Correct Option: (C) $\frac{1}{K}$

Step 1 (Formula / Governing Law & Given Data): [0.25 Mark]

According to the law of conservation of energy, at the distance of closest approach ($d$), the initial kinetic energy ($K$) of the alpha particle is entirely converted into electrostatic potential energy ($U$) of the system (alpha particle and gold nucleus).

$$K = U = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{d}$$

Where $q_1 = 2e$ (charge of alpha particle) and $q_2 = Ze$ (charge of gold nucleus).

Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]

Substitute the charges into the energy conservation equation:

$$K = \frac{1}{4\pi\epsilon_0} \cdot \frac{(2e)(Ze)}{d}$$

Rearranging the equation to solve for the distance of closest approach ($d$):

$$d = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{K}$$

Since $\frac{1}{4\pi\epsilon_0}$, $2$, $Z$, and $e^2$ are constants for a given experiment, we can express the relationship as:

$$d \propto \frac{1}{K}$$

Step 3 (Final Concluded Answer): [0.25 Mark]

The distance of closest approach $d$ is inversely proportional to the initial kinetic energy $K$ of the alpha particle. Therefore, the correct option is (C).

💡 EL10 Concept Insight: Think of the alpha particle like a ball rolling up a hill; the faster you roll it (higher kinetic energy), the higher it climbs before stopping, meaning it gets closer to the top (the nucleus) if the hill is steeper, but here, the "hill" is the electric repulsion, so more speed allows it to overcome that repulsion and get much closer to the center.