Two ions of masses 4

$\mathrm{r}=\frac{\mathrm{P}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mk}}}{\mathrm{qB}}$

Given they have same kinetic energy

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

$\frac{r_{1}}{r_{2}}=\frac{\sqrt{4}}{2} \times \frac{3}{\sqrt{16}}=\frac{3}{4}$

$\mathrm{r}_{2}=\frac{4 \mathrm{r}_{1}}{3}$( $r_{2}$ is for hearier ion and $r_{1}$ is for lighter ion)

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

$\theta \rightarrow$ Deflection

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

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

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