Select the correct statement(s) about three-dimensional `hcp` system.

Options

A. The number of atoms in `hcp` unit cell is six.
B. The volume of `hcp` unit cell is `24sqrt2r^(3)`.
C. The empty space in `hcp` unit cell is `26%`.
D. The base area of `hcp` unit is `6sqrt3r^(2)`.

Answer

To solve the question regarding the three-dimensional hexagonal close-packed (hcp) system, we will analyze each of the given statements one by one.

### Step-by-Step Solution:

1. **Number of Atoms in hcp Unit Cell:**
- In a hexagonal close-packed (hcp) structure, there are 6 atoms at the corners of the top and bottom hexagonal faces, and 2 atoms at the base center. Additionally, there are 3 atoms located within the unit cell.
- The contribution of the corner atoms is \( \frac{1}{6} \) for each of the 12 corner atoms (6 from the top and 6 from the bottom).
- The contribution from the 2 base center atoms is \( \frac{1}{2} \) each.
- The total number of atoms can be calculated as:
\[ \text{Total atoms} = \left(12 \times \frac{1}{6}\right) + \left(2 \times \frac{1}{2}\right) + 3 = 2 + 1 + 3 = 6 \]
- **Conclusion:** The first statement is correct: the number of atoms in the hcp unit cell is 6.

2. **Volume of hcp Unit Cell:**
- The volume \( V \) of the hcp unit cell can be calculated using the formula:
\[ V = \text{Area of base} \times \text{Height} \]
- The area of the hexagonal base can be derived from the area of equilateral triangles. The area of the hexagon is:
\[ \text{Area} = 6 \times \left(\frac{\sqrt{3}}{4} a^2\right) = \frac{3\sqrt{3}}{2} a^2 \]
- The height \( h \) of the hcp unit cell is related to the edge length \( a \) by:
\[ \frac{h}{a} = \frac{\sqrt{8}}{3} \implies h = \frac{\sqrt{8}}{3} a \]
- Substituting these into the volume formula gives:
\[ V = \frac{3\sqrt{3}}{2} a^2 \times \frac{\sqrt{8}}{3} a = 4\sqrt{2} a^3 \]
- Given that \( a = 2r \), we can substitute to find:
\[ V = 24\sqrt{2} r^3 \]
- **Conclusion:** The second statement is correct: the volume of the hcp unit cell is \( 24\sqrt{2} r^3 \).

3. **Empty Space in hcp Unit Cell:**
- The packing efficiency can be calculated using the formula:
\[ \text{Packing Efficiency} = \frac{Z \times \frac{4}{3} \pi r^3}{\text{Volume of unit cell}} \]
- Where \( Z = 6 \) for hcp. Substituting the values:
\[ \text{Packing Efficiency} = \frac{6 \times \frac{4}{3} \pi r^3}{24\sqrt{2} r^3} = \frac{8\pi}{24\sqrt{2}} = \frac{\pi}{3\sqrt{2}} \approx 0.74 \]
- The empty space percentage is:
\[ \text{Empty Space} = 1 - \text{Packing Efficiency} \approx 1 - 0.74 = 0.26 \text{ or } 26\% \]
- **Conclusion:** The third statement is correct: the empty space in the hcp unit cell is 26%.

4. **Base Area of hcp Unit Cell:**
- The base area can be calculated as:
\[ \text{Base Area} = 6 \times \left(\frac{\sqrt{3}}{4} (2r)^2\right) = 6 \times \left(\frac{\sqrt{3}}{4} \times 4r^2\right) = 6\sqrt{3} r^2 \]
- **Conclusion:** The fourth statement is correct: the base area of the hcp unit cell is \( 6\sqrt{3} r^2 \).

### Final Conclusion:
All four statements about the three-dimensional hcp system are correct.