the cold end. This is a possible interpretation of the whole apparatus, rather than an extra diagram specified by the question. The enthalpy calculation needs only the stated inlet and outlet properties; it does not assume one equilibrium vessel containing both phases at 14 K. In particular, $H(p_0)-H_{\ell }$ is an outlet-stream enthalpy difference, not the latent heat of vaporization at one common temperature.

Let L be the vessel initially containing argon and R the vessel initially containing helium. Both have volume $V$ and fixed temperature $T_0$, and the aperture has area $A_h$. The incoming and outgoing rates of each species must be considered separately. For species $s$,Thus its density difference relaxes exponentially with time constantThe two-vessel effusive mixing solution for the four densities isUsing atomic masses of approximately $40$ and $4$ atomic mass units gives ${\tau }_{\mathrm{He}}/{\tau }_{\mathrm{Ar}}\simeq 1/\sqrt{10}\simeq 0.32$. Helium therefore transfers faster. Each species eventually reaches $n_0/2$ in each vessel, but its transient is different. In particular, the total density temporarily rises in the initially argon vessel and falls in the initially helium vessel, before both return to $n_0$.

For unequal maintained temperatures $T_L,T_R$, replace the rate equation byWith each species' total density sum still $n_0$, its final densities and relaxation time areEvery species has the same final density ratio $\sqrt{T_R/T_L}$ and partial-pressure ratio $\sqrt{T_L/T_R}$. Thus there is no permanent composition separation in this ideal steady state, although the rates remain mass dependent. A maintained hot side has lower density and higher pressure. Equal particle rates do not imply thermal equilibrium: atoms arriving from the hotter side carry more energy, so external heating and cooling are required to hold different temperatures. The model assumes thin-hole effusion, local equilibrium within each vessel and negligible aperture volume.

For a classical dilute ideal gas in thermal equilibrium, the probability of a single-particle velocity is proportional to the Boltzmann factor for kinetic energy, $e^{-m{v}^2/(2k_{B}T)}$. Velocities with speed between $v$ and $v+dv$ fill a spherical shell of velocity-space volume $4\pi v^{2}dv$. The normalized Maxwell speed distribution is thereforeThe factor $v^2$ counts states, rather than representing an extra Boltzmann energy factor.

Set $\alpha =m/(2k_{B}T)$ and use $I_j={\int }_0^{\infty }{v}^j{e}^{-\alpha v^2}dv$. Normalization cancels in ratios, so the given recurrence givesConsequentlyThe root mean square speed is larger than the mean speed, consistent with the positive variance of the speed distribution.

Equilibrium radiation and cavity frequencies. A blackbody absorbs all incident radiation at every frequency and direction. In thermal equilibrium its emission spectrum depends only on its temperature. Radiation inside an isothermal absorbing cavity, sampled through a small aperture that hardly disturbs equilibrium, realizes black-body radiation to a good approximation. Microscopic absorption and emission establish the equilibrium distribution; a perfectly reflecting wall is the convenient idealization used to count the modes once that distribution is specified.

In a rectangular electromagnetic cavity, the wave equation and perfect conductor wall boundary conditions select component wavenumbers $k_x=\pi l/L_x$, $k_y=\pi m/L_y$, $k_z=\pi n/L_z$. Combining the vacuum dispersion relation $\omega =c|\mathbf{k}|$ and $\nu =\omega /(2\pi )$ givesFor example, electric-field components can have the spatial dependenceFor three positive indices, Gauss's law requires $k_x{A}_x+k_y{A}_y+k_z{A}_z=0$, leaving two independent polarizations of an electromagnetic wave.

Index qualification. The original paper describes the indices as positive integers. That describes the three-nonzero-index family but omits valid electromagnetic modes with exactly one zero index, which have one polarization of an electromagnetic wave. For instance, with $n=0$, $l,m\ge 1$, only $E_z$ above remains; it is normal to the top and bottom faces and vanishes at the side walls. Two zero indices give no nontrivial cavity field. The omitted one-zero-index family does not affect the leading three-dimensional mode count, but is essential to the thin-cavity limit below. Retaining the paper's strict positivity restriction for every mode would instead give a thickness gap and would not yield its requested two-dimensional power law.

Three-dimensional mode count. A lattice cell in positive $\mathbf{k}$ space has volume ${\pi }^3/V$, where $V=L_x{L}_y{L}_z$. In a continuum approximation, a shell of radius $k$ and thickness $dk$ contains its positive octant, multiplied by two polarizations:Using $k=2\pi \nu /c$ givesFor a finite cavity the exact density of states is discrete; this formula is its bulk continuum limit, appropriate when many modes contribute and wall corrections are small.

Mean photon number. A mode is a quantum harmonic oscillator with thermal excitation energies $sh\nu$, $s=0,1,2,\dots$. The zero-point energy does not enter its thermal occupation. Put $q=e^{-h\nu /(k_{B}T)}$. Its partition function and mean occupation areThis is the Bose-Einstein distribution with zero photon chemical potential: equilibrium absorption and emission do not conserve photon number.

Thermal energy. Multiply the density of states by the mean excitation energy $h\nu ⟨s⟩$ and integrate:The substitution $x=h\nu /(k_{B}T)$ isolates all the dependence on cavity dimensions and temperature:Expanding $1/(e^x-1)={\sum }_{r=1}^{\infty }{e}^{-rx}$ gives an integral ${\sum }_{r}6/r^4=6\zeta (4)={\pi }^4/15$. HenceThis is the thermal energy only, excluding the zero-point contribution. The three-dimensional continuum regime requires $k_{B}T$ large compared with the geometric level spacings in all three directions.

Spectral sketch. The spectral energy density per unit ordinary frequency is Planck's law,At small frequency it rises as $8\pi k_{B}T{\nu }^2/c^3$, while at large frequency it decays exponentially. Raising temperature increases the spectrum at every fixed positive frequency, moves the peak to larger frequency in proportion to $T$, and increases the integrated area in proportion to $T^4$. A plot per unit wavelength has a different peak position because the integration variable changes.

Thin-cavity radiation. Let $A=L_x{L}_y$. The relevant low-temperature window isNonzero thickness indices are then thermally suppressed, while many in-plane modes are populated. The $n=0$ family has one polarization of an electromagnetic wave. Its allowed in-plane wavenumbers form a lattice of cell area ${\pi }^2/A$; a quarter of a circular shell therefore containsThus the dimensional crossover of thermal cavity radiation givesThe integral is $2\zeta (3)$ by the same geometric-series argument. A model with a different number of independent polarizations of an electromagnetic wave changes the prefactor but not the $T^3$ scaling. For a cavity with fixed finite $L_x,L_y$, cooling still further eventually freezes its lowest in-plane mode as well: the strict $T\to 0$ thermal energy is exponentially suppressed. The requested area law is the intermediate quasi-two-dimensional regime, not an arbitrarily low-temperature continuum limit.

Origin of diffusion. In a material near local thermal equilibrium, microscopic carriers transfer energy from hotter regions to colder regions. On distances long compared with their mean free path and times long compared with relaxation times, the local constitutive law is Fourier's law,Combining this temperature-gradient-driven flux with local conservation of energy produces a diffusion equation. This description assumes diffusive thermal conduction; it is not generally valid for ballistic transport, and flow would add advective transport.

Energy balance with heating. Choose any fixed volume $V$, let $c$ be its uniform volumetric heat capacity and $H$ its heat generation rate per unit volume. The rate of increase of internal energy equals heat generation minus outward heat flux:Since $V$ is arbitrary, for constant thermal conductivityThe ratio $\kappa /c$ is the thermal diffusivity in ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Here $c$ is in $\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$, $\kappa$ in $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and $H$ in $\mathrm{W}{\mathrm{m}}^{-3}$, so both right-hand terms have units $\mathrm{K}{\mathrm{s}}^{-1}$. If thermal conductivity varies with position or temperature, the divergence $\nabla \cdot (\kappa \nabla T)$ must be retained rather than replacing it by $\kappa {\nabla }^{2}T$.

Temperature as an equilibrium relation. Classical thermodynamics begins with the zeroth law of thermodynamics: thermal equilibrium is transitive. A thermometer can therefore assign a common empirical temperature to systems in mutual thermal equilibrium. Different thermometer properties initially give different monotone empirical scales; a resistance reading by itself does not define the absolute scale.

The Second law of thermodynamics provides a substance-independent thermodynamic scale. A reversible engine operating between reservoirs has a heat ratio depending only on their temperatures. One chooses absolute temperature so that $Q_h/Q_c=T_h/T_c$, with $Q_h,Q_c$ the positive heat magnitudes. Its reversible efficiency is $1-T_c/T_h$. Equivalently, reversible heat defines entropy through $dS=\delta Q_{\mathrm{rev}}/T$. For an equilibrium simple compressible system,Classical thermodynamics uses this relation without needing a microscopic account of entropy. A dilute-gas thermometer realizes an empirical scale approaching the thermodynamic scale after corrections for nonideal behavior.

Statistical meaning of the same derivative. In the microcanonical description, an equilibrium energy shell containing $\Omega (U,V,N)$ accessible states has $S=k_B\ln \Omega$. Its statistical temperature is consequentlyTo see why this is an equilibrium temperature, let two weakly coupled systems exchange a small energy while their total energy stays fixed. The combined multiplicity is ${\Omega }_A(U_A){\Omega }_B(U_{\mathrm{tot}}-U_A)$. Maximizing its logarithm giveswhich is equality of the two inverse temperatures. Stability requires the total entropy maximum, not only stationarity. The overwhelmingly dominant macrostate in a large system therefore realizes the thermodynamic equilibrium condition.

A large reservoir gives a small subsystem the canonical probabilities $p_i\propto e^{-\beta E_i}$, with $\beta =1/(k_{B}T)$. In that ensemble, $F=-k_{B}T\ln Z$ and $S=-k_B{\sum }_i{p}_i\ln p_i$ agree with macroscopic thermodynamic relations in their usual equilibrium regime. The equipartition identity for a classical dilute monatomic gas, $⟨p^2/(2m)⟩=3k_{B}T/2$, is a useful operational connection between microscopic motion and temperature, but is not a universal definition: quantum modes need not obey equipartition, and a system away from equilibrium need not have one temperature at all.

For finite systems one must specify the entropy convention and energy resolution; surface and fluctuation corrections can distinguish definitions that agree in the ordinary thermodynamic limit. Bounded-energy population-inverted systems can also have a negative entropy derivative and hence a negative statistical temperature. That possibility lies outside an ordinary positive-temperature ideal-gas thermometer; it does not mean temperature is merely an arbitrary average kinetic energy.

Obtaining the energy–temperature conversion. The Boltzmann constant connects entropy measured in $\mathrm{J}{\mathrm{K}}^{-1}$ with the dimensionless logarithm of a state count. Historically it could be determined from the ideal gas law $pV=N{k}_{B}T$: measure the molar gas constant $R$ and independently determine the Avogadro constant $N_A$, then obtain $k_B=R/N_A$. The independent particle-number determination matters; using $R$ alone would not fix energy per particle.

An electrical method uses Johnson-Nyquist noise. A resistor of resistance $R_e$ at known temperature has, in the classical range $hf\ll k_{B}T$, a one-sided open-circuit voltage-noise spectrum $S_V(f)=4k_{B}T{R}_e$. Over a calibrated flat bandwidth $\Delta f$,A practical measurement uses an equilibrium resistor, a calibrated low-noise amplifier and a known transfer function, subtracts or independently characterizes amplifier noise, and replaces a nominal bandwidth by the actual effective noise bandwidth. Varying temperature and fitting the mean-square voltage is more informative than treating one amplified voltage sample as the thermal variance. Other primary methods relate temperature to sound speed in a dilute gas or to calibrated particle diffusion and mobility.

There is a historically relevant unit-definition qualification: by the date of this exam, the revised SI had already taken effect on 20 May 2019. The kelvin is now defined by fixingThe same experiments now realize or compare thermodynamic temperatures using this fixed constant, whereas historically they measured the conversion constant on an independently defined temperature scale. The physical distinction between equilibrium temperature and microscopic state counting is unchanged.^[ref]

The expression is the canonical partition function for a system in thermal equilibrium with a large heat bath at fixed temperature $T$. The system exchanges energy with the bath, while its particle number and external mechanical parameters, such as volume, are fixed. The bath is large enough that its temperature is not appreciably changed by this exchange.

The system has allowed energy levels ${ϵ}_j$. The index $j$ labels distinct energy values, and the integer degeneracy factor $g_j$ counts distinct quantum states with that same energy. $k_B$ is the Boltzmann constant, and $\exp [-{ϵ}_j/(k_{B}T)]$ is a state's Boltzmann factor. The dimensionless sum $Z$ normalizes the probabilities:Thus the degeneracy multiplies the probability of occupying the level, not the probability assigned to each individual state within it.

The canonical model is not restricted to a classical ideal gas. It applies to a quantum rotor, for example, provided the sum counts all of that system's allowed states with their proper degeneracies. For a many-particle system, ${ϵ}_j$ would be the energy of a many-particle state, with any quantum-statistical restrictions already built into the state count.

(a) Treat the lake as a heat bath at fixed temperature $T_0=288.15\mathrm{K}$ and the block's heat capacity as constant. Its initial temperature is $T_i=473.15\mathrm{K}$, and $C=mc=355\mathrm{J}{\mathrm{K}}^{-1}$. Although the cooling is irreversible, entropy is a state function, so evaluate the block's change along an imagined reversible thermodynamic process:The lake receives heat $Q=C(T_i-T_0)=65675\mathrm{J}$, and thereforeAdding the two contributions givesThe positive result is consistent with the Second law of thermodynamics for heat transfer across a finite temperature difference. The calculation uses $c=710\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ exactly as printed in both the original PDF and the transcription; it does not replace the supplied value by a tabulated value for lead.

(b) The block starts and eventually finishes at the lake's temperature, so its net entropy change is approximately zero. Neglecting buoyancy, air resistance and the final immersion depth, the lost gravitational potential energy $mgh$ becomes internal energy in the lake after the motion and waves have dissipated. The lake remains at $T_0$, givingTemporary heating of the block or temporary storage in kinetic energy does not change this final energy and entropy balance once thermal equilibrium is restored.

Let $N_u,N_l$ be the upper- and lower-level populations and $g_u,g_l$ their degeneracy factors. At thermal equilibrium, the Boltzmann distribution givesNo degeneracy factors are supplied, so the numerical answers use equal degeneracy factors. Stating the ratio as upper population divided by lower population removes an otherwise arbitrary inversion.

(a) At $T=500\mathrm{K}$,The two populations are nearly equal because the level separation is much smaller than the thermal energy $k_{B}T$.

(b) At $T=2\mathrm{mK}=0.002\mathrm{K}$,Almost every atom occupies the lower level. If the levels have unequal degeneracy factors, each boxed ratio must be multiplied by $g_u/g_l$; their individual values cannot be inferred from the paper.

Each independent cavity mode is a quantum harmonic oscillator of angular frequency $\omega$. Its occupation numbers $n=0,1,\dots$ represent the numbers of photons in that mode. At thermal equilibrium, photons can be absorbed and emitted by the walls, so their number is not conserved and the photon chemical potential is zero. The Boltzmann distribution of the oscillator excitations then givesThe expression supplied in the paper is the thermal excitation energy. The oscillator's temperature-independent zero-point energy is excluded from the emitted thermal radiation.

Putting $ϵ=\hslash \omega$ and $dϵ=\hslash d\omega$ into the mode count from (a) gives the number of modes per unit cavity volume:Multiplying by the mean thermal energy in each mode yields Planck's law in the angular-frequency convention:The two photon polarizations are already included in the density of states; no further factor of two is needed. $u(T,\omega )d\omega$ has units $\mathrm{J}{\mathrm{m}}^{-3}$, whereas $u(T,\omega )$ has units $\mathrm{J}\mathrm{s}{\mathrm{m}}^{-3}$.

The equilibrium photon gas is isotropic, so the energy density per unit solid angle in a narrow angular-frequency interval is $u(T,\omega )d\omega /(4\pi )$. For rays making angle $\theta$ with an outward surface normal, a surface element $dA$ is crossed in time $dt$ by rays initially in a volume $cdtdA\cos \theta$. Only the outward hemisphere contributes. ThusThe hemispherical restriction supplies a factor $1/2$ and the directional projection supplies another $1/2$. A small aperture in a cavity therefore emits this spectral power per unit area. A blackbody at the same temperature emits the same amount: otherwise radiation exchange between it and the aperture would violate thermal equilibrium.

Integrating Planck's law over all positive angular frequencies gives the total emitted heat flux densitySet $x=\hslash \omega /(k_{B}T)$. ThenThe remaining integral is independent of temperature, proving $M\propto T^4$. Its value follows by expanding $(e^x-1{)}^{-1}={\sum }_{m=1}^{\infty }{e}^{-mx}$ and integrating termwise:Consequently the Stefan–Boltzmann law isThis is emitted power alone. A body in a radiation environment also absorbs incoming radiation, which must be included when calculating net radiative cooling.

The fluctuation-dissipation theorem relates spontaneous fluctuations in thermal equilibrium to the dissipative part of the linear response to a weak conjugate force. For an overdamped trapped coordinate, the Langevin equation isThe fluctuation-dissipation relation for a Langevin particle fixes the noise strength from the same drag coefficient $\gamma$ that dissipates motion. With a drive proportional to $e^{-i\omega t}$, the linear response function is $\chi (\omega )=1/(K-i\gamma \omega )$. The two-sided angular-frequency power spectral density then obeysIts integral $\int S_{xx}d\omega /(2\pi )=k_{B}T/K$ agrees with the equipartition theorem. Without a restoring force, the same noise-drag relation gives the diffusion-mobility relation $D=k_{B}T/\gamma$.

In thermal calibration of an optical trap, record a bead's spontaneous displacement fluctuations and compare their spectrum with its response to a weak calibrated oscillatory force. The overdamped harmonic-trap power spectrum and the dissipative response must be consistent with the same temperature and drag. The equilibrium and weak-drive conditions are essential: an actively driven bath need not obey this equilibrium relation.

An intrinsic semiconductor has negligible electrically active impurity doping, and thermal excitation across its band gap produces equal electron and hole densities $n=p=n_i$. In an extrinsic sample, semiconductor doping supplies donor impurities or acceptor impurities, so one carrier species dominates over a range of temperatures. Doping does not remove the possibility of low-temperature dopant freeze-out or high-temperature intrinsic behavior.

Assume parabolic isotropic band edges, spin degeneracy two, a single valley, thermal equilibrium and nondegenerate carriers. Write $E_F$ for the Fermi level, and measure conduction-band energy by $E=E_c+ϵ$. Counting momentum states gives the density of states per unit volumeThe low-occupation Maxwell–Boltzmann statistics givesThe substitution $x=ϵ/(k_{B}T)$ and ${\int }_0^{\infty }\sqrt{x}{e}^{-x}dx=\sqrt{\pi }/2$ give the coefficient. Similarly $p=N_v{e}^{(E_v-E_F)/(k_{B}T)}$ and $N_v=2[m_h^{\ast }{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. Multiplication eliminates $E_F$, proving the semiconductor mass-action lawHere the supplied effective masses are treated as the masses entering the effective density of states. A detailed multivalley silicon model would require the appropriate valley degeneracy and density-of-states masses consistently; adding an extra factor without changing the stated model would alter the requested expression.

A pentavalent arsenic impurity supplies a weakly bound excess electron. In the hydrogenic donor state approximation, dielectric screening reduces the Coulomb attraction and the band effective mass replaces the free mass. Hence the donor binding energy isThis is the effective-mass estimate, without a material-specific central-cell correction.

For the temperature plot use donor density $N_D=10^{15}{\mathrm{cm}}^{-3}$ and no acceptors. The charge neutrality with partial donor ionization model puts $E_D=E_c-E_b$ and includes the twofold spin degeneracy of an occupied donor:These equations give three regimes. At low temperature, $p$ is negligible and strong dopant freeze-out gives $n\simeq \sqrt{N_D{N}_c/2}{e}^{-E_b/(2k_{B}T)}$. At intermediate temperature, the donors are almost ionized and intrinsic holes remain rare, so $n\simeq N_D$. At high temperature, $n\simeq p\simeq n_i\propto T^{3/2}{e}^{-E_g/(2k_{B}T)}$. Thus the log-density plot against inverse temperature has a steep intrinsic segment on the left, a nearly horizontal extrinsic plateau, and a freeze-out decline on the right; the $T^{3/2}$ and $T^{3/4}$ factors give slight curvature.

To quantify “approximately constant”, require $0.9N_D\le n\le 1.1N_D$. At $300\mathrm{K}$, the specified model gives $N_c\simeq 4.12\times 10^{18}$, $N_v\simeq 8.61\times 10^{18}$ and $n_i\simeq 2.33\times 10^9{\mathrm{cm}}^{-3}$. The full neutrality equations give the ten-percent plateau intervalThe lower boundary is set by donor ionization and the upper boundary by intrinsic electron-hole production. The large ratio $N_c/N_D$ allows substantial donor ionization even when $k_{B}T<E_b$; using only $k_{B}T\gg E_b$ would incorrectly exclude much of the plateau. The quoted interval depends on the chosen tolerance and the constant-gap band model, rather than being an exact measured range for arsenic-doped silicon.

Band edges and optical transitions. A direct band gap has the conduction minimum and valence maximum at the same crystal momentum. A near-edge photon then connects the two almost vertically on an energy-versus-wavevector plot, because its momentum is small on the Brillouin zone scale. An indirect band gap has the extrema at different momenta, so an additional momentum transfer, commonly supplied by phonon absorption or emission, is required for a near-edge radiative transition. This reduces its probability relative to an allowed direct transition.

Near the zone centre, an isotropic direct-gap model has $E_c(k)=E_c+{\hslash }^2{k}^2/(2m_e^{\ast })$ and downward valence dispersions $E_{v,j}(k)=E_v-{\hslash }^2{k}^2/(2m_{h,j}^{\ast })$. The heavy-hole and light-hole bands are the upper branches with large and small positive hole masses: the heavy branch is flatter. In common cubic semiconductors these arise from the $j=3/2$ manifold, while spin-orbit coupling places the $j=1/2$ split-off valence band lower by ${\Delta }_{\mathrm{so}}$. Its thermally generated hole population is suppressed by approximately $e^{-{\Delta }_{\mathrm{so}}/(k_{B}T)}$ relative to upper-valence holes, apart from density-of-states factors. Neglect is justified when ${\Delta }_{\mathrm{so}}\gg k_{B}T$; it is not a universal exclusion of this band at arbitrary temperature.

For an indirect material such as Si, the conduction minimum need not be at $\Gamma$. In applying the carrier-density formulae, $E_c$ must mean the actual minimum of the included conduction valley. The supplied scalar effective masses and densities of states constitute a simplified band-edge model; they do not specify Si's full valley anisotropy or multiplicity.

Carrier populations. Let $f(E)$ be the Fermi-Dirac distribution. In the nondegenerate conduction tail, $f(E)\simeq e^{-(E-\mu )/(k_{B}T)}$. Set $x=E-E_c$ and integrate the given density of states:The integral is $\Gamma (3/2)(k_{B}T{)}^{3/2}=\sqrt{\pi }(k_{B}T{)}^{3/2}/2$. Defining the effective density of states givesThe factor two is already included in the supplied density of states; it must not be added a second time.

A hole is an unoccupied valence state. For $E$ sufficiently below $\mu$, $1-f(E)\simeq e^{(E-\mu )/(k_{B}T)}$. Substitution of $x=E_v-E$ gives in exactly the same wayTheir product eliminates the chemical potential:This is the semiconductor mass-action law at thermal equilibrium. In an intrinsic semiconductor, charge neutrality additionally givesIn a doped semiconductor the product still applies in this nondegenerate equilibrium model, but the two populations need not be equal.

Intrinsic chemical potential. Taking logarithms of $n=p$ yields the intrinsic-semiconductor chemical potentialFor unequal band degeneracies, use the first form with the full $N_c,N_v$. For the supplied mass ratio $m_h^{\ast }/m_e^{\ast }=1/2$, the shift from midgap at $300\mathrm{K}$ is approximately $-13.4\mathrm{meV}$, toward the valence edge.

Three temperature regimes. The original plot runs from high temperature on the left to low temperature on the right. At high temperature, thermal excitation across the gap dominates and $n\simeq p\simeq n_i$ rises rapidly with temperature. In the middle, donors and acceptors are effectively ionized, the material is extrinsic, and $n\simeq N_D-N_A$ makes the plateau. At low temperature, dopant freeze-out binds increasing numbers of electrons to donor levels, reducing the mobile-electron population as the sample cools. Its activation scale is a donor binding energy, not the full band gap. Compensation and incomplete ionization affect that regime.

Gap estimate from the original figure. On the dashed intrinsic branch, $n_i=C{T}^{3/2}{e}^{-E_g/(2k_{B}T)}$. Hence band-gap extraction from intrinsic carrier activation uses the slope of $\ln (n_i/T^{3/2})$ versus $1/T$. With $x=1000\mathrm{K}/T$ and $k_B$ expressed in eV/K, two points giveThe very compressed intrinsic branch is only approximately readable: $n_i\sim 10^{15}\mathrm{c}{\mathrm{m}}^{-3}$ near $x\sim 1.8$ and $n_i\sim 10^{13}\mathrm{c}{\mathrm{m}}^{-3}$ near $x\sim 2.5$. These give $E_g\sim 1.0\mathrm{eV}$, rather than a precision determination. Neglecting the thermal prefactor gives about $1.13\mathrm{eV}$ from those same rough points. Thus a gap of order $1$–$1.1\mathrm{eV}$ is justified by this schematic plot; its resolution does not support distinguishing a few hundredths of an electronvolt. The declining freeze-out branch must not be used to infer the full gap.

Room-temperature densities. At $300\mathrm{K}$, the supplied masses giveIn the fully ionized extrinsic plateau, put $N=N_D-N_A=10^{15}\mathrm{c}{\mathrm{m}}^{-3}$. Charge neutrality and the mass-action law implySince $n_i\ll N$, $n\simeq 10^{15}\mathrm{c}{\mathrm{m}}^{-3}$ is insensitive to the imprecise gap. The minority density is much more sensitive:Using $E_g=1.0\mathrm{eV}$ gives $n_i\simeq 5.8\times 10^{10}\mathrm{c}{\mathrm{m}}^{-3}$ and $p\simeq 3.3\times 10^6\mathrm{c}{\mathrm{m}}^{-3}$. Using $1.1\mathrm{eV}$ instead gives $n_i\simeq 8.3\times 10^9\mathrm{c}{\mathrm{m}}^{-3}$ and $p\simeq 7.0\times 10^4\mathrm{c}{\mathrm{m}}^{-3}$. These are explicitly conditional estimates within the stated mass model. The plot alone cannot fix a precise minority density because the uncertain gap occurs inside an exponential.