Given they have same kinetic energy

${r} \propto \frac{\sqrt{{m}}}{{q}}$

$\frac{{r}_{1}}{{T}_{2}}=\frac{\sqrt{4}}{2} \times \frac{3}{\sqrt{16}}=\frac{3}{4}$

${I}_{2}=\frac{4 {r}_{1}}{3}\left[{r}_{2}\right.$ is for hearier ion and ${r}_{1}$ is for lighter ion)

$\sin \theta=\frac{{d}}{{R}}$

$\theta \rightarrow$ Deflection

$\theta \propto \frac{1}{{R}}$

$({R} \rightarrow$ Radius of path)

$\because {R}_{2}>{R}_{1} \Rightarrow \theta_{2}<\theta_{1}$