The problem asks for the percentage of empty space (void space) in a hexagonal close-packed (HCP) unit cell.

Step 1: Understand Packing Efficiency and Empty Space

The percentage of empty space in a unit cell is related to the packing efficiency (PE) by the formula: $\text{Empty Space } \% = (1 - \text{Packing Efficiency}) \times 100\%$ Packing Efficiency is defined as the ratio of the volume occupied by atoms within the unit cell to the total volume of the unit cell: $\text{Packing Efficiency (PE)} = \frac{\text{Volume occupied by atoms in unit cell}}{\text{Total volume of the unit cell}} = \frac{Z \times V_{\text{atom}}}{V_{\text{unit cell}}}$ where $Z$ is the effective number of atoms per unit cell, $V_{\text{atom}}$ is the volume of a single atom, and $V_{\text{unit cell}}$ is the volume of the unit cell.

Step 2: Calculate the effective number of atoms (Z) per unit cell

In an HCP unit cell:

• There are 12 atoms at the corners (6 on the top face, 6 on the bottom face). Each corner atom is shared by 6 unit cells. Contribution from corners = $12 \times \frac{1}{6} = 2$ atoms.
• There are 2 atoms at the center of the two hexagonal faces (top and bottom). Each is shared by 2 unit cells. Contribution from face centers = $2 \times \frac{1}{2} = 1$ atom.
• There are 3 atoms entirely within the body of the unit cell. Contribution from body atoms = $3 \times 1 = 3$ atoms.

So, the total effective number of atoms per unit cell is: $Z = 2 + 1 + 3 = 6 \text{ atoms}$

Step 3: Calculate the volume occupied by atoms

Assuming each atom is a sphere of radius 'r', the volume of one atom is $V_{\text{atom}} = \frac{4}{3}\pi r^3$. The total volume occupied by atoms in the unit cell is: $V_{\text{occupied}} = Z \times V_{\text{atom}} = 6 \times \frac{4}{3}\pi r^3 = 8\pi r^3$

Step 4: Calculate the volume of the HCP unit cell

The volume of the hexagonal prism unit cell is given by the area of its base multiplied by its height ($c$).

• Base Area: The base is a regular hexagon with side length 'a'. In a close-packed structure, the atoms at the corners are in contact. Therefore, the side length of the hexagon 'a' is equal to twice the atomic radius, $a = 2r$. The area of a regular hexagon with side 'a' is $6 \times \frac{\sqrt{3}}{4}a^2$. $\text{Base Area} = 6 \times \frac{\sqrt{3}}{4}(2r)^2 = 6 \times \frac{\sqrt{3}}{4} \times 4r^2 = 6\sqrt{3}r^2$
• Height (c): The height 'c' of the HCP unit cell is related to the atomic radius 'r' by the ideal c/a ratio for close packing, which is $\sqrt{\frac{8}{3}}$. $c = a \sqrt{\frac{8}{3}} = (2r)\sqrt{\frac{8}{3}} = 2r \frac{2\sqrt{2}}{\sqrt{3}} = 4r\sqrt{\frac{2}{3}}$
• Volume of unit cell: $V_{\text{unit cell}} = (\text{Base Area}) \times c = (6\sqrt{3}r^2) \times \left(4r\sqrt{\frac{2}{3}}\right) = 24r^3 \sqrt{3} \sqrt{\frac{2}{3}} = 24r^3\sqrt{2}$

Step 5: Calculate the Packing Efficiency (PE)

Now we can calculate the packing efficiency: $\text{PE} = \frac{V_{\text{occupied}}}{V_{\text{unit cell}}} = \frac{8\pi r^3}{24\sqrt{2} r^3} = \frac{\pi}{3\sqrt{2}}$ Using the values $\pi \approx 3.14159$ and $\sqrt{2} \approx 1.41421$: $\text{PE} \approx \frac{3.14159}{3 \times 1.41421} \approx \frac{3.14159}{4.24263} \approx 0.7405$ So, the packing efficiency is approximately 74%.

Step 6: Calculate the percentage of empty space

$\text{Empty Space } \% = (1 - \text{PE}) \times 100\% = (1 - 0.7405) \times 100\% = 0.2595 \times 100\% \approx 26\%$ The empty space in the HCP unit cell is 26%.

Conclusion

Comparing this result with the given options: A: 74% is the packing efficiency, not the empty space. B: 47.6% corresponds to the empty space in a simple cubic lattice. C: 32% corresponds to the empty space in a body-centered cubic (BCC) lattice. D: 26% matches our calculated value for the empty space in an HCP lattice.