In scattering experiment, find the distance of closest approach, if a `6 MeV alpha - particle` is used

Options

A. `3.2 xx 10^(-16) m`
B. `2 xx 10^(-14) m`
C. `4.6 xx 10^(-15) m`
D. `3.2 xx 10^(-15)` m

Answer

To find the distance of closest approach for a 6 MeV alpha particle in a scattering experiment, we will use the principle that the kinetic energy of the alpha particle is converted into potential energy at the closest approach. Here’s how to solve the problem step by step:

### Step 1: Understand the Problem
The distance of closest approach occurs when the kinetic energy of the alpha particle is completely converted into electrostatic potential energy due to the repulsion from the nucleus of the target atom (in this case, we can assume it's a gold nucleus).

### Step 2: Write the Energy Equations
The kinetic energy (KE) of the alpha particle can be expressed as:
\[ KE = 6 \text{ MeV} = 6 \times 10^6 \text{ eV} \]

The potential energy (PE) at the closest approach is given by the formula:
\[ PE = \frac{1}{4 \pi \epsilon_0} \frac{Z_1 Z_2 e^2}{r} \]
where:
- \( Z_1 \) is the atomic number of the gold nucleus (which is 79),
- \( Z_2 \) is the charge number of the alpha particle (which is 2 since it consists of 2 protons),
- \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \) C),
- \( r \) is the distance of closest approach.

### Step 3: Set Kinetic Energy Equal to Potential Energy
At the closest approach, we set KE equal to PE:
\[ 6 \times 10^6 \text{ eV} = \frac{1}{4 \pi \epsilon_0} \frac{(79)(2)(1.6 \times 10^{-19})^2}{r} \]

### Step 4: Substitute Known Values
The electrostatic constant \( \epsilon_0 \) is approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2 \). Thus, we can rewrite the equation:
\[ 6 \times 10^6 = \frac{(9 \times 10^9) \cdot (79)(2)(1.6 \times 10^{-19})^2}{r} \]

### Step 5: Solve for \( r \)
Rearranging the equation to solve for \( r \):
\[ r = \frac{(9 \times 10^9) \cdot (79)(2)(1.6 \times 10^{-19})^2}{6 \times 10^6} \]

### Step 6: Calculate the Values
Calculating the numerator:
- \( 79 \times 2 = 158 \)
- \( (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \)
- Thus, \( (9 \times 10^9) \cdot (158) \cdot (2.56 \times 10^{-38}) \)

Now, calculate:
\[ 9 \times 10^9 \times 158 \times 2.56 \times 10^{-38} \approx 3.63 \times 10^{-17} \]

Now, divide by \( 6 \times 10^6 \):
\[ r \approx \frac{3.63 \times 10^{-17}}{6 \times 10^6} \approx 6.05 \times 10^{-24} \text{ m} \]

### Step 7: Final Result
Thus, the distance of closest approach \( r \) is approximately:
\[ r \approx 6.05 \times 10^{-14} \text{ m} \]