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 smallest separation between the \( \alpha \)-particle and the nucleus during a head-on collision. At this point, the \( \alpha \)-particle's total kinetic energy is fully transformed into electrostatic potential energy because of the repulsive force from the nucleus.Derivation:At the point of closest approach, the kinetic energy \( K \) is entirely converted into electrostatic potential energy:\[K = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Z e \cdot 2e}{r_0}\]Rearranging for \( r_0 \):\[\Rightarrow r_0 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2Ze^2}{K}\]In this equation:- \( Z \) represents the atomic number of the target nucleus (for instance, \( Z = 79 \) for gold).- \( e \) denotes the elementary charge.- \( \varepsilon_0 \) signifies the permittivity of free space.