= Solution
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 $\pi^2/3$, gives
$$
C_{el}=\frac92R\frac{k_B T}{E_F}.
$$
Equating it to the stated low-temperature <Debye heat capacity> yields the nonzero crossover
$$
\boxed{T=\left[\frac{15}{8\pi^4}\frac{k_B}{E_F}\theta_D^3\right]^{1/2}=0.911\,\mathrm K.}
$$
We used $k_B=8.61733\times10^{-5}\,\mathrm{eV\,K^{-1}}$ 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 $g(E_F)$.
Back to article page
