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 unti celll of alluminium is ?

Options

A. Body centred cubic cell
B. Face-centred cubic cell
C. End-centred cubic cell
D. Simple cubic cell

Explanation

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

### Step 1: Write down the formula for density
The formula for the density (\( \rho \)) of a crystal is given by:

\[ \rho = \frac{Z \cdot m}{N_a \cdot a^3} \]

Where:
- \( Z \) = number of atoms per unit cell
- \( m \) = molar mass of the element (in grams)
- \( N_a \) = Avogadro's number (\( 6.022 \times 10^{23} \) mol\(^{-1}\))
- \( a \) = edge length of the unit cell (in cm)

### Step 2: Convert the edge length from picometers to centimeters
The edge length of aluminum is given as 405 pm (picometers). We need to convert this to centimeters:

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

### Step 3: Substitute the known values into the density formula
We know:
- Density (\( \rho \)) = 2.70 g/cm³
- Molar mass of aluminum (\( m \)) = 26.98 g
- Avogadro's number (\( N_a \)) = \( 6.022 \times 10^{23} \) mol\(^{-1}\)
- Edge length (\( a \)) = \( 405 \times 10^{-10} \) cm

Now we can rearrange the density formula to solve for \( Z \):

\[ Z = \frac{\rho \cdot N_a \cdot a^3}{m} \]

### Step 4: Calculate \( a^3 \)
First, calculate \( a^3 \):

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

### Step 5: Substitute the values into the equation for \( Z \)
Now substitute the values into the equation for \( Z \):

\[ Z = \frac{2.70 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 6.64 \times 10^{-29} \, \text{cm}^3}{26.98 \, \text{g}} \]

### Step 6: Perform the calculation
Calculating the numerator:

\[ 2.70 \cdot 6.022 \times 10^{23} \cdot 6.64 \times 10^{-29} \approx 1.07 \times 10^{-5} \]

Now divide by the molar mass:

\[ Z = \frac{1.07 \times 10^{-5}}{26.98} \approx 4 \]

### Step 7: Determine the type of unit cell
The value of \( Z = 4 \) indicates that the unit cell is a face-centered cubic (FCC) structure since FCC has 4 atoms per unit cell.

### Conclusion
Thus, the type of unit cell of aluminum is **Face-Centered Cubic (FCC)**.

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