Percentage of Empty Space in Hexagonal Close Packed Cell

To find the empty space in a hexagonal close-packed (HCP) unit cell, we first calculate the packing efficiency (the fraction of the total volume occupied by the constituent spheres).

1. Number of atoms per unit cell ($Z$):
   In an HCP unit cell:

   • $12$ corner atoms each shared by $6$ unit cells: $12 \times \frac{1}{6} = 2$
   • $2$ face-centered atoms each shared by $2$ unit cells: $2 \times \frac{1}{2} = 1$
   • $3$ atoms completely inside the body of the unit cell: $3 \times 1 = 3$

   $\text{Total number of atoms, } Z = 2 + 1 + 3 = 6$

2. Volume occupied by the atoms ($V_{\text{atoms}}$):
   Each atom is considered a sphere of radius $r$:
   $V_{\text{atoms}} = Z \times \left(\frac{4}{3}\pi r^3\right) = 6 \times \frac{4}{3}\pi r^3 = 8\pi r^3$

3. Total volume of the HCP unit cell ($V_{\text{cell}}$):

   • The base of the unit cell is a regular hexagon with side length $a = 2r$.
     $\text{Area of base} = 6 \times \left(\frac{\sqrt{3}}{4} a^2\right) = 6 \times \frac{\sqrt{3}}{4}(2r)^2 = 6\sqrt{3}r^2$
   • The height of the unit cell is $c = 4r\sqrt{\frac{2}{3}}$.

   $\text{Volume of the cell } (V_{\text{cell}}) = \text{Base Area} \times c = (6\sqrt{3}r^2) \times \left(4r\sqrt{\frac{2}{3}}\right) = 24\sqrt{2}r^3$

4. Packing Efficiency:
   $\text{Packing Efficiency} = \frac{V_{\text{atoms}}}{V_{\text{cell}}} \times 100\% = \frac{8\pi r^3}{24\sqrt{2}r^3} \times 100\% = \frac{\pi}{3\sqrt{2}} \times 100\% \approx 74\%$

5. Percentage of empty space:
   $\text{Empty space} = 100\% - \text{Packing Efficiency} = 100\% - 74\% = 26\%$

Therefore, the correct option is (D).