Calculate the impact parameter of a 5 MeV particle scattered by `90^(@)`, when it approach a gold nucleus (Z=79).

Explanation

To calculate the impact parameter \( B \) of a 5 MeV particle scattered by \( 90^\circ \) when it approaches a gold nucleus (with atomic number \( Z = 79 \)), we can use the formula:

\[ B = \frac{Z e^2 \cot(\theta/2)}{4 \pi \epsilon_0 K} \]

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),
- \( \theta \) is the scattering angle (in this case, \( \theta = 90^\circ \)),
- \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)),
- \( K \) is the kinetic energy of the alpha particle in joules.

### Step-by-Step Solution:

1. **Convert the kinetic energy from MeV to Joules:**
\[ K = 5 \, \text{MeV} = 5 \times 1.6 \times 10^{-13} \, \text{J} = 8 \times 10^{-13} \, \text{J} \]

2. **Calculate \( \cot(\theta/2) \):**
Since \( \theta = 90^\circ \):
\[ \cot(90^\circ/2) = \cot(45^\circ) = 1 \]

3. **Substitute the values into the formula for \( B \):**
\[ B = \frac{Z e^2 \cot(\theta/2)}{4 \pi \epsilon_0 K} \]
\[ B = \frac{79 \times (1.6 \times 10^{-19})^2 \times 1}{4 \pi \times (8.85 \times 10^{-12}) \times (8 \times 10^{-13})} \]

4. **Calculate \( e^2 \):**
\[ e^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \]

5. **Calculate the denominator:**
\[ 4 \pi \epsilon_0 K = 4 \pi \times (8.85 \times 10^{-12}) \times (8 \times 10^{-13}) \approx 8.88 \times 10^{-24} \, \text{C}^2/\text{N m}^2 \]

6. **Substitute the values into the equation for \( B \):**
\[ B = \frac{79 \times 2.56 \times 10^{-38}}{8.88 \times 10^{-24}} \approx \frac{2.0224 \times 10^{-36}}{8.88 \times 10^{-24}} \approx 2.27 \times 10^{-14} \, \text{m} \]

### Final Result:
\[ B \approx 2.27 \times 10^{-14} \, \text{m} \]