Question

Draw the graph showing the variation of scattered particles detected (\( N \)) with the scattering angle (\( \theta \)) in the Geiger-Marsden experiment. Write two conclusions that you can draw from this graph. Obtain the expression for the distance of closest approach in this experiment.

Hint:

Rutherford’s experiment showed that atoms have a small dense nucleus and are mostly empty space.

Answer 1

Geiger-Marsden Experiment (Rutherford Scattering) 
- The experiment measured the number of alpha particles (\( N \)) scattered at different angles (\( \theta \)). 
- The observed scattering pattern led to significant conclusions about atomic structure. 
Graph: Variation of \( N \) with \( \theta \) 

 Conclusions from the Graph
1. Most alpha particles pass undeflected, meaning the atom is mostly empty space.
2. Few particles are scattered at large angles, implying a small, dense, positively charged nucleus. Expression for Distance of Closest Approach The distance of closest approach (\( r_0 \)) is the minimum separation between the alpha particle and the nucleus before it stops and reverses.
 - At the point of closest approach, the initial kinetic energy of the alpha particle is converted into electrostatic potential energy: \[ \frac{1}{2} m v^2 = \frac{1}{4\pi\epsilon_0} \frac{Z e \cdot 2e}{r_0} \] 

Solving for \( r_0 \): \[ r_0 = \frac{1}{4\pi\epsilon_0} \frac{2 Z e^2}{\frac{1}{2} m v^2} \] \[ r_0 = \frac{4 \pi \epsilon_0 \cdot 2 Z e^2}{m v^2} \] Thus, the distance of closest approach is: \[ r_0 = \frac{2 Z e^2}{4 \pi \epsilon_0 \cdot \frac{1}{2} m v^2} \] 

This represents the minimum distance between the alpha particle and the nucleus before repulsion stops its motion.