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Using $dx/dT=-x/T$ in the internal energy givesThis is a Schottky anomaly with unequal degeneracies, with lower-level state degeneracy two and upper-level state degeneracy one. Its limiting forms areIt vanishes at low temperature because exciting the upper orientation is exponentially unlikely. It also vanishes at high temperature because the populations have nearly saturated and further heating hardly changes the internal energy. Between these limits it has one smooth maximum. The original plot shows its shape; numerical maximization gives $k_{\mathrm{B}}{T}_{\max }/\delta \simeq 0.449$ and $C_{\max }/(N{k}_{\mathrm{B}})\simeq 0.241$.

For the same atoms in a three-dimensional volume $V$, the one-atom translational partition function is $q_{\mathrm{trans}}^{(3)}=V/{\Lambda }^3$. Comparing the same $N$ atoms at the same temperature cancels the common $N!$ in their canonical partition functions. Therefore the molar adsorption confinement free energy, with the sign defined as adsorbed minus free, isThe logarithm is dimensionless: $V/\sigma$ and $\Lambda$ are both lengths. For a macroscopic gas depth $V/\sigma \gg \Lambda$, confinement raises the translational Helmholtz free energy. Equivalently, the translational state count is reduced by a factor $\sigma \Lambda /V$ per atom. The question supplies neither a volume nor an adsorption binding energy, so this answer is symbolic and describes the confinement cost, not a numerical net adsorption free energy.

For example, a binding energy $D>0$ per atom would supply a separate $-N_{\mathrm{A}}D$ to $\Delta F_m$, together with any normal-motion internal contribution retained by a more detailed model. Adsorption can then be favourable despite the loss of free translational phase space. Merely calling the change a loss must not reverse the sign of the explicitly defined $F_{\mathrm{ads}}-F_{\mathrm{gas}}$.

At low temperature, retain $J=0$ in the even sum and $J=1$ in the odd sum. Their weights are respectively $1$ and $3e^{-2{\theta }_{\mathrm{rot}}/T}$ before the nuclear-spin multiplicity, soEquilibrium hydrogen then becomes almost entirely parahydrogen. At high temperature, many even and odd rotational levels contribute, and their spatial sums have the same leading value. HenceThis is normal hydrogen. Its familiar high-temperature ratio is a nuclear-spin multiplicity ratio, not the $9:1$ ratio of the two lowest levels. Maintaining conversion is essential: a sample cooled without an effective conversion mechanism can retain its earlier composition instead of following $K(T)$. NBS thermodynamic data distinguish equilibrium composition from normal hydrogen and compare the state-sum prediction with spectroscopic and thermal evidence.^[ref]

For a frozen ortho-para mixture, the mole fractions do not change during the heat measurement, and the molar heat capacity is the composition-weighted average. A high-temperature composition retained on cooling has $x_o=3/4$, $x_p=1/4$, soFor a mixture that continues to interconvert and remain in equilibrium near the specified temperature, use the combined rotational sum and differentiate that sum:ConsequentlyThe equilibrium composition itself is $K=3(0.970)/1.165\simeq 2.4979$, giving $x_o\simeq 0.71411$. Its changing composition supplies a further positive heat capacity: warming converts some lower-energy para molecules into higher-energy ortho molecules. One can isolate this term asFor clarity, if a mixture is first equilibrated at this temperature and then its composition is frozen during measurement, the conversion term is absent: its rotational value is $0.58789R$ and its total value is $2.08789R$. This is distinct from the equilibrium heat capacity of interconverting states even though the instantaneous composition is the same. An experiment must specify conversion activity and equilibration time; compare sufficiently slow temperature steps with a conversion catalyst against a composition-preserving sample, while keeping the same thermal background and checking that the sample has reached the stated temperature.

Apply part (b) to each supplied rotational sum. Multiplying the odd sum by its constant nuclear-spin factor three does not change $q^{\prime }/q$ or $q^{\prime \prime }/q$. At the specified temperature the dimensionless rotational contributions areThus the rotational heat capacities are $1.425R$ for parahydrogen and $0.253R$ for orthohydrogen. For the total gas heat capacities, including translation and neglecting vibration,The difference is substantial because the permitted rotational ladders differ. The para ladder starts at $J=0$ and next reaches $J=2$, separated by $6k_{\mathrm{B}}{\theta }_{\mathrm{rot}}$; its excited population is changing appreciably here. The ortho ladder starts at $J=1$ and next reaches $J=3$, a larger separation of $10k_{\mathrm{B}}{\theta }_{\mathrm{rot}}$, so much of the ortho population remains in its lowest rotational manifold. A rotational contribution greater than $R$ is possible in the intermediate quantum regime and is not a violation of high-temperature equipartition.

Metallic atomic densities place very many electrons in a small volume. The Pauli exclusion principle prevents all these fermions occupying the lowest spatial state, so filling the available states produces a Fermi energy of order electronvolts. At room temperature, $k_{\mathrm{B}}T\simeq 0.026\mathrm{eV}$, much smaller than that scale. Equivalently, ordinary metals are at $T\ll T_F=E_F/k_{\mathrm{B}}$, where $T_F$ is the Fermi temperature. Their electrons are quantum degenerate even at ordinary laboratory temperatures.

For the free electron model, the Fermi energy is the highest occupied one-electron energy at zero temperature, measured relative to the bottom of the free-electron band. The occupied wavevectors fill a disk in two dimensions or a sphere in three dimensions. The radius is the Fermi wavevector $k_F$, withEach allowed wavevector has two spin states. At nonzero temperature, the Fermi-Dirac distribution rounds the occupation boundary; the zero-temperature Fermi energy remains a useful density-defined scale and should not be defined as the thermal energy $k_{\mathrm{B}}T$.

An isolated spin-degenerate lowest energy band accommodates two electrons per atom when there is one atom per primitive cell. Starting from one per atom, the estimate for filling the band is thereforeThis corresponds to an additional areal carrier count of about $4.77\times 10^{19}{\mathrm{m}}^{-2}$. If a genuine band gap separates the filled first band from the next band, there are no partially occupied band states at low temperature, so ordinary metallic conduction is lost.

Simply touching an edge is insufficient. In the circular approximation first contact would occur at $k_F=\pi /a$, corresponding to $n_e{a}^2=\pi /2\simeq 1.57$ electrons per cell; that is a partly filled band and need not be insulating. Near zone boundaries, the periodic potential reshapes the free-electron contours. The two-electron filling estimate assumes a global gap without overlapping bands, rather than inferring an insulator merely from one local avoided crossing.

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.

A typical intrinsic semiconductor has a nearly full valence band ending at $E_v$ and a nearly empty conduction band starting at $E_c$, separated byAt finite temperature, the equilibrium Fermi level lies near the middle of the gap. In the nondegenerate parabolic-band approximation,where $N_c,N_v$ are the two effective density of states values. Equal densities of states put it exactly at midgap; unequal masses or valley multiplicities shift it slightly. The left panel gives the requested energy-level sketch, with doped cases for comparison.

The extra electron is bound to an approximately singly charged donor through a dielectric-screened Coulomb attraction. In the isotropic hydrogenic donor state model, replace the vacuum electron mass by $m^{\ast }$ and the vacuum permittivity by ${ϵ}_r{ϵ}_0$:At low temperature, the donor is mainly in its $1s$ state. Allowed electric-dipole transitions to $np$ states haveAssign the three printed lines to $n=2,3,4$. Their independently inferred binding energies areThey are mutually close, giving $R^{\ast }\simeq 10.63\mathrm{meV}$. The hydrogenic impurity absorption series accumulates at the donor-ionization limit. Above about $10.6\mathrm{meV}$, absorption can place the electron in the conduction-band continuum, explaining the continuous spectrum.

Using the supplied $m^{\ast }=0.2m_e$ and $R_H=13.61\mathrm{eV}$,The corresponding effective radius isThese are the parameters inferred from the printed lines within the stipulated scalar model, rather than substituted tabulated material constants. Real impurity spectra can depart from a single hydrogenic ladder through anisotropy and short-distance impurity effects.^[ref]

There are three useful temperature regimes.

At low temperature, dopant freeze-out keeps most donor electrons bound. Warming ionizes donors and increases conductivity through the rising mobile-electron concentration. For an uncompensated dilute donor system, negligible holes and spin degeneracy two, charge neutrality with partial donor ionization givesDeep in freeze-out, $n\ll N_D$, soThe activation scale is set by the shallow donor binding, rather than the whole band gap.

In the extrinsic regime, donors are largely ionized and $n\simeq N_D$ is nearly constant. The temperature dependence of conductivity then mainly follows the electron mobility. Increasing phonon scattering can make conductivity fall as temperature rises, even though donors remain ionized; impurity scattering can give a different mobility trend at lower temperature.

At sufficiently high temperature, thermally generated electron–hole pairs overwhelm the dopant population. The intrinsic, gap-activated increase in conductivity returns. The boundaries depend on donor concentration, compensation and scattering, so no one exponential describes the entire doped-sample curve.

Arsenic has five valence electrons, one more than silicon. As a substitutional donor impurity, it supplies a weakly bound extra electron which is largely thermally released into the conduction band at room temperature, leaving an ionized positive donor.

Consequently the mobile electron density increases far above the intrinsic value, so the conductivity increases substantially even at ppm dopant levels. The crystal remains macroscopically neutral: fixed donor charge compensates the added mobile charge.

Warming thermally ionizes neutral donors, depleting the occupied bound ground states from which these absorption lines start. Increased scattering can also broaden the lines. Once the donors are predominantly ionized, their bound-to-bound absorption becomes weak or disappears.

The inferred binding provides the natural thermal scale:The first excitation energy similarly corresponds to about $92.5\mathrm{K}$. An order-of-magnitude estimate is therefore a temperature of order $10^2\mathrm{K}$. This is not a sharp onset: significant ionization can start below that scale because free carriers have a large translational density of states. The precise loss of line strength requires donor density and compensation, as in part (ii), rather than just equating one energy with $k_{\mathrm{B}}T$.

Gallium has three valence electrons, one fewer than silicon, and acts as an acceptor impurity. A neutral acceptor can be described as a negatively charged centre binding an electron hole. Thermal release of these holes gives a p-type semiconductor.

At low temperature, the bound hole can absorb far-IR photons in transitions between acceptor levels. The simplest scalar hydrogenic acceptor state model predictsso one expects a discrete bound-hole absorption series followed by a photoionization continuum. In that idealization, the first lines would be at $3R_A^{\ast }/4$, $8R_A^{\ast }/9$ and $15R_A^{\ast }/16$, with threshold $R_A^{\ast }$.

The hole effective mass is not supplied, so the electron mass $0.2m_e$ cannot be reused to give numerical Ga line positions. In a real valence band, multiple hole bands, anisotropic masses and spin-orbit effects can split and complicate the series; acceptor spectra are generally less accurately described by one scalar hydrogenic ladder.^[ref]

At fixed temperature,Thus magnetic-field-dependent Langmuir adsorption is even in $B$ and increases with $|B|$. At fixed nonzero gas pressure, stronger field increases surface coverage; in an isotherm plot, the half-coverage pressure $p_{1/2}=1/b$ moves to lower pressure. The saturation coverage remains one because the number of sites is unchanged.

For weak fields, with ${\theta }_0=b(0)p/[1+b(0)p]$,There is no linear small-field change. At large $|x|$, $\cosh x\sim e^{|x|}/2$, and at fixed $p>0$ the coverage approaches one. The field lowers the free energy of the surface spin ensemble relative to the unsplit bulk; simply averaging the two surface energies to $-E_b$ would miss this equilibrium effect.

For fixed degeneracies and ionization energy,Therefore $K_c$ increases with temperature in this model. Ionization costs positive energy, so it is suppressed at low temperature by its Boltzmann factor. At higher temperature that penalty becomes less important.

Ionization also creates an additional independently translating particle and more ways to distribute the particles and their momenta. This favours the ionized mixture through its translational and mixing entropy, consistent with the $T^{3/2}$ prefactor and increasing ionization at high temperature.

For each coordinate, the horizontal graph position iswith $\tilde{\omega }$ in inverse centimetres. Read $S/R$ from the supplied curve at these positions. Each $A_1$ frequency contributes one coordinate and each $E$ frequency contributes two. The graphical estimates areThus the listed frequency groups contribute approximately $0$, $1.4$, $4.7$, $0.33$, $2.0$ and $13.8\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$, respectively, andThe nine coordinates are consistent with $3N-6=9$ for a five-atom nonlinear molecule. The lowest-frequency doubly degenerate bend dominates the vibrational entropy; the C–H stretch is effectively frozen at this temperature.

As a numerical check on the graphical reading, the harmonic expression $S/R=x/(e^x-1)-\ln (1-e^{-x})$, with $x=1/r$, gives a total $22.10\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$. This agrees with the precision of the supplied graph.

At $500\mathrm{K}$, the same approximation givesAs temperature rises, the relative advantage from a fixed gap difference becomes smaller, even while the intrinsic carrier densities of both phases increase.

For an intrinsic, nondegenerate semiconductor, charge neutrality and the Boltzmann approximation to the Fermi-Dirac distribution give $n_i=\sqrt{N_C{N}_V}\exp [-E_g/(2kT)]$. Assume the two polymorphs have equal effective-density-of-states prefactors and retain the quoted band gaps at the temperature being estimated. Their carrier-density ratio is thenwith $k_B$ expressed in eV per kelvin. At $77\mathrm{K}$,Both absolute intrinsic densities are extremely small at this temperature; a large ratio does not imply that either material is a good intrinsic conductor. Defects or unintended dopants can dominate real “undoped” specimens.