An electron and a proton enter a magnetic field at right angles to the field with the same kinetic energy

Options

A. the electron trajectory will be less curved than the proton trajectory
B. the proton trajectory will be less curved than the electron trajectory
C. both trajectories will be equally curved
D. both particles move in straight lines

Answer

To solve the problem, we need to analyze the motion of an electron and a proton entering a magnetic field at right angles with the same kinetic energy. Here’s the step-by-step solution:

### Step 1: Understand the Given Information
We know that both the electron and the proton have the same kinetic energy (KE). The kinetic energy can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

### Step 2: Set Up the Equation for Kinetic Energy
For the electron and proton, we can write:
\[ \frac{1}{2} m_e v_e^2 = \frac{1}{2} m_p v_p^2 \]
where \( m_e \) and \( m_p \) are the masses of the electron and proton, and \( v_e \) and \( v_p \) are their respective velocities.

### Step 3: Rearrange the Equation
Cancelling the \(\frac{1}{2}\) from both sides gives us:
\[ m_e v_e^2 = m_p v_p^2 \]
From this, we can derive:
\[ \frac{m_e}{m_p} = \frac{v_p^2}{v_e^2} \]
Taking the square root of both sides, we find:
\[ \frac{v_e}{v_p} = \sqrt{\frac{m_p}{m_e}} \]

### Step 4: Analyze the Masses
Since the mass of the proton \( m_p \) is much greater than the mass of the electron \( m_e \) (approximately 1836 times), we conclude:
\[ v_e > v_p \]
This means the electron has a higher velocity than the proton.

### Step 5: Determine the Radius of Circular Motion
In a magnetic field, the radius \( r \) of the circular path of a charged particle is given by:
\[ r = \frac{mv}{qB} \]
where \( q \) is the charge of the particle and \( B \) is the magnetic field strength.

### Step 6: Substitute for Radius
Substituting the kinetic energy relation into the radius formula, we can express it as:
\[ r = \frac{mv}{qB} \]
Using \( KE = \frac{1}{2} mv^2 \), we can express \( mv \) as:
\[ mv = \sqrt{2m \cdot KE} \]
Thus, the radius becomes:
\[ r = \frac{\sqrt{2m \cdot KE}}{qB} \]

### Step 7: Compare the Radii
Since both particles have the same kinetic energy, we can compare their radii:
- For the electron:
\[ r_e = \frac{\sqrt{2m_e \cdot KE}}{q_e B} \]
- For the proton:
\[ r_p = \frac{\sqrt{2m_p \cdot KE}}{q_p B} \]

### Step 8: Conclusion
Since \( m_e < m_p \) and \( v_e > v_p \), the radius of the electron's path will be smaller than that of the proton's path. Thus, the trajectory of the electron will be less curved than that of the proton.

### Final Answer
The electron trajectory will be less curved than the proton trajectory.