Ionized hydrogen atoms and `alpha-`particle with moments enters perpendicular to a constant megnetic field. B. The ratio of their radii of their paths `r_(H): r_(alpha)` be :

Options

A. `1:4`
B. `2:1`
C. `1:2`
D. `4:1`

Explanation

To solve the problem of finding the ratio of the radii of the paths of ionized hydrogen atoms and alpha particles in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion in a Magnetic Field**:
When a charged particle moves perpendicular to a magnetic field, it experiences a magnetic force that causes it to move in a circular path. The radius of this circular path (r) can be expressed as:
\[ r = \frac{mv}{qB} \]
where:
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle,
- \( q \) is the charge of the particle,
- \( B \) is the magnetic field strength.

2. **Identifying the Particles**:
- For ionized hydrogen (H), the charge \( q \) is equal to the elementary charge \( e \) (since it has one proton).
- For an alpha particle (α), which consists of 2 protons and 2 neutrons, the charge \( q \) is \( 2e \).

3. **Calculating the Radius for Each Particle**:
- For the hydrogen atom:
\[ r_H = \frac{mv_H}{eB} \]
- For the alpha particle:
\[ r_{\alpha} = \frac{mv_{\alpha}}{2eB} \]

4. **Finding the Ratio of Radii**:
To find the ratio \( \frac{r_H}{r_{\alpha}} \):
\[ \frac{r_H}{r_{\alpha}} = \frac{\frac{mv_H}{eB}}{\frac{mv_{\alpha}}{2eB}} \]
This simplifies to:
\[ \frac{r_H}{r_{\alpha}} = \frac{mv_H}{eB} \cdot \frac{2eB}{mv_{\alpha}} = \frac{2v_H}{v_{\alpha}} \]

5. **Assuming Equal Velocities**:
If we assume that both particles enter the magnetic field with the same velocity \( v \) (i.e., \( v_H = v_{\alpha} \)), then:
\[ \frac{r_H}{r_{\alpha}} = \frac{2v}{v} = 2 \]
Therefore, the ratio of the radii is:
\[ r_H : r_{\alpha} = 2 : 1 \]

### Final Answer:
The ratio of the radii of the paths of ionized hydrogen atoms to alpha particles is \( 2 : 1 \).

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