The fraction of volume occupied by atoms in a face centered cubic unit cell is:

Options

A. (a) 0.32
B. (b) 0.48
C. (c) 0.68
D. (d) 0.74

Answer

To find the fraction of volume occupied by atoms in a face-centered cubic (FCC) unit cell, we can follow these steps:

### Step 1: Determine the number of atoms in the FCC unit cell
In an FCC unit cell:
- There are 8 corner atoms, and each corner atom contributes \( \frac{1}{8} \) of its volume to the unit cell.
- There are 6 face-centered atoms, and each face-centered atom contributes \( \frac{1}{2} \) of its volume to the unit cell.

Calculating the total number of atoms (Z):
\[ Z = \text{(Number of corner atoms)} \times \text{(Contribution per corner atom)} + \text{(Number of face-centered atoms)} \times \text{(Contribution per face-centered atom)} \]
\[ Z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]
Thus, there are 4 atoms per FCC unit cell.

### Step 2: Calculate the volume of the atoms
The atoms in the FCC unit cell can be approximated as spheres. The volume \( V \) of a single sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, the total volume of the 4 atoms is:
\[ \text{Total volume of atoms} = Z \times V = 4 \times \frac{4}{3} \pi r^3 = \frac{16}{3} \pi r^3 \]

### Step 3: Calculate the volume of the unit cell
The volume \( V_{\text{cube}} \) of the cubic unit cell is given by:
\[ V_{\text{cube}} = a^3 \]
where \( a \) is the edge length of the cube.

### Step 4: Relate the edge length \( a \) to the radius \( r \)
In an FCC structure, the face diagonal of the cube is equal to four times the radius of the atoms:
\[ \text{Face diagonal} = \sqrt{2}a = 4r \]
From this, we can solve for \( a \):
\[ a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r \]

### Step 5: Substitute \( a \) into the volume of the cube
Now substituting \( a \) into the volume of the cube:
\[ V_{\text{cube}} = (2\sqrt{2}r)^3 = 8 \cdot 2\sqrt{2} \cdot r^3 = 16\sqrt{2}r^3 \]

### Step 6: Calculate the packing fraction
The packing fraction (the fraction of volume occupied by the atoms) is given by:
\[ \text{Packing fraction} = \frac{\text{Total volume of atoms}}{\text{Volume of cube}} = \frac{\frac{16}{3} \pi r^3}{16\sqrt{2}r^3} \]
Simplifying this:
\[ \text{Packing fraction} = \frac{\frac{16}{3} \pi}{16\sqrt{2}} = \frac{\pi}{3\sqrt{2}} \]

### Step 7: Numerical approximation
Calculating the numerical value:
\[ \frac{\pi}{3\sqrt{2}} \approx 0.74 \]

### Final Answer
The fraction of volume occupied by atoms in a face-centered cubic unit cell is approximately **0.74**.

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