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Replacement of a lattice ion by another of the same formal charge. charge neutrality alone does not ensure unlimited solubility; size and coordination preferences also matter.

For an ideal infinite filament with zero line charge density and current $I$ in a laboratory frame, the line-integrated four-current is $(0,I)$ in temporal/axial components. An axial boost of speed $u$ gives the title density and current. The laboratory magnetic field ${\mu }_{0}I\hat{\mathbf{\phi }}/(2\pi r)$ therefore becomes both an electric field from ${\lambda }^{\prime }$ and a magnetic field from $I^{\prime }$. Charge neutrality is frame-dependent for a current-carrying system. A model built from oppositely charged moving and resting populations makes this result explicit; the charge of each fixed material particle set remains invariant.

An unoccupied oxygen site in a reference crystal structure. Removing an oxide ion gives an effective charge $+2e$ relative to the perfect lattice; the charge of the actual missing ion is $-2e$. A physical sample must satisfy overall charge neutrality through other defects, electronic carriers or changed cation valences.

Substitutional arsenic has one more valence electron than germanium, so it acts as a donor impurity. Thermal ionization of its shallow bound state supplies mobile conduction-band electrons and leaves fixed positive donor charge. The resulting n-type semiconductor can have $n_e\gg n_i$ even when only a small fraction of host atoms are replaced. Its electrical conductivity then rises because the mobile carrier count increases; the opposite donor charge preserves macroscopic charge neutrality.

This argument assumes that the donors are substantially ionized at the measurement temperature and that the added impurity scattering does not overwhelm the carrier increase. The screened bound-donor picture and its departures from a simple hydrogenic model are described in institutional semiconductor spectroscopy notes.^[ref]

Assume dilute nondegenerate electrons and electron holes, parabolic three-dimensional band edges with the stated scalar band effective masses, spin multiplicity two, one effective valley for each edge, and thermal equilibrium. Take the unchanged host ground state to have unit activity and neglect bound excitons. The effective density of states values areFor a Fermi level $\mu$ between the band edges,The factors of $\mu$ cancel on multiplication. With $E_g=E_c-E_v$, the semiconductor mass-action law isThis density-form equilibrium constant has units ${\mathrm{m}}^{-6}$. A dimensionless thermodynamic constant uses a specified reference number density $n^{\circ }$, giving $K=K_n/(n^{\circ }{)}^2$. The dot representing the host ground state is not a freely translating gas species requiring a further gas partition function. For an intrinsic semiconductor, charge neutrality gives $n_e=n_h=n_i=\sqrt{K_n}$.

Retain the inferred mobilities and assume one ionized donor per added arsenic atom, negligible compensation, the same band gap, and the nondegenerate semiconductor mass-action law. Relative to the intrinsic sample, a conductivity factor $f$ requiresFor complete donor ionization, charge neutrality gives $N_D=n_e-n_h$, soThe host atom density, using $M_{\mathrm{Ge}}\simeq 72.63\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, isThus the required atomic fraction isapproximately one donor per $5.8\times 10^4$ host atoms. The electron density must rise to about $2000n_i$, not $1000n_i$, because the original intrinsic sample conducted through both electrons and holes. This is an estimate under unchanged mobility: impurity scattering, incomplete ionization, compensation and band-edge changes can alter it. Here $n_e/N_c\simeq 0.12$, so the nondegenerate treatment is an approximation rather than an arbitrarily dilute limit.

In a nondegenerate semiconductor, the Fermi-Dirac distribution reduces near the conduction edge to $f(E)\simeq e^{-(E-E_F)/k_{\mathrm{B}}T}$. The corresponding dilute hole probability near the valence edge givesAn intrinsic, neutral sample has $n=p=n_i$. Multiplying the two expressions therefore yieldsEach excitation creates an electron-hole pair; charge neutrality is why the carrier concentration has half the gap in its activation exponent. The electrical conductivity isThus $\sigma \propto e^{-E_g/(2k_{\mathrm{B}}T)}$ to leading exponential accuracy. The prefactor contains the mobilities and the density-of-states temperature dependence; for fixed parabolic masses, $\sqrt{N_c{N}_v}\propto T^{3/2}$.

Measure conductivity in the intrinsic regime and plot $\ln \sigma$ against $1/T$. If the prefactor is constant or its temperature dependence has been corrected, a slope $m$ givesUse the intrinsic branch rather than a donor freeze-out or extrinsic branch. A temperature-dependent gap or mobility changes the apparent activation energy, so a precision extraction must account for those prefactors.

Let $\alpha$ be the degree of ionization. Conservation of nuclei and charge neutrality, with no pre-existing electrons or other ions, giveSubstitution into the definition of $K_c$ givesThe physical root lies between zero and one:The second form avoids subtracting nearly equal terms at large $y$. The limits are $\alpha \simeq \sqrt{y}$ when $y\ll 1$ and $\alpha \simeq 1-1/y$ when $y\gg 1$. The dependence on initial number density matters: $K_c$ alone, without $c_{\mathrm{tot}}$ and the standard-state convention, does not specify the ionized fraction.

The finite-temperature Fermi level lies at the middle of the gap in the stipulated symmetric intrinsic model:Here $E_v$ is the valence-band maximum and $E_c$ the conduction-band minimum. Thermal promotion creates one conduction electron and one electron hole, so charge neutrality requires $n=p$. Equal constant densities of states in broad, symmetric bands give equal populations when the distances $E_c-\mu$ and $\mu -E_v$ are equal.

This is a statement about the chemical potential, not an available electron state in the gap. Unequal band densities of states would shift the intrinsic Fermi level away from exact midgap; the equality follows from the symmetry assumed here.

With the supplied fixed band gap and effective masses, the intrinsic carrier concentration isNumerically,For uncompensated, approximately ionized donor impurities, charge neutrality and the semiconductor mass-action law requireThere is no abrupt extrinsic-to-intrinsic transition. An order-of-magnitude lower crossover density is thereforeRobust extrinsic dominance requires $N_D\gg n_i$; at the nominal equality, $n=1.618N_D$ and $p=0.618N_D$, so intrinsic contributions are already appreciable. For a quantitative tolerance $p/n\le \eta$, the minimum ionized dopant density is $n_i(1-\eta )/\sqrt{\eta }$. For example, $\eta =0.1$ would require $N_D\ge 2.85n_i$ at the upper temperature.

Using the nominal crossover estimate and the complete-ionization approximation at room temperature, $n\simeq N_D$ andFor the twofold neutral-donor degeneracy, a consistency calculation gives about $91.3\%$ ionization at $300\mathrm{K}$, raising this minority density to $1.30\times 10^{15}{\mathrm{m}}^{-3}$, about $9.6\%$ above the complete-ionization estimate. Stronger extrinsic dominance uses more donors and correspondingly fewer electron holes. Actual germanium has temperature-dependent band parameters and multiple valleys; the numerical estimate consistently uses the fixed effective data supplied rather than adding an unsupplied multiplicity.

For a one-dimensional free-electron gas of length $L$, allowed wave vectors have spacing $2\pi /L$. Including two spin states and occupied wave vectors from $-k_F$ to $+k_F$,The ground-state kinetic energy is found by integrating rather than assigning every electron the Fermi energy:Charge neutrality gives $N_e=ZN$ because each positive ion has charge $+Ze$. Thus $K_e=ZN{E}_F/3$. Since the ion separation is $2R$, $L\simeq 2RN$ and $n_e=Z/(2R)$. These are one-dimensional number densities and state counts; using a three-dimensional $k_F\propto n^{1/3}$ relation here would change the model. The results concern the ground-state Fermi-gas contribution, neglecting electron interactions beyond the supplied electrostatic energy.

Use the supplied effective electrostatic term and enforce charge neutrality while varying the chain length: $R=Z/(2n_e)$, with $N$ and $Z$ fixed. The total energy per positive ion becomesThe positive quadratic term supplies degeneracy pressure against compression. Differentiating gives the neutral one-dimensional metal equilibrium density,where $a_B=4\pi {ϵ}_0{\hslash }^2/(m_e{e}^2)$ is the Bohr radius. The second derivative $Z{\pi }^2{\hslash }^2/(12m_e)>0$ verifies a minimum. Within this specified toy energy, the minimizing electron density is independent of $Z$; instead the equilibrium ion spacing is $2R^{\ast }=Z/n_e^{\ast }$.

At the minimum the electrostatic contribution is minus twice the kinetic contribution, so ${\epsilon }_{\ast }=-Z{E}_F^{\ast }/3$. These follow from the given electrostatic approximation, not from a full self-consistent calculation of a real one-dimensional metal. In particular, additional Coulomb terms cannot be appended without the corresponding neutral-background subtraction that defines the supplied finite energy.

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.

Take uncompensated donor impurities and initially approximate them as ionized. Charge neutrality then gives $n-p=N_d$, while the semiconductor mass-action law gives $np=n_i^2$. Solving the quadratic without prematurely discarding minority carriers,Since $N_d\gg n_i$, the hole density isUse the conduction population to locate the chemical potential:It moves toward the conduction band but remains about $10.4k_{B}T$ below it, so the mobile electrons are nondegenerate. Despite the donor binding energy being comparable with $k_{B}T$, the large number of available conduction states makes ionization almost complete at this dilute doping. For a singly occupied two-spin donor, $N_d^{+}/N_d=[1+2e^{(\mu -E_D)/(k_{B}T)}{]}^{-1}$, with $E_D=E_c-0.02584\mathrm{eV}$. The inferred value is $0.99984$, confirming the approximation to much better than one percent. Including this small correction raises the hole density only to about $4.5824\times 10^{10}{\mathrm{m}}^{-3}$.

Near a parabolic valence band maximum $E_v$, the missing-electron effective mass is $m_h^{\ast }>0$. The same two-dimensional state counting givesMultiplication eliminates the chemical potential, giving the semiconductor mass-action lawFor an intrinsic semiconductor, charge neutrality imposes $n=p=n_i$, soThe intrinsic chemical potential is not exactly midgap when the two effective masses differ: ${\mu }_i=(E_c+E_v)/2+(k_{B}T/2)\log (m_h^{\ast }/m_e^{\ast })$. The exponential suppression still justifies the assumed nondegenerate regime here.

With equal constant electron mobility and hole mobility $\mathcal{M}$, the sheet electrical conductivity is $\sigma =e\mathcal{M}(n+p)$. In the intrinsic regime, $n+p\simeq 2n_i\propto T{e}^{-E_g/(2k_{B}T)}$, soIt decreases steeply with increasing $1/T$ on the high-temperature side. There is a small logarithmic curvature rather than a perfectly straight activation line.

In the fully ionized extrinsic regime, charge neutrality and the semiconductor mass-action law give $n-p=n_d$, $np=n_i^2$, henceThis gives a nearly horizontal region passing through $\sigma (200\mathrm{K})=0.01{\Omega }^{-1}$. At lower temperatures semiconductor carrier freeze-out decreases the ionized donor density and bends the curve downward again.

To locate that bend realistically, include donor charge-state degeneracy. For a neutral donor with two spin states and an empty donor with one state, the ionized density isSolve $n-n_i^2/n=n_d^{+}$ for the mobile density. In the asymptotic freeze-out regime, where $p$ is negligible and $n\ll n_d$,Its activation scale $E_B/2\simeq 3.14\mathrm{meV}$ is much smaller than the intrinsic scale $E_g/2=180\mathrm{meV}$.

Within the specified interval the low-temperature downturn is modest: $N_c(30\mathrm{K})\simeq 2.81\times 10^{10}{\mathrm{cm}}^{-2}$, and the partial-ionization equation gives $n(30\mathrm{K})\simeq 0.65n_d$. The sample is not yet deep in the asymptotic freeze-out limit at 30 K. Calibrating the mobility to the stated 200 K value gives representative valuesThe donor statistical weights were not explicitly supplied, so they are a stated assumption for these illustrative curve details; a one-state donor approximation changes the low-temperature rounding but not the three physical regimes or their activation scales.

Equal effective masses and equal spin and valley degeneracies give electron-hole symmetry. In the intrinsic semiconductor, charge neutrality requires equal numbers of conduction band electrons and valence band holes. It therefore pins the chemical potential to the band touching point, $\mu =0$. The Fermi-Dirac distribution givesMeasured relative to the filled valence band at zero temperature, the thermal internal energy includes both species. Removing a valence band electron at $-E$ costs the positive hole energy $E$, just as adding a conduction band electron at $E$ does. HenceDifferentiate at fixed area and fixed total electronic particle number, allowing creation of electron-hole pairs:There is no conflict with the equipartition theorem. The normalization $N_c$ counts only the excited conduction band electrons, while the heat capacity includes both electrons and holes; moreover, $N_c\propto T$ is not fixed. In fact $\Delta U/N_c={\pi }^2{k}_{B}T/(6\ln 2)$, and differentiating $N_c(T)$ supplies another equal contribution to the heat capacity. Finally, $\mu =0$ gives occupation $1/2$ at the touching point, so the excitation gas is not in the classical dilute limit even though $k_{B}T$ exceeds the vanishing band gap.

Assume uncompensated donor impurities, negligible intrinsic electron holes, and a nondegenerate conduction band. Then charge neutrality equates the conduction electron number density to the ionized-donor number density:The low-temperature condition relative to the band gap supports neglect of intrinsic carriers; one also needs $n_i\ll n$ and the Boltzmann approximation for the conduction electrons. Combining part (b)(i) with the effective density of states result givesEliminate the chemical potential by writing $e^{\beta (\mu -{ϵ}_D)}=(n/N_c)e^{\beta ({ϵ}_c-{ϵ}_D)}$. With the donor binding energy $E_i={ϵ}_c-{ϵ}_D>0$,Writing $A(T)=N_c{e}^{-\beta E_i}$, the physical root of the resulting quadratic equation isThis lies between zero and $N_D$. At semiconductor carrier freeze-out, $A\ll N_D$, it becomes $n\simeq \sqrt{N_D{N}_c}{e}^{-E_i/(2k_{B}T)}$; at essentially complete ionization, $A\gg N_D$, it tends to $N_D$. Including donor charge-state degeneracy would replace $A$ by $A{g}_{\mathrm{empty}}/g_{\mathrm{full}}$.

There are three regimes. At the lowest temperatures, most donor impurities retain their electrons; thermal ionization gives semiconductor carrier freeze-out. The result from part (b) isThis produces the approximately straight descending low-temperature tail on the logarithmic scale plot against $1/T$. The activation energy in the exponential is $E_i/2$, because the chemical potential adjusts with donor ionization.

At intermediate temperatures, the donor impurities are essentially fully ionized while intrinsic excitation remains small. Charge neutrality gives $n\simeq N_D$, producing the extrinsic plateau, roughly from $20$ to $40\mathrm{K}$ in the figure.

At higher temperatures, excitation from the valence band to the conduction band creates electron-hole pairs. The intrinsic carrier concentration grows as $n_i\propto T^{3/2}{e}^{-E_g/(2k_{B}T)}$ for constant parabolic band effective masses and constant band gap. With ionized donor impurities, charge neutrality and the semiconductor mass-action law giveThus $n\simeq n_i+N_D/2$ when $n_i\gg N_D$. This explains the renewed rapid rise and the eventual intrinsic branch; that branch is governed by the band gap, rather than the donor binding energy.