An electron and a proton enter region of uniform magnetic field in a direction at right angles to the field with the same kinetic energy. They describe circular paths of radius `r_(e)` and `r_(p)` ?respectively. Then what will be the relation between `r_(e)` and `r_(p)`?

Options

A. `r_(e)=r_(p)`
B. `r_(e)ltr_(p)`
C. `r_(e)gtr_(p)`
D. `r_(e)` may be less than or greater than `r_(p)` depending on the direction of the magnetic field

Answer

To solve the problem, we need to derive the relationship between the radii of the circular paths of an electron and a proton moving in a magnetic field with 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 at right angles to the field, it experiences a magnetic force that acts as a centripetal force. The magnetic force \( F \) on a charged particle is given by:
\[ F = QVB \sin \theta \]
where \( Q \) is the charge, \( V \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field. Since \( \theta = 90^\circ \), we have:
\[ F = QVB \]

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

3. **Equating Forces**:
Setting the magnetic force equal to the centripetal force, we have:
\[ QVB = \frac{mv^2}{r} \]

4. **Solving for Radius**:
Rearranging the equation to solve for the radius \( r \):
\[ r = \frac{mv}{QB} \]

5. **Kinetic Energy Relation**:
The kinetic energy \( KE \) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
Since both the electron and proton have the same kinetic energy, we can express the velocity in terms of kinetic energy:
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

6. **Substituting Velocity into Radius Formula**:
Substituting \( v \) back into the radius formula:
\[ r = \frac{m \sqrt{\frac{2 \cdot KE}{m}}}{QB} = \frac{\sqrt{2 \cdot KE \cdot m}}{QB} \]

7. **Comparing Radii of Electron and Proton**:
Let \( r_e \) be the radius for the electron and \( r_p \) be the radius for the proton. Thus:
\[ r_e = \frac{\sqrt{2 \cdot KE \cdot m_e}}{Q_e B} \]
\[ r_p = \frac{\sqrt{2 \cdot KE \cdot m_p}}{Q_p B} \]
Since the charge \( Q \) is the same for both (electron and proton), we can simplify:
\[ r_e = \frac{\sqrt{2 \cdot KE \cdot m_e}}{Q B} \]
\[ r_p = \frac{\sqrt{2 \cdot KE \cdot m_p}}{Q B} \]

8. **Finding the Relation**:
Since \( KE \) and \( Q \) are the same for both particles, we can compare the radii:
\[ \frac{r_e}{r_p} = \frac{\sqrt{m_e}}{\sqrt{m_p}} \]
Given that the mass of the proton \( m_p \) is greater than the mass of the electron \( m_e \) (i.e., \( m_p > m_e \)), we conclude:
\[ r_e < r_p \]

### Conclusion:
Thus, the relationship between the radii is:
\[ r_e < r_p \]