The density of a face centred cubic element (atomic mass = 60.2 amu) is 6.25 gm `cm^(-3)`, calculate the edge length of the unit cell.

Explanation

To calculate the edge length of a face-centered cubic (FCC) unit cell given the atomic mass and density, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Atomic mass (M) = 60.2 amu
- Density (ρ) = 6.25 g/cm³
- Number of atoms per unit cell for FCC (Z) = 4
- Avogadro's number (NA) = 6.022 × 10²³ mol⁻¹

2. **Convert Atomic Mass to grams per mole:**
- Since 1 amu = 1 g/mol, the atomic mass in grams per mole is:
\[ M = 60.2 \text{ g/mol} \]

3. **Use the Density Formula:**
- The formula for density (ρ) in terms of the unit cell parameters is:
\[ \rho = \frac{Z \cdot M}{A^3 \cdot N_A} \]
- Rearranging this formula to solve for the edge length (A):
\[ A^3 = \frac{Z \cdot M}{\rho \cdot N_A} \]
- Therefore, we can express A as:
\[ A = \left(\frac{Z \cdot M}{\rho \cdot N_A}\right)^{1/3} \]

4. **Substitute the Known Values:**
- Substitute Z = 4, M = 60.2 g/mol, ρ = 6.25 g/cm³, and NA = 6.022 × 10²³ mol⁻¹ into the equation:
\[ A^3 = \frac{4 \cdot 60.2}{6.25 \cdot 6.022 \times 10^{23}} \]

5. **Calculate the Right Side:**
- Calculate the numerator:
\[ 4 \cdot 60.2 = 240.8 \]
- Calculate the denominator:
\[ 6.25 \cdot 6.022 \times 10^{23} = 3.76 \times 10^{24} \text{ g/cm}^3 \text{ mol}^{-1} \]
- Now, calculate A³:
\[ A^3 = \frac{240.8}{3.76 \times 10^{24}} = 6.39 \times 10^{-23} \text{ cm}^3 \]

6. **Find the Cube Root:**
- Now take the cube root to find A:
\[ A = \left(6.39 \times 10^{-23}\right)^{1/3} \approx 4.02 \times 10^{-8} \text{ cm} \]

7. **Convert to Picometers:**
- To convert cm to picometers (1 cm = 10¹² pm):
\[ A \approx 4.02 \times 10^{-8} \text{ cm} \times 10^{12} \text{ pm/cm} = 402 \text{ pm} \]

### Final Answer:
The edge length of the unit cell is approximately **402 pm**.