1. Use energy conservation at the distance of closest approach

When the $\alpha$-particle moves directly toward a heavy nucleus, its initial kinetic energy is converted into electrostatic potential energy at the closest approach.

Given kinetic energy of the $\alpha$-particle: $K = \frac{1}{2}mv^2$

If $r$ is the distance of closest approach, then the electrostatic potential energy is $U = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r}$ because the charge on the $\alpha$-particle is $+2e$.

At closest approach, $\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{r}$

2. Solve for $r$

Rearranging, $r = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{\frac{1}{2}mv^2}$

So, $r \propto \frac{Ze^2}{mv^2}$

Hence, $r \propto \frac{1}{mv^2}$ for fixed $Z$ and $e$.

3. Check each option

• A: $v^2$
  Incorrect, since $r \propto \dfrac{1}{v^2}$.

• B: $\dfrac{1}{m}$
  Correct as an individual proportionality, because $r \propto \dfrac{1}{m}$.

• C: $\dfrac{1}{v^2}$
  Correct, because $r \propto \dfrac{1}{v^2}$.

• D: $\dfrac{1}{Ze}$
  Incorrect, since $r \propto Z$ (and actually $\propto e^2$), not inversely.

4. Conclusion

The distance of closest approach is proportional to both $\dfrac{1}{m}$ and $\dfrac{1}{v^2}$. So if forced to choose from the given options in a single-correct format, option C is certainly correct, but B is also proportionality-wise correct.

Thus, the physics gives: $r \propto \frac{1}{mv^2}$

So among the given options, C is correct, though the question is not ideally framed for single-correct because B also follows.