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

### Step 1: Understand the FCC structure
In a face-centered cubic (FCC) lattice, atoms are located at each corner of the cube and at the center of each face. Each unit cell contains 4 atoms.

### Step 2: Relate the edge length to the atomic radius
In an FCC lattice, the relationship between the edge length (A) and the atomic radius (R) can be expressed as:
\[ A = 2\sqrt{2}R \]
This is because the face diagonal of the cube (which connects two corner atoms through the center atom on the face) is equal to \( 4R \) and can also be expressed as \( \sqrt{2}A \).

### Step 3: Solve for the atomic radius (R)
Rearranging the equation gives us:
\[ R = \frac{A}{2\sqrt{2}} \]

### Step 4: Substitute the given edge length
Now, substitute the given edge length \( A = 407 \, \text{pm} \):
\[ R = \frac{407 \, \text{pm}}{2\sqrt{2}} \]

### Step 5: Calculate the radius
Calculating this gives:
\[ R = \frac{407}{2 \times 1.414} \]
\[ R = \frac{407}{2.828} \]
\[ R \approx 144.3 \, \text{pm} \]

### Step 6: Calculate the diameter (d)
The diameter (d) of the gold atom is twice the radius:
\[ d = 2R \]
\[ d = 2 \times 144.3 \, \text{pm} \]
\[ d \approx 288.6 \, \text{pm} \]

### Final Answer
Thus, the diameter of the gold atom is approximately **288.6 pm**.

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