Assuming that atoms are in a crystal structure and arranged as close-packed spheres, what is the ratio of the volume of the atoms to the volume available for the simple cubic structure? Assume a one-atom basis.

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Solution: Each corner atom in a cubic unit cell is shared by a total number of eight unit cells so that the corner atom contributes only \( \frac{1}{8}\) of its effective part to a unit cell. Since there are in all 8 corners their total contribution is equal to \( \frac{8}{8}\) = 1.User Loaded Image | CSIT Guide

Therefore, a number of atoms per unit cell = 1.

From figure,

a = 2r

or, r = a/2

Therefore, volume occupied by the atom in the unit cell = \( \frac{4}{3} \pi r^{3}\) = \( \frac{4}{3} \pi \left( \frac{a}{2} \right)^3\)

Volume of unit cell = a3.

Thus, packing fraction = \( \frac{\frac{4}{3}\pi \left ( \frac{a}{2} \right)^3}{a^3} \) = \( \frac{\pi}{6} \) = 0.52 = 52%.

Therefore, 52% space of unit cell is occupied by the atoms. Here the atoms are loosely packed. Only Polonium at a certain temperature is known to exhibit such a structure. For example, KCL which has alternate icons of K and CL also behaves like a simple lattice as regards scattering of X-rays because the two icons are almost identical.

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