Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S3 / 1

The de Broglie hypothesis associates a wave phase with a material particle. Its momentum and energy setThis is not a literal extended classical lump of matter oscillating at every point. A wavefunction gives amplitudes for measurements, and a localized particle is represented by a superposition of momentum waves. Electron diffraction, for example, exhibits interference at angles set by the same wavelength relation.

Using the given one-component equipartition estimate, $p^2=M{k}_{B}T$. Substituting into the de Broglie relation givesComparison with a confinement length or interparticle spacing $d$ gives the criterion $T\lesssim h^2/(M{k}_B{d}^2)$. Numerical factors depend on the convention: the statistical thermal de Broglie wavelength is usually $h/\sqrt{2\pi M{k}_{B}T}$, while a three-component root-mean-square momentum introduces a factor $\sqrt{3}$. The question's specified one-component convention is used consistently here. Once degeneracy becomes important, classical equipartition itself ceases to describe the electrons accurately.

Set $\lambda \simeq d$ with the convention of part b. ThenThus ordinary solid-state temperatures are far below this rough overlap scale. The electrons need quantum statistics and quantum states throughout that regime. The number is an order-of-magnitude wavelength criterion, not an exact Fermi temperature: electron density, electronic band structure, effective mass and the numerical convention all matter.

At the same temperature, momenta increase as the square root of mass, so wavelengths decrease as its inverse square root. For a nucleus of mass $M_N\simeq A{m}_p$,At the electron overlap threshold, ${\lambda }_e=d$, the nuclear wavelength is about $7.0\mathrm{pm}/\sqrt{A}$, far below the $0.3$ nm spacing. Nuclear exchange effects are consequently much less important than electron exchange effects. This statement about nuclear overlap does not mean lattice vibrations are always classical.

At room temperature, electrons require quantum mechanics, including their occupied orbitals or bands and Fermi statistics. Heavy nuclei can often be treated as distinguishable particles near classical equilibrium lattice sites, because their exchange wavelengths are short. Their coupled displacements form phonons, however, and the occupation of a vibrational mode is classical only if $k_{B}T\gg \hslash \omega$ for that mode. Even at room temperature some high-frequency modes remain quantum mechanical.

At $1$ K, quantum lattice vibrations, zero-point motion and the scarcity of thermally excited phonons are essential. Many heavy nuclei still do not have overlapping exchange wavefunctions; quantum vibration and quantum exchange are distinct criteria. For the same spacing, the nuclear overlap threshold from the rough estimate is $T_N\simeq 211/A$ K, illustrating why the nuclear mass cannot be omitted from a blanket low-temperature claim.