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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.

Assume thermal equilibrium, a three-dimensional nondegenerate semiconductor, one parabolic conduction extremum and one parabolic valence extremum, with two spin states in each. Use the supplied effective masses as scalar density-of-states parameters. With conduction energy $E=E_c+\epsilon$, state counting gives the per-volume density of statesSince $E_c-\mu \gg k_{B}T$, replace the Fermi-Dirac distribution by its Boltzmann factor. Using $x=\epsilon /(k_{B}T)$ and the supplied integral,The electron hole occupation is $1-f(E)$ in the valence band. Applying the same integration below $E_v$ givesTheir product cancels the chemical potential, giving the semiconductor mass-action lawThus, for masses expressed in ordinary mass units,The labels $B,C$ in this last box are exponents, unrelated to the magnetic field or Curie constant above. Valley degeneracy or multiple valence bands must be included in the effective density of states, or absorbed into suitable density-of-states effective masses, not counted twice. Real silicon has anisotropic multiple valleys, so its conductivity mass is not automatically its density-of-states mass; the numerical answers below implement the paper's supplied scalar-mass model.^[ref]

The formula is not an identity for arbitrarily degenerate carriers. For a concrete counterexample, take $\mu -E_c=10k_{B}T$ while holes remain dilute. Fermi occupation gives $n/N_c\simeq 24.1$, whereas the Boltzmann replacement would give $e^{10}\simeq 22026$; the predicted product then differs by about three orders of magnitude. Separate electron and hole quasi-Fermi levels out of equilibrium would also prevent the cancellation used above.

For an undoped neutral intrinsic semiconductor, $n=p$. Equating the two expressions from a and taking logarithms givesand hence the intrinsic-semiconductor chemical potentialTherefore $\boxed{\alpha =\frac{3}{4}\ln (m_h^{\ast }/m_e^{\ast })}$. The logarithm is temperature-independent for fixed parabolic effective masses. The chemical potential is exactly at midgap only if the two effective density of states values coincide; for $m_h^{\ast }>m_e^{\ast }$ it moves above midgap. Additional valley multiplicities would add their logarithmic contribution.

A substitutional phosphorus atom supplies one more valence electron than the silicon atom it replaces. In the shallow hydrogenic donor state model, the remaining positive impurity attracts that extra electron through the screened Coulomb potentialRescaling the hydrogen atom binding energy by the effective mass and dielectric screening gives the ionization energyThe donor level lies at $E_c-E_D^{\mathrm{bind}}$; its binding magnitude is positive. The associated effective Bohr radius of a hydrogenic donor is $a_B^{\ast }={ϵ}_r(m_e/m_e^{\ast })a_0\simeq 2.38\mathrm{nm}$, consistent with a shallow state extended over several lattice spacings.

At $1000\mathrm{K}$, $k_{B}T=0.08617\mathrm{eV}$, approximately $3.3$ times this binding energy. Dilute donors are consequently expected to be largely ionized, rather than frozen out. The actual ionized fraction also depends on the chemical potential, available conduction states and donor charge-state degeneracy; comparing $k_{B}T$ to the binding energy alone is not a normalized ionization probability. At sufficiently high temperature, intrinsic electron-hole production can dominate the total carriers even though donors are ionized. Band anisotropy and short-range impurity corrections limit the numerical hydrogenic estimate.

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 $4.5824\times 10^{10}{\mathrm{m}}^{-3}$.

The chemical potential is the change in Helmholtz free energy when particles are added at fixed temperature and volume:Equivalently, $dU=TdS-PdV+\mu dN$ identifies it as the derivative of internal energy at fixed entropy and volume. In equilibrium, systems that exchange particles have equal chemical potentials.

The Fermi energy is the zero-temperature chemical potential, or, for independent electrons, the highest occupied single-particle energy in the zero-temperature limit. Its numerical value depends on the chosen energy reference; the Fermi energy need not be measured from the bottom of a band unless that convention is specified.

Monovalency gives one electron per atom, so the two-spin-state band is half filled. At zero temperature, occupying $-k_F<k<k_F$ gives $2(Na/2\pi )(2k_F)=N$. HenceRelative to the band minimum, the Fermi energy is $2t$.

The density of states obeys $g({ϵ}_0+x)=g({ϵ}_0-x)$. At $\mu ={ϵ}_0$, the Fermi-Dirac distribution satisfiesPairing the energies above and below the band centre therefore leaves exactly half of the two available states per atom occupied, at every positive temperature. Increasing $\mu$ strictly increases the mean particle number, so this solution at fixed half filling is unique. ConsequentlyThis particle-hole symmetry of a half-filled bipartite band argument uses the symmetric tight binding band and fixed half filling; it is not a general temperature-independence property of chemical potential in metals with additional bands or an asymmetric density of states.

The reciprocal primitive vectors are $(2\pi /a_x,0)$ and $(0,2\pi /a_y)$, so the first Brillouin zone isIt is wider in the $k_x$ direction because $a_x<a_y$.

The geometry alone does not fix the hopping amplitudes or a unique Fermi surface. One consistent tight-binding model isrepresenting larger hopping between the closer atoms. It remains a nearest-neighbour bipartite band at half filling, so its chemical potential is ${ϵ}_0$. Its Fermi surface consists of the two sheetsThese meet the top and bottom zone edges and continue periodically into the neighbouring zones. The right panel of the figure gives an explicit example with $t_x=2t_y$ and $a_x=0.65a_y$. The central region between the sheets is occupied and has half the zone area. If the hopping amplitudes instead remained equal, a geometrically stretched diamond would also satisfy the stated information. The metric inequality $a_x<a_y$ alone does not determine which of these shapes occurs.

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.

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.

Let $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant, $T$ the absolute temperature, and $\mu$ the common equilibrium chemical potential. The conduction band minimum and valence band maximum are ${ϵ}_c$ and ${ϵ}_v$; their separation is the band gap $E_g={ϵ}_c-{ϵ}_v$. The positive parameters $m_e,m_h$ in the supplied density of states functions are the band effective masses of electrons and electron holes, not necessarily the bare electron mass.

The electron occupation is the Fermi-Dirac distribution $f(ϵ)=[e^{\beta (ϵ-\mu )}+1{]}^{-1}$. The appropriate statistical factors are $f_c=f$ for electrons and $f_v=1-f$ for electron holes. In a nondegenerate semiconductor, ${ϵ}_c-\mu \gg k_{B}T$ and $\mu -{ϵ}_v\gg k_{B}T$, so the Boltzmann approximation givesFor the conduction band, set $u=ϵ-{ϵ}_c$ and then $\xi =\beta u$. The supplied integral yieldsFor the valence band, set $u={ϵ}_v-ϵ$ to obtainThe effective density of states factors $N_c,N_v$ include the two spin states already present in the supplied density of states functions. Multiplication cancels the chemical potential and gives the semiconductor mass-action lawHere $n_i=\sqrt{N_c{N}_v}{e}^{-E_g/(2k_{B}T)}$ is the intrinsic carrier concentration. Equality $n=p=n_i$ additionally requires an electrically neutral intrinsic semiconductor. The product law also applies to an equilibrium doped semiconductor while both populations remain nondegenerate. The derivation assumes parabolic energy bands, the stated spin degeneracy, and a single common chemical potential.

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.

The conduction and valence dispersions are two cones touching at the Dirac point, with slopes $\pm \hslash v_F$. Here $\mathbf{k}$ is measured relative to one valley centre. Each cone has a density of states proportional to $|E|$, which vanishes at the touching energy. Gate-induced carriers satisfy approximately $n=C_g(V_g-V_D)/e$ per unit area. Negative and positive offsets from $V_D$ produce holes and electrons, respectively. Away from neutrality a simple charge carrier mobility model gives $\sigma \simeq e\mu |n|$, so resistance decreases on either side of a maximum near $V_D$. Disorder, electron–hole puddles, contacts and quantum transport round the peak; the cone dispersion alone fixes neither its height nor its width, nor does it require exact electron–hole symmetry in a real device.

A direct derivation of the magnetic spectrum starts from the single-valley Dirac Hamiltonian operator $H=v_F({\sigma }_x{\pi }_x+{\sigma }_y{\pi }_y)$. Introduce oscillator operators using the magnetic length ${\ell }_B=\sqrt{\hslash /(e|B|)}$. Up to exchanging sublattices or valleys,Squaring gives adjacent oscillator number operators, henceThe positive expression printed in the source describes the electron branch only; the negative-energy partners must also be included. The same square-root dependence follows from quantizing circular orbit area, with the Dirac Berry phase cancelling the usual half-integer offset.

Each spin and valley contributes $Ae|B|/h$ states in area $A$. Including two spins and two valleys gives $N_{\mathrm{LL}}=4Ae|B|/h$, or areal degeneracy $4e|B|/h$. The zero-energy level is half-filled at charge neutrality: the carrier density counts departures from that state.

For $n=10^{16}{\mathrm{m}}^{-2}$, the filling factor and fourfold graphene Hall sequence areIn a plateau, ${\rho }_{xx}\simeq 0$ and $|{\rho }_{xy}|=h/({\nu }_N{e}^2)$. The Hall sign reverses between electron and hole doping; the sketch uses its magnitude to avoid tensor-index sign conventions. Representative plateau-centre fields are $20.68$, $6.893$, $4.136$ and $2.954\mathrm{T}$ for $\nu =2,6,10,14$. Between plateaux, crossing extended states in broadened levels gives longitudinal peaks and Hall steps. Their ideal filling-centre positions have $\nu =4j$, namelyThus peaks near $10.34$, $5.17$ and $3.45\mathrm{T}$ separate the first prominent plateaux as field increases. Many weak oscillations accumulate toward zero field and are damped there. Finite-width plateaux require localized states between extended-state energies; an exactly clean Landau spectrum does not by itself prescribe plateau widths. The illustration uses an explicitly schematic broadening, not inferred experimental widths.

At $6.5\mathrm{T}$, $\nu =6.36$, close to the $\nu =6$ plateau. With $v_F=10^6\mathrm{m}{\mathrm{s}}^{-1}$, $E_1=92.5\mathrm{meV}$ and $E_2=130.8\mathrm{meV}$. The relevant adjacent gap near the chemical potential is therefore $\Delta =E_2-E_1\simeq 38.3\mathrm{meV}$, rather than the gap from the zero level. Resolving levels requires quantum-lifetime broadening $\hslash /{\tau }_q\lesssim \Delta$ and thermal broadening appreciably smaller than $\Delta$, giving the order-of-magnitude limitsThe upper temperature is an energy-scale estimate, not a guaranteed sharp-feature temperature. For oscillation amplitudes the thermal damping factor is $X/\sinh X$, $X=2{\pi }^2{k}_{B}T/(\hslash {\omega }_c^{\ast })$, with $\hslash {\omega }_c^{\ast }=\hslash eB{v}_F^2/E_F\simeq 36.7\mathrm{meV}$ and $E_F=\hslash v_F\sqrt{\pi n}$. Requiring $X\lesssim 1$ gives the more conservative scale $T\lesssim 22\mathrm{K}$ for little thermal damping. Visible weaker features may persist above that scale; their actual cutoff depends on resolution, disorder and temperature-dependent scattering. The transport lifetime and quantum lifetime also need not coincide.

At $B_z=8\mathrm{T}$,The cyclotron spacing is $\hslash e(8\mathrm{T})/m^{\ast }$, smaller than at 10 T, and each resolved level's degeneracy per area is also reduced by $4/5$. Thus the same carrier density occupies more levels. This ratio is determined without the missing graph, but it does not specify whether the chemical potential lies in a localized tail or an extended-state centre.

TODO — missing density/plateau assignment: a numerical index and chemical-potential marker at 8 T require the unavailable density information. For example the first illustrative density in part (i) gives ${\nu }_8\simeq 2.5$, while $N_S=3.87\times 10^{15}{\mathrm{m}}^{-2}$ instead gives ${\nu }_8\simeq 2$. These produce different schematic placements even though both satisfy every surviving sample parameter. Unresolved spin degeneracy and disorder add further qualifications; no unique 8 T label should be invented.

At $B_z=10\mathrm{T}$, the nominal filling isA labelled density-of-states picture has disorder-broadened peaks at these orbital and spin energies, localized tails and extended states near their centres. A chemical potential in the localized region corresponds to a Hall plateau and small longitudinal resistivity; passing through an extended-state centre corresponds to a transition.

TODO — missing density/plateau assignment: there is no basis for choosing one of those positions at 10 T without the missing Hall graph. For example $N_S=4.84\times 10^{15}{\mathrm{m}}^{-2}$ gives nominal ${\nu }_{10}\simeq 2$, whereas $N_S=6.04\times 10^{15}{\mathrm{m}}^{-2}$ gives ${\nu }_{10}\simeq 2.5$. In an ideal resolved-level picture the first can put the chemical potential between filled levels, while the second leaves a level partly filled. Both are compatible with the surviving current and width, demonstrating why the requested placement is undetermined.

As magnetic field grows from zero, a clean two-dimensional electron gas first has an ordinary finite longitudinal resistivity, then develops Shubnikov–de Haas oscillations as broadened Landau levels pass the chemical potential. Thermal and disorder broadening obscure the low-field oscillations; their visibility grows when the level spacing exceeds those broadening scales. In a well-developed integer quantum Hall regime, ${\rho }_{xx}$ has minima close to zero on Hall plateaux and maxima during transitions between them, when extended states cross the chemical potential. Detailed peak heights and envelopes depend on disorder and scattering and are not fixed by quantization alone.

At fixed density the nominal filling is $\nu (B)=N_{S}h/(eB)$, so level-filling features are approximately periodic in $1/B$, with period $e/(N_{S}h)$ for resolved spin levels or $2e/(N_{S}h)$ for an unresolved spin pair. Increasing $B$ lowers the filling; spin splitting can create additional odd-index features. The original schematic below shows these relations without pretending to reconstruct the missing data.

TODO — missing source graph: the actual $0$–$12\mathrm{T}$ positions and integer plateau labels require $N_S$ or surviving plateau fields from part (b). Neither is provided independently. Thus a uniquely labelled field-axis sketch for this sample cannot be completed. The following DOS diagram distinguishes extended centres and localized tails generically; its chemical-potential markers are illustrative and do not stand for the requested 8 T or 10 T assignments.

A perfectly clean two-dimensional system has sharply degenerate Landau levels and integer filling only at isolated fields for fixed density. Real disorder broadens each level into a band, with states localized in the tails and extended states near its centre. When the chemical potential moves through a localized region, changing field can change the occupation of those localized states without changing the number of conducting extended channels. The Hall conductance then remains $p{e}^2/h$ and the Hall voltage at fixed current remains $I_{x}h/(p{e}^2)$ over a finite field interval.

In the same regime, separated edge channels and localized bulk states suppress dissipative backscattering, so ${\rho }_{xx}$ is very small. At a transition the chemical potential reaches the next band of extended states; longitudinal conduction rises and the Hall conductance changes to another integer. This is a mobility-gap explanation: a plateau does not require the density of states itself to vanish, nor does it require the carrier density to grow linearly with field. Low temperature, sufficient mobility and suitable contacts are needed to resolve it.

This physical explanation is determined by the surviving question text. Particular measured plateau heights and widths cannot be quoted because the source graph is absent, as recorded in part (b).

The lower curve is a valence band and the upper curve is a conduction band, separated by an indirect band gap. The readable upper curve has a minimum just inside the X boundary and slightly below the plotted chemical potential. It is therefore partially occupied, while the visible valence states lie below the chemical potential and are filled at low temperature. The sample is electron-doped, with electron pockets around the six symmetry-related conduction minima of silicon.

The full high-symmetry path and part of the lower-left dispersion are absent from the held image. Source gap: those missing path labels and curve segments cannot be read or reconstructed from this copy. The identification of the conventional six silicon valleys and the valence maximum at $\Gamma$ uses the usual silicon band model, rather than purported readings of missing pixels.

Two methods for calculating such an electronic band structure are density functional theory, solving a self-consistent periodic electronic problem, and a tight-binding model, diagonalizing a hopping Hamiltonian built from localized atomic orbitals. Neither is a measurement of the plotted dispersion relation.

For $\mathbf{B}\parallel [111]$, its squared projection onto the long axis of every valley is $1/3$. All six valleys therefore have the same cyclotron effective mass:At the fixed photon energy,Thus the two-peak spectrum is replaced by one peak near $1.4\mathrm{T}$. The source does not determine the linewidth or absolute oscillator strength for the new direction; the sketch below shows only a normalized illustrative line at the computed position.