Past exam of the quantum condensed matter course of the University of Cambridge / 2016 / 1

In the three-dimensional Debye model, the lattice has $3N$ normal modes for $N$ atoms. Replacing their dispersion by a linear acoustic dispersion up to a cutoff gives the Debye heat capacityAt low temperature the upper limit can be replaced by infinity. Since the integral then equals $4{\pi }^4/15$,At high temperature each harmonic mode contributes $k_B$ by the equipartition theorem, giving the Dulong-Petit lawFor one mole replace $N{k}_B$ by $R$. The curve starts with zero slope, rises as $T^3$, and levels off toward the classical limit. This is the lattice contribution; a metal's electronic heat capacity is a separate low-temperature term.

The Debye wavevector follows by counting three polarizations inside the cutoff sphere:With the supplied speed of sound treated as the effective Debye speed, ${\omega }_D=v{k}_D$ andHere $k_D\simeq 1.52\times 10^{10}{\mathrm{m}}^{-1}$. Real crystals have different longitudinal and transverse sound speeds; the one-speed estimate uses the stated value, rather than inferring those unprovided branches.

Use physical fields proportional to the real part of $e^{-i\omega t}$. The Drude frequency response isThus the sketch is an even Lorentzian line shape, with maximum ${\sigma }_0^2$ at zero frequency, half that value at $|\omega |={\tau }^{-1}$, and an ${\omega }^{-2}$ tail. The positive-frequency half is shown in the preceding figure. Squaring the modulus differs from plotting either $|\sigma |$ or the real part of $\sigma$.

For a single carrier species, the plasma frequency and electrical conductivity satisfyThereforeThe corresponding relaxation time is $\tau \simeq 3.57\times 10^{-14}\mathrm{s}$. This is an angular frequency; the ordinary frequency is $\boxed{f_{1/2}={\omega }_{1/2}/(2\pi )\simeq 4.46\times 10^{12}\mathrm{Hz}}$. The units of electrical conductivity used in the calculation are ${\mathrm{\Omega }}^{-1}{\mathrm{m}}^{-1}$, consistently with the SI plasma-frequency relation.

For nondegenerate Electrons in a parabolic conduction band,when the density-of-states effective mass and valley multiplicity are fixed. The intrinsic carrier extrapolation across a crossover must therefore specify how the activation energy changes with temperature.

The usual intended estimate treats the quoted $0.9\mathrm{eV}$ as a constant intrinsic activation energy between the two temperatures, and identifies the crossover by $n_i(T_c)\simeq n_{\mathrm{ext}}$. ThenFor $T_1=350\mathrm{K}$, $T_c=280\mathrm{K}$ and $k_B=8.6173\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$, the exponent is $-7.46$ and the prefactor is $0.8^{3/2}=0.7155$. HenceEquivalently, a fixed-gap intrinsic model with $E_g/2=0.9\mathrm{eV}$ gives this result. A crossover is gradual: if $n_{\mathrm{ext}}$ denotes the net ionized donor density and $n_i(T_c)=n_{\mathrm{ext}}$, charge neutrality gives the actual Electron population $n(T_c)=(1+\sqrt{5})n_{\mathrm{ext}}/2$, rather than exactly $n_{\mathrm{ext}}$.

There are two limits to interpreting the numerical estimate literally. First, the printed data specify $E_c-\mu$ only at $350\mathrm{K}$, without specifying $E_g$ or the valence density of states. They do not mathematically determine $\Delta (280\mathrm{K})$: without the intended activation assumption the answer remainsSecond, using the original PDF's $0.9\mathrm{eV}$ and density implies $N_c(350)\simeq 2.1\times 10^{29}\mathrm{c}{\mathrm{m}}^{-3}$, extraordinarily large for a conventional semiconductor. This confirms that the numerical data should not be treated as a realistic material determination. No unprinted decimal-point correction has been assumed.