Percentage of free space in cubic close packed struchure and in body centred structure are respectively.

Options

A. `30% and 26%`
B. `26% and 32%`
C. `32% and 48%`
D. `48% and 26%`

Answer

To determine the percentage of free space in cubic close-packed (CCP) and body-centered cubic (BCC) structures, we need to calculate the packing efficiency for each structure and then find the free space percentage.

### Step 1: Calculate Packing Efficiency for CCP (Cubic Close-Packed)

1. **Identify the number of atoms in the unit cell**:
- For CCP (or FCC), the number of atoms per unit cell is 4.

2. **Determine the edge length (A) in terms of atomic radius (R)**:
- The edge length \( A \) for CCP is given by \( A = 2\sqrt{2}R \).

3. **Calculate the volume of the unit cell**:
- The volume of the unit cell \( V_{cell} \) is \( A^3 = (2\sqrt{2}R)^3 = 16\sqrt{2}R^3 \).

4. **Calculate the volume occupied by one atom**:
- The volume of one atom (sphere) is \( V_{atom} = \frac{4}{3}\pi R^3 \).

5. **Calculate the total volume occupied by all atoms in the unit cell**:
- Total volume occupied by atoms = Number of atoms × Volume of one atom = \( 4 \times \frac{4}{3}\pi R^3 = \frac{16}{3}\pi R^3 \).

6. **Calculate packing efficiency**:
- Packing efficiency \( PE = \frac{\text{Total volume occupied by atoms}}{\text{Volume of unit cell}} \times 100 \)
- \( PE = \frac{\frac{16}{3}\pi R^3}{16\sqrt{2}R^3} \times 100 \)
- Simplifying gives \( PE = \frac{\pi}{3\sqrt{2}} \times 100 \).

7. **Numerical value of packing efficiency**:
- \( PE \approx 74\% \).

8. **Calculate free space**:
- Free space = \( 100\% - PE = 100\% - 74\% = 26\% \).

### Step 2: Calculate Packing Efficiency for BCC (Body-Centered Cubic)

1. **Identify the number of atoms in the unit cell**:
- For BCC, the number of atoms per unit cell is 2.

2. **Determine the edge length (A) in terms of atomic radius (R)**:
- The edge length \( A \) for BCC is given by \( A = \frac{4}{\sqrt{3}}R \).

3. **Calculate the volume of the unit cell**:
- The volume of the unit cell \( V_{cell} \) is \( A^3 = \left(\frac{4}{\sqrt{3}}R\right)^3 = \frac{64}{3\sqrt{3}}R^3 \).

4. **Calculate the volume occupied by one atom**:
- Volume of one atom (sphere) is \( V_{atom} = \frac{4}{3}\pi R^3 \).

5. **Calculate the total volume occupied by all atoms in the unit cell**:
- Total volume occupied by atoms = Number of atoms × Volume of one atom = \( 2 \times \frac{4}{3}\pi R^3 = \frac{8}{3}\pi R^3 \).

6. **Calculate packing efficiency**:
- Packing efficiency \( PE = \frac{\text{Total volume occupied by atoms}}{\text{Volume of unit cell}} \times 100 \)
- \( PE = \frac{\frac{8}{3}\pi R^3}{\frac{64}{3\sqrt{3}}R^3} \times 100 \).
- Simplifying gives \( PE \approx 68.04\% \).

7. **Calculate free space**:
- Free space = \( 100\% - PE = 100\% - 68.04\% \approx 31.96\% \) (approximately 32%).

### Final Answer
- The percentage of free space in cubic close-packed structure is **26%** and in body-centered structure is **32%**.