Crystal Structures — HCP Unit Cell Atom Count Consider the close-packed hexagonal (HCP) space lattice. The statement “In a unit cell of an HCP lattice, there are twenty-four atoms” is:

Introduction / Context:
Understanding how many atoms are effectively contained in a crystallographic unit cell is foundational for calculating packing factors, densities, and interpreting diffraction. The hexagonal close-packed (HCP) structure has a specific conventional unit cell that differs from the cubic cells students often memorize for FCC/BCC.

Given Data / Assumptions:

• We refer to the conventional HCP unit cell, not an arbitrary supercell.
• Standard counting rules apply: corner atoms contribute 1/8 when shared by eight cells, face-centered contributions depend on sharing, and interior atoms count fully.

Concept / Approach:
The conventional HCP unit cell effectively contains 6 atoms: 12 corner atoms contribute 1/6 each in the hexagonal prism description (or 1/8 in alternative notations with appropriate sharing), 2 atoms are fully inside (at 1/3 and 2/3 of the height), and additional fractional contributions from face/edge centers sum appropriately to yield 6. This atom count is consistent with the well-known HCP packing factor of about 0.74 (same as FCC) and standard density calculations for metals like Mg, Ti, and Zn.

Step-by-Step Solution:
Recall: HCP has ABAB stacking with 2 atoms per primitive cell; the conventional hexagonal cell contains 3 primitive cells.Thus, atoms per conventional cell = 2 * 3 = 6.Therefore, the statement claiming 24 atoms in the unit cell is incorrect.Any apparent higher count arises from miscounting shared atoms or using a nonminimal supercell.

Verification / Alternative check:
Standard materials texts list HCP effective atoms per conventional cell as 6, with c/a ≈ 1.633 for ideal packing and identical packing efficiency to FCC.

Why Other Options Are Wrong:

• “Correct” contradicts the established 6-atom count for the conventional HCP cell.

Common Pitfalls:
Confusing primitive versus conventional cells or forgetting to apply fractional contributions for shared lattice points properly.

Final Answer:
Incorrect