 Solution

Quantum statistics matter when thermal energy is comparable with a characteristic vibrational level spacing. The chain's natural angular frequency is ${\omega }_0=\sqrt{\kappa /m}=5.7471\times 10^{12}{\mathrm{s}}^{-1}$, givingFactors of order unity depend on which vibrational scale is used. The largest chain phonon angular frequency is $2{\omega }_0$, corresponding to $87.8\mathrm{K}$. An isolated two-atom bond has reduced mass $m/2$ and angular frequency $\sqrt{2}{\omega }_0$, corresponding to $62.1\mathrm{K}$. Thus quantum corrections become appreciable on the scale of several tens of kelvin, and all modes are classical only when $T$ is comfortably above the largest relevant vibrational scale. Below it, high-frequency phonons are progressively frozen out and equipartition fails.

For the isolated bond, the zero-temperature root-mean-square relative displacement isIt is small enough for a local bond expansion. The classical weakly anharmonic estimate in part d therefore has a useful parametric windowwhere the lower condition means comfortably above the vibrational scale, rather than a sharp threshold. The model's upper bound concerns local anharmonicity and binding, not a prediction of the melting temperature.