The atomic radius of palladium is 1.375 A. The unit cell of palladium is a face-centred cube. Calculate the density of palladium.

Answer

To calculate the density of palladium (Pd) given its atomic radius and the type of unit cell, follow these steps:

### Step 1: Identify the parameters
- Atomic radius (R) of palladium = 1.375 Å = 1.375 × 10⁻¹⁰ m = 1.375 × 10⁻⁸ cm
- Atomic weight (m) of palladium = 106 g/mol
- For a face-centered cubic (FCC) structure, the number of atoms per unit cell (Z) = 4.

### Step 2: Calculate the edge length (a) of the unit cell
In an FCC unit cell, the relationship between the atomic radius (R) and the edge length (a) is given by:
\[ 4R = \sqrt{2}a \]
Rearranging this gives:
\[ a = \frac{4R}{\sqrt{2}} \]

Substituting the value of R:
\[ a = \frac{4 \times 1.375 \times 10^{-8} \text{ cm}}{\sqrt{2}} \]
\[ a = \frac{5.5 \times 10^{-8} \text{ cm}}{1.414} \]
\[ a \approx 3.89 \times 10^{-8} \text{ cm} \]

### Step 3: Calculate the volume of the unit cell
The volume (V) of the cubic unit cell is given by:
\[ V = a^3 \]
Substituting the value of a:
\[ V = (3.89 \times 10^{-8} \text{ cm})^3 \]
\[ V \approx 5.92 \times 10^{-23} \text{ cm}^3 \]

### Step 4: Calculate the density (D) of palladium
The density is calculated using the formula:
\[ D = \frac{Z \times m}{N_A \times V} \]
Where:
- \( N_A \) = Avogadro's number = \( 6.022 \times 10^{23} \text{ mol}^{-1} \)

Substituting the values:
\[ D = \frac{4 \times 106 \text{ g/mol}}{6.022 \times 10^{23} \text{ mol}^{-1} \times 5.92 \times 10^{-23} \text{ cm}^3} \]
\[ D = \frac{424 \text{ g}}{3.53 \text{ cm}^3} \]
\[ D \approx 8.49 \text{ g/cm}^3 \]

### Final Answer
The density of palladium is approximately **8.49 g/cm³**.

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