If an electron and a proton having same momenta enter perpendicular to a magnetic field, then

Options

A. curved path of electron and proton will have same radius of curvature (ignoring the sense of revolution)
B. they will move undeflected
C. curved path of electron is more curved than that of the proton
D. path of proton is more curved

Explanation

To solve the problem, we need to analyze the motion of an electron and a proton when they enter a magnetic field perpendicularly with the same momentum. We will derive the radius of their circular paths and compare them.

### Step-by-Step Solution:

1. **Understanding the Motion in a Magnetic Field**:
When charged particles like electrons and protons enter a magnetic field perpendicularly, they experience a magnetic force that causes them to move in a circular path. The magnetic force acting on a charged particle is given by the formula:
\[ F = Q \cdot v \cdot B \]
where \( F \) is the magnetic force, \( Q \) is the charge of the particle, \( v \) is its velocity, and \( B \) is the magnetic field strength.

**Hint**: Remember that the direction of the magnetic force is perpendicular to both the velocity of the particle and the magnetic field.

2. **Equating the Forces**:
For circular motion, the magnetic force provides the centripetal force required to keep the particle in circular motion. The centripetal force can be expressed as:
\[ F = \frac{mv^2}{R} \]
where \( m \) is the mass of the particle, \( v \) is its velocity, and \( R \) is the radius of the circular path.

**Hint**: Recognize that both forces (magnetic and centripetal) are equal when the particle moves in a circular path.

3. **Setting the Equations Equal**:
By setting the magnetic force equal to the centripetal force, we have:
\[ Q \cdot v \cdot B = \frac{mv^2}{R} \]

4. **Solving for the Radius \( R \)**:
Rearranging the equation to solve for the radius \( R \):
\[ R = \frac{mv}{Q \cdot B} \]

5. **Substituting Momentum**:
The momentum \( p \) of a particle is defined as:
\[ p = mv \]
Therefore, we can express \( R \) in terms of momentum:
\[ R = \frac{p}{Q \cdot B} \]

6. **Comparing the Electron and Proton**:
Since both the electron and the proton have the same momentum \( p \), we can analyze their radii:
- For the electron, charge \( Q_e = -1.6 \times 10^{-19} \, C \) (negative charge).
- For the proton, charge \( Q_p = +1.6 \times 10^{-19} \, C \) (positive charge).

The radius for both particles can be expressed as:
\[ R_e = \frac{p}{|Q_e| \cdot B} \quad \text{and} \quad R_p = \frac{p}{|Q_p| \cdot B} \]

Since \( |Q_e| = |Q_p| \), we find that:
\[ R_e = R_p \]

7. **Conclusion**:
Both the electron and proton will have the same radius of curvature in the magnetic field despite their opposite charges. Thus, they will follow circular paths of the same radius.

### Final Answer:
The electron and proton will have the same radius of curvature when they enter a magnetic field perpendicularly with the same momentum.