An electron and a proton have equal kinetic energies. They enter in a magnetic field perpendicularly, Then

Options

A. Both will follow a circular path with same radius
B. Both will follow a helical path
C. Both will follow a parabolic path
D. All the statements are false

Answer

To solve the problem, we need to analyze the motion of an electron and a proton when they enter a magnetic field perpendicularly, given that they have equal kinetic energies.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
The kinetic energy (KE) of a particle is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

2. **Given Condition**:
We know that the electron and proton have equal kinetic energies:
\[ KE_e = KE_p \]
Therefore,
\[ \frac{1}{2} m_e v_e^2 = \frac{1}{2} m_p v_p^2 \]
where \( m_e \) and \( m_p \) are the masses of the electron and proton, respectively, and \( v_e \) and \( v_p \) are their velocities.

3. **Magnetic Force and Circular Motion**:
When a charged particle moves in a magnetic field perpendicularly, it experiences a magnetic force that causes it to move in a circular path. The radius \( r \) of the circular path is given by:
\[ r = \frac{mv}{qB} \]
where \( q \) is the charge of the particle and \( B \) is the magnetic field strength.

4. **Relating Radius to Kinetic Energy**:
Since both particles have the same kinetic energy, we can express their velocities in terms of their kinetic energies:
\[ v_e = \sqrt{\frac{2KE}{m_e}} \quad \text{and} \quad v_p = \sqrt{\frac{2KE}{m_p}} \]
Substituting these into the radius formula gives:
\[ r_e = \frac{m_e \sqrt{\frac{2KE}{m_e}}}{q_e B} = \frac{\sqrt{2KE \cdot m_e}}{q_e B} \]
\[ r_p = \frac{m_p \sqrt{\frac{2KE}{m_p}}}{q_p B} = \frac{\sqrt{2KE \cdot m_p}}{q_p B} \]

5. **Comparing the Radii**:
Since the charge \( q \) of the electron and proton are equal (\( q_e = q_p \)), the radius can be compared as:
\[ r_e \propto \sqrt{m_e} \quad \text{and} \quad r_p \propto \sqrt{m_p} \]
Given that the mass of the proton \( m_p \) is much greater than the mass of the electron \( m_e \) (approximately 1836 times), it follows that:
\[ r_e < r_p \]

6. **Conclusion**:
The electron will have a smaller radius of curvature compared to the proton when both enter the magnetic field perpendicularly. Therefore, both particles will not follow the same circular path.

### Final Answer:
- The correct conclusion is that both will not follow circular paths with the same radius. The radius of the electron's path is much smaller than that of the proton.