Ionized hydrogen atoms and $\alpha$ -particles with same momenta enter perpendicular to a constant magnetic field, $B$. The ratio of their radii of their paths $r_H:r_{\alpha }$ will be:

The radius ($r$) of the circular path of a charged particle in a perpendicular magnetic field is:
$$r=\frac{p}{qB}$$
where $p$ is the momentum, $q$ is the charge, and $B$ is the magnetic field strength.Given that $p$ and $B$ are the same for both particles:
$$r\propto \frac{1}{q}$$
For an ionized hydrogen atom (proton), charge $q_H=e$.For an $\alpha$ -particle, charge $q_{\alpha }=2e$.Therefore, the ratio is:
$$\frac{r_H}{r_{\alpha }}=\frac{q_{\alpha }}{q_H}=\frac{2e}{e}=2$$
So, $r_H:r_{\alpha }=2:1$.