Question

Define the ‘distance of closest approach’. An \( \alpha \)-particle of kinetic energy \( K \) is bombarded on a thin gold foil. Derive an expression for the ‘distance of closest approach’.

Hint:

Always use conservation of energy in head-on nuclear collisions: convert kinetic energy to potential energy at the closest approach point.

Answer 1

Definition: The \textit{distance of closest approach} \( r_0 \) is the minimum distance between the \( \alpha \)-particle and the nucleus during head-on collision. At this point, the entire kinetic energy of the \( \alpha \)-particle is converted into electrostatic potential energy due to repulsion from the nucleus. Derivation: At the distance of closest approach, kinetic energy \( K \) is converted into electrostatic potential energy: \[ K = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Z e \cdot 2e}{r_0} \] \[ \Rightarrow r_0 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2Ze^2}{K} \] Where: - \( Z \) is the atomic number of the target nucleus (e.g., for gold \( Z = 79 \)), - \( e \) is the elementary charge, - \( \varepsilon_0 \) is the permittivity of free space.