A beam of alpha paricles is incident on a target of lead. `A` particular alpha paticles comes in 'head- on' to a particular lead nucleus and stops `6.50 xx 10^(-14)` m away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has `82` protons, remains at rest. The mass of alpha particle is `6.64 xx 10^(-27)kg`
Calculate the electrostatic potential energy at the instant when the alpha particle stops?

Options

A. `36.3MeV`
B. `45.0MeV`
C. `3.63MeV`
D. `40.0MeV`

Explanation

To calculate the electrostatic potential energy at the instant when the alpha particle stops, we can follow these steps:

### Step 1: Identify the charges involved
The alpha particle has a charge of \( +2e \) (since it consists of 2 protons), and the lead nucleus has a charge of \( +82e \) (since it has 82 protons).

### Step 2: Use the formula for electrostatic potential energy
The electrostatic potential energy \( U \) between two point charges is given by the formula:
\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]
where:
- \( k \) is Coulomb's constant, approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
- \( q_1 \) is the charge of the alpha particle
- \( q_2 \) is the charge of the lead nucleus
- \( r \) is the distance between the charges

### Step 3: Calculate the charges in coulombs
Using \( e = 1.6 \times 10^{-19} \, \text{C} \):
- Charge of the alpha particle, \( q_1 = 2e = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \, \text{C} \)
- Charge of the lead nucleus, \( q_2 = 82e = 82 \times 1.6 \times 10^{-19} = 1.312 \times 10^{-17} \, \text{C} \)

### Step 4: Substitute values into the formula
The distance \( r \) is given as \( 6.5 \times 10^{-14} \, \text{m} \). Now substituting the values into the potential energy formula:
\[ U = \frac{(9 \times 10^9) \cdot (3.2 \times 10^{-19}) \cdot (1.312 \times 10^{-17})}{6.5 \times 10^{-14}} \]

### Step 5: Calculate the potential energy
Calculating the numerator:
\[ 9 \times 10^9 \cdot 3.2 \times 10^{-19} \cdot 1.312 \times 10^{-17} \approx 3.77 \times 10^{-26} \]
Now, dividing by \( r \):
\[ U \approx \frac{3.77 \times 10^{-26}}{6.5 \times 10^{-14}} \approx 5.8 \times 10^{-13} \, \text{J} \]

### Step 6: Convert to electron volts
To convert joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ U \approx \frac{5.8 \times 10^{-13}}{1.6 \times 10^{-19}} \approx 3.625 \times 10^6 \, \text{eV} = 3.625 \, \text{MeV} \]

### Final Answer
Thus, the electrostatic potential energy at the instant when the alpha particle stops is approximately:
\[ \boxed{3.63 \, \text{MeV}} \]