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The conventional face-centered cubic lattice cell contains four atoms, soThe preceding free-electron and Debye expressions giveAt low temperature the electronic and lattice heat-capacity crossover in a monovalent metal is found fromThus the electronic term dominates for $0<T\lesssim 3.4$ K within the normal-metal model. Both low-temperature approximations are well controlled at that crossover. At exactly zero temperature both heat capacities vanish; below sufficiently small finite-size energy scales a continuum model also needs revision.

Room temperature is not $T\ll {\theta }_D$: using the cubic formula at 300 K would incorrectly predict $C_{\mathrm{ph}}\simeq 142n{k}_B$, far above the harmonic high-temperature limit $3n{k}_B$. Instead use the Dulong-Petit law as an estimate:Keeping the finite cutoff in the Debye heat capacity givesAt 300 K this is $2.8006n{k}_B$, whereas $C_e=0.017994n{k}_B$, giving $\boxed{C_e/C_{\mathrm{ph}}=6.42\times 10^{-3}}$. Either appropriate room-temperature estimate shows that the lattice contribution dominates. The electronic approximation remains excellent because $300\ll T_F$.

For the electronic and lattice heat-capacity crossover in a monovalent metal, to obtain a numerical estimate from $E_F$ alone, assume monovalent potassium is a three-dimensional free electron gas with a parabolic band. Including both spins, the molar density of states is $g(E_F)=3N_A/(2E_F)$. Using the supplied approximate coefficient $3$, rather than silently replacing it by ${\pi }^2/3$, givesEquating it to the stated low-temperature Debye heat capacity yields the nonzero crossoverWe used $k_B=8.61733\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$ and $E_F=2\mathrm{eV}$. This is self-consistently far below ${\theta }_D=100\mathrm{K}$ and the Fermi temperature. Using the exact Sommerfeld coefficient instead gives $0.954\mathrm{K}$; that is a change of approximation, not a discrepancy in the algebra. Without a band/DOS assumption, the supplied Fermi energy does not by itself determine $g(E_F)$.