Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 2 / A9

In a face-centered cubic lattice, nearest neighbours touch along a face diagonal. Four radii span that diagonal, so $\sqrt{2}a=4r$ andThusThe primitive vectors enclose volume $a^3/4$, containing one atom; the conventional cubic cell of volume $a^3$ contains four. Keeping these cell conventions distinct avoids a factor-of-four error in the electron density.

With one valence electron per atom, the electron number density is $n=4/a^3$. For a spin-degenerate, three-dimensional free electron model at zero temperature,Substitution givesCopper has the larger Fermi wavevector and Fermi energy because its smaller cell gives a larger electron density. The energies are measured from the bottom of the free-electron dispersion, not from vacuum. Treating these real metals as monovalent free-electron systems is the model assumption, not a statement that their filled d bands are absent.

The reciprocal-space volume inside a spherical Fermi surface is ${\mathcal{V}}_{\mathrm{F}}=4\pi k_{\mathrm{F}}^3/3$. The first Brillouin zone has volume $(2\pi {)}^3$ divided by the primitive real-space cell volume:Since $k_{\mathrm{F}}^3=3{\pi }^2(4/a^3)$,Numerically,The half-zone volume reflects one electron per primitive cell and two spin states per Bloch wavevector: a completely filled single band would require two electrons per cell. Volume alone does not establish containment inside the polyhedral zone. Here the nearest reciprocal-lattice points are of type $(111)$, so their zone faces are a distance $\pi \sqrt{3}/a$ from the origin. The free-electron radius obeysso these ideal monovalent spheres fit inside the first zone. Real band structure can distort a Fermi surface into zone-face contacts or necks; that is additional physics beyond the free sphere.

At zero temperature, states are occupied up to $E_{\mathrm{F}}$. Write the density of states as $g(E)=C\sqrt{E}$, including any fixed volume and spin factors in $C$. ThenwhileDividing eliminates $C$:The mean is positive even at absolute zero because the Pauli exclusion principle fills a range of momenta; electrons cannot all occupy the zero-momentum state.

At fixed $N$, the Fermi energy scales as $E_{\mathrm{F}}\propto (N/V{)}^{2/3}$, so $\partial E_{\mathrm{F}}/\partial V=-(2/3)E_{\mathrm{F}}/V$. From $U=3N{E}_{\mathrm{F}}/5$,This is the nonrelativistic electron degeneracy pressure. Because $n{E}_{\mathrm{F}}\propto V^{-5/3}$,The units are energy per volume, hence pressure. Converting the Fermi energies to joules gives the ideal Fermi gas bulk modulusThese are free-electron model values. The complete mechanical bulk modulus of a real metal also involves the ionic lattice, electronic bands and interactions; the model's positive electron pressure does not by itself explain the equilibrium crystal volume.

For fixed atom number, volume scales as the cube of the touching-sphere radius. Compressing silver to the stated gold lattice size therefore requiresa volume decrease of about $0.415\%$. If the bulk modulus is constant,Using the silver value derived above givesFor this small compression, $\Delta p\simeq -B\Delta V/V$ gives almost the same estimate. This is an additional pressure relative to the initial standard-pressure state, using the question's free-electron model modulus rather than an empirical silver modulus.