To find the density of the unit cell for a metal that crystallizes in a closed packed structure, we can follow these steps:
### Step-by-Step Solution:
1. **Identify the Structure**:
The problem states that the metal has a closed packed structure where spheres in the first layer are in contact with six neighbors, and the spheres of the second and third layers are placed in the depressions of the layers below. This indicates a cubic close-packed (CCP) structure.
2. **Determine the Number of Atoms in the Unit Cell**:
In a CCP structure, there are 4 atoms per unit cell. This is because:
- There are 8 corner atoms, each contributing 1/8th to the unit cell (8 * 1/8 = 1).
- There are 6 face-centered atoms, each contributing 1/2 to the unit cell (6 * 1/2 = 3).
- Total = 1 + 3 = 4 atoms per unit cell.
3. **Use the Density Formula**:
The density (ρ) of the unit cell can be calculated using the formula:
\[
\rho = \frac{n \cdot M}{V}
\]
where:
- \( n \) = number of atoms in the unit cell (4 for CCP),
- \( M \) = molar mass (atomic weight) of the metal (given as 197 u),
- \( V \) = volume of the unit cell in cm³.
4. **Calculate the Volume of the Unit Cell**:
The side length of the unit cell (a) is given as 4.07 Å. To convert this to centimeters:
\[
a = 4.07 \, \text{Å} = 4.07 \times 10^{-8} \, \text{cm}
\]
The volume \( V \) of the unit cell is:
\[
V = a^3 = (4.07 \times 10^{-8} \, \text{cm})^3 = 6.77 \times 10^{-24} \, \text{cm}^3
\]
5. **Calculate the Density**:
Now substituting the values into the density formula:
\[
\rho = \frac{4 \cdot 197 \, \text{g/mol}}{6.77 \times 10^{-24} \, \text{cm}^3}
\]
To convert the volume from cm³ to moles, we need to multiply by Avogadro's number (\( N_A = 6.022 \times 10^{23} \)):
\[
V = 6.77 \times 10^{-24} \, \text{cm}^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1} = 4.07 \times 10^{-1} \, \text{mol}
\]
Now substituting back into the density formula:
\[
\rho = \frac{4 \cdot 197}{4.07 \times 10^{-1}} \approx 19.7 \, \text{g/cm}^3
\]
### Final Result:
The density of the unit cell is approximately \( \rho \approx 19.7 \, \text{g/cm}^3 \).
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