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

[!TLDR] BCC has 8 corner atoms (1/8 each) plus 1 unshared body-centre atom. [!ANSWER] $Z = \left(8 \times \tfrac{1}{8}\right) + 1 = 2$ atoms per bcc unit cell.

A body-centred cubic unit cell has atoms at the $8$ corners, exactly as in the simple cubic case, plus one additional atom sitting at the geometric centre of the cube's body. The corner contribution is unchanged: $8 \times \tfrac{1}{8} = 1$. The body-centre atom, however, lies entirely inside this one unit cell — it is not shared with any neighbouring cell — so it contributes the full atom, $1$. Adding the two contributions: $Z = 1 + 1 = 2$. So a bcc unit cell contains twice as many atoms as a simple cubic cell of the same nominal size, which is part of why bcc packs more efficiently (68.0% versus 52.4%). [!ANSWER] $Z = 2$ atoms per bcc unit cell: $1$ from the $8$ shared corners plus $1$ full atom at the body centre.

Add the corner contribution (8 corners x 1/8) to the body-centre contribution (1 unshared atom, full weight).

Forgetting that the body-centre atom is not shared with any other cell and mistakenly giving it a fractional weight.