# Least Center-to-Center Distance in Alpha Particle Scattering When an alpha particle approaches a stationary nucleus, its **closest approach** (minimum center-to-center distance) is determined by conservation of energy. The alpha particle's initial kinetic energy is entirely converted into electrostatic potential energy at the point of closest approach. ## Key Points - **Alpha particle charge:** \( q_{\alpha} = 2e \) - **Target nucleus charge:** \( q_{\text{nucleus}} = Ze \) (where \( Z \) is the atomic number) - **Proton charge:** \( e = 1.602 \times 10^{-19}~\mathrm{C} \) - **Initial kinetic energy of alpha:** \( K \) - **Distance of closest approach:** \( r_{\min} \) ## Energy Conservation At closest approach: \[ \text{Initial kinetic energy} = \text{Electrostatic potential energy} \] \[ K = \frac{1}{4\pi\epsilon_} \frac{(2e)(Ze)}{r_{\min}} \] where \( \epsilon_ = 8.854 \times 10^{-12}~\mathrm{C}^2/\mathrm{N}\cdot\mathrm{m}^2 \). ## Solving for \( r_{\min} \): \[ r_{\min} = \frac{1}{4\pi\epsilon_} \frac{2Ze^2}{K} \] ### Substitute constants: \[ \frac{1}{4\pi\epsilon_} = 8.988 \times 10^9~\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}^2 \] So: \[ r_{\min} = \frac{(8.988 \times 10^9)\, 2Z (1.602 \times 10^{-19})^2}{K} \] ## Final Formula \[ \boxed{ r_{\min} = \frac{2Ze^2}{4\pi\epsilon_ K} } \] - \( Z \): Atomic number of the target nucleus - \( e \): Proton charge (\(1.602 \times 10^{-19}~\mathrm{C}\)) - \( K \): Alpha particle's initial kinetic energy (in joules) ## Example If you provide \( Z \) and \( K \), you can directly compute \( r_{\min} \) using the above formula. --- **Note:** This calculation assumes a head-on collision (impact parameter = ) and a stationary nucleus. The result is the center-to-center distance at closest approach.