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Show that π/6 of the available volume is occupied by hard spheres in contact in a simple cubic arrangement.

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The cross-section of the portions of four spheres each of radius r touching each other and lying in a cell of edge a is shown in Fig. 5.1. The volume of each sphere lying within the cell is 1/4 × 4/3πr3 or π/3 r3. Volume of four spheres lying within the cell is 4π/3r3. Volume of the cell is a3 or (2r)3. Therefore, the available volume occupied by hard spheres in the simple cubic structure is 4πr3/3/(2r)3 or π/6 .

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