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 can follow these steps:

### Step 1: Understand the formula for density
The density (d) of a unit cell can be expressed using the formula:
\[ d = \frac{Z \cdot m}{N_a \cdot a^3} \]
where:
- \( Z \) = number of atoms per unit cell
- \( m \) = molar mass of the substance (in grams per mole)
- \( N_a \) = Avogadro's number (\( 6.022 \times 10^{23} \, \text{mol}^{-1} \))
- \( a \) = edge length of the unit cell (in cm)

### Step 2: Convert the given values
Given:
- Atomic mass \( m = 108 \, \text{g/mol} \)
- Density \( d = 10.5 \, \text{g/cm}^3 \)
- Edge length \( a = 409 \, \text{pm} = 409 \times 10^{-10} \, \text{cm} \)

### Step 3: Substitute the values into the density formula
Rearranging the density formula to find \( Z \):
\[ Z = \frac{d \cdot N_a \cdot a^3}{m} \]

### Step 4: Calculate \( a^3 \)
First, calculate \( a^3 \):
\[ a^3 = (409 \times 10^{-10} \, \text{cm})^3 = 6.839 \times 10^{-29} \, \text{cm}^3 \]

### Step 5: Substitute into the equation for \( Z \)
Now substituting the values into the rearranged density formula:
\[ Z = \frac{10.5 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 6.839 \times 10^{-29} \, \text{cm}^3}{108 \, \text{g/mol}} \]

### Step 6: Perform the calculation
Calculating the numerator:
\[ 10.5 \cdot 6.022 \times 10^{23} \cdot 6.839 \times 10^{-29} \approx 4.31 \]
Now, divide by the molar mass:
\[ Z \approx \frac{4.31}{108} \approx 4 \]

### Step 7: Identify the crystal structure
With \( Z = 4 \), we can compare this with known structures:
- **Face-Centered Cubic (FCC)** has \( Z = 4 \)
- **Body-Centered Cubic (BCC)** has \( Z = 2 \)
- **Simple Cubic (SC)** has \( Z = 1 \)

Since the calculated \( Z \) value is 4, the structure of the crystal lattice is **Face-Centered Cubic (FCC)**.

### Final Answer
The structure of the crystal lattice would be **Face-Centered Cubic (FCC)**.

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