Let's solve the problem step by step: --- ## **Given:** - Initial speed of electron, \( v_i = 4.44 \times 10^5 \) m/s - Proton is at rest and considered fixed. - We want the distance between the electron and proton when the electron's speed is \( 2v_i \). --- ## **Step 1: Conservation of Energy** Total energy (kinetic + electrostatic potential) is conserved: \[ K_i + U_i = K_f + U_f \] ### **Initial State (electron very far from proton):** - Kinetic energy: \( K_i = \frac{1}{2}mv_i^2 \) - Potential energy: \( U_i = \) (since \( r \rightarrow \infty \)) ### **Final State (distance \( r \) from proton):** - New speed: \( v_f = 2v_i \) - Kinetic energy: \( K_f = \frac{1}{2}m(2v_i)^2 = 2m v_i^2 \) - Potential energy: \( U_f = -\frac{ke^2}{r} \) Where: - \( m \) = mass of electron = \( 9.11 \times 10^{-31} \) kg - \( k \) = Coulomb's constant = \( 8.99 \times 10^9 \) N·m²/C² - \( e \) = elementary charge = \( 1.60 \times 10^{-19} \) C --- ## **Step 2: Write the Energy Conservation Equation** \[ \frac{1}{2}mv_i^2 + = 2mv_i^2 - \frac{ke^2}{r} \] --- ## **Step 3: Solve for \( r \)** Rearrange: \[ \frac{1}{2}mv_i^2 - 2mv_i^2 = -\frac{ke^2}{r} \] \[ -\frac{3}{2}mv_i^2 = -\frac{ke^2}{r} \] \[ \frac{3}{2}mv_i^2 = \frac{ke^2}{r} \] \[ r = \frac{ke^2}{\frac{3}{2}mv_i^2} \] \[ r = \frac{2ke^2}{3mv_i^2} \] --- ## **Step 4: Plug in Values** \[ r = \frac{2 \times (8.99 \times 10^9) \times (1.60 \times 10^{-19})^2}{3 \times (9.11 \times 10^{-31}) \times (4.44 \times 10^5)^2} \] Calculate numerator: - \( (1.60 \times 10^{-19})^2 = 2.56 \times 10^{-38} \) - \( 2 \times 8.99 \times 10^9 \times 2.56 \times 10^{-38} \) - \( = 2 \times 8.99 \times 2.56 \times 10^{9 - 38} \) - \( = 2 \times 8.99 \times 2.56 \times 10^{-29} \) - \( = 2 \times 22.9744 \times 10^{-29} \) - \( = 45.9488 \times 10^{-29} \) - \( = 4.59488 \times 10^{-28} \) Calculate denominator: - \( 4.44 \times 10^5 \) squared = \( 19.7136 \times 10^{10} \) - \( 3 \times 9.11 \times 10^{-31} \times 19.7136 \times 10^{10} \) - \( = 3 \times 9.11 \times 19.7136 \times 10^{-31 + 10} \) - \( = 3 \times 179.634096 \times 10^{-21} \) - \( = 538.902288 \times 10^{-21} \) - \( = 5.38902288 \times 10^{-19} \) Now, \[ r = \frac{4.59488 \times 10^{-28}}{5.38902288 \times 10^{-19}} \] \[ = .8526 \times 10^{-9} \text{ m} \] \[ = 8.53 \times 10^{-10} \text{ m} \] --- ## **Final Answer** \[ \boxed{8.53 \times 10^{-10} \text{ m}} \] This is the distance between the two particles when the electron's speed is \( 2v_i \).