Hydrogen ion and singly ionized helium atom are accelerated,

(3) According to work energy theorem, gain in kinetic energy is equal to work done in displacement of charge.

$\therefore \quad \frac{1}{2} m v^{2}=q \Delta V$

Here, $\Delta V=$ potential difference between two positions of charge $q$.

For same $q$ and $\Delta V$.

$v \propto \frac{1}{\sqrt{m}}$

Mass of hydrogen ion $\mathrm{m}_{\mathrm{H}}=1$

Mass of helium ion $\mathrm{m}_{\mathrm{He}}=4$

$\therefore \quad \frac{v_{\mathrm{H}}}{v_{\mathrm{He}}}=\sqrt{\frac{4}{1}}=2: 1$