A metallic element crystallizes into a lattice contained sequence of layers `ABABAB….` Any packing of sphere leaves out voilds in the lattice.The percentage by volume of this lattice as empty space is

Options

A. `74%`
B. `68%`
C. `52.4%`
D. `26%`

Explanation

To find the percentage by volume of a metallic lattice that is empty space, we can follow these steps:

### Step 1: Understand the Structure
The metallic element crystallizes in an ABABAB... arrangement, which indicates a hexagonal close-packed (HCP) structure. In this structure, spheres (atoms) are packed closely together, but there are still voids (empty spaces) present.

### Step 2: Calculate the Volume of the Unit Cell
The volume of a hexagonal unit cell can be calculated using the formula:
\[ V_{cell} = \text{Base Area} \times \text{Height} \]

#### Base Area Calculation
The base of the hexagonal unit cell consists of equilateral triangles. The area of one triangle can be calculated as:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
For an equilateral triangle with side length \( A \):
- Base = \( A \)
- Height = \( A \sin(60^\circ) = A \frac{\sqrt{3}}{2} \)

Thus, the area of the base becomes:
\[ \text{Base Area} = \frac{3\sqrt{3}}{2} \left(\frac{A^2}{4}\right) = \frac{3\sqrt{3}}{4} A^2 \]

#### Height Calculation
The height \( H \) of the unit cell is the distance between two closest packed layers, which is \( 2 \sqrt{2} R \) where \( R \) is the radius of the atoms.

### Step 3: Substitute Values to Find Volume
Now, substituting the base area and height into the volume formula:
\[ V_{cell} = \text{Base Area} \times H = \left(\frac{3\sqrt{3}}{4} A^2\right) \times (2\sqrt{2} R) \]

### Step 4: Relate Side Length to Atomic Radius
For close-packed structures, the relationship between the side length \( A \) and the atomic radius \( R \) is:
\[ A = 2R \]
Substituting this into the volume formula gives:
\[ V_{cell} = \frac{3\sqrt{3}}{4} (2R)^2 \times (2\sqrt{2} R) = \frac{3\sqrt{3}}{4} \times 4R^2 \times 2\sqrt{2} R = 6\sqrt{6} R^3 \]

### Step 5: Calculate the Volume of Atoms in the Unit Cell
The total number of atoms in the unit cell can be calculated as follows:
- 12 atoms at the corners contribute \( \frac{1}{6} \) each
- 2 atoms on the face contribute \( \frac{1}{2} \) each
- 3 atoms in the body center contribute fully

Calculating the total contribution:
\[ \text{Total atoms} = 12 \times \frac{1}{6} + 2 \times \frac{1}{2} + 3 = 2 + 1 + 3 = 6 \]

The volume of the atoms in the unit cell is:
\[ V_{atoms} = 6 \times \frac{4}{3} \pi R^3 = 8\pi R^3 \]

### Step 6: Calculate Packing Efficiency
The packing efficiency is given by:
\[ \text{Packing Efficiency} = \frac{V_{atoms}}{V_{cell}} \]
Substituting the volumes we calculated:
\[ \text{Packing Efficiency} = \frac{8\pi R^3}{6\sqrt{6} R^3} = \frac{8\pi}{6\sqrt{6}} \]

### Step 7: Calculate Percentage of Empty Space
The percentage of empty space in the lattice is:
\[ \text{Empty Space} = 100\% - \text{Packing Efficiency} \]
Substituting the packing efficiency:
\[ \text{Empty Space} = 100\% - \left(\frac{8\pi}{6\sqrt{6}} \times 100\%\right) \]

### Final Calculation
After calculating the packing efficiency, we find that the packing efficiency is approximately 74%. Therefore, the percentage of empty space is:
\[ \text{Empty Space} \approx 100\% - 74\% = 26\% \]

### Conclusion
The percentage by volume of the lattice that is empty space is approximately **26%**.

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