An `alpha` particle of `K.E. 10^(-12)J` exibits back scattering form a gold nucleus Z=79. What can be the maximum possible radius of the gold nucleus?

Explanation

To find the maximum possible radius of the gold nucleus when an alpha particle exhibits back scattering, we can use the concept of the closest approach. The formula for the closest approach distance \( r_0 \) for an alpha particle approaching a nucleus is given by:

\[ r_0 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Z \cdot e^2}{K.E.} \]

Where:
- \( Z \) is the atomic number of the nucleus (for gold, \( Z = 79 \)),
- \( e \) is the charge of an electron (\( e = 1.6 \times 10^{-19} \) C),
- \( K.E. \) is the kinetic energy of the alpha particle,
- \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)).

Given:
- \( K.E. = 10^{-12} \, \text{J} \)

### Step-by-Step Solution

1. **Identify the values**:
- \( Z = 79 \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( K.E. = 10^{-12} \, \text{J} \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)

2. **Substitute the values into the formula**:
\[ r_0 = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{79 \cdot (1.6 \times 10^{-19})^2}{10^{-12}} \]

3. **Calculate the numerator**:
- First calculate \( (1.6 \times 10^{-19})^2 \):
\[ (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \]
- Now calculate \( 79 \cdot 2.56 \times 10^{-38} \):
\[ 79 \cdot 2.56 \times 10^{-38} = 2.0224 \times 10^{-36} \, \text{C}^2 \]

4. **Calculate the denominator**:
- Calculate \( 4 \pi (8.85 \times 10^{-12}) \):
\[ 4 \pi (8.85 \times 10^{-12}) \approx 1.112 \times 10^{-10} \, \text{C}^2/\text{N m}^2 \]

5. **Combine the results**:
\[ r_0 = \frac{2.0224 \times 10^{-36}}{1.112 \times 10^{-10}} \cdot 10^{12} \]

6. **Calculate \( r_0 \)**:
\[ r_0 \approx \frac{2.0224 \times 10^{-36} \times 10^{12}}{1.112 \times 10^{-10}} \approx 3.64 \times 10^{-14} \, \text{m} \]

### Final Answer
The maximum possible radius of the gold nucleus is approximately:
\[ r_0 \approx 3.64 \times 10^{-14} \, \text{m} \]