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There are four cation sites per conventional unit cell, hence $0.4$ Gd substitutions per cell on average. Two substitutions of Gd${}^{3+}$ for Ce${}^{4+}$ require one oxygen vacancy for charge neutrality, so the cell contains $0.2$ vacancies. Equivalently, the composition is $\mathrm{C}{\mathrm{e}}_{0.9}\mathrm{G}{\mathrm{d}}_{0.1}{\mathrm{O}}_{1.95}$. ThereforeThis uses the supplied lattice parameter literally and treats it as the conventional fluorite-cell edge.

In a large volume, one one-particle state occupies momentum-space volume $h^3/V$. A spherical shell of radius $p$ has volume $4\pi p^{2}dp$, so for spin degeneracy $g$ the density of states is $4\pi gV{p}^{2}dp/h^3$. With ultra-relativistic dispersion $E=cp$ and zero chemical potential, the Bose-Einstein distribution givesThe convergent dimensionless integral is ${\pi }^4/15$, obtained by expansion into exponentials and summing $\sum n^{-4}$. Therefore $ϵ\propto T^4$ and $\alpha =4$. Photons are massless bosons with two polarizations; equilibrium emission and absorption do not conserve photon number and force ${\mu }_{\gamma }=0$.

For recombination, chemical equilibrium imposes ${\mu }_p+{\mu }_e={\mu }_H+{\mu }_{\gamma }={\mu }_H$, including rest energies in the chemical potentials. Divide the three nonrelativistic number-density expressions to obtainHere $m_p{c}^2+m_e{c}^2-m_H{c}^2=I$. Neglect the relative proton/hydrogen mass difference and take the ground-state degeneracy ratio $g_p{g}_e/g_H=1$ (electron and proton spins give $2\cdot 2$ and hydrogen gives $4$). Charge neutrality gives $n_p=n_e$. This yields the stated Saha equation

With $n_e=X_e{n}_B$ and $n_H=(1-X_e)n_B$, the ratio asked for isEquivalently the prefactor is $16\pi \zeta (3)\eta (2\pi {)}^{-3/2}[kT/(m_e{c}^2){]}^{3/2}$. Because photons vastly outnumber baryons, even a small high-energy tail can keep hydrogen ionized. Neutralization therefore requires exponential suppression of that tail far below $kT=I$, explaining the much lower recombination temperature. The equilibrium formula is an estimate; the actual expanding-universe recombination kinetics eventually depart from Saha equilibrium.

Ignoring helium, charge neutrality gives $n_e=n_p$, while baryon conservation gives $n_B=n_p+n_H$. Defining the hydrogen ionization fraction $X_e=n_e/n_B$ yields $n_p=n_e=X_e{n}_B$ and $n_H=(1-X_e)n_B$. Substituting in the abundance relation givesUse $n_B=\eta n_{\gamma }$ and $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$ to obtain the Saha equation:If the positive right side is denoted by $A(T)$, the physical root isThis form avoids subtracting nearly equal numbers when $A$ is small, and makes the limits transparent: $X_e\simeq 1-A$ for $A\ll 1$, whereas $X_e\simeq A^{-1/2}$ for $A\gg 1$. It is an equilibrium prediction; a kinetic cosmological recombination treatment is needed after reactions cease to keep up with expansion.

Write $n_b=n_p+n_H$ for the cosmological baryon density in this hydrogen-only model, and $x_e=n_p/n_b$ for the hydrogen ionization fraction. By charge neutrality, $n_e=n_p$, so the Saha ionization equation becomesThe baryon-to-photon ratio is $\eta =n_b/n_{\gamma }$, and the equilibrium photon number density is $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$. At an order-one hydrogen ionization fraction, neglecting factors of order one therefore givesReplacing $T$ inside the slowly varying logarithm by $B$ givesWith $B=13.6\mathrm{eV}$, $M_e=5\times 10^5\mathrm{eV}$ and $\eta =10^{-10}$, this is $T\simeq 0.351\mathrm{eV}\simeq 4.1\times 10^3\mathrm{K}$. Solving the preceding implicit approximation instead gives $T\simeq 0.306\mathrm{eV}\simeq 3.5\times 10^3\mathrm{K}$. Both are reasonable estimates of the recombination temperature at the accuracy requested. The small baryon abundance delays hydrogen recombination: even when $T\ll B$, the high-energy tail of the abundant photons can still ionize the much rarer hydrogen atoms.

Let $f(\mathbf{p})d^{3}p$ be particle number per unit volume in a momentum element, including the spin-state count in $f$. This is the local momentum part of a phase-space distribution function, normalized by the number density $n=\int f(\mathbf{p})d^{3}p$. For an isotropic distribution with particle energy $E(p)=\sqrt{m^2{c}^4+p^2{c}^2}$, the kinetic pressure of an isotropic gas is the average momentum flux:The kinetic energy density, explicitly excluding rest mass, isFor $p\ll mc$, $E-m{c}^2=p^2/(2m)$ and $\mathbf{v}=\mathbf{p}/m$, giving $P=2u_{\mathrm{kin}}/3$. For $p\gg mc$, $E\simeq pc$ and $v\simeq c$, giving $P\simeq u_{\mathrm{kin}}/3$. ThusNeither relation assumes a Maxwellian distribution. At intermediate momenta neither constant ratio is exact, and anisotropic distributions require a pressure tensor rather than this scalar pressure.

For a classical Maxwell-Boltzmann distribution, $f=n(2\pi m{k}_{B}T{)}^{-3/2}\exp [-p^2/(2m{k}_{B}T)]$. Three Gaussian component integrals give $⟨p^2⟩=3m{k}_{B}T$, so the ideal gas hasIn a fully ionized mixture, sum over independent species to obtain $P_{\mathrm{gas}}=\rho k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$, where the mean molecular weight counts ions and free electrons. This excludes partial-ionization and interaction corrections. A dilute classical ultrarelativistic gas still has $P=n{k}_{B}T$ but $u_{\mathrm{kin}}\simeq 3n{k}_{B}T$; its momentum distribution is proportional to $e^{-pc/(k_{B}T)}$ instead of the nonrelativistic Gaussian.

For a thermal photon gas, the two polarization states and zero photon chemical potential give the Planck photon distributionUse $x=pc/(k_{B}T)$ and ${\int }_0^{\infty }{x}^3/(e^x-1)dx={\pi }^4/15$. The radiation constant and photon equation of state are thereforeLikewise $n_{\gamma }=2\zeta (3)(k_{B}T/\hslash c{)}^3/{\pi }^2$. Unlike a gas with a fixed particle number, photons do not have pressure proportional to baryonic mass density.

For fully degenerate electrons, the Fermi-Dirac distribution becomes a filled momentum sphere with two spin states. Counting them gives the Fermi momentumHere ${\mu }_e$ is the mean molecular weight per electron. The equation of state of a cold electron gas follows by integrating momentum flux up to $p_F$:Its two limits areThe corresponding kinetic energy densities are $3n_e[p_F^2/(2m_e)]/5$ and $3n_e{p}_{F}c/4$, respectively. Thus in the high-density relativistic limit the electron pressure scales as ${\rho }^{4/3}$ and is nearly independent of temperature. This is the ideal noninteracting, fixed-composition electron result, not a universal equation of state at nuclear densities where captures, interactions and the composition change.

The boundaries on a stellar equation-of-state regime diagram concern the electron component. Define the electron relativistic density threshold and thermal electron relativistic threshold byA degenerate electron gas is nonrelativistic well below ${\rho }_{\ast }$ and ultrarelativistic well above it: this is an approximately vertical division on a log-density plot. A nondegenerate electron gas instead becomes thermally relativistic near $T_{\ast }$: this is an approximately horizontal division. The regimes have broad crossovers rather than a discontinuity at either line.

The exact electron Fermi temperature, subtracting electron rest energy, isFor electrons with negligible thermal pairs, strong degeneracy requires $T\ll T_F$, whereas $T\gg T_F$ gives a nondegenerate gas. In the pair-rich regime the actual electron and Positron distributions must instead be treated with their chemical potentials, as discussed below. The two limiting degeneracy boundaries areThese slopes $2/3$ and $1/3$ explain the bent degeneracy boundary on the logarithmic graph. A thermal-wavelength test $n_e[2\pi {\hslash }^2/(m_e{k}_{B}T){]}^{3/2}\sim 2$ gives the same nonrelativistic density/temperature scaling, with an order-one definition of the crossover.

The radiation-to-gas pressure boundary in the nondegenerate fully ionized regime isRadiation dominates above this line. Once electrons are strongly degenerate, compare radiation pressure with $P_e(\rho )$ instead of continuing the ideal-gas comparison into that regime. The radiation-to-degeneracy pressure boundary is $T=[3P_e(\rho )/a_{\mathrm{rad}}{]}^{1/4}$, with logarithmic slopes $5/12$ and $1/3$ in the nonrelativistic and ultrarelativistic limits. Degeneracy of the electrons and dominance of their pressure are distinct criteria.

For the pair curve, distinguish baryonic net electrons from thermally created electrons and Positrons. Chemical equilibrium with photons requires opposite electron/Positron chemical potentials when the one-particle energy includes rest energy, as in the distributions below. At low temperature and low degeneracy the zero-chemical-potential density per charge species isCharge neutrality gives $n_{-}-n_{+}=n_b=\rho /({\mu }_e{m}_u)$, while the nondegenerate equilibrium product is $n_{-}{n}_{+}=n_0^2$. HencePairs become important when $n_0$ is comparable with or exceeds $n_b$, approximately the electron-positron thermal pair abundance curve ${\rho }_{\mathrm{pair}}(T)={\mu }_e{m}_u{n}_0(T)$. Below its density at a fixed temperature, pairs dominate over the net charge electrons. The exponential makes the low-temperature portion steep in a log-log plot. For ${\mu }_e=2$, the full zero-potential integral gives approximate pair-marker densities $1.80\times 10^3\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $10^9\mathrm{K}$ and $7.52\times 10^5\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$ at $3\times 10^9\mathrm{K}$. At relativistic temperature use the full zero-chemical-potential Fermi-Dirac distribution instead; it gives $n_0\to [3\zeta (3)/(2{\pi }^2)](k_{B}T/\hslash c{)}^3$, so the high-temperature pair curve approaches slope three in log-density versus log-temperature. Strong net-electron degeneracy suppresses Positrons and requires the full chemical-potential-dependent distribution.

The original diagram uses fully ionized helium, ${\mu }_e=2$ and ${\mu }_{\mathrm{mol}}=4/3$, only to set numerical locations. It plots the full $T_F$ curve and the zero-chemical-potential pair integral, rather than extending their asymptotes into the crossover. The pair boundary is deliberately an approximate abundance marker, not a phase transition; at high temperature the nonrelativistic ion approximation and fixed-composition model also have limits. For $T\gg T_{\ast }$ with abundant nondegenerate pairs, the combined electron/Positron energy density is $7a_{\mathrm{rad}}{T}^4/4$, so photons plus pairs have $u=11a_{\mathrm{rad}}{T}^4/4$ and $P=u/3$. Real stellar matter adds Coulomb effects, partial ionization, nuclear reactions and, at sufficiently high density, nuclear-matter physics beyond the ideal regime map.

Let $Y_{\mathrm{He}}\simeq 0.25$ be the present primordial helium mass fraction. If $n_{\mathrm{He}}$ is the helium-nucleus density and $n_H$ the hydrogen-nucleus density, thenSince $Y_{\mathrm{He}}=4n_{\mathrm{He}}/n_b$,This retains the correction from neutrons bound in helium. Charge neutrality requires one electron per proton, including bound electrons, so with the baryon-to-photon ratio $\eta =n_b/n_{\gamma }\simeq 6\times 10^{-10}$,

Pair equilibrium with negligible chemical potentials cannot continue once its predicted electron abundance falls below the residual abundance required by charge neutrality. Put $x=m_e/T$ and equate part (b), to logarithmic accuracy, to the order-$10^{-10}$ result of part (c):Taking a logarithm and retaining the numerical factor singled out in the hint givesThe solution is $x\simeq 28$, soThus the zero-chemical-potential electron-positron equilibrium description must fail before roughly $20\mathrm{keV}$; thereafter the small asymmetric electron population survives.

Chemical equilibrium for $p+e^{-}\rightleftharpoons H+\gamma$ gives ${\mu }_p+{\mu }_e={\mu }_H$, because the photon chemical potential vanishes. Applying the nonrelativistic number-density formula to all three massive species and taking the ratio givesUse charge neutrality $n_p=n_e$, the Hydrogen binding energy $E_{\mathrm{bind}}=m_p{c}^2+m_e{c}^2-m_H{c}^2$, $m_H\simeq m_p$, and the ground-state statistical weights $g_H=g_p{g}_e$. ThenThe approximations are a dilute, nondegenerate ideal gas in thermal and chemical equilibrium, nonrelativistic electrons, protons and hydrogen, negligible excited hydrogen states, and negligible corrections from the proton-to-hydrogen mass ratio. This relation is the Saha ionization equation.

Chemical equilibrium for $p+e\leftrightarrow H$ givesand charge neutrality gives $n_p=n_e$. Insert the nonrelativistic Maxwell-Boltzmann number densities, use $m_H{c}^2=m_p{c}^2+m_e{c}^2-E_{\mathrm{bind}}$, and neglect the proton correction in the reduced mass. The spin degeneracies cancel in the ground-state approximation. One obtains Saha equationEquivalently, its reciprocal is the ionization ratio. The exponential favors neutral hydrogen as the temperature falls, while the translational phase-space factor favors ionization at high temperature.

In the nonrelativistic regime,The assumption $E-\mu \gg k_{B}T$ puts the Fermi-Dirac distribution in its Maxwell-Boltzmann limit, soTherefore the nonrelativistic Maxwell--Boltzmann number density is

For $p+e^{-}\leftrightarrow H+\gamma$, chemical equilibrium and the vanishing photon chemical potential implyApplying the number-density formula to each massive species and eliminating the chemical potentials givesUse charge neutrality, $n_p\simeq n_e$, the mass approximation $m_H\simeq m_p$, and the convention in the question that suppresses the order-one internal-degeneracy ratio $g_e{g}_p/g_H$. With the Hydrogen binding energywe obtain the Saha ionization equationThe assumptions are thermal equilibrium and chemical equilibrium before cosmological recombination, a nonrelativistic nondegenerate gas, zero photon chemical potential, charge neutrality, negligible proton-electron mass correction in the translational prefactor, and the stated degeneracy-factor approximation.

The oxygen-pressure exponents can be explained by a change in the dominant charge state of metal vacancies, with holes carrying the p-type current. For a divalent metal oxide, a doubly negative metal vacancy is $V_{\mathrm{M}}^{\prime \prime }$ in Kröger–Vink notation. A conventional oxygen-incorporation reaction isTreating regular oxygen-site activity as constant, the law of mass action gives $K_2=[V_{\mathrm{M}}^{\prime \prime }]p_h^2/p_{{\mathrm{O}}_2}^{1/2}$. If these are the compensating defects, charge neutrality gives $p_h=2[V_{\mathrm{M}}^{\prime \prime }]$, hence $p_h^3\propto p_{{\mathrm{O}}_2}^{1/2}$ and $\sigma \propto p_{{\mathrm{O}}_2}^{1/6}$, assuming nearly constant hole mobility.

For singly negative vacancies, useNow $p_h=[V_{\mathrm{M}}^{\prime }]$ and $p_h^2\propto p_{{\mathrm{O}}_2}^{1/2}$, so $\sigma \propto p_{{\mathrm{O}}_2}^{1/4}$. The charge-state equilibrium $V_{\mathrm{M}}^{\prime }\rightleftharpoons V_{\mathrm{M}}^{\prime \prime }+h^{\bullet }$ gives $[V_{\mathrm{M}}^{\prime \prime }]/[V_{\mathrm{M}}^{\prime }]=K_{\mathrm{ion}}/p_h$: the singly charged fraction can therefore dominate as oxygen pressure and hole concentration increase. This is a consistent dilute-defect model, not a unique experimental identification of every defect. Changes in mobility, defect association or other compensating defects require independent tests.

A Lorentz four-vector is a set of four components transforming as $A^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{A}^{\nu }$ under a Lorentz transformation. Examples include the four-position and four-velocity. A Lorentz invariant is a quantity unchanged by that transformation. With metric signature $(+---)$, the scalar product $A_{\mu }{B}^{\mu }=A^0{B}^0-\mathbf{A}\cdot \mathbf{B}$ is invariant. The components of a four-vector are generally not individually invariant.

The proton's total electric charge is taken to be a scalar: it does not acquire a factor $\gamma$ when the proton moves. Experimental evidence comes from sensitive charge neutrality tests of matter. Hydrogen molecules and helium atoms remain neutral despite different binding and internal electron and nuclear motions. If constituent charge scaled with kinetic energy as relativistic mass does, the same cancellation of fixed numbers of proton and electron charges would generally fail in different bound systems.

In a gas-efflux test of charge neutrality, gas escapes from an electrically isolated vessel connected to an electrometer. Any net charge per escaping molecule would leave a measurable opposite vessel charge, after residual ions and charged contaminants are removed. The hydrogen/helium efflux experiment supports velocity-independent additive charge. ^[ref]

This is evidence for charge invariance, not merely conservation of electric charge: the latter concerns constancy during a process, while invariance compares descriptions or internal motions without rescaling charge by a boost factor. Net-matter neutrality constrains combinations of constituent charges; it should not be described as an exact proof covering every possible velocity-dependent microscopic model.

A completely filled valence band carries no net current: its occupied crystal-momentum states have velocities that cancel. Near a valence-band maximum, an almost full band is conveniently described by the few missing electrons, or electron holes. Removing an electron of charge $-e_0$, with $e_0>0$, changes the charge relative to the filled band by $+e_0$. This is a quasiparticle description of electronic motion, not a mobile positive ion or a Positron.

For an isotropic parabolic maximum, ${ϵ}_v(\mathbf{k})=E_v-{\hslash }^2{k}^2/(2m_h^{\ast })$. Assign the missing-electron state momentum ${\mathbf{k}}_h=-{\mathbf{k}}_e$ and hole energyIt has positive effective mass $m_h^{\ast }$. Its velocity equals the velocity of the missing electron state, and its positive-charge current replaces the negative of that state's electronic current. Semiclassical motion is thus $\hslash {\dot{\mathbf{k}}}_h=+e_0(\mathbf{E}+{\mathbf{v}}_h\times \mathbf{B})$. With momentum relaxation time ${\tau }_h$, the low-field mobility is ${\mu }_h=e_0{\tau }_h/m_h^{\ast }$ andIf electrons also contribute, $\sigma =e_0(p{\mu }_h+n{\mu }_e)$: the two species' longitudinal conductivities add.

A Hall bar separates carrier concentration from mobility. Apply current density $j_x$, magnetic field $B_z$, and measure transverse electric field $E_y$. For one isotropic hole population, transverse force balance gives $E_y=v_x{B}_z$, soFor thickness $t$, the Hall-voltage magnitude is $|V_H|=IB/(e_{0}pt)$. A longitudinal four-terminal resistance measurement supplies $\sigma$ independently. Use ohmic contacts, known conducting thickness, controlled temperature and low enough current to avoid self-heating. Reversing magnetic field removes even-in-field longitudinal offsets from contact misalignment; reversing current also helps reject thermoelectric offsets. These are central controls, rather than merely taking one transverse voltage reading.^[ref]

The simple inference assumes one carrier type and a known Hall scattering factor. More generally $R_H=r_H/(e_{0}p)$, so $\sigma R_H$ measures Hall mobility rather than automatically the drift mobility. With two carrier populations at weak field,Its sign alone need not identify the more numerous carrier. A Hall experiment also cannot determine $m_h^{\ast }$ and ${\tau }_h$ separately: their ratio sets mobility. cyclotron resonance supplies an independent mass measurement by sweeping frequency or magnetic field through $\omega =e_{0}B/m_c$ and detecting resonant absorption. The measured cyclotron effective mass can depend on band anisotropy and need not equal every curvature mass; the isotropic parabolic model identifies them. Its linewidth provides a scattering-time estimate only with a specified damping model.

semiconductor doping generates holes by adding acceptor impurities. In silicon, a substitutional group-III atom such as boron provides one fewer valence electron than silicon. A shallow acceptor level lies slightly above the valence-band edge: accepting a valence electron makes the dopant negatively ionized and leaves a mobile hole in the valence band. The impurity stays fixed in the lattice. charge neutrality and the nondegenerate equilibrium mass-action law giveFor fully ionized, uncompensated acceptors in the extrinsic regime, $p\simeq N_A$ and $n\simeq n_i^2/N_A$; with fully ionized donor compensation, $p\simeq N_A-N_D$ while holes remain dominant. The Fermi level moves towards the valence band. At low temperature, semiconductor carrier freeze-out suppresses mobile electron holes; at high temperature intrinsic electron-hole generation can dominate. Increasing dopant density also changes scattering, so a conductivity increase alone cannot distinguish increased carrier density from changed mobility. Temperature-dependent Hall and conductivity measurements separate these effects, subject to the carrier and scattering qualifications above.

Doping changes the mobile carrier population by introducing controlled impurity levels. In an intrinsic semiconductor, the filled valence band is separated from the nearly empty conduction band by a band gap $E_g$. Thermal excitation produces an electron-hole pair, so the electron and hole number densities obey $n=p=n_i$. A electron hole represents a missing valence-band electron and behaves as a mobile positive charge carrier. For nondegenerate bands in thermal equilibrium,Here $N_C,N_V$ are the effective density of states, $E_C,E_V$ the band edges, and $E_F$ denotes the chemical potential, conventionally called the Fermi level. The semiconductor mass-action law holds in this nondegenerate equilibrium regime.

In a tetrahedrally bonded elemental semiconductor such as silicon, a substitutional group-V atom has an extra outer electron relative to the host. It is a donor impurity: a shallow level just below $E_C$ can ionise, releasing an electron to the conduction band and leaving a fixed positively charged donor. The resulting n-type semiconductor has electrons as majority carriers and holes as minority carriers. Conversely a group-III substitution supplies one fewer bonding electron. An acceptor impurity can take an electron from the valence band, leaving a mobile electron hole and a fixed negative acceptor; the resulting p-type semiconductor has holes as majority carriers.

The impurity state is shallow because the electron or hole responds to a screened Coulomb potential and an effective mass, rather than the bare atomic potential. A simple hydrogenic donor estimate iswhere ${ϵ}_r$ is the relative permittivity. This is a useful scaling model, with real band anisotropy and central-cell effects omitted. Dopants are fixed in the lattice; their ionisation changes mobile carrier numbers without making a uniformly doped bulk sample carry a large net electric charge.

The charge neutrality condition isWhen shallow dopants are almost fully ionised and $N_D-N_A\gg n_i$, an n-type region has $n\simeq N_D-N_A$ and $p\simeq n_i^2/n$. A p-type region similarly has $p\simeq N_A-N_D$ and $n\simeq n_i^2/p$. Thus compensation by the opposite impurity type reduces the majority-carrier density; the total dopant concentration is not generally the mobile carrier density. At sufficiently high impurity density the degenerate semiconductor regime requires the full Fermi-Dirac distribution and these exponential approximations fail.

There are three useful temperature regimes. At low temperature, dopant freeze-out leaves many shallow levels un-ionised. In the extrinsic regime, the dopants are largely ionised and set the majority-carrier density. At high enough temperature, intrinsic pair creation dominates and $n,p$ approach $n_i$. The Fermi level moves towards the conduction band with donor doping and towards the valence band with acceptor doping. The electrical conductivity isDoping therefore controls conductivity and its carrier type, but ionised impurities also scatter carriers and can reduce charge carrier mobility. Carrier density and mobility must both be considered.

A p-n junction forms when p-type and n-type regions are joined within a semiconductor. Initially the concentration gradients drive diffusion current: electrons diffuse from the n region to the p region and holes in the opposite direction. Recombination removes mobile majority carriers near the join, exposing fixed positive donors on the n side and fixed negative acceptors on the p side. This forms a depletion region and an electric field directed from n to p. Its drift current opposes diffusion. At equilibrium the net current vanishes and the Fermi level is uniform, although the band edges vary spatially through band bending.

For an abrupt nondegenerate junction with fully ionised dopants, its built-in potential isUnder the depletion approximation, the space-charge densities are $-e{N}_A$ on a p-side width $w_p$ and $+e{N}_D$ on an n-side width $w_n$. Integrating Poisson equation and imposing zero field outside gives $N_A{w}_p=N_D{w}_n$ andMost of the depletion width therefore extends into the more lightly doped side. The schematic shows the impurity levels before contact and the resulting space charge and potential after contact; there is no externally extractable equilibrium battery voltage from the built-in field alone.

A forward bias, with p positive relative to n, reduces the barrier from $V_{\mathrm{bi}}$ towards $V_{\mathrm{bi}}-V$ and injects minority carriers across the junction. In the ideal low-injection, diffusion-controlled regime this produces the Shockley diode equation, $I=I_s[e^{eV/(k_{B}T)}-1]$. Reverse bias increases the barrier and the depletion region; only a small thermally generated minority-carrier current flows before breakdown. Reverse-biased depletion regions also give a voltage-dependent junction capacitance. Real devices have recombination, leakage, series resistance and breakdown mechanisms beyond the ideal law. Controlled doping creates majority and minority carriers, and their junction produces a depletion barrier that makes rectification possible.

Take a p-n junction made from one semiconductor, with the p-type semiconductor on the left and the n-type semiconductor on the right. At thermal equilibrium the chemical potential $\mu =E_F$ is horizontal throughout. Before contact the chemical potential lies nearer the valence band on the p side and nearer the conduction band on the n side; after contact the band edges bend to permit one common value.

In this orientation both the conduction band edge $E_C$ and the valence band edge $E_V$ decrease from p to n by $e{V}_{\mathrm{bi}}$. For a homojunction their separation $E_g=E_C-E_V$ is constant. The acceptor impurity level $E_A$ lies just above $E_V$ where acceptors are present, and the donor impurity level $E_D$ lies just below $E_C$ where donors are present. These impurity levels follow the local band bending; they do not form one additional horizontal band across the device.

The shaded depletion region has few mobile electrons or electron holes and contains fixed negative ionized acceptor impurities on the p side and positive ionized donor impurities on the n side. Thus its electric field points from n to p. Its neutral outer regions have negligible electric field, so the band edges become flat there. In the depletion approximation, charge neutrality requires $N_A{x}_p=N_D{x}_n$; the symmetric drawing illustrates $N_A=N_D$ rather than assuming that equality is universal. Diffusion and field-driven drift current cancel at equilibrium: no applied voltage does not mean zero internal electric field or zero built-in electric potential difference.

graphene is a two-dimensional honeycomb lattice with two sublattices. Nearest-neighbour hopping gives two bands $E_{\pm }(\mathbf{k})=\pm t|{\sum }_{j=1}^3{e}^{i\mathbf{k}\cdot {\mathbf{\delta }}_j}|$. They touch at two inequivalent valleys. Expanding near valley $\tau =\pm 1$ givesThe two-component spinor describes sublattice amplitudes, not real spin. This is a massless two-dimensional Dirac Hamiltonian with material velocity $v_F$ in place of $c$. The emergent band quasiparticles are relativistic in this mathematical sense; the electron's vacuum rest mass is not set to zero. Real spin and valley each contribute a degeneracy of two in the ideal low-energy model.

A single-walled carbon nanotube is a rolled graphene sheet with carbon nanotube chiral vector ${\mathbf{C}}_h=n{\mathbf{a}}_1+m{\mathbf{a}}_2$. Periodicity requires $\mathbf{k}\cdot {\mathbf{C}}_h=2\pi \ell$, selecting lines through graphene momentum space and producing one-dimensional transverse electronic modes. Near a valley,A quantization line through one of the Dirac valleys in graphene gives a metallic branch; ideal nanotube zone folding gives the criterion $n-m\equiv 0(\mathrm{mod}3)$. Otherwise the nearest line misses the valley and opens $E_g=2\hslash v_F|q_{\perp }|$. The transverse momentum gives a carbon-nanotube Dirac mass, with $m_{\mathrm{eff}}=\hslash |q_{\perp }|/v_F$. For a large-radius ideal semiconducting tube, $|q_{\perp }|\simeq 2/(3d)$ and $E_g\simeq 4\hslash v_F/(3d)$, showing the inverse-diameter gap. Curvature, strain, spin-orbit coupling and interactions can change the simple classification; nominally metallic non-armchair tubes can acquire small gaps. One-dimensional transverse electronic mode thresholds give Van Hove singularities, unlike graphene's low-energy density of states proportional to $|E|$.

A gate changes occupation and the Fermi level in both systems. Ideal monolayer graphene can be tuned through charge neutrality from holes to electrons, but an ordinary scalar gate does not open a bulk band gap. Including spin and valley,The gate relation includes geometric and quantum capacitances, so one should not equate gate voltage to Fermi energy without the capacitance model. A semiconducting carbon nanotube can instead be depleted when its Fermi level lies in a gap and turned on in electron or hole bands. Contact Schottky barriers can strongly influence the observed nanotube field effect. A truly gapless ideal metallic tube cannot be switched off simply by moving the Fermi level through its band crossing. These gate responses reflect band structure and contacts, not a change of wrapping indices.

Semiconductor doping introduces electrically active impurities. In gallium arsenide, silicon has one more valence electron than gallium but one fewer than arsenic. Silicon on a gallium site can therefore supply a shallow donor electron, while silicon on an arsenic site can accept an electron and leave a mobile hole. It is an amphoteric dopant, so its net effect depends on site occupation and growth conditions.^[ref]

For isotropic parabolic bands, shift the energy origin in the given free-electron density of states to the appropriate band edge and use its effective mass:and, for valence electron states,The stated normalization includes the two spin states and assumes one effective valley. In a nondegenerate semiconductor, the conduction occupation is approximately $e^{-(E-\mu )/(k_{B}T)}$ and the probability of a valence hole is approximately $e^{(E-\mu )/(k_{B}T)}$. Substituting $x=(E-{ϵ}_c)/(k_{B}T)$ in the first integral, and the corresponding hole energy in the second, giveswhere the effective density of states values are $N_c=2[m_e{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$ and $N_v=2[m_h{k}_{B}T/(2\pi {\hslash }^2){]}^{3/2}$. The factor $\sqrt{\pi }/2$ comes from the supplied Gaussian-weighted integral. Multiplying cancels the shared thermodynamic equilibrium chemical potential and proves the semiconductor mass-action law:For the given sample $N_D=2\times 10^{21}{\mathrm{m}}^{-3}$ and $N_A=10^{21}{\mathrm{m}}^{-3}$. With complete donor ionization and occupied acceptors, charge neutrality gives $n-p=N_D-N_A=10^{21}{\mathrm{m}}^{-3}$. At $293\mathrm{K}$ the given masses and gap giveSolving the compensated semiconductor equations, or using $n_i\ll N_D-N_A$, yieldsThe very small minority density is consistent with the wide gap and the strongly n-type balance. Also $n/N_c\simeq 2.6\times 10^{-3}\ll 1$ and $p/N_v\ll 1$, checking the nondegenerate approximation. The acceptor levels are occupied in this n-type regime; complete acceptor ionization in the compensation equation is appropriate to the model.