Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is `0.407 x 10^(-9)`m?

Options

A. 0.115 nm
B. 0.144 nm
C. 0.235 nm
D. 0.156 nm

Answer

To find the atomic radius of gold (Au) which crystallizes in a face-centered cubic (FCC) unit cell, we can follow these steps:

### Step 1: Identify the given values
- Atomic mass of gold: 197 u (not needed for this calculation)
- Edge length of the unit cell (a): \(0.407 \times 10^{-9}\) m

### Step 2: Use the formula for atomic radius in a face-centered cubic unit cell
In a face-centered cubic (FCC) structure, the relationship between the edge length (a) and the atomic radius (r) is given by the formula:
\[ r = \frac{a}{2\sqrt{2}} \]

### Step 3: Substitute the edge length into the formula
Substituting the given edge length into the formula:
\[ r = \frac{0.407 \times 10^{-9} \text{ m}}{2\sqrt{2}} \]

### Step 4: Calculate the denominator
First, calculate \(2\sqrt{2}\):
\[ 2\sqrt{2} \approx 2 \times 1.414 \approx 2.828 \]

### Step 5: Complete the calculation
Now, substitute this value back into the equation:
\[ r = \frac{0.407 \times 10^{-9}}{2.828} \approx 0.144 \times 10^{-9} \text{ m} \]
Converting this to nanometers:
\[ r \approx 0.144 \text{ nm} \]

### Step 6: Final answer
Thus, the atomic radius of gold is approximately:
\[ \boxed{0.144 \text{ nm}} \]

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