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

Explanation

To find the distance of closest approach of a 7.7 MeV alpha particle to a gold nucleus (Z = 79) in a Geiger-Marsden experiment, we can follow these steps:

### Step 1: Understand the Concept
The distance of closest approach occurs when the kinetic energy of the alpha particle is completely converted into electric potential energy at the point of closest approach.

### Step 2: Use Conservation of Energy
According to the conservation of energy:
\[ K.E. = U \]
Where:
- \( K.E. \) is the kinetic energy of the alpha particle.
- \( U \) is the electric potential energy at the distance of closest approach.

### Step 3: Write the Equations
The kinetic energy of the alpha particle can be expressed as:
\[ K.E. = \text{(Energy in Joules)} = 7.7 \, \text{MeV} = 7.7 \times 1.6 \times 10^{-13} \, \text{J} \]
\[ K.E. = 1.232 \times 10^{-12} \, \text{J} \]

The electric potential energy \( U \) at the distance \( d \) is given by:
\[ U = \frac{1}{4\pi \epsilon_0} \cdot \frac{2Ze^2}{d} \]
Where:
- \( Z = 79 \) (for gold),
- \( e = 1.6 \times 10^{-19} \, \text{C} \) (charge of a proton),
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step 4: Set the Energies Equal
Setting the kinetic energy equal to the potential energy:
\[ 1.232 \times 10^{-12} = \frac{1}{4\pi \epsilon_0} \cdot \frac{2 \cdot 79 \cdot (1.6 \times 10^{-19})^2}{d} \]

### Step 5: Solve for \( d \)
Rearranging the equation to solve for \( d \):
\[ d = \frac{1}{4\pi \epsilon_0} \cdot \frac{2 \cdot 79 \cdot (1.6 \times 10^{-19})^2}{1.232 \times 10^{-12}} \]

### Step 6: Substitute Values
Substituting the known values:
- \( \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( e^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \).

Thus,
\[ d = \frac{9 \times 10^9 \cdot 2 \cdot 79 \cdot 2.56 \times 10^{-38}}{1.232 \times 10^{-12}} \]

### Step 7: Calculate \( d \)
Calculating the value:
1. Calculate \( 2 \cdot 79 \cdot 2.56 \times 10^{-38} = 404.48 \times 10^{-38} \).
2. Multiply by \( 9 \times 10^9 \):
\[ 9 \times 10^9 \cdot 404.48 \times 10^{-38} = 3.639 \times 10^{-28} \]
3. Finally, divide by \( 1.232 \times 10^{-12} \):
\[ d = \frac{3.639 \times 10^{-28}}{1.232 \times 10^{-12}} \approx 2.95 \times 10^{-14} \, \text{m} \]

### Step 8: Convert to Femtometers
Since \( 1 \, \text{femtometer} = 10^{-15} \, \text{m} \):
\[ d \approx 29.5 \, \text{fm} \]

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

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