Calcium crystallises in a face-centered cubic unit cell with a 0.556 nm and density `1.4848 g//cm^3`. Percentage of Schottky defects in the crystal is:

Options

A. 0.03
B. 0.02
C. 0.038
D. 0.04

Explanation

To solve the problem of determining the percentage of Schottky defects in calcium, which crystallizes in a face-centered cubic (FCC) unit cell, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data**:
- Lattice parameter (edge length, a) = 0.556 nm = \(0.556 \times 10^{-7}\) cm
- Density (\(d\)) = 1.4848 g/cm³
- Molar mass of calcium (Ca) = 40 g/mol

2. **Calculate the volume of the unit cell**:
\[ V = a^3 = (0.556 \times 10^{-7} \text{ cm})^3 = 0.556^3 \times 10^{-21} \text{ cm}^3 \]
\[ V = 0.172 \times 10^{-21} \text{ cm}^3 = 1.72 \times 10^{-22} \text{ cm}^3 \]

3. **Calculate the number of atoms per unit cell in FCC**:
In a face-centered cubic unit cell, the number of atoms per unit cell is:
\[ \text{Number of atoms} = 4 \]

4. **Use the density formula**:
The density of the unit cell can be expressed as:
\[ d = \frac{\text{mass of atoms in unit cell}}{\text{volume of unit cell}} \]
The mass of the atoms in the unit cell can be calculated as:
\[ \text{mass} = \text{number of atoms} \times \frac{\text{molar mass}}{N_A} \]
where \(N_A\) is Avogadro's number (\(6.022 \times 10^{23}\) mol⁻¹).

Substituting the values:
\[ \text{mass} = 4 \times \frac{40 \text{ g/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}} = \frac{160}{6.022 \times 10^{23}} \text{ g} \]

5. **Set up the equation for density**:
\[ d = \frac{160 \times (1 - x)}{6.022 \times 10^{23} \times V} \]
Substituting the volume calculated:
\[ 1.4848 = \frac{160 \times (1 - x)}{6.022 \times 10^{23} \times 1.72 \times 10^{-22}} \]

6. **Solve for \(1 - x\)**:
Rearranging gives:
\[ 1.4848 \times 6.022 \times 10^{23} \times 1.72 \times 10^{-22} = 160 \times (1 - x) \]
\[ 1 - x = \frac{1.4848 \times 6.022 \times 10^{23} \times 1.72 \times 10^{-22}}{160} \]

Calculate the left-hand side:
\[ 1 - x = \frac{1.4848 \times 6.022 \times 1.72}{160} \approx 0.96 \]

7. **Calculate \(x\)**:
\[ x = 1 - 0.96 = 0.04 \]

8. **Calculate the percentage of Schottky defects**:
\[ \text{Percentage of Schottky defects} = x \times 100 = 0.04 \times 100 = 4\% \]

### Final Answer:
The percentage of Schottky defects in the crystal is **4%**.

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