A proton and an alpha particle enter the same magnetic field which is perpendicular to their velocity. If they have same kinetic energy then ratio of radii of their circular path is

Options

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

Answer

To solve the problem, we need to find the ratio of the radii of the circular paths of a proton and an alpha particle moving in the same magnetic field, given that they have the same kinetic energy.

### Step-by-Step Solution:

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

2. **Kinetic Energy of the Particles**:
The kinetic energy \( KE \) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
Since both the proton and the alpha particle have the same kinetic energy, we can write:
\[ KE_p = KE_{\alpha} \]
This implies:
\[ \frac{1}{2} m_p v_p^2 = \frac{1}{2} m_{\alpha} v_{\alpha}^2 \]
Simplifying this, we get:
\[ m_p v_p^2 = m_{\alpha} v_{\alpha}^2 \]

3. **Mass and Charge of the Particles**:
- Mass of the proton \( m_p = m \)
- Mass of the alpha particle \( m_{\alpha} = 4m \) (since an alpha particle consists of 2 protons and 2 neutrons)
- Charge of the proton \( q_p = e \)
- Charge of the alpha particle \( q_{\alpha} = 2e \)

4. **Finding the Ratio of Velocities**:
From the kinetic energy equation:
\[ m v_p^2 = 4m v_{\alpha}^2 \]
Dividing both sides by \( m \) (mass of the proton):
\[ v_p^2 = 4 v_{\alpha}^2 \]
Taking the square root:
\[ \frac{v_p}{v_{\alpha}} = 2 \]

5. **Finding the Ratio of Radii**:
Now, we can find the ratio of the radii \( r_p \) and \( r_{\alpha} \):
\[ \frac{r_p}{r_{\alpha}} = \frac{m_p v_p / (q_p B)}{m_{\alpha} v_{\alpha} / (q_{\alpha} B)} \]
Since \( B \) cancels out:
\[ \frac{r_p}{r_{\alpha}} = \frac{m_p v_p}{m_{\alpha} v_{\alpha}} \cdot \frac{q_{\alpha}}{q_p} \]
Substituting the known values:
\[ \frac{r_p}{r_{\alpha}} = \frac{m \cdot v_p}{4m \cdot v_{\alpha}} \cdot \frac{2e}{e} \]
Simplifying:
\[ \frac{r_p}{r_{\alpha}} = \frac{v_p}{4 v_{\alpha}} \cdot 2 = \frac{2 v_p}{4 v_{\alpha}} = \frac{v_p}{2 v_{\alpha}} \]
Now substituting \( \frac{v_p}{v_{\alpha}} = 2 \):
\[ \frac{r_p}{r_{\alpha}} = \frac{2}{2} = 1 \]

### Final Answer:
The ratio of the radii of their circular paths is:
\[ \frac{r_p}{r_{\alpha}} = 1 \]