Ionized hydrogen atoms and $\alpha$-particles with the same momenta enter perpendicular to a constant magnetic

Question

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

Accepted Answer

1. **Radius Formula:** The radius ($r$) of the circular path of a charged particle moving perpendicular to a uniform magnetic field ($B$) is given by the equation:
   $$r = \frac{mv}{qB} = \frac{p}{qB}$$
   where $m$ is mass, $v$ is velocity, $q$ is charge, and $p = mv$ is momentum .
2. **Proportionality:** Since the momentum ($p$) and the magnetic field ($B$) are the same for both particles, the radius is inversely proportional to the charge ($q$).
   $$r \propto \frac{1}{q}$$
3. **Particle Charges:**
   - Ionized Hydrogen atom ($H^+$) has a charge $q_H = +e$.
   - $\alpha$-particle ($He^{2+}$) has a charge $q_\alpha = +2e$.
4. **Calculation:**
   $$\frac{r_H}{r_\alpha} = \frac{q_\alpha}{q_H} = \frac{2e}{e} = \frac{2}{1}$$
   Thus, the ratio is 2:1.

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