Step 1: Set up with the kinetic energy:
Call the initial kinetic energy $K = \frac{1}{2}mv^2$. Rutherford's picture says all of $K$ becomes potential energy at the turning point.
Step 2: Potential energy expression:
For a charge $q_1 = 2e$ and $q_2 = Ze$ separated by $r_0$, $U = k \cdot 2e \cdot Ze / r_0$, where $k = 1/(4\pi\epsilon_0)$.
Step 3: Rearrange:
Setting $K = U$ gives \[ r_0 = \frac{2kZe^2}{K} \] Here $2kZe^2$ is constant for a given target, so $r_0 \propto 1/K$.
Step 4: Replace K by v:
Since $K \propto v^2$, we get $r_0 \propto 1/v^2 = v^{-2}$.
A faster alpha particle gets closer to the nucleus, and it does so quickly, in inverse proportion to $v^2$.
Step 5: Options:
Only the second option matches. The fourth option is wrong since $r_0$ increases, not decreases, with $Ze$.
Final Answer:
Hence $r_0$ varies as $v^{-2}$. \[\boxed{r_0 \propto v^{-2}}\]