Copper crystallizes in a cubic lattice structure. Atomic radius of copper is 128 pm and its atomic mass is 63.5. The density of copper is

Options

A. `10.71 g cm^(-3)`
B. `4.93 g cm^(-3)`
C. `8.9 g cm^(-3)`
D. `11.2 g cm^(-3)`

Explanation

To find the density of copper, we will follow these steps:

### Step 1: Determine the number of atoms per unit cell (Z)
Copper crystallizes in a face-centered cubic (FCC) structure. In an FCC unit cell, there are:
- 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom to the unit cell.

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

### Step 2: Calculate the edge length (A) of the unit cell
The relationship between the atomic radius (R) and the edge length (A) in an FCC structure is given by:
\[ A\sqrt{2} = 4R \]
Given the atomic radius \( R = 128 \, \text{pm} \):
\[ A = \frac{4R}{\sqrt{2}} = \frac{4 \times 128 \, \text{pm}}{\sqrt{2}} = \frac{512}{\sqrt{2}} \approx 362 \, \text{pm} \]

### Step 3: Convert the edge length to centimeters
To calculate the density, we need the edge length in centimeters:
\[ A = 362 \, \text{pm} = 362 \times 10^{-12} \, \text{m} = 362 \times 10^{-10} \, \text{cm} \]

### Step 4: Calculate the volume of the unit cell (A³)
\[ A^3 = (362 \times 10^{-10} \, \text{cm})^3 = 4.74 \times 10^{-29} \, \text{cm}^3 \]

### Step 5: Calculate the mass of the unit cell
The mass (m) of one atom of copper is given by:
\[ m = \frac{\text{atomic mass}}{N_A} = \frac{63.5 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} \approx 1.055 \times 10^{-22} \, \text{g} \]
The mass of the unit cell (which contains Z atoms) is:
\[ \text{Mass of unit cell} = Z \times m = 4 \times 1.055 \times 10^{-22} \, \text{g} \approx 4.22 \times 10^{-22} \, \text{g} \]

### Step 6: Calculate the density (D)
The density is given by the formula:
\[ D = \frac{\text{mass of unit cell}}{\text{volume of unit cell}} = \frac{4.22 \times 10^{-22} \, \text{g}}{4.74 \times 10^{-29} \, \text{cm}^3} \approx 8900 \, \text{g/cm}^3 \]

### Final Answer
The density of copper is approximately \( 8900 \, \text{g/cm}^3 \).

---