Geiger-Marsden Experiment: Analysis and Derivation

Step 1 (Graph and Conclusions): [1 Mark]

The graph of the number of scattered particles ($N$) versus the scattering angle ($\theta$) shows an exponential decrease as the angle increases.

• Conclusion 1: Most $\alpha$-particles pass through the gold foil with very little or no deflection, indicating that the atom is mostly empty space.
• Conclusion 2: A very small fraction of $\alpha$-particles undergo large-angle scattering, indicating the presence of a small, dense, positively charged nucleus at the center.

Step 2 (Governing Law for Distance of Closest Approach): [1 Mark]

At the distance of closest approach ($r_0$), the entire initial kinetic energy ($K$) of the $\alpha$-particle is converted into electrostatic potential energy ($U$) as the particle momentarily comes to rest before reversing its direction.

By the Law of Conservation of Energy:

$$K = U$$ $$\frac{1}{2}mv^2 = \frac{1}{4\pi\epsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$$

Step 3 (Mathematical Derivation and Final Expression): [1 Mark]

Rearranging the equation to solve for $r_0$:

$$r_0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{\frac{1}{2}mv^2}$$ $$r_0 = \frac{4Ze^2}{4\pi\epsilon_0 mv^2}$$

Where:

• $Z$ = Atomic number of the target nucleus.
• $e$ = Elementary charge.
• $m$ = Mass of the $\alpha$-particle.
• $v$ = Initial velocity of the $\alpha$-particle.