= Solution
{c}
Square the stated <Lindemann melting criterion> and insert the high-temperature displacement:
$$
\frac{\hbar^2T_m}{m_{\mathrm{Ar}}k_BT_E^2}=(0.05d)^2.
$$
Thus, with $m_{\mathrm{Ar}}=(0.040\,\mathrm{kg\,mol^{-1}})/N_A$,
$$
\boxed{T_m=0.05^2d^2\frac{m_{\mathrm{Ar}}k_BT_E^2}{\hbar^2}\simeq1.00\times10^2\,\mathrm K.}
$$
Here $T_m/T_E\simeq1.66$, 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.
Back to article page
