The unit cell of aluminium is a cube with an edge length of 410 pm. The density of aluminium is 3 gm `cm^(-3)`. What is the type of unit cell of aluminium crystals ?

Options

A. simple cubic
B. body - centred cubic
C. face - centred cubic
D. hexagonal

Answer

To determine the type of unit cell of aluminum crystals, we need to follow these steps:

### Step 1: Convert the edge length from picometers to centimeters.
Given:
- Edge length (A) = 410 pm

To convert picometers to centimeters:
\[ A = 410 \, \text{pm} = 410 \times 10^{-12} \, \text{m} = 410 \times 10^{-10} \, \text{cm} = 4.10 \times 10^{-8} \, \text{cm} \]

### Step 2: Use the density of aluminum to find the number of atoms per unit cell (Z).
Given:
- Density (ρ) = 3 g/cm³
- Molar mass of aluminum (M) = 27 g/mol
- Avogadro's number (Nₐ) = \(6.022 \times 10^{23} \, \text{mol}^{-1}\)

We use the formula for density:
\[ \rho = \frac{Z \cdot M}{N_a \cdot A^3} \]
Rearranging the formula to find Z:
\[ Z = \frac{\rho \cdot N_a \cdot A^3}{M} \]

### Step 3: Calculate \(A^3\).
\[ A^3 = (4.10 \times 10^{-8} \, \text{cm})^3 = 6.87 \times 10^{-24} \, \text{cm}^3 \]

### Step 4: Substitute the values into the Z formula.
Now substitute the values into the Z formula:
\[ Z = \frac{3 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 6.87 \times 10^{-24} \, \text{cm}^3}{27 \, \text{g/mol}} \]

### Step 5: Calculate Z.
Calculating the numerator:
\[ 3 \cdot 6.022 \times 10^{23} \cdot 6.87 \times 10^{-24} \approx 3.11 \]
Now divide by the molar mass:
\[ Z \approx \frac{3.11}{27} \approx 4 \]

### Step 6: Determine the type of unit cell.
From the calculated value of Z:
- Z = 4 corresponds to a face-centered cubic (FCC) unit cell.

### Conclusion:
The type of unit cell of aluminum crystals is **Face-Centered Cubic (FCC)**.

---