Q.Calculate the number of atoms per unit cell in a face-centred cubic (fcc) lattice, accounting separately for the corner atoms and the face-centre atoms.

[!TLDR] FCC has 8 corner atoms (1/8 each) plus 6 face-centre atoms (1/2 each, one per face). [!ANSWER] $Z = \left(8 \times \tfrac{1}{8}\right) + \left(6 \times \tfrac{1}{2}\right) = 1 + 3 = 4$ atoms per fcc unit cell.

A face-centred cubic unit cell has atoms at the $8$ corners, contributing $8 \times \tfrac{1}{8} = 1$, exactly as before, plus one additional atom at the centre of each of the cube's $6$ faces. Each face is shared between exactly $2$ unit cells (the cell on either side of that face), so each face-centre atom contributes $\tfrac{1}{2}$ of an atom; with $6$ faces, this gives $6 \times \tfrac{1}{2} = 3$. Adding the corner and face contributions: $Z = 1 + 3 = 4$. An fcc unit cell therefore contains four atoms — twice as many as bcc and four times as many as simple cubic — consistent with fcc having the highest packing efficiency (74.0%) of the three cubic arrangements. [!ANSWER] $Z = 4$ atoms per fcc unit cell: $1$ from the $8$ shared corners plus $3$ from the $6$ shared face-centre atoms.

Add the corner contribution (8 corners x 1/8) to the face-centre contribution (6 faces x 1/2, since each face is shared between 2 cells).

Using the body-centre sharing rule (weight 1) for face-centre atoms instead of the correct weight of 1/2 for a face-shared position.