# Alpha Particle and Target Nucleus: Closest Approach ## Problem Statement - **Alpha particle**: 2 protons, initial kinetic energy \(K = 1.1\, \text{pJ} = 1.1 \times 10^{-12}\, \text{J}\) - **Target nucleus**: 87 protons (likely francium, \(Z = 87\)) - **Proton charge**: \(e = 1.602 \times 10^{-19}\,\text{C}\) - **Assumption**: The target nucleus is stationary. Find the **minimum distance** between the centers (closest approach), assuming a head-on collision and that the alpha particle stops at that point (all kinetic energy converted to potential energy). --- ## Step 1: Potential Energy at Closest Approach The **electrostatic potential energy** between two point charges at distance \( r \): \[ U = \frac{1}{4\pi \varepsilon_} \frac{q_1 q_2}{r} \] Where: - \( q_1 = \) charge of alpha particle \( = 2e \) - \( q_2 = \) charge of nucleus \( = 87e \) - \( \varepsilon_ = 8.854 \times 10^{-12}\,\text{C}^2/(\text{N} \cdot \text{m}^2) \) At closest approach, all kinetic energy is converted: \[ K = U \implies K = \frac{1}{4\pi \varepsilon_} \frac{(2e)(87e)}{r} \] --- ## Step 2: Solve for \( r \) \[ K = \frac{1}{4\pi \varepsilon_} \frac{174 e^2}{r} \implies r = \frac{1}{4\pi \varepsilon_} \frac{174 e^2}{K} \] --- ## Step 3: Insert Values - \( e = 1.602 \times 10^{-19}\,\text{C} \) - \( K = 1.1 \times 10^{-12}\,\text{J} \) - \( \frac{1}{4\pi \varepsilon_} = 8.988 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2 \) \[ r = (8.988 \times 10^9) \frac{174 (1.602 \times 10^{-19})^2}{1.1 \times 10^{-12}} \] First, calculate \( e^2 \): \[ (1.602 \times 10^{-19})^2 = 2.566 \times 10^{-38}\,\text{C}^2 \] Now, plug in the numbers: \[ r = (8.988 \times 10^9) \times \frac{174 \times 2.566 \times 10^{-38}}{1.1 \times 10^{-12}} \] Calculate numerator: \[ 174 \times 2.566 \times 10^{-38} = 446.484 \times 10^{-38} = 4.46484 \times 10^{-36} \] Now: \[ r = (8.988 \times 10^9) \times \frac{4.46484 \times 10^{-36}}{1.1 \times 10^{-12}} \] Compute \( \frac{4.46484 \times 10^{-36}}{1.1 \times 10^{-12}} = 4.058036 \times 10^{-24} \) So, \[ r = (8.988 \times 10^9) \times (4.058036 \times 10^{-24}) \] \[ r = 36.478 \times 10^{-15}\,\text{m} \] \[ r = 3.6 \times 10^{-14}\,\text{m} \] --- ## Final Answer ### **The closest center-to-center distance is:** \[ \boxed{3.6 \times 10^{-14}\,\text{m}} \] --- ### **Summary Table** | Quantity | Value | |----------------|---------------------------| | \(K\) | \(1.1 \times 10^{-12}\) J | | \(e\) | \(1.602 \times 10^{-19}\) C| | \(Z_{nucleus}\)| 87 | | \(r_{min}\) | \(3.6 \times 10^{-14}\) m | --- **Note:** This is the theoretical closest approach, assuming no nuclear forces and a point-like nucleus. **Image alt text**: Diagram showing an alpha particle approaching a stationary nucleus, reaching a minimum separation where all kinetic energy has converted to electrostatic potential energy.