Effect on the Lattice Parameter

3. Effect on the Lattice Parameter

Let us now consider the effects of the increase in the surface-to-volume ratio as the object size decreases. We first analyze the simple case of a liquid sphere of diameter 2R. Due to curvature, a pressure difference ΔP is generated between the inside and outside of the sphere. In hydrostatic equilibrium, this is given by the Laplace equation:

ΔP dV = γ dA

where dV is the volume change corresponding to a change dA in surface area. For a sphere, this becomes:

ΔP = 2γ / R

For a spherical solid, the specific surface energy must be replaced by the surface stress tensor g_ij. Considering a simple cubic structure (isotropic case), we have:

g = γ + dγ/dA

Surface stress accounts for deformation effects, unlike surface energy which only accounts for surface creation.

The compressibility is defined as:

χ = -ΔV / VΔP

where V is the atomic volume of the solid, which can also be written as a³, where a is the lattice parameter.

Combining the above equations, the relative change in lattice parameter is obtained as:

Δa / a = -2g / 3χR

Observation

This relation indicates that as particle size decreases, the lattice parameter contracts due to surface-induced pressure. This effect becomes significant at the nanoscale.

Fig. 1.2: Contraction of the lattice parameter with decreasing particle size.

Experimental studies confirm that the lattice parameter decreases as a function of the inverse of particle diameter. This contraction has been observed using techniques such as electron energy loss spectroscopy (EELS) and X-ray absorption (EXAFS).

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