To solve this problem, we need to understand the concept of the distance of closest approach in the context of an $\alpha$-particle bombarding a heavy nucleus.$\\$The distance of closest approach is the minimum distance between the $\alpha$-particle and the nucleus during the collision.$\\$When an $\alpha$-particle with mass $m$ and velocity $v$ approaches a nucleus of charge $Ze,$ it experiences a repulsive Coulomb force.$\\$The initial kinetic energy of the $\alpha$-particle is given by:$\\ K.E. = \frac{1}{2} m v^2 \\$At the distance of closest approach, all the kinetic energy is converted into electrostatic potential energy:$\\ \frac{1}{2} m v^2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{(2e)(Ze)}{r} \\$Here, $r$ is the distance of closest approach, and $2e$ is the charge of the $\alpha$-particle.$\\$Rearranging the equation to solve for $r: \\ r = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{(2e)(Ze)}{\frac{1}{2} m v^2} \\ r = \frac{4 \pi \varepsilon_0 \cdot 2Ze^2}{m v^2} \\$From this expression, we can see that the distance of closest approach $r$ is inversely proportional to $m. \\$Therefore, the distance of closest approach depends on $m$ as $\frac{1}{m}. \\$This corresponds to Option 3.