A proton (mass m and charge `+e`) and an `alpha`-particle (mass `4` m and charge `+2e`) are projected with the same kinetic energy at right angles to the uniform magnetic field. Which one of the following statements will be true?

Options

A. The `alpha`-particle will be bent in a circular path with a small radius that for the proton
B. The radius of the path of the `alpha`-particle will be greater than that of the proton
C. The `alpha`-particle and the proton will be bent in a circular path with the same radiuys
D. The `alpha`-particle and the proton will go through the field in a straight line

Answer

To solve the problem, we need to analyze the motion of a proton and an alpha particle in a magnetic field when they are projected with the same kinetic energy at right angles to the field.

### Step-by-Step Solution:

1. **Understanding the Motion in a Magnetic Field**:
When a charged particle moves in a magnetic field, it experiences a magnetic force that causes it to move in a circular path. The radius of this circular path depends on the mass of the particle, its charge, and its velocity.

2. **Formula for Radius of Circular Motion**:
The radius \( r \) of the circular path of a charged particle in a magnetic field is given by:
\[ r = \frac{mv}{qB} \]
where:
- \( m \) = mass of the particle
- \( v \) = velocity of the particle
- \( q \) = charge of the particle
- \( B \) = magnetic field strength

3. **Kinetic Energy Relation**:
The kinetic energy \( K \) of a particle is given by:
\[ K = \frac{1}{2} mv^2 \]
From this, we can express the velocity \( v \) in terms of kinetic energy:
\[ v = \sqrt{\frac{2K}{m}} \]

4. **Substituting Velocity into the Radius Formula**:
Substituting the expression for \( v \) into the radius formula gives:
\[ r = \frac{m \sqrt{\frac{2K}{m}}}{qB} = \frac{\sqrt{2Km}}{qB} \]

5. **Calculating Radius for Proton**:
For the proton (mass \( m \), charge \( +e \)):
\[ r_p = \frac{\sqrt{2Km}}{eB} \]

6. **Calculating Radius for Alpha Particle**:
For the alpha particle (mass \( 4m \), charge \( +2e \)):
\[ r_{\alpha} = \frac{\sqrt{2K(4m)}}{2eB} = \frac{2\sqrt{2Km}}{2eB} = \frac{\sqrt{2Km}}{eB} \]

7. **Comparing the Radii**:
From the above calculations, we see:
\[ r_p = \frac{\sqrt{2Km}}{eB} \quad \text{and} \quad r_{\alpha} = \frac{\sqrt{2Km}}{eB} \]
This shows that the radius of the circular path for both the proton and the alpha particle is the same.

8. **Conclusion**:
Since both particles have the same radius of curvature in the magnetic field, we conclude that the correct statement is:
- The alpha particle and the proton will be bent in a circular path with the same radius.

### Final Answer:
The correct option is that the alpha particle and the proton will be bent in a circular path with the same radius.

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