An element having atomic mass 107.9 u had FCC lattice. The edge length of its unit cell is `408.6 ` pm. Calculate density of the unit cell. [Given , `N_A = 6.022 xx 10^(23)]`.

Explanation

To calculate the density of the unit cell of the element with a face-centered cubic (FCC) lattice, we will follow these steps:

### Step 1: Identify the parameters
- Atomic mass (m) = 107.9 u
- Edge length (a) = 408.6 pm = 408.6 × 10^(-12) m
- Avogadro's number (N_A) = 6.022 × 10^(23) mol^(-1)

### Step 2: Calculate the number of atoms in the unit cell (Z)
For an FCC lattice, the number of atoms per unit cell (Z) is 4.

### Step 3: Convert atomic mass to kg
To calculate density, we need the mass in kilograms.
1 u = 1.660539 × 10^(-27) kg

Thus,
\[ m = 107.9 \, \text{u} \times 1.660539 \times 10^{-27} \, \text{kg/u} = 1.794 \times 10^{-25} \, \text{kg} \]

### Step 4: Calculate the volume of the unit cell
The volume (V) of the cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of edge length:
\[ V = (408.6 \times 10^{-12} \, \text{m})^3 = 6.814 \times 10^{-29} \, \text{m}^3 \]

### Step 5: Calculate the density (ρ)
Density (ρ) is given by the formula:
\[ \rho = \frac{Z \times m}{V} \]
Substituting the values:
\[ \rho = \frac{4 \times 1.794 \times 10^{-25} \, \text{kg}}{6.814 \times 10^{-29} \, \text{m}^3} \]
Calculating the density:
\[ \rho = \frac{7.176 \times 10^{-25} \, \text{kg}}{6.814 \times 10^{-29} \, \text{m}^3} \approx 105.3 \, \text{kg/m}^3 \]

### Final Answer
The density of the unit cell is approximately **105.3 kg/m³**.

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