Packing refers to the arrangement of constituent units in such a way that the forces of attraction among the constituent particles is the maximum and the contituents occupy the maximum available space. In two dimensions, there are hexagonal close packing and cubic close packing. In three dimentions, there are hexagonal, cubic as well as body centred close packings.
The empty space left in ccp packing is:

Options

A. `26%`
B. `74%`
C. `52.4%`
D. `80%`

Answer

To find the empty space left in cubic close packing (ccp), also known as face-centered cubic (fcc) packing, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Structure**:
- In cubic close packing (ccp), the atoms are arranged at the corners and the centers of each face of the cube. This means there are atoms at 8 corners and 6 face centers.

2. **Calculating the Number of Atoms**:
- Each corner atom contributes \( \frac{1}{8} \) of its volume to the unit cell (since each corner is shared by 8 unit cells), and there are 8 corners.
- Each face-centered atom contributes \( \frac{1}{2} \) of its volume to the unit cell (since each face is shared by 2 unit cells), and there are 6 faces.
- Therefore, the total number of atoms per unit cell (Z) is:
\[ Z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

3. **Finding the Edge Length (a)**:
- The face diagonal of the cube contains 3 atomic radii (3r), where r is the radius of the atoms.
- The face diagonal can also be expressed in terms of the edge length (a) of the cube using Pythagoras theorem:
\[ \text{Face diagonal} = a\sqrt{2} \]
- Setting these equal gives:
\[ 3r = a\sqrt{2} \implies a = \frac{3r}{\sqrt{2}} \]

4. **Calculating the Volume of the Unit Cell**:
- The volume of the unit cell (V_cell) is given by:
\[ V_{\text{cell}} = a^3 = \left(\frac{3r}{\sqrt{2}}\right)^3 = \frac{27r^3}{2\sqrt{2}} \]

5. **Calculating the Volume of the Atoms in the Unit Cell**:
- The volume of one atom (sphere) is:
\[ V_{\text{atom}} = \frac{4}{3}\pi r^3 \]
- The total volume of 4 atoms in the unit cell is:
\[ V_{\text{total atoms}} = 4 \times \frac{4}{3}\pi r^3 = \frac{16}{3}\pi r^3 \]

6. **Calculating the Fraction of Volume Occupied**:
- The fraction of the volume occupied by the atoms in the unit cell is:
\[ \text{Fraction occupied} = \frac{V_{\text{total atoms}}}{V_{\text{cell}}} = \frac{\frac{16}{3}\pi r^3}{\frac{27r^3}{2\sqrt{2}}} = \frac{32\pi\sqrt{2}}{81} \]

7. **Calculating the Percentage of Volume Occupied**:
- To find the percentage of volume occupied:
\[ \text{Percentage occupied} = \left(\frac{32\pi\sqrt{2}}{81}\right) \times 100 \]
- This evaluates to approximately 74%.

8. **Calculating the Empty Space**:
- The percentage of empty space in the ccp packing is:
\[ \text{Percentage empty space} = 100\% - \text{Percentage occupied} = 100\% - 74\% = 26\% \]

### Final Answer:
The empty space left in cubic close packing (ccp) is **26%**.

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