A proton and an electron are moving with the same de-Broglie wavelength (consider the non-relativistic case). Then

Options

A. In a magnetic field both the particles describe circles of equal radius
B. Both the particle have the same momentum
C. The speed of the proton and the electron are in the ratio `m_(e)//m_(p)`, where `m_(e)` is the electron mass and `m_(p)`, the proton mass
D. The product of mass and kinetic energy is the same for both particles

Explanation

To solve the problem, we need to analyze the situation where a proton and an electron are moving with the same de-Broglie wavelength. We will use the de-Broglie wavelength formula and the principles of motion in a magnetic field to derive the necessary relationships.

### Step-by-Step Solution:

1. **Understanding de-Broglie Wavelength**:
The de-Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.

2. **Setting Up the Equation**:
Given that the de-Broglie wavelengths of the proton (\( \lambda_p \)) and the electron (\( \lambda_e \)) are equal:
\[ \lambda_e = \lambda_p \]
This leads to:
\[ \frac{h}{m_e v_e} = \frac{h}{m_p v_p} \]
Simplifying this gives:
\[ m_e v_e = m_p v_p \quad \text{(Equation 1)} \]

3. **Analyzing Motion in a Magnetic Field**:
When a charged particle moves in a magnetic field, it describes a circular path. The radius of the circular path (\( R \)) is given by:
\[ R = \frac{mv}{qB} \]
where \( q \) is the charge of the particle and \( B \) is the magnetic field strength.

4. **Calculating Radii for Electron and Proton**:
For the electron:
\[ R_e = \frac{m_e v_e}{qB} \]
For the proton:
\[ R_p = \frac{m_p v_p}{qB} \]
Since \( q \) and \( B \) are the same for both particles, we can compare the radii:
\[ R_e = R_p \quad \Rightarrow \quad \frac{m_e v_e}{qB} = \frac{m_p v_p}{qB} \]
This confirms that the radii are equal.

5. **Momentum Comparison**:
From Equation 1, we have:
\[ m_e v_e = m_p v_p \]
This indicates that the momentum of the electron and proton is equal.

6. **Velocity Ratio**:
Rearranging Equation 1 gives:
\[ \frac{v_p}{v_e} = \frac{m_e}{m_p} \]
This shows the ratio of their velocities.

7. **Kinetic Energy Comparison**:
The kinetic energy (\( KE \)) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
For the electron:
\[ KE_e = \frac{1}{2} m_e v_e^2 \]
For the proton:
\[ KE_p = \frac{1}{2} m_p v_p^2 \]
Multiplying the kinetic energy by mass gives:
\[ m_e KE_e = \frac{1}{2} m_e^2 v_e^2 \]
\[ m_p KE_p = \frac{1}{2} m_p^2 v_p^2 \]
Since \( m_e v_e = m_p v_p \), squaring both sides leads to:
\[ m_e^2 v_e^2 = m_p^2 v_p^2 \]
This shows that the products of mass and kinetic energy are equal.

### Conclusion:
All options in the question are correct:
1. The radii of the circular paths are equal.
2. The momenta are equal.
3. The speed ratio is \( \frac{m_e}{m_p} \).
4. The product of mass and kinetic energy is the same for both particles.