Let's address each part of the question step by step, including explanations and a summary at the end. --- ### (a) **Smallest Distance (Closest Approach) Calculation** #### **Step 1: Set Up Energy Conservation** **Explanation:** When the alpha particle moves toward the gold nucleus, its initial kinetic energy is converted into electric potential energy as it slows down. At the point of closest approach, all kinetic energy has been converted, and the particle momentarily stops before reversing direction. - **Initial kinetic energy:** \( KE = \frac{1}{2}mv^2 \) where \( m = 6.64 \times 10^{-27} \) kg (mass of alpha particle), \( v = 2.00 \times 10^7 \) m/s. - **Electrostatic potential energy at distance \( r \):** \( PE = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r} \), where \( q_1 = +2e \), \( q_2 = +79e \), and \( \epsilon_0 = 8.85 \times 10^{-12} \) F/m. Set initial kinetic energy equal to potential energy at the closest approach: \[ \frac{1}{2}mv^2 = \frac{1}{4\pi\epsilon_0}\frac{(2e)(79e)}{r} \] #### **Step 2: Solve for the Smallest Distance \( r \)** **Explanation:** Rearrange the equation and solve for \( r \): \[ r = \frac{1}{4\pi\epsilon_0} \frac{(2e)(79e)}{\frac{1}{2}mv^2} \] Insert values: - \( e = 1.60 \times 10^{-19} \) C - \( m = 6.64 \times 10^{-27} \) kg - \( v = 2.00 \times 10^7 \) m/s - \( \frac{1}{4\pi\epsilon_0} = 8.99 \times 10^9 \) N·m²/C² Calculate numerator: \[ (2e)(79e) = 2 \times 79 \times (1.60 \times 10^{-19})^2 = 158 \times 2.56 \times 10^{-38} = 404.48 \times 10^{-38} = 4.0448 \times 10^{-36} \] Calculate denominator: \[ \frac{1}{2}mv^2 = 0.5 \times 6.64 \times 10^{-27} \times (2.00 \times 10^7)^2 = 0.5 \times 6.64 \times 4 \times 10^{14 - 27} = 0.5 \times 6.64 \times 4 \times 10^{-13} \] \[ = 13.28 \times 2 \times 10^{-13} = 26.56 \times 10^{-13} = 2.656 \times 10^{-12} \] Now substitute: \[ r = \frac{8.99 \times 10^9 \times 4.0448 \times 10^{-36}}{2.656 \times 10^{-12}} \] \[ = \frac{36.370 \times 10^{-27}}{2.656 \times 10^{-12}} \] \[ = \frac{36.370}{2.656} \times 10^{-27+12} \] \[ = 13.71 \times 10^{-15} \text{ m} \] --- ### (b) **Effect if Gold Nucleus Moves (Qualitative)** **Explanation:** If the gold nucleus is allowed to move, not all kinetic energy of the alpha particle will be available to convert into potential energy because some energy will be transferred to the gold nucleus (conservation of momentum). As a result, the alpha particle would not get as close to the nucleus; the distance of closest approach would increase because less energy is available to overcome the repulsive force. --- ### (c) **Dimensional Analysis of Closest Approach Formula** **Step 1: Predict Units** **Explanation:** The smallest distance \( r \) should have units of meters (\( m \)). Check the units in the formula: \[ r = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{\frac{1}{2}mv^2} \] \[ \left[ \frac{\text{N}\cdot\text{m}^2}{\text{C}^2} \cdot \frac{\text{C}^2}{\text{kg}\cdot\frac{\text{m}^2}{\text{s}^2}} \right] \] \[ \left[ \frac{\text{N}\cdot\text{m}^2}{\text{kg}\cdot\frac{\text{m}^2}{\text{s}^2}} \right] \] Since \( 1\, \text{N} = 1\, \text{kg}\cdot\text{m}/\text{s}^2 \): \[ \left[ \frac{\text{kg}\cdot\text{m}/\text{s}^2 \cdot \text{m}^2}{\text{kg}\cdot\text{m}^2/\text{s}^2} \right] = \left[ \frac{\text{kg}\cdot\text{m}^3/\text{s}^2}{\text{kg}\cdot\text{m}^2/\text{s}^2} \right] = [\text{m}] \] **Step 2: Compare to Prediction** **Explanation:** The formula's units reduce to meters, matching the expected unit for distance. --- ### **Final Answer Summary** - **(a) The smallest distance between the alpha particle and the nucleus (assuming a stationary nucleus) is approximately \( 1.37 \times 10^{-14} \) m.** - **(b) If the gold nucleus is allowed to move, the closest approach distance increases because some kinetic energy is transferred to the nucleus.** - **(c) Dimensional analysis confirms that the final formula for distance yields units of meters, as expected for a length.**