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`
What initial kinetic energy (in joule and in MeV)did the alpha particle have?

Options

A. `36.3`
B. `0.36`
C. `3.63`
D. `2.63`

Explanation

To solve the problem, we will calculate the initial kinetic energy of the alpha particle using the concept of electrostatic potential energy at the point of closest approach to the lead nucleus. We will follow these steps:

### Step 1: Understand the scenario
An alpha particle approaches a lead nucleus and stops at a distance of \(6.50 \times 10^{-14} \, \text{m}\) from the center of the nucleus. The lead nucleus has 82 protons. The initial kinetic energy of the alpha particle will be converted into electrostatic potential energy at the point of closest approach.

### Step 2: Use the conservation of energy principle
The initial kinetic energy (KE_initial) of the alpha particle will be equal to the electrostatic potential energy (PE) at the closest approach:
\[ KE_{\text{initial}} = PE \]

### Step 3: Calculate the electrostatic potential energy
The electrostatic potential energy between two point charges is given by the formula:
\[ PE = \frac{k \cdot q_1 \cdot q_2}{r} \]
where:
- \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\) (Coulomb's constant),
- \(q_1\) is the charge of the alpha particle,
- \(q_2\) is the charge of the lead nucleus,
- \(r\) is the distance of closest approach.

### Step 4: Determine the charges
The charge of the alpha particle (\(q_1\)) is:
\[ q_1 = 2 \times (1.6 \times 10^{-19} \, \text{C}) = 3.2 \times 10^{-19} \, \text{C} \]
The charge of the lead nucleus (\(q_2\)) is:
\[ q_2 = 82 \times (1.6 \times 10^{-19} \, \text{C}) = 1.312 \times 10^{-17} \, \text{C} \]

### Step 5: Substitute the values into the potential energy formula
Now substituting the values into the potential energy formula:
\[ PE = \frac{(9 \times 10^9) \cdot (3.2 \times 10^{-19}) \cdot (1.312 \times 10^{-17})}{6.50 \times 10^{-14}} \]

### Step 6: Calculate the potential energy
Calculating the above expression:
\[ PE = \frac{(9 \times 10^9) \cdot (3.2 \times 10^{-19}) \cdot (1.312 \times 10^{-17})}{6.50 \times 10^{-14}} \approx 4.81 \times 10^{-13} \, \text{J} \]

### Step 7: Convert to MeV
To convert joules to mega electron volts (MeV), we use the conversion factor \(1 \, \text{J} = 6.242 \times 10^{12} \, \text{MeV}\):
\[ KE_{\text{initial}} \, \text{(in MeV)} = 4.81 \times 10^{-13} \, \text{J} \times 6.242 \times 10^{12} \, \text{MeV/J} \approx 3.00 \, \text{MeV} \]

### Final Answer
- Initial kinetic energy in joules: \(4.81 \times 10^{-13} \, \text{J}\)
- Initial kinetic energy in MeV: \(3.00 \, \text{MeV}\)

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