An element X (At. Wt. =24) forms FCC lattice. If the edge length of lattice is `4 xx 10^(-8)` cm and the observed density is `2.4 xx 10^(3 kg//m^(3)`. Then the percentage occupancy of lattice point by element X is : `(N_(A) = 6 x 10^(23))`

Options

A. 96
B. 98
C. 99.9
D. None of these

Answer

To solve the problem step-by-step, we will calculate the theoretical density of the FCC lattice and then find the percentage occupancy of the lattice points by element X.

### Step 1: Determine the number of effective atoms (Z) in the FCC lattice.
In a Face-Centered Cubic (FCC) lattice:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom to the unit cell.

Calculating Z:
\[ Z = \left(8 \times \frac{1}{8}\right) + \left(6 \times \frac{1}{2}\right) = 1 + 3 = 4 \]

### Step 2: Use the formula for density to calculate the theoretical density (\( \rho \)).
The formula for density is given by:
\[ \rho = \frac{Z \times M}{N_A \times a^3} \]
Where:
- \( Z = 4 \) (from Step 1)
- \( M = 24 \, \text{g/mol} \) (atomic weight of element X)
- \( N_A = 6 \times 10^{23} \, \text{mol}^{-1} \) (Avogadro's number)
- \( a = 4 \times 10^{-8} \, \text{cm} \) (edge length)

First, convert the edge length from cm to m:
\[ a = 4 \times 10^{-8} \, \text{cm} = 4 \times 10^{-10} \, \text{m} \]

Now calculate \( a^3 \):
\[ a^3 = (4 \times 10^{-10})^3 = 64 \times 10^{-30} \, \text{m}^3 \]

Now substitute the values into the density formula:
\[ \rho = \frac{4 \times 24}{6 \times 10^{23} \times 64 \times 10^{-30}} \]
\[ = \frac{96}{6 \times 64 \times 10^{-7}} = \frac{96}{384 \times 10^{-7}} = \frac{1}{4} \times 10^{7} \, \text{g/m}^3 \]
Converting grams to kilograms:
\[ = \frac{1}{4} \times 10^{7} \, \text{g/m}^3 = 2.5 \times 10^{3} \, \text{kg/m}^3 \]

### Step 3: Calculate the percentage occupancy.
The observed density is given as \( 2.4 \times 10^{3} \, \text{kg/m}^3 \).

Now calculate the percentage occupancy:
\[ \text{Percentage Occupancy} = \left( \frac{\text{Observed Density}}{\text{Theoretical Density}} \right) \times 100 \]
\[ = \left( \frac{2.4 \times 10^{3}}{2.5 \times 10^{3}} \right) \times 100 \]
\[ = \left( \frac{2.4}{2.5} \right) \times 100 = 96\% \]

### Final Answer:
The percentage occupancy of the lattice point by element X is **96%**.

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