Gold has a fcc lattice with edge length 407 pm. The diameter of the gold atom is

Options

A. 303.1 pm
B. 287.8 pm
C. 352.5 pm
D. 576.6 pm

Answer

To find the diameter of a gold atom in a face-centered cubic (FCC) lattice with an edge length of 407 picometers (pm), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Edge Length**:
The edge length (a) of the FCC lattice is given as 407 pm.

2. **Understand the Relationship in FCC Lattice**:
In a face-centered cubic (FCC) lattice, the relationship between the edge length (a) and the atomic radius (r) is given by the formula:
\[ 4r = a\sqrt{2} \]
This relationship arises because the face diagonal of the cube contains 4 atomic radii.

3. **Rearranging the Formula**:
To find the radius (r), we can rearrange the formula:
\[ r = \frac{a\sqrt{2}}{4} \]

4. **Substituting the Edge Length**:
Now, substitute the edge length (a = 407 pm) into the equation:
\[ r = \frac{407 \times \sqrt{2}}{4} \]

5. **Calculating the Radius**:
First, calculate \(\sqrt{2}\):
\[ \sqrt{2} \approx 1.414 \]
Now substitute this value:
\[ r = \frac{407 \times 1.414}{4} \approx \frac{575.878}{4} \approx 143.9695 \text{ pm} \]

6. **Finding the Diameter**:
The diameter (D) of the gold atom is twice the radius:
\[ D = 2r = 2 \times 143.9695 \approx 287.939 \text{ pm} \]

7. **Rounding the Result**:
Rounding this to three significant figures gives:
\[ D \approx 287.9 \text{ pm} \]

### Final Answer:
The diameter of the gold atom is approximately **287.9 pm**.