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Step-by-Step Solution
Step 1: State the formula for the radius of the circular path
When a charged particle moves perpendicular to a uniform magnetic field, the radius of its circular path is given by:
$r = \frac{p}{q B}$
where
$p$ is the linear momentum of the particle,
$q$ is the charge of the particle,
$B$ is the magnetic field.
Step 2: Express the radius for the hydrogen ion
An ionized hydrogen atom (essentially a proton) has charge $e$. With momentum $p$ and charge $e$, the radius of its path is:
$r_H = \frac{p}{e\, B}$
Step 3: Express the radius for the alpha particle
An alpha particle ($\alpha$) has charge $2e$. With the same momentum $p$ and a charge of $2e$, the radius of its path is:
$r_\alpha = \frac{p}{2e\, B}$
Step 4: Determine the ratio of the radii
Taking the ratio $r_H : r_\alpha$, we get:
$\frac{r_H}{r_\alpha}
= \frac{\frac{p}{e\, B}}{\frac{p}{2e\, B}}
= \frac{p}{e\, B} \times \frac{2e\, B}{p}
= 2 : 1$
This shows that the radius of the path of the ionized hydrogen (proton) is twice that of the alpha particle under the given conditions.
