Given mass number of gold `= 197`, Density of gold `= 19.7 g cm^(-3)`. The radius of the gold atom is approximately:

Options

A. `1.5 xx 10^(-8) m`
B. `1.5 xx 10^(-9) m`
C. `1.5 xx 10^(-10) m`
D. `1.5 xx 10^(-12) m`

Explanation

To find the radius of a gold atom given its mass number and density, we can follow these steps:

### Step-by-Step Solution:

1. **Given Values**:
- Mass number of gold (A) = 197
- Density of gold (D) = 19.7 g/cm³

2. **Convert Mass Number to Mass**:
The mass of a single gold atom can be calculated using the mass number. The mass of one atom in grams can be found using the formula:
\[ m = A \times 1.66 \times 10^{-27} \text{ kg} \]
Since we need the mass in grams, we convert kilograms to grams:
\[ m = 197 \times 1.66 \times 10^{-27} \text{ kg} \times 1000 \text{ g/kg} = 3.27 \times 10^{-25} \text{ g} \]

3. **Density Formula**:
The density (D) is defined as mass (m) divided by volume (V):
\[ D = \frac{m}{V} \]
Rearranging this gives us:
\[ V = \frac{m}{D} \]

4. **Volume of a Sphere**:
The volume of a sphere (which we can approximate for the atom) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Setting the two expressions for volume equal gives:
\[ \frac{4}{3} \pi r^3 = \frac{m}{D} \]

5. **Solve for Radius**:
Rearranging for \( r^3 \):
\[ r^3 = \frac{3m}{4\pi D} \]
Now substituting the values we have:
\[ r^3 = \frac{3 \times (3.27 \times 10^{-25} \text{ g})}{4\pi \times (19.7 \text{ g/cm}^3)} \]

6. **Calculate \( r^3 \)**:
First, calculate the denominator:
\[ 4\pi \times 19.7 \approx 247.5 \text{ g/cm}^3 \]
Now substituting this into the equation:
\[ r^3 = \frac{3 \times 3.27 \times 10^{-25}}{247.5} \approx 3.97 \times 10^{-27} \text{ cm}^3 \]

7. **Finding the Radius**:
To find \( r \), take the cube root:
\[ r \approx \sqrt[3]{3.97 \times 10^{-27}} \approx 1.56 \times 10^{-9} \text{ cm} = 1.56 \text{ Å} \]

### Final Answer:
The radius of the gold atom is approximately \( 1.56 \text{ Å} \).