In the head-on collision of two alpha particles $\alpha_1$ and $\alpha_2$ with the gold nucleus, the closest approaches are 31.4 fermi and 94.2 fermi respectively. Then the ratio of the energy possessed by the alpha particles $\alpha_2 / \alpha_1$ is:

The distance of closest approach $r_0$ for an alpha particle of kinetic energy $K$ colliding head-on with a nucleus of atomic number $Z$ is given by the formula $r_0 = \dfrac{1}{4 \pi \epsilon_0} \dfrac{2Ze^2}{K}$.

From this expression, it is evident that $r_0 \propto \dfrac{1}{K}$, which implies $K \propto \dfrac{1}{r_0}$.

Given the closest approaches for the two alpha particles are $r_1 = 31.4$ fermi and $r_2 = 94.2$ fermi, the ratio of their kinetic energies $K_2 / K_1$ is given by:

$\dfrac{K_2}{K_1} = \dfrac{r_1}{r_2}$

Substituting the given values:

$\dfrac{K_2}{K_1} = \dfrac{31.4}{94.2} = \dfrac{1}{3}$

Answer: $1: 3$