With the help of a labelled diagram,show that there are $4$ octahedral voids per unit cell in a cubic close-packed $(CCP)$ structure.

(N/A) In a cubic close-packed $(CCP)$ or face-centered cubic $(FCC)$ structure,the number of octahedral voids is equal to the number of atoms present in the unit cell.
$1$. The number of atoms per unit cell in $CCP$ is $4$.
$2$. One octahedral void is present at the body center of the cube. This void is entirely within the unit cell and is formed by $6$ atoms (one at each face center).
$3$. Additionally,there are $12$ octahedral voids located at the edges of the cube. Each edge-centered void is shared by $4$ adjacent unit cells. Therefore,the contribution of each edge-centered void to a single unit cell is $1/4$.
$4$. Total number of octahedral voids = (Number of voids at body center $\times 1$) + (Number of voids at edge centers $\times 1/4$)
$5$. Total number of octahedral voids = $(1 \times 1) + (12 \times 1/4) = 1 + 3 = 4$.
Thus,there are $4$ octahedral voids per unit cell in a $CCP$ structure.