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Question 43

The total number of octahedral void(s) per atom present in a cubic close packed structure is:

Solution

In a Cubic Close Packed (ccp) or Face-Centered Cubic (fcc) structure:

Atoms per unit cell ($$Z$$): 4 (1 corner + 3 face centers).

Octahedral voids per unit cell ($$N$$): 4 (1 at the body center and 3 at the edge centers).

$$\text{Octahedral voids per atom} = \frac{\text{Total Voids}}{\text{Total Atoms}} = \frac{4}{4} = 1 \text{}$$

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