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)

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