The unit cell of aluminium is a cube with an edge length of 405 pm. The density of aluminium is 2.7 g/cc. The type of lattice Al has is

Options

A. bcc
B. fcc
C. sc
D. end centred

Explanation

To determine the type of lattice that aluminum has, we will follow these steps:

### Step 1: Convert the edge length to centimeters
Given the edge length of the unit cell of aluminum is 405 pm (picometers), we need to convert this to centimeters.

\[ \text{Edge length (a)} = 405 \, \text{pm} = 405 \times 10^{-10} \, \text{cm} \]

### Step 2: Use the density formula to find the number of atoms per unit cell (Z)
The formula for density (d) in terms of the number of atoms per unit cell (Z), molar mass (M), and volume of the unit cell (a³) is given by:

\[ d = \frac{Z \cdot M}{a^3 \cdot N_0} \]

Where:
- \(d\) = density (2.7 g/cm³)
- \(M\) = molar mass of aluminum (27 g/mol)
- \(N_0\) = Avogadro's number (\(6.02 \times 10^{23}\) mol⁻¹)
- \(a\) = edge length of the unit cell (405 pm = \(405 \times 10^{-10}\) cm)

### Step 3: Rearranging the formula to solve for Z
Rearranging the formula to find Z:

\[ Z = \frac{d \cdot a^3 \cdot N_0}{M} \]

### Step 4: Substitute the values into the equation
Substituting the values into the equation:

\[ Z = \frac{(2.7 \, \text{g/cm}^3) \cdot (405 \times 10^{-10} \, \text{cm})^3 \cdot (6.02 \times 10^{23} \, \text{mol}^{-1})}{27 \, \text{g/mol}} \]

### Step 5: Calculate \(a^3\)
Calculating \(a^3\):

\[ a^3 = (405 \times 10^{-10})^3 = 6.64 \times 10^{-29} \, \text{cm}^3 \]

### Step 6: Calculate Z
Now substituting \(a^3\) back into the equation for Z:

\[ Z = \frac{(2.7) \cdot (6.64 \times 10^{-29}) \cdot (6.02 \times 10^{23})}{27} \]

Calculating this step-by-step:

1. Calculate the numerator:
\[ 2.7 \cdot 6.64 \times 10^{-29} \cdot 6.02 \times 10^{23} \approx 1.07 \times 10^{-5} \]

2. Now divide by the molar mass:
\[ Z = \frac{1.07 \times 10^{-5}}{27} \approx 3.96 \approx 4 \]

### Step 7: Determine the type of lattice
From the calculated value of Z, we can determine the type of lattice:
- For Face-Centered Cubic (FCC), Z = 4
- For Body-Centered Cubic (BCC), Z = 2
- For Simple Cubic, Z = 1

Since we found \(Z = 4\), aluminum has a Face-Centered Cubic (FCC) lattice.

### Final Answer
The type of lattice that aluminum has is **Face-Centered Cubic (FCC)**.

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