The unit cell of aluminium is a cube with an edge length of 405 pm. The density of aluminium is `2.70 g cm^(-3)` . What type of unit cell of aluminium is ?

Options

A. fcc
B. simple cubic
C. bcc
D. None of these

Answer

To determine the type of unit cell of aluminum based on the given data, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Edge length of aluminum unit cell, \( a = 405 \, \text{pm} = 405 \times 10^{-10} \, \text{cm} \)
- Density of aluminum, \( D = 2.70 \, \text{g/cm}^3 \)
- Molar mass of aluminum, \( M = 26 \, \text{g/mol} \)
- Avogadro's number, \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \)

2. **Convert Edge Length to Cubic Centimeters:**
- Since \( a = 405 \, \text{pm} = 405 \times 10^{-10} \, \text{cm} \), we can calculate \( a^3 \):
\[ a^3 = (405 \times 10^{-10})^3 \, \text{cm}^3 \]

3. **Use the Density Formula:**
- The formula for density in terms of unit cell parameters is given by:
\[ D = \frac{Z \cdot M}{N_A \cdot a^3} \]
where \( Z \) is the number of atoms per unit cell.

4. **Rearrange the Formula to Solve for \( Z \):**
\[ Z = \frac{D \cdot N_A \cdot a^3}{M} \]

5. **Substitute the Values:**
- First, calculate \( a^3 \):
\[ a^3 = (405 \times 10^{-10})^3 = 6.63 \times 10^{-29} \, \text{cm}^3 \]
- Now, substitute the values into the rearranged formula:
\[ Z = \frac{2.70 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{atoms/mol} \cdot 6.63 \times 10^{-29} \, \text{cm}^3}{26 \, \text{g/mol}} \]

6. **Calculate \( Z \):**
\[ Z = \frac{2.70 \cdot 6.022 \times 10^{23} \cdot 6.63 \times 10^{-29}}{26} \]
- Performing the calculation gives:
\[ Z \approx 4 \]

7. **Determine the Type of Unit Cell:**
- Since \( Z = 4 \), this indicates that aluminum has a face-centered cubic (FCC) structure.

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

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