Gold (atomic radius = 0.150nm) crystallises in a face centred unit cell. What is the length of the side of the cell ?

Explanation

To find the length of the side of the face-centered cubic (FCC) unit cell for gold, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the FCC Structure**: In a face-centered cubic (FCC) unit cell, atoms are located at each of the eight corners of the cube and at the center of each of the six faces.

2. **Identify the Atomic Radius**: The atomic radius of gold (Au) is given as \( r = 0.150 \) nm.

3. **Determine the Relationship Between Atomic Radius and Edge Length**: In an FCC unit cell, the face diagonal can be expressed in terms of the atomic radius. The face diagonal of the cube can be represented as:
\[ \text{Face Diagonal} = 4r \]
where \( r \) is the atomic radius.

4. **Relate Face Diagonal to Edge Length**: The face diagonal (d) of a cube with edge length \( a \) can also be expressed using the Pythagorean theorem:
\[ d = a\sqrt{2} \]

5. **Set the Two Expressions for Face Diagonal Equal**: Since both expressions represent the same diagonal, we can set them equal:
\[ 4r = a\sqrt{2} \]

6. **Solve for Edge Length \( a \)**: Rearranging the equation to find \( a \):
\[ a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r \]

7. **Substitute the Value of \( r \)**: Now substitute the given value of the atomic radius:
\[ a = 2\sqrt{2} \times 0.150 \, \text{nm} \]

8. **Calculate the Edge Length**:
- First, calculate \( 2\sqrt{2} \):
\[ 2\sqrt{2} \approx 2 \times 1.414 \approx 2.828 \]
- Then multiply by the atomic radius:
\[ a \approx 2.828 \times 0.150 \approx 0.4242 \, \text{nm} \]

### Final Answer:
The length of the side of the face-centered cubic unit cell for gold is approximately \( 0.4242 \, \text{nm} \).

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