At `100^@C`, copper (Cu) has FCC unit cell structure with cell edge length of x Å. What is the approximate density of `Cu` (in `g cm^(-3))` at this temperature ? [Atomic Mass of Cu = 63.55 u]

Options

A. `105/(x^3)`
B. `422/(x^3)`
C. `205/(x^3)`
D. `211/(x^3)`

Explanation

To find the density of copper (Cu) at 100°C, we will use the formula for density in terms of the unit cell structure. Copper has a face-centered cubic (FCC) structure. Here are the steps to calculate the density:

### Step-by-step Solution:

1. **Identify the parameters:**
- Atomic mass of Cu (M) = 63.55 u
- Number of atoms per FCC unit cell (Z) = 4
- Avogadro's number (N_A) = \(6 \times 10^{23} \, \text{atoms/mol}\)
- Edge length of the unit cell (a) = x Å = \(x \times 10^{-8} \, \text{cm}\)

2. **Convert the edge length to centimeters:**
- Since 1 Å = \(10^{-8} \, \text{cm}\), the edge length in centimeters is:
\[ a = x \times 10^{-8} \, \text{cm} \]

3. **Calculate the volume of the unit cell (V):**
- The volume of the cubic unit cell is given by:
\[ V = a^3 = (x \times 10^{-8})^3 = x^3 \times 10^{-24} \, \text{cm}^3 \]

4. **Use the density formula:**
- The density (D) is given by the formula:
\[ D = \frac{Z \times M}{N_A \times V} \]
- Substituting the values:
\[ D = \frac{4 \times 63.55}{6 \times 10^{23} \times (x^3 \times 10^{-24})} \]

5. **Simplify the expression:**
- This simplifies to:
\[ D = \frac{4 \times 63.55 \times 10^{24}}{6 \times 10^{23} \times x^3} \]
- Further simplifying:
\[ D = \frac{4 \times 63.55}{6} \times \frac{10^{24}}{10^{23} \times x^3} = \frac{4 \times 63.55}{6} \times \frac{10}{x^3} \]

6. **Calculate the numerical value:**
- Calculate \( \frac{4 \times 63.55}{6} \):
\[ \frac{4 \times 63.55}{6} \approx 42.23 \]
- Thus, the density becomes:
\[ D \approx \frac{422.3}{x^3} \, \text{g/cm}^3 \]

### Final Result:
The approximate density of copper (Cu) at 100°C is:
\[ D \approx \frac{422}{x^3} \, \text{g/cm}^3 \]