If an element (at. Mass =50) crystallise in fc lattie ,with a= 0.50 nm . What is the denstiy of unit cell if it conatins `0.25%` Schottuy defects (use `N_(A) = 6xx 10^(23)`)?

Options

A. `2.0 g//"cc"`
B. `2.66 g//"cc"`
C. `3.06 g//"cc"`
D. None of these

Answer

To find the density of a unit cell of an element that crystallizes in a face-centered cubic (FCC) lattice with a given atomic mass and edge length, while accounting for Schottky defects, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Atomic mass (m) = 50 g/mol
- Edge length (a) = 0.50 nm = \(0.50 \times 10^{-7}\) cm
- Schottky defects = 0.25%
- Avogadro's number (\(N_A\)) = \(6 \times 10^{23}\) mol\(^{-1}\)

2. **Determine the Number of Atoms in the FCC Unit Cell:**
- In an FCC unit cell, the total number of atoms (Z) is 4.

3. **Calculate the Effective Number of Atoms Considering Schottky Defects:**
- The effective number of atoms after accounting for Schottky defects can be calculated as:
\[ \text{Effective atoms} = Z \times \left(1 - \frac{\text{Schottky defects}}{100}\right) \]
- Substituting the values:
\[ \text{Effective atoms} = 4 \times \left(1 - \frac{0.25}{100}\right) = 4 \times 0.9975 = 3.99 \]

4. **Convert Edge Length to cm and Calculate Volume of the Unit Cell:**
- Convert the edge length from nm to cm:
\[ a = 0.50 \text{ nm} = 0.50 \times 10^{-7} \text{ cm} \]
- Calculate the volume of the unit cell (V):
\[ V = a^3 = (0.50 \times 10^{-7})^3 = 1.25 \times 10^{-21} \text{ cm}^3 \]

5. **Calculate the Density of the Unit Cell:**
- The density (\(D\)) can be calculated using the formula:
\[ D = \frac{Z \times m}{N_A \times V} \]
- Substituting the values:
\[ D = \frac{3.99 \times 50}{6 \times 10^{23} \times 1.25 \times 10^{-21}} \]
- Calculate the density:
\[ D = \frac{199.5}{7.5 \times 10^{2}} = 2.66 \text{ g/cm}^3 \]

### Final Answer:
The density of the unit cell containing 0.25% Schottky defects is approximately **2.66 g/cm³**.