The unit cell of a metallic element of atomic mass (`108 g //"mole"`) and density `10.5 g//cm^(3)` is a cube with edge length of 409 pm. The structure of the crystal lattice would be:

Options

A. fcc
B. bcc
C. hcp
D. Simple Cube

Answer

To determine the structure of the crystal lattice for the given metallic element, we will follow these steps:

### Step 1: Write down the given data
- Atomic mass (m) = 108 g/mol
- Density (D) = 10.5 g/cm³
- Edge length (a) = 409 pm = 409 x 10^-10 cm (since 1 pm = 10^-12 m and 1 cm = 10^-2 m)

### Step 2: Use the formula for density of a unit cell
The formula for density (D) in terms of the number of atoms in the unit cell (Z), molar mass (m), and edge length (a) is given by:

\[ D = \frac{Z \cdot m}{a^3 \cdot N_A} \]

Where:
- \(N_A\) = Avogadro's number = \(6.022 \times 10^{23} \text{ mol}^{-1}\)

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

\[ Z = \frac{D \cdot a^3 \cdot N_A}{m} \]

### Step 4: Calculate \(a^3\)
Convert the edge length from picometers to centimeters and then calculate \(a^3\):

\[ a = 409 \text{ pm} = 409 \times 10^{-10} \text{ cm} \]
\[ a^3 = (409 \times 10^{-10})^3 = 6.84 \times 10^{-29} \text{ cm}^3 \]

### Step 5: Substitute the values into the equation for Z
Now substitute the values into the rearranged formula:

\[ Z = \frac{10.5 \text{ g/cm}^3 \cdot 6.84 \times 10^{-29} \text{ cm}^3 \cdot 6.022 \times 10^{23} \text{ mol}^{-1}}{108 \text{ g/mol}} \]

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

\[ 10.5 \cdot 6.84 \times 10^{-29} \cdot 6.022 \times 10^{23} \approx 4.34 \times 10^{-5} \]

Now, divide by the molar mass:

\[ Z \approx \frac{4.34 \times 10^{-5}}{108} \approx 4.02 \approx 4 \]

### Step 7: Determine the crystal structure
Since the value of Z (number of atoms per unit cell) is 4, the crystal structure corresponds to a face-centered cubic (FCC) lattice.

### Conclusion
The structure of the crystal lattice for the given metallic element is **FCC (Face-Centered Cubic)**.

---