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 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 potential energy due to the electrostatic repulsion between the positively charged alpha particle and the nucleus. At this point, the kinetic energy is zero.

### Step 2: Write the Energy Conservation Equation
At the distance of closest approach (let's denote it as \( D \)), the kinetic energy (KE) of the alpha particle is equal to the potential energy (PE) due to electrostatic repulsion:
\[ KE = PE \]
\[ 6 \text{ MeV} = \frac{k \cdot q_1 \cdot q_2}{D} \]

### Step 3: Convert MeV to Joules
We need to convert the kinetic energy from MeV to Joules for our calculations.
1 MeV = \( 1.6 \times 10^{-13} \) Joules, so:
\[ 6 \text{ MeV} = 6 \times 1.6 \times 10^{-13} \text{ J} = 9.6 \times 10^{-13} \text{ J} \]

### Step 4: Identify the Charges
The alpha particle has a charge of \( +2e \) (where \( e = 1.6 \times 10^{-19} \) C), and the gold nucleus has a charge of \( +79e \). Thus:
\[ q_1 = 2e = 2 \times 1.6 \times 10^{-19} \text{ C} \]
\[ q_2 = 79e = 79 \times 1.6 \times 10^{-19} \text{ C} \]

### Step 5: Substitute Values into the Equation
Now, substituting the values into the potential energy equation:
\[ 9.6 \times 10^{-13} = \frac{(9 \times 10^9) \cdot (2 \times 1.6 \times 10^{-19}) \cdot (79 \times 1.6 \times 10^{-19})}{D} \]

### Step 6: Solve for D
Rearranging the equation to solve for \( D \):
\[ D = \frac{(9 \times 10^9) \cdot (2 \times 1.6 \times 10^{-19}) \cdot (79 \times 1.6 \times 10^{-19})}{9.6 \times 10^{-13}} \]

### Step 7: Calculate the Values
Calculating the numerator:
\[ = 9 \times 10^9 \cdot 2 \cdot 1.6 \times 10^{-19} \cdot 79 \cdot 1.6 \times 10^{-19} \]
\[ = 9 \times 10^9 \cdot 2 \cdot 79 \cdot (1.6 \times 1.6) \times 10^{-38} \]
\[ = 9 \times 10^9 \cdot 158.4 \cdot 2.56 \times 10^{-38} \]
\[ = 9 \times 10^9 \cdot 405.504 \times 10^{-38} \]
\[ = 3.649536 \times 10^{-28} \text{ J m} \]

Now substituting back to find \( D \):
\[ D = \frac{3.649536 \times 10^{-28}}{9.6 \times 10^{-13}} = 3.79 \times 10^{-15} \text{ m} \]

### Final Answer
The distance of closest approach \( D \) is approximately:
\[ D \approx 3.79 \times 10^{-14} \text{ m} \]