In a head-on collision between an `alpha`-particle and a gold nucleus, the minimum distance of approach is `3.95 xx 10^(4)m`. Calculate the energy of the `alpha`-particle.

Answer

To calculate the energy of the alpha particle in a head-on collision with a gold nucleus, we can use the concept of electrostatic potential energy. The formula for the potential energy (U) at the minimum distance of approach (R0) between two charged particles is given by:

\[ U = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Z_1 Z_2 e^2}{R_0} \]

Where:
- \( Z_1 \) and \( Z_2 \) are the atomic numbers of the two particles (for an alpha particle, \( Z_1 = 2 \) and for gold, \( Z_2 = 79 \)).
- \( e \) is the elementary charge (\( e \approx 1.6 \times 10^{-19} \) C).
- \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)).
- \( R_0 \) is the minimum distance of approach, given as \( 3.95 \times 10^{-14} \, \text{m} \).

### Step-by-Step Calculation:

1. **Identify the Constants**:
- \( Z_1 = 2 \) (alpha particle)
- \( Z_2 = 79 \) (gold nucleus)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)
- \( R_0 = 3.95 \times 10^{-14} \, \text{m} \)

2. **Substitute the Values into the Formula**:
\[ U = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{(2)(79)(1.6 \times 10^{-19})^2}{3.95 \times 10^{-14}} \]

3. **Calculate the Denominator**:
\[ 4 \pi \epsilon_0 \approx 4 \times 3.14 \times 8.85 \times 10^{-12} \approx 1.11 \times 10^{-10} \, \text{C}^2/\text{N m}^2 \]

4. **Calculate the Numerator**:
\[ (2)(79)(1.6 \times 10^{-19})^2 = 158 \cdot (2.56 \times 10^{-38}) \approx 4.05 \times 10^{-36} \]

5. **Combine the Results**:
\[ U = \frac{4.05 \times 10^{-36}}{3.95 \times 10^{-14} \cdot 1.11 \times 10^{-10}} \]
\[ U \approx \frac{4.05 \times 10^{-36}}{4.39 \times 10^{-24}} \approx 9.23 \times 10^{-13} \, \text{J} \]

6. **Convert Joules to Electron Volts**:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
\[ U \approx \frac{9.23 \times 10^{-13}}{1.6 \times 10^{-19}} \approx 5775 \, \text{eV} \]

### Final Answer:
The energy of the alpha particle is approximately **5775 eV**.