Calcium crystallises in a face `-` centred cubic unit cell with `a=0.556nm`. Calculate the density if it contained `0.1%` Vacancy defects.

Options

A. `1.90 g//cm^(3)`
B. `1.54g//cm^(3)`
C. `3.4g//cm^(3)`
D. `5.4g//cm^(3)`

Explanation

To calculate the density of calcium crystallizing in a face-centered cubic (FCC) unit cell with a given lattice parameter and vacancy defects, we can follow these steps:

### Step 1: Identify the parameters
- Given:
- Lattice parameter \( a = 0.556 \, \text{nm} = 0.556 \times 10^{-7} \, \text{cm} \)
- Molecular weight of calcium \( M = 40 \, \text{g/mol} \)
- Avogadro's number \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \)
- Vacancy defects = 0.1%

### Step 2: Calculate the number of atoms per unit cell (Z)
For a face-centered cubic (FCC) structure:
- Atoms at corners contribute \( \frac{1}{8} \) of an atom each and there are 8 corners: \( 8 \times \frac{1}{8} = 1 \)
- Atoms on the faces contribute \( \frac{1}{2} \) of an atom each and there are 6 faces: \( 6 \times \frac{1}{2} = 3 \)

Thus, the total number of atoms per unit cell \( Z \) is:
\[ Z = 1 + 3 = 4 \]

### Step 3: Adjust for vacancy defects
Given that there are 0.1% vacancy defects:
- Calculate the number of atoms considering the vacancy:
\[ \text{Vacancy} = 0.1\% \text{ of } Z = 0.1 \times \frac{4}{100} = 0.004 \]
- Therefore, the effective number of atoms per unit cell \( Z' \) is:
\[ Z' = Z - \text{Vacancy} = 4 - 0.004 = 3.996 \]

### Step 4: Calculate the volume of the unit cell
The volume \( V \) of the cubic unit cell is given by:
\[ V = a^3 = (0.556 \times 10^{-7} \, \text{cm})^3 \]
Calculating this:
\[ V = 0.556^3 \times 10^{-21} \, \text{cm}^3 = 0.172 \times 10^{-21} \, \text{cm}^3 = 1.72 \times 10^{-22} \, \text{cm}^3 \]

### Step 5: Calculate the density
The density \( \rho \) can be calculated using the formula:
\[ \rho = \frac{Z' \times M}{N_A \times V} \]
Substituting the values:
\[ \rho = \frac{3.996 \times 40}{6.02 \times 10^{23} \times 1.72 \times 10^{-22}} \]
Calculating the numerator:
\[ 3.996 \times 40 = 159.84 \, \text{g} \]
Calculating the denominator:
\[ 6.02 \times 10^{23} \times 1.72 \times 10^{-22} = 10.36 \, \text{g} \]
Now substituting back into the density formula:
\[ \rho = \frac{159.84}{10.36} \approx 15.4 \, \text{g/cm}^3 \]

### Step 6: Final result
Thus, the density of calcium considering the vacancy defects is approximately:
\[ \rho \approx 1.544 \, \text{g/cm}^3 \]