In a ccp type structure, if half of atoms are removed from face centered atoms, then percentage void in unit cell is approximately

Options

A. `54%`
B. `46.25%`
C. `63%`
D. `37%`

Answer

To solve the problem of finding the percentage void in a CCP (Cubic Close Packing) type structure after removing half of the face-centered atoms, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the CCP Structure**:
In a CCP structure, atoms are located at the corners and the face centers of the cube. There are 8 corner atoms and 6 face-centered atoms.

2. **Calculate the Number of Atoms in the Unit Cell**:
- Corner atoms contribute: \(8 \text{ corners} \times \frac{1}{8} = 1 \text{ atom}\)
- Face-centered atoms contribute: \(6 \text{ faces} \times \frac{1}{2} = 3 \text{ atoms}\)
- Total atoms in a CCP unit cell = \(1 + 3 = 4 \text{ atoms}\)

3. **Removing Half of the Face-Centered Atoms**:
- If half of the face-centered atoms are removed, the contribution from the face-centered atoms becomes \(3 \text{ atoms} \div 2 = 1.5 \text{ atoms}\).
- The contribution from the corner atoms remains the same, which is \(1 \text{ atom}\).
- Therefore, the new total number of atoms in the unit cell = \(1 + 1.5 = 2.5 \text{ atoms}\).

4. **Calculate the Volume of the Unit Cell**:
- The volume of the cube (unit cell) is given by \(a^3\), where \(a\) is the edge length.
- For a CCP structure, the relationship between the edge length \(a\) and the radius \(r\) of the atoms is given by \(4r = \sqrt{2}a\), which simplifies to \(a = 2\sqrt{2}r\).

5. **Calculate the Volume Occupied by Atoms**:
- The volume occupied by one atom is \(\frac{4}{3}\pi r^3\).
- The total volume occupied by \(2.5\) atoms is \(2.5 \times \frac{4}{3}\pi r^3\).

6. **Calculate Packing Efficiency**:
- Packing efficiency is given by:
\[ \text{Packing Efficiency} = \left( \frac{\text{Volume occupied by atoms}}{\text{Volume of unit cell}} \right) \times 100 \]
- Substituting the values:
\[ \text{Packing Efficiency} = \left( \frac{2.5 \times \frac{4}{3}\pi r^3}{(2\sqrt{2}r)^3} \right) \times 100 \]
- Simplifying the volume of the unit cell:
\[ (2\sqrt{2}r)^3 = 8 \times 2\sqrt{2}^3 \times r^3 = 16\sqrt{2}r^3 \]
- Thus, the packing efficiency becomes:
\[ \text{Packing Efficiency} = \left( \frac{10\pi r^3/3}{16\sqrt{2}r^3} \right) \times 100 \]
- This simplifies to:
\[ \text{Packing Efficiency} = \left( \frac{10\pi}{48\sqrt{2}} \right) \times 100 \]

7. **Calculate the Percentage of Void**:
- The percentage of voids in the unit cell is given by:
\[ \text{Percentage Void} = 100 - \text{Packing Efficiency} \]
- If the packing efficiency is approximately \(46.4\%\), then:
\[ \text{Percentage Void} = 100 - 46.4 = 53.6\% \]

8. **Final Answer**:
The percentage void in the unit cell is approximately **54%**.