Abstract

A new explanation of why the crystal structure of the rare gas solids, Ne, Ar, Kr and Xe is f. c. c. rather than h. c. p. is offered. The magnitude of the relative energy difference,∆= (Ef. c. c.–Eh. c. p.)/Et. c. c., is estimated and it is shown that the effect is numerically large enough in all these solids (∆≳ + 1 x 10–3) to overcome the small preference of two-body interatomic potentials for the h. c. p. structure (∆≃ – 10–4). The effect is much weaker in helium and so the h. c. p. structure of solid helium emerges naturally as a consequence of the two-body potential. The explanation depends on the modification of the (long-range) van der Waals energy by the (short-range) overlap of atomic excited states with the neighbouring atoms in the crystal. The resulting crystal field in the f. c. c. and h. c. p. structures splits excited d-states by different amounts. The f. c. c. structure is favoured because the energy split is wider in f. c. c. (which is centrosymmetric) than in h. c. p. (which does not have a centre of symmetry at the atomic sites); the resulting van der Waals attractive energy is thereby greater in f. c. c. An alternative approach is also developed, which uses the band states of the crystal as a starting-point, and yields a similar result. We expect that, if good enough band structure calculations of h. c. p. rare gas solids were available, the best way to estimate the value of∆would be to calculate the van der Waals energy in the solid in terms of band structure energies for the excited states and gas phase values for the dipole matrix elements. Preliminary estimates of the size of the effect, based on currently available band structure data, suggest that∆ranges from approximately 12 x 10–4for Ne to 27 x 10–4for Xe; these values are quite sufficient to explain the stability of the f. c. c. structure.

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