In a geiger - marsden experiment. Find the distance of closest approach to the nucleus of a 7.7 me v `alpha`- particle before it comes momentarily to rest and reverses its direction. (z for gold nucleus = 79) .

Options

A. 30 fm
B. 20 fm
C. 40 fm
D. 10 fm

Explanation

To find the distance of closest approach of a 7.7 MeV alpha particle to a gold nucleus (Z = 79), we will use the concept of conservation of energy. The kinetic energy of the alpha particle will be converted into electric potential energy at the point of closest approach.

### Step-by-Step Solution:

1. **Understand the Energy Conservation Principle**:
The initial kinetic energy (KE) of the alpha particle will be equal to the electric potential energy (PE) at the distance of closest approach (D).
\[ KE = PE \]

2. **Write the Expression for Kinetic Energy**:
The kinetic energy of the alpha particle can be expressed in joules. Given that the energy is 7.7 MeV, we convert this to joules:
\[ KE = 7.7 \, \text{MeV} = 7.7 \times 10^6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.232 \times 10^{-12} \, \text{J} \]

3. **Write the Expression for Electric Potential Energy**:
The electric potential energy (PE) between two charged particles is given by:
\[ PE = \frac{1}{4 \pi \epsilon_0} \cdot \frac{(2e)(Ze)}{D} \]
where \(e\) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)), \(Z\) is the atomic number of the nucleus (79 for gold), and \(D\) is the distance of closest approach.

4. **Set the Kinetic Energy Equal to Potential Energy**:
\[ 1.232 \times 10^{-12} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{(2e)(Ze)}{D} \]

5. **Substitute Known Values**:
The value of \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).
\[ 1.232 \times 10^{-12} = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{(2 \times 1.6 \times 10^{-19})(79 \times 1.6 \times 10^{-19})}{D} \]

6. **Calculate the Right Side**:
First, calculate the numerator:
\[ 2e \cdot Ze = 2 \times (1.6 \times 10^{-19}) \times (79 \times 1.6 \times 10^{-19}) = 2 \times 1.6^2 \times 79 \times 10^{-38} \]
\[ = 2 \times 2.56 \times 79 \times 10^{-38} = 404.48 \times 10^{-38} \, \text{C}^2 \]

7. **Calculate the Constant**:
Calculate \( \frac{1}{4 \pi \epsilon_0} \):
\[ \frac{1}{4 \pi (8.85 \times 10^{-12})} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \]

8. **Combine the Results**:
Now, substituting back into the equation:
\[ 1.232 \times 10^{-12} = 9 \times 10^9 \cdot \frac{404.48 \times 10^{-38}}{D} \]

9. **Solve for D**:
Rearranging gives:
\[ D = \frac{9 \times 10^9 \cdot 404.48 \times 10^{-38}}{1.232 \times 10^{-12}} \]
Calculate the right side:
\[ D \approx \frac{3.639 \times 10^{-28}}{1.232 \times 10^{-12}} \approx 2.95 \times 10^{-14} \, \text{m} \approx 30 \, \text{fm} \]

### Final Answer:
The distance of closest approach to the nucleus is approximately \(30 \, \text{fm}\).