Q.How many atoms are present per unit cell in a simple cubic (primitive) lattice? Show the corner-atom contribution that leads to your answer.

[!TLDR] Each of the 8 corner atoms in a simple cubic cell is shared among 8 unit cells, contributing 1/8 each. [!ANSWER] $Z = 8 \times \tfrac{1}{8} = 1$ atom per simple cubic unit cell.

In a simple cubic (primitive) lattice, atoms sit only at the $8$ corners of the cube. Each corner is a meeting point of $8$ different unit cells stacked around it in three dimensions, so any one corner atom is shared equally among those $8$ cells and contributes only $\tfrac{1}{8}$ of an atom to any single cell. Summing the contribution of all $8$ corners: $Z = 8 \times \tfrac{1}{8} = 1$. So exactly one atom's worth of material genuinely belongs to one simple cubic unit cell, even though visually the cell appears to have $8$ atoms drawn at its corners. [!ANSWER] $Z = 1$ atom per unit cell, from $8$ corner atoms each contributing $\tfrac{1}{8}$.

Identify the only occupied position (corners), determine how many cells share a corner (8), and multiply the number of corners by the sharing fraction.

Counting all 8 visible corner atoms as belonging fully to the cell (giving the wrong answer $Z=8$) instead of applying the $1/8$ sharing fraction.