Mentallic Gold crystallise in fcc lattice and the length of cubic unit cell is 407 pm.
(Given : Atomic mass of Gold =`197 ,N_(A)=6xx10^(23)`)
The density if it have `0.2%` Schottky defect is `(" in gm"//cm^(3))`

Options

A. `4.86`
B. `9.72`
C. `19.48`
D. `19.44`

Answer

To find the density of metallic gold with 0.2% Schottky defects, we will follow these steps:

### Step 1: Determine the number of effective atoms per unit cell (Z)
Gold crystallizes in a face-centered cubic (FCC) lattice. In an ideal FCC unit cell, the number of atoms (Z) is calculated as follows:

- There are 8 corner atoms, each contributing \( \frac{1}{8} \) to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) to the unit cell.

So, the total number of atoms in an ideal FCC unit cell is:
\[ Z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

However, since there is a 0.2% Schottky defect, we need to adjust this number:
\[ \text{Effective Z} = Z - (0.2\% \text{ of } Z) = 4 - (0.002 \times 4) = 4 - 0.008 = 3.992 \]

### Step 2: Convert the edge length from picometers to centimeters
The edge length \( a \) is given as 407 pm. We need to convert this to centimeters:
\[ a = 407 \text{ pm} = 407 \times 10^{-12} \text{ m} = 407 \times 10^{-10} \text{ cm} \]

### Step 3: Calculate the volume of the unit cell
The volume \( V \) of the cubic unit cell is given by:
\[ V = a^3 = (407 \times 10^{-10} \text{ cm})^3 \]

Calculating this gives:
\[ V = 6.77 \times 10^{-29} \text{ cm}^3 \]

### Step 4: Calculate the mass of the unit cell
The mass \( m \) of the unit cell can be calculated using the formula:
\[ m = Z \times \text{Atomic mass of gold} = 3.992 \times 197 \text{ g/mol} \]

Calculating this gives:
\[ m = 787.064 \text{ g/mol} \]

### Step 5: Convert the mass to grams per unit cell
To convert this mass to grams per unit cell, we use Avogadro's number \( N_A \):
\[ \text{Mass per unit cell} = \frac{m}{N_A} = \frac{787.064 \text{ g/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}} = 1.308 \times 10^{-22} \text{ g} \]

### Step 6: Calculate the density
Finally, we can calculate the density \( \rho \) using the formula:
\[ \rho = \frac{\text{Mass per unit cell}}{\text{Volume of unit cell}} = \frac{1.308 \times 10^{-22} \text{ g}}{6.77 \times 10^{-29} \text{ cm}^3} \]

Calculating this gives:
\[ \rho \approx 19.34 \text{ g/cm}^3 \]

### Final Answer
The density of metallic gold with 0.2% Schottky defects is approximately \( 19.34 \text{ g/cm}^3 \).

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