The unit of an element of atomic mass 96 and density `10.3 g cm^(-3)` is a cube with edge length of 314 pm. Find the structure of the crystal lattice (simple cubic, FCC or BCC) (Avogadro's constant , `N_0 = 6.023 xx 10^(23) mol^(-1))`

Explanation

To determine the structure of the crystal lattice (simple cubic, FCC, or BCC) for the given element with an atomic mass of 96, density of 10.3 g/cm³, and edge length of 314 pm, we can follow these steps:

### Step 1: Convert the edge length from picometers to centimeters
The edge length \( a \) is given as 314 pm. To convert this to centimeters:
\[ a = 314 \, \text{pm} = 314 \times 10^{-12} \, \text{m} = 314 \times 10^{-10} \, \text{cm} \]

### Step 2: Use the density formula
The density \( D \) of a crystal can be expressed using the formula:
\[ D = \frac{z \cdot M}{a^3 \cdot N_A} \]
Where:
- \( D \) = density (g/cm³)
- \( z \) = number of atoms per unit cell
- \( M \) = molar mass (g/mol)
- \( a \) = edge length (cm)
- \( N_A \) = Avogadro's number (6.023 x 10²³ mol⁻¹)

### Step 3: Substitute the known values into the formula
Given:
- \( D = 10.3 \, \text{g/cm}^3 \)
- \( M = 96 \, \text{g/mol} \)
- \( a = 314 \times 10^{-10} \, \text{cm} \)
- \( N_A = 6.023 \times 10^{23} \, \text{mol}^{-1} \)

Substituting these values into the density formula:
\[ 10.3 = \frac{z \cdot 96}{(314 \times 10^{-10})^3 \cdot (6.023 \times 10^{23})} \]

### Step 4: Calculate \( a^3 \)
Calculating \( a^3 \):
\[ a^3 = (314 \times 10^{-10})^3 = 3.09 \times 10^{-28} \, \text{cm}^3 \]

### Step 5: Rearrange the equation to solve for \( z \)
Rearranging the equation:
\[ z = \frac{10.3 \cdot (314 \times 10^{-10})^3 \cdot (6.023 \times 10^{23})}{96} \]

### Step 6: Calculate \( z \)
Substituting the values:
\[ z = \frac{10.3 \cdot 3.09 \times 10^{-28} \cdot 6.023 \times 10^{23}}{96} \]
Calculating this gives:
\[ z \approx 2 \]

### Step 7: Determine the crystal structure
The value of \( z \) indicates the number of atoms per unit cell:
- For Simple Cubic (SC), \( z = 1 \)
- For Body-Centered Cubic (BCC), \( z = 2 \)
- For Face-Centered Cubic (FCC), \( z = 4 \)

Since we found \( z = 2 \), the crystal structure of the element is **Body-Centered Cubic (BCC)**.