In a hexagonal close packed (hcp) structure of spheres, the fraction of the volume occupied by the sphere is A. In a cubic close packed structure the fraction is B. The relation for A and B is:

Options

A. A=B
B. A < B
C. A > B
D. A is equal to the fraction in a simple cubic lattice.

Answer

To find the relation between the volume fractions \( A \) and \( B \) for hexagonal close packed (hcp) and cubic close packed (ccp) structures respectively, we will calculate the packing fractions for both structures step by step.

### Step 1: Calculate the packing fraction for Hexagonal Close Packed (hcp) Structure

1. **Identify the arrangement**: In an hcp structure, the stacking of atoms follows an AB-AB pattern.
2. **Count the atoms**:
- **Corner atoms**: There are 12 corner atoms, each contributing \( \frac{1}{6} \) of an atom to the unit cell.
\[ \text{Contribution from corner atoms} = 12 \times \frac{1}{6} = 2 \]
- **Body-centered atoms**: There are 3 body-centered atoms, each contributing fully to the unit cell.
\[ \text{Contribution from body-centered atoms} = 3 \]
- **Face-centered atoms**: There are 2 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom to the unit cell.
\[ \text{Contribution from face-centered atoms} = 2 \times \frac{1}{2} = 1 \]
- **Total atoms (Z)**:
\[ Z = 2 + 3 + 1 = 6 \]

3. **Calculate the volume of the atoms**: The volume of one atom (sphere) is given by:
\[ V_{\text{atom}} = \frac{4}{3} \pi r^3 \]
Therefore, the total volume of the atoms in the unit cell is:
\[ V_{\text{total atoms}} = Z \times V_{\text{atom}} = 6 \times \frac{4}{3} \pi r^3 = 8 \pi r^3 \]

4. **Calculate the volume of the unit cell**: The volume of the hexagonal unit cell can be calculated as:
\[ V_{\text{cell}} = 24 \sqrt{2} r^3 \]

5. **Calculate the packing fraction (A)**:
\[ A = \frac{V_{\text{total atoms}}}{V_{\text{cell}}} = \frac{8 \pi r^3}{24 \sqrt{2} r^3} = \frac{8 \pi}{24 \sqrt{2}} = \frac{\pi}{3 \sqrt{2}} \approx 0.74 \]

### Step 2: Calculate the packing fraction for Cubic Close Packed (ccp) Structure

1. **Identify the arrangement**: In a ccp (or face-centered cubic, FCC) structure, atoms are located at the corners and the centers of the faces of the cube.
2. **Count the atoms**:
- **Corner atoms**: There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
\[ \text{Contribution from corner atoms} = 8 \times \frac{1}{8} = 1 \]
- **Face-centered atoms**: There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom to the unit cell.
\[ \text{Contribution from face-centered atoms} = 6 \times \frac{1}{2} = 3 \]
- **Total atoms (Z)**:
\[ Z = 1 + 3 = 4 \]

3. **Calculate the volume of the atoms**: The total volume of the atoms in the unit cell is:
\[ V_{\text{total atoms}} = Z \times V_{\text{atom}} = 4 \times \frac{4}{3} \pi r^3 = \frac{16}{3} \pi r^3 \]

4. **Calculate the volume of the unit cell**: The edge length \( a \) of the cube can be related to the radius \( r \) of the atoms:
\[ a = \frac{4}{\sqrt{2}} r \]
Therefore, the volume of the cubic unit cell is:
\[ V_{\text{cell}} = a^3 = \left(\frac{4}{\sqrt{2}} r\right)^3 = \frac{64}{2\sqrt{2}} r^3 = 32 \sqrt{2} r^3 \]

5. **Calculate the packing fraction (B)**:
\[ B = \frac{V_{\text{total atoms}}}{V_{\text{cell}}} = \frac{\frac{16}{3} \pi r^3}{32 \sqrt{2} r^3} = \frac{16 \pi}{96 \sqrt{2}} = \frac{\pi}{6 \sqrt{2}} \approx 0.74 \]

### Conclusion: Relation between A and B

From the calculations:
- Packing fraction for hcp: \( A = 0.74 \)
- Packing fraction for ccp: \( B = 0.74 \)

Thus, the relation between \( A \) and \( B \) is:
\[ A = B \]