Atoms per conventional cubic cell in the diamond structure Silicon and germanium crystallize in the diamond structure. How many atoms are contained in a conventional cubic cell of volume a^3 (where a is the cubic lattice parameter)?

Introduction / Context:
The diamond crystal structure (adopted by Si and Ge) can be described as two interpenetrating face-centered cubic (fcc) lattices offset by one-quarter of a body diagonal. Counting the atoms per conventional cell is a classic crystallography exercise, important for computing densities and understanding basis atoms per lattice point.

Given Data / Assumptions:

• Conventional cubic cell of edge length a.
• Diamond structure = fcc Bravais lattice + two-atom basis.
• Each atom at a corner is shared by 8 cells; each face-centered atom is shared by 2 cells; basis duplication must be accounted for correctly.

Concept / Approach:

Start from fcc: there are 4 atoms per conventional fcc cell (8 corners × 1/8 + 6 faces × 1/2 = 1 + 3 = 4). The diamond structure places one atom at each fcc lattice point and adds a second atom of the basis displaced by (a/4, a/4, a/4). Thus, the number of atoms doubles compared with fcc: 4 × 2 = 8 atoms per conventional cell.

Step-by-Step Counting:

Verification / Alternative check:

Mass density ρ = (Z * M) / (N_A a^3), where Z = 8 for diamond cubic, is consistent with experimental values for Si/Ge when using their molar masses and lattice constants.

Why Other Options Are Wrong:

• 4: corresponds to simple fcc without the extra basis atom.
• 16 or 32: overcount by including multiple equivalent positions or multiple cells.
• 2: refers to the basis per lattice point, not atoms per conventional cell.

Common Pitfalls:

Mixing up basis size with atoms per unit cell; forgetting the sharing fractions for corner and face atoms in fcc counting.

Final Answer:

8