The mass of a proton is 1847 times that of an electron A electron and a proton are injected into a unifrom electric field at right angle to the direction of the field with the same initial K.E.

Options

A. the electron trajectory will be les lcurved than the proton trajectory.
B. both the trajectories will be straight
C. the proton trajectory will be less curved than the electron trajectroy
D. both the trajectories will be equally curved

Explanation

To solve the problem step by step, we need to analyze the motion of both the electron and the proton in a uniform electric field, considering their masses and kinetic energies.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Mass of proton (m_p) = 1847 × Mass of electron (m_e)
- Both particles are injected into a uniform electric field at right angles to the direction of the field.
- Both have the same initial kinetic energy (KE).

2. **Kinetic Energy Relation**:
- The kinetic energy (KE) is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- Since both the electron and the proton have the same kinetic energy, we can express their velocities in terms of their masses:
\[ KE_e = \frac{1}{2} m_e v_e^2 \]
\[ KE_p = \frac{1}{2} m_p v_p^2 \]
- Setting these equal gives:
\[ \frac{1}{2} m_e v_e^2 = \frac{1}{2} m_p v_p^2 \]
- Rearranging gives:
\[ m_e v_e^2 = m_p v_p^2 \]

3. **Substituting Mass of Proton**:
- Substitute \( m_p = 1847 m_e \):
\[ m_e v_e^2 = 1847 m_e v_p^2 \]
- Dividing both sides by \( m_e \) (assuming \( m_e \neq 0 \)):
\[ v_e^2 = 1847 v_p^2 \]
- This implies:
\[ v_e = \sqrt{1847} v_p \]

4. **Acceleration in Electric Field**:
- The force on a charged particle in an electric field (E) is given by:
\[ F = qE \]
- The acceleration (a) can be calculated using Newton's second law:
\[ a = \frac{F}{m} = \frac{qE}{m} \]

5. **Trajectory Calculation**:
- The vertical displacement (y) in terms of time (t) can be expressed as:
\[ y = \frac{1}{2} a t^2 \]
- Substituting for acceleration:
\[ y = \frac{1}{2} \left(\frac{qE}{m}\right) t^2 \]

6. **Time Relation**:
- The time (t) can also be expressed in terms of horizontal distance (x) and initial horizontal velocity (v):
\[ t = \frac{x}{v} \]
- Substituting this into the equation for y gives:
\[ y = \frac{1}{2} \left(\frac{qE}{m}\right) \left(\frac{x}{v}\right)^2 \]

7. **Final Trajectory Expression**:
- Since \( v \) is related to kinetic energy, we can express \( v^2 \) in terms of kinetic energy (K):
\[ v^2 = \frac{2K}{m} \]
- Substituting this into the equation for y leads to:
\[ y = \frac{qE}{4K} x^2 \]

8. **Conclusion**:
- The trajectory of both particles depends on \( q \), \( E \), and \( K \), but not on mass. Since both particles have the same charge and are in the same electric field with the same kinetic energy, their trajectories will be equally curved.

### Final Answer:
Both trajectories will be equally curved.