Aluminium metal (atomic weight = 27 g) crystallises in the cubic system with edge length `4.0 Å`. The density of metal is 27.16 amu/`Å^(3)`. Determine the unit cell type and calculate the radius of the Aluminium metal.

Answer

To solve the problem, we will follow these steps:

### Step 1: Given Data
- Atomic weight of Aluminium (M) = 27 g/mol
- Edge length of the cubic unit cell (a) = 4.0 Å
- Density (ρ) = 27.16 amu/ų

### Step 2: Convert Density to g/cm³
Since 1 amu = 1 g/mol and 1 ų = 10⁻²⁴ cm³, we can convert the density:
\[ \rho = 27.16 \text{ amu/Å}^3 = 27.16 \text{ g/mol} \times \frac{1 \text{ mol}}{10^{-24} \text{ cm}^3} = 27.16 \times 10^{24} \text{ g/cm}^3 \]
This is approximately equal to \(2.716 \text{ g/cm}^3\).

### Step 3: Calculate the Number of Atoms per Unit Cell (Z)
Using the formula for density:
\[ \rho = \frac{Z \cdot M}{N_A \cdot a^3} \]
Where:
- \(N_A\) = Avogadro's number = \(6.022 \times 10^{23} \text{ mol}^{-1}\)
- \(a\) = edge length in cm = \(4.0 \text{ Å} = 4.0 \times 10^{-8} \text{ cm}\)

Rearranging to find Z:
\[ Z = \frac{\rho \cdot N_A \cdot a^3}{M} \]
Substituting the values:
\[ Z = \frac{2.716 \text{ g/cm}^3 \cdot 6.022 \times 10^{23} \text{ mol}^{-1} \cdot (4.0 \times 10^{-8} \text{ cm})^3}{27 \text{ g/mol}} \]

Calculating \(a^3\):
\[ (4.0 \times 10^{-8})^3 = 64 \times 10^{-24} \text{ cm}^3 = 6.4 \times 10^{-23} \text{ cm}^3 \]

Now substituting:
\[ Z = \frac{2.716 \cdot 6.022 \times 10^{23} \cdot 6.4 \times 10^{-23}}{27} \]

Calculating Z:
\[ Z \approx 4 \]

### Step 4: Determine Unit Cell Type
Since \(Z = 4\), Aluminium crystallizes in a Face-Centered Cubic (FCC) structure.

### Step 5: Calculate the Radius of Aluminium Atom (R)
For an FCC unit cell, the relationship between the edge length (a) and the atomic radius (R) is given by:
\[ a = 4R/\sqrt{2} \]
Rearranging gives:
\[ R = \frac{a \sqrt{2}}{4} \]
Substituting \(a = 4.0 \text{ Å}\):
\[ R = \frac{4.0 \sqrt{2}}{4} = \sqrt{2} \text{ Å} \approx 1.414 \text{ Å} \]

### Final Answers
- Unit Cell Type: Face-Centered Cubic (FCC)
- Radius of Aluminium Atom: \( \sqrt{2} \text{ Å} \approx 1.414 \text{ Å} \)

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