Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 2

Both in-plane orientations minimize the one-site energy. Since every adsorption site is occupied, the sites label independent choices even though the molecules are chemically identical. The ground state of the whole system therefore hasThe ground-state degeneracy is extensive: independently choosing either in-plane orientation at each site costs no energy. There is no extra vacancy or free-translational contribution in this model.

Write $\beta =(k_{\mathrm{B}}T{)}^{-1}$ and $x=\beta \delta$. The one-site partition function includes both degenerate lower states,The occupied adsorption sites are distinguishable positions, so the canonical partition function is a product, without an additional $N!$ divisor:Differentiating $\ln Z_N$ gives the internal energy,The final factor is the population of the perpendicular orientation. The shared binding offset $-\epsilon$ shifts all energy levels equally and does not determine their relative populations.

At low temperature, the upper orientation has negligible Boltzmann weight, whereas at high temperature all three orientations become equally probable. ConsequentlyThe high-temperature limit is a finite arithmetic mean of three available one-site energy levels. It is not an unbounded translational or rotational equipartition contribution: the prescribed site model has only three states per molecule. Their common binding offset can keep this limiting internal energy negative without any inconsistency.

For $k_{\mathrm{B}}T\gg \delta$, the Boltzmann factors of the orientations become almost equal, giving $3^N$ nearly equiprobable microstates. The Boltzmann entropy of that distribution approachesThe exact limiting value is reached only as $T\to \infty$ for a nonzero splitting. The useful crossover scale is $T\sim \delta /k_{\mathrm{B}}$; one needs $T\gg \delta /k_{\mathrm{B}}$ for a good saturation approximation. The binding offset $\epsilon$ is not an additional population scale here because all sites remain occupied and it shifts every orientation equally.

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 $C_{\max }/(N{k}_{\mathrm{B}})\simeq 0.241$.

Only the two lower orientations remain populated, giving $2^N$ equiprobable microstates. ThusThis residual entropy follows from the stipulated independent-site ground-state degeneracy. The zero-entropy version of the third law of thermodynamics presupposes a unique ordered ground state; that assumption does not hold here. Intermolecular interactions or a small splitting of the two in-plane states could remove this degeneracy, but they are excluded from the given model.

Let $\Lambda =h/\sqrt{2\pi m{k}_{\mathrm{B}}T}$ be the thermal de Broglie wavelength. The two free momentum components of one adsorbed atom give the phase-space integralEach momentum integral is a Gaussian integral, soThe second expression is the classical ideal-gas partition function for mobile, indistinguishable surface atoms. Its validity requires a dilute nondegenerate gas, $N{\Lambda }^2/\sigma \ll 1$, and neglects interactions and excited motion normal to the surface. The $N!$ appears here because particles move between positions, unlike the fully occupied fixed-site model of A6.

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 $F_{\mathrm{ads}}-F_{\mathrm{gas}}$.

A planar rigid rotor has angular eigenfunctions $e^{im\varphi }$ and single-valuedness imposes $m=0,\pm 1,\pm 2,\dots$. Its energy depends on $m^2$, so if the level is labelled by $J=|m|$, thenThe two states at $J>0$ represent opposite directions of angular momentum. If the signed quantum number itself is called $J$, each signed state is counted once instead; one must not both sum over both signs and multiply by two again.

For an unrestricted three-dimensional linear rigid rotor, $J=0,1,2,\dots$ has $M=-J,\dots ,J$ and state degeneracy $2J+1$, with $E_J={\hslash }^{2}J(J+1)/(2I)$. Planar confinement removes one rotational coordinate and changes both the spectrum and degeneracy. These are the spatial counts before nuclear spin statistics. For homonuclear hydrogen, the allowed even/odd spatial states must also be paired with the appropriate parahydrogen or orthohydrogen nuclear states, as considered explicitly in A8.

Define ${\theta }_{\mathrm{rot}}={\hslash }^2/(2I{k}_{\mathrm{B}})$. Summing the spatial states counted in part (c),When $T\gg {\theta }_{\mathrm{rot}}$ the closely spaced terms can be replaced by a Gaussian integral:Here labelled means that the two ends of the molecular axis were distinguished in the angular count. In the usual high-temperature molecular convention that removes equivalent orientations and factors out nuclear-spin multiplicity, divide by the rotational symmetry number $s=2$ for homonuclear hydrogen:The full exact spin isomers of hydrogen count is instead the even-$m$ spatial sum plus three times the odd-$m$ sum; its high-temperature limit is $2\sqrt{\pi T/{\theta }_{\mathrm{rot}}}$, including the four nuclear-spin states. These conventions differ only by temperature-independent factors in this limit. All give $U_{\mathrm{rot},\mathrm{m}}=RT/2$ and $C_{V,m}^{\mathrm{rot}}=R/2$, consistent with one quadratic rotational degree of freedom. By comparison, the free linear rigid rotor has $q_{\mathrm{rot}}\propto T$ and $C_{V,m}^{\mathrm{rot}}=R$.

The two protons are identical spin-one-half fermions, so the total state is antisymmetric under exchange. In the electronic and vibrational ground state, an even rotational $J$ has a symmetric spatial wavefunction and therefore uses the antisymmetric nuclear-spin singlet state; odd $J$ uses the symmetric triplet state. Thus parahydrogen has one nuclear-spin state, and orthohydrogen has three. Each rotational level also has spatial state degeneracy $2J+1$.

The translational partition functions and all common electronic/vibrational factors cancel in the ratio of equilibrium populations. Using the same zero of energy for the two spin species givesSubtracting the $J=1$ energy separately from all odd levels without restoring that offset in the equilibrium ratio would incorrectly remove the low-temperature preference for parahydrogen. The nuclear spin statistics account for the factor three, independently of the spatial factor $2J+1$.

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 fixed number of non-interacting molecules, $Z_N=q^N/N!$ in the dilute ideal gas limit. The temperature-independent factorial drops out of the internal energy derivative:Differentiate once more at fixed volume to obtainPrimes denote fixed-volume derivatives. If $q$ factorizes into translational, rotational and other independent contributions, their heat capacities add. In particular, the three-dimensional translational factor contributes $3R/2$, while a constant spin degeneracy factor contributes zero. The formula also follows from the positive canonical energy variance, $C_V=\mathrm{Var}(E)/(k_{\mathrm{B}}{T}^2)$.

For a high-temperature linear rigid rotor, $q_{\mathrm{rot}}=T/{\theta }_{\mathrm{rot}}$ up to temperature-independent symmetry/spin factors. Therefore $q^{\prime }/q=1/T$ and $q^{\prime \prime }=0$, givingThis agrees with the equipartition theorem for two quadratic rotational coordinates. It is the rotational contribution; adding free three-dimensional translation gives $C_{V,m}=5R/2$ when vibrations remain unexcited. A constant rotational symmetry number does not affect this derivative.

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.

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.

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 $k_{\mathrm{B}}T$.

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.

Let $\mathcal{A}$ be the total two-dimensional area. There are $\mathcal{A}/(2\pi {)}^2$ wavevector states per reciprocal-space area per spin. Including spin degeneracy two and filling a disk givesFor areal number density $n_e=N_e/\mathcal{A}$ this yieldsThe disk is centred at zero wavevector, and its boundary is a fermi circle. Omitting the two spin states would incorrectly double the energy inferred for a given number density.

Using $M_{\mathrm{Au}}\simeq 196.97\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ and the Avogadro constant, the areal atom count isOne atom occupies each square cell of area $a^2$, so the inferred square lattice spacing isThis uses the given deposited mass and the assumed single square layer; it does not substitute a bulk gold lattice parameter for the measured areal count.

One mobile electron per gold atom givesThis is an areal number density; no layer thickness is required and it should not be converted into a volume density for the two-dimensional model.

Substitution in part (a) givesandThe conversion uses $1\mathrm{eV}=1.60218\times 10^{-19}\mathrm{J}$. These are the predictions of the stipulated two-dimensional free electron model, with the vacuum electron mass and one carrier per inferred square cell.

The reciprocal lattice is generated byIts points are spaced by $2\pi /a$ along both reciprocal axes. Perpendicular bisectors between the origin and its four nearest reciprocal neighbours enclose the first Brillouin zone,The original drawing labels both the reciprocal-lattice spacing and the nearer zone boundaries. Its shaded disk is the occupied region for one electron per real-space cell.

For one electron per square cell, $n_e=1/a^2$, soNumerically, $\pi /a\simeq 2.17\times 10^{10}{\mathrm{m}}^{-1}$ exceeds $k_F\simeq 1.73\times 10^{10}{\mathrm{m}}^{-1}$, so the fermi circle lies entirely inside the first Brillouin zone.

The state-counting explanation is that a spin-degenerate band has two electron states per primitive cell. One electron per cell therefore occupies half of that band. In the circular free-electron approximation the occupied reciprocal-space area isA disk of that area fits inside this square: its radius is less than the distance to an edge. Half filling alone is not a shape-independent theorem that every possible Fermi surface must avoid every zone boundary; the square geometry and circular free-electron dispersion make the statement true here.

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.

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 $n_e=n_h=n_i=\sqrt{K_n}$.

The equal supplied masses give equal effective density of states. At the specified temperature,Using the same energy units in the exponential,For equal electron and hole mobilities, both carriers contribute to the intrinsic electrical conductivity:The factor two is essential. A dimensional check givesusing $\mathrm{A}=\mathrm{C}{\mathrm{s}}^{-1}$.

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]

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.

Two derivatives of the free-electron dispersion giveThe band effective mass becomes the ordinary electron mass when no periodic-potential modification of the quadratic dispersion relation is present.

Differentiating the tight binding dispersion produced by the Hückel method givesHenceUnlike a free particle, the band effective mass depends on wavevector and can change sign. At an inflection point $\cos (ka)=0$, the inverse mass vanishes and the curvature-based mass diverges; a quadratic approximation then ceases to describe the dispersion locally. For the usual $\beta <0$ convention, the bottom at $k=0$ has positive mass and the top at $k=\pi /a$ has negative electron mass. The constant $\alpha$ has no effect on curvature.

In a weakly doped n-type semiconductor, the mobile electrons occupy states near the bottom of the conduction band. Their local band curvature is positive, so their band effective mass is positive and approximately constant sufficiently near that minimum. For the usual $\beta <0$ convention, expand about $k=0$:Stronger overlap gives larger $|\beta |$, greater curvature and a smaller mass within this model. Positive $\beta$ simply places the minimum at the zone boundary and again gives a positive edge mass. A negative electron mass near a band maximum instead motivates the positive-mass electron hole description; it is not the usual conduction-edge situation for an n-type semiconductor. No numerical comparison with $m_e$ follows without values of $a$ and $\beta$.