Past exam of the condensed matter physics course of the University of Oxford / 2019 / 2

The Einstein model of a solid replaces the vibrations of $N$ identical atoms by $3N$ independent quantum harmonic oscillators with a common frequency ${\omega }_E$. Each atom has three displacement directions; the restoring force is harmonic, coupling between different atoms is replaced by a fixed local restoring potential, and anharmonicity and frequency dispersion are neglected. The crystal volume and ${\omega }_E$ are held fixed when calculating the vibrational heat capacity at constant volume.

The one-oscillator partition function isConsequently, with $T_E=\hslash {\omega }_E/k_B$,Differentiating at fixed volume givesThe zero-point energy does not contribute to this derivative. The result tends to the Dulong-Petit law $3N{k}_B$ at high temperature and vanishes exponentially at low temperature; the latter is a limitation of the single-frequency model for a real acoustic spectrum.

Apply the given small-$x$ expansion with $x=T_E/T$ to the internal energy, including its zero-point energy:The temperature-independent terms cancel. Taking the derivative gives the Einstein high-temperature heat-capacity expansionThusThere is no $1/T$ correction to the heat capacity even though the internal energy has a $1/T$ correction.

Use the ratio of the two measured coefficients so that the sample size cancels:HenceThe units of the $7500$ coefficient are $\mathrm{J}\mathrm{K}$, so that its division by $T^2$ has the same heat capacity units as the leading $25\mathrm{J}{\mathrm{K}}^{-1}$ term.

The leading coefficient fixes the atom count:With molar mass $40\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, the sample mass isOne atom belongs to each primitive unit cell of monatomic argon on a face-centered cubic lattice, but the conventional cube contains four atoms. If its side is $a$, $N=4V/a^3$, and nearest neighbors lie across half a face diagonal. Thereforeusing $V=21\times 10^{-6}{\mathrm{m}}^3$. The corresponding conventional lattice constant is $a\simeq 5.18\text{A˚}$. Treating the source's one-atom cell as the conventional cube would omit the factor four and give a wrong nearest-neighbor distance.

For one Cartesian displacement, the harmonic potential is $m_{\mathrm{Ar}}{\omega }_E^2{x}^2/2$. At $T\gg T_E$, the equipartition theorem assigns $k_{B}T/2$ to this quadratic potential term. HenceThis is a one-coordinate mean squared displacement, not the total three-dimensional displacement variance, which is three times larger in the isotropic model.

For comparison, the exact Einstein mean-square displacement isUsing $\coth y\simeq 1/y$ recovers the requested high-temperature result. Its nonzero zero-temperature limit is the displacement contribution of the zero-point energy.

Square the stated Lindemann melting criterion and insert the high-temperature displacement:Thus, with $m_{\mathrm{Ar}}=(0.040\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1})/N_A$,Here $T_m/T_E\simeq 1.66$, so the use of the high-temperature approximation is only moderately justified. It is an estimate, not a controlled $T/T_E\to \infty$ limit. Retaining the exact Einstein mean-square displacement in the same heuristic criterion gives about $97\mathrm{K}$, indicating the size of that particular correction. The empirical threshold $0.05d$ and the independent-oscillator model introduce additional uncertainty.