An element with cell edge of 288 pm has a density of 7.2 g cm-3. What type of structure does the element have if it’s atomic mass M=51.8 g mol-1?

Options

A. Body-Centred Cubic (BCC)
B. Face-Centred Cubic (FCC)
C. Simple Cubic
D. Hexagonal Closed Packing (HCP)

Answer

To determine the type of structure of the element with the given parameters, we will follow these steps:

### Step 1: Identify the given values
- Cell edge (a) = 288 pm = 288 × 10^-12 m
- Density (ρ) = 7.2 g/cm³ = 7.2 × 10^3 kg/m³ (since 1 g/cm³ = 1000 kg/m³)
- Atomic mass (M) = 51.8 g/mol = 51.8 × 10^-3 kg/mol
- Avogadro's number (N_A) = 6.02 × 10^23 mol^-1

### Step 2: Convert the cell edge to cubic meters
To find the volume of the unit cell (V), we use the formula:
\[ V = a^3 \]
\[ V = (288 \times 10^{-12} \text{ m})^3 \]

### Step 3: Calculate the volume of the unit cell
Calculating the volume:
\[ V = (288 \times 10^{-12})^3 = 2.38 \times 10^{-31} \text{ m}^3 \]

### Step 4: Use the density formula to find Z
The formula for density in terms of the number of atoms per unit cell (Z) is:
\[ \rho = \frac{Z \times M}{V \times N_A} \]
Rearranging this formula to find Z gives:
\[ Z = \frac{\rho \times V \times N_A}{M} \]

### Step 5: Substitute the values into the equation
Substituting the known values:
\[ Z = \frac{(7.2 \times 10^3 \text{ kg/m}^3) \times (2.38 \times 10^{-31} \text{ m}^3) \times (6.02 \times 10^{23} \text{ mol}^{-1})}{(51.8 \times 10^{-3} \text{ kg/mol})} \]

### Step 6: Calculate Z
Calculating the numerator:
\[ \text{Numerator} = (7.2 \times 10^3) \times (2.38 \times 10^{-31}) \times (6.02 \times 10^{23}) \]
\[ = 1.01 \times 10^{-7} \text{ kg} \]

Now calculating Z:
\[ Z = \frac{1.01 \times 10^{-7}}{51.8 \times 10^{-3}} \]
\[ Z \approx 1.95 \approx 2 \]

### Step 7: Determine the structure type
Since Z = 2, the element has a body-centered cubic (BCC) structure.

### Conclusion
The element has a BCC structure.

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