Aluminium(a.w=27) crystalises in a cubic unit cell with edge length (a)=100pm, with density 'd'=180g/cm, then type of unit cell is

Options

A. SCC
B. BCC
C. FCC
D. HCP

Answer

To determine the type of unit cell for Aluminium (Al) given its edge length and density, we can follow these steps:

### Step 1: Understand the Formula for Density
The density (\(d\)) of a crystalline solid can be expressed using the formula:
\[ d = \frac{Z \cdot M}{N_a \cdot a^3} \]
where:
- \(Z\) = number of atoms per unit cell
- \(M\) = molar mass of the substance (in grams per mole)
- \(N_a\) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))
- \(a\) = edge length of the unit cell (in cm)

### Step 2: Convert Given Values
1. **Convert edge length from pm to cm**:
\[ a = 100 \, \text{pm} = 100 \times 10^{-12} \, \text{m} = 100 \times 10^{-10} \, \text{cm} = 1 \times 10^{-8} \, \text{cm} \]

2. **Use the molar mass of Aluminium**:
\[ M = 27 \, \text{g/mol} \]

3. **Density is given**:
\[ d = 180 \, \text{g/cm}^3 \]

### Step 3: Substitute Values into the Density Formula
Substituting the known values into the density formula:
\[ 180 = \frac{Z \cdot 27}{6.022 \times 10^{23} \cdot (1 \times 10^{-8})^3} \]

### Step 4: Calculate \(a^3\)
Calculate \(a^3\):
\[ a^3 = (1 \times 10^{-8})^3 = 1 \times 10^{-24} \, \text{cm}^3 \]

### Step 5: Rearrange the Density Equation to Solve for \(Z\)
Rearranging the equation gives:
\[ Z = \frac{180 \cdot 6.022 \times 10^{23} \cdot 1 \times 10^{-24}}{27} \]

### Step 6: Calculate \(Z\)
Now, calculate \(Z\):
1. Calculate the numerator:
\[ 180 \cdot 6.022 \times 10^{23} \cdot 1 \times 10^{-24} = 10.8436 \, \text{(approximately)} \]
2. Now divide by \(27\):
\[ Z = \frac{10.8436}{27} \approx 4 \]

### Step 7: Identify 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 for Aluminium is **Face-Centered Cubic (FCC)**.

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