A proton, an electron, and a Helium nucleus, have the same Kinetic energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let `r_(p), r_(e)` and `r_(He)` be their respective radii, then

Options

A. `r_(e)gtr_(p)gtr_(He)`
B. `r_(e)ltr_(p)lt4_(He)`
C. `r_(e)gtr_(p)=r_(He)`
D. `r_(e)ltr_(p)=r_(He)`

Explanation

To solve the problem, we need to find the relationship between the radii of the circular orbits of a proton, an electron, and a helium nucleus (alpha particle) when they have the same kinetic energy and are subjected to a magnetic field.

### Step-by-Step Solution:

1. **Understanding the Forces**:
The magnetic force acting on a charged particle moving in a magnetic field provides the centripetal force required for circular motion. The magnetic force \( F_m \) is given by:
\[ F_m = qvB \]
where \( q \) is the charge of the particle, \( v \) is its velocity, and \( B \) is the magnetic field strength.

2. **Centripetal Force**:
The centripetal force \( F_c \) required for circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the particle, \( v \) is its velocity, and \( r \) is the radius of the circular path.

3. **Setting Forces Equal**:
For a particle in circular motion in a magnetic field, we set the magnetic force equal to the centripetal force:
\[ qvB = \frac{mv^2}{r} \]

4. **Solving for Radius**:
Rearranging the above equation gives us the expression for the radius \( r \):
\[ r = \frac{mv}{qB} \]

5. **Relating Velocity to Kinetic Energy**:
The kinetic energy \( K \) of the particle is given by:
\[ K = \frac{1}{2} mv^2 \]
From this, we can express \( v \) in terms of \( K \):
\[ v = \sqrt{\frac{2K}{m}} \]

6. **Substituting Velocity into Radius Equation**:
Substituting \( v \) into the radius equation:
\[ r = \frac{m \sqrt{\frac{2K}{m}}}{qB} = \frac{\sqrt{2Km}}{qB} \]

7. **Calculating Radii for Each Particle**:
Now we can write the expressions for the radii of the proton, electron, and helium nucleus:
- For the proton:
\[ r_p = \frac{\sqrt{2K m_p}}{q_p B} \]
- For the electron:
\[ r_e = \frac{\sqrt{2K m_e}}{q_e B} \]
- For the helium nucleus (alpha particle), which has a mass \( m_{He} = 4m_p \) and charge \( q_{He} = 2e \):
\[ r_{He} = \frac{\sqrt{2K m_{He}}}{q_{He} B} = \frac{\sqrt{2K \cdot 4m_p}}{2eB} = \frac{2\sqrt{2K m_p}}{2eB} = \frac{\sqrt{2K m_p}}{eB} \]

8. **Comparing the Radii**:
Now we can compare the radii:
- \( r_p \) for the proton:
\[ r_p = \frac{\sqrt{2K m_p}}{eB} \]
- \( r_e \) for the electron:
\[ r_e = \frac{\sqrt{2K m_e}}{eB} \]
- \( r_{He} \) for the helium nucleus:
\[ r_{He} = \frac{\sqrt{2K m_p}}{eB} \]

Since \( m_p \) (mass of the proton) is much greater than \( m_e \) (mass of the electron), we can conclude:
\[ r_{He} = r_p > r_e \]

### Final Result:
Thus, the relationship between the radii is:
\[ r_{He} = r_p > r_e \]