A proton, a deutron and `alpha`-particle, whose kinetic energies are same, enter perpendicularly a uniform magnetic field. Compare the radii of their circualr paths.

Options

A. `1:1:sqrt2`
B. `sqrt2:1:1`
C. `1:sqrt2:1`
D. `1:2:sqrt2`

Explanation

To solve the problem of comparing the radii of the circular paths of a proton, a deuteron, and an alpha particle entering a uniform magnetic field with the same kinetic energy, we can follow these steps:

### 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 acts as a centripetal force, causing it to move in a circular path. The radius \( R \) of this circular path can be derived from the balance of forces.

2. **Centripetal Force and Magnetic Force**:
The centripetal force required for circular motion is given by:
\[ F_c = \frac{mv^2}{R} \]
The magnetic force acting on a charged particle moving in a magnetic field is given by:
\[ F_m = qvB \]
Setting these two forces equal gives:
\[ \frac{mv^2}{R} = qvB \]

3. **Solving for the Radius \( R \)**:
Rearranging the equation for \( R \) gives:
\[ R = \frac{mv}{qB} \]

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

5. **Substituting for \( v \) in the Radius Equation**:
Substituting \( v \) back into the equation for \( R \):
\[ R = \frac{m \sqrt{\frac{2K}{m}}}{qB} = \frac{\sqrt{2Km}}{qB} \]

6. **Comparing the Radii**:
Since the kinetic energy \( K \) is the same for all three particles, we can express the radii in terms of their mass \( m \) and charge \( q \):
\[ R \propto \frac{\sqrt{m}}{q} \]

7. **Mass and Charge of Each Particle**:
- **Proton**: \( m_p = 1.67 \times 10^{-27} \, \text{kg}, \, q_p = 1.6 \times 10^{-19} \, \text{C} \)
- **Deuteron**: \( m_d \approx 3.34 \times 10^{-27} \, \text{kg}, \, q_d = 1.6 \times 10^{-19} \, \text{C} \)
- **Alpha Particle**: \( m_{\alpha} \approx 6.64 \times 10^{-27} \, \text{kg}, \, q_{\alpha} = 3.2 \times 10^{-19} \, \text{C} \)

8. **Calculating the Ratios**:
- For the proton: \( R_p \propto \frac{\sqrt{m_p}}{q_p} \)
- For the deuteron: \( R_d \propto \frac{\sqrt{m_d}}{q_d} \)
- For the alpha particle: \( R_{\alpha} \propto \frac{\sqrt{m_{\alpha}}}{q_{\alpha}} \)

Now substituting the values:
- \( R_p \propto \frac{\sqrt{m_p}}{q_p} \)
- \( R_d \propto \frac{\sqrt{2m_p}}{q_p} \)
- \( R_{\alpha} \propto \frac{\sqrt{4m_p}}{2q_p} \)

9. **Final Ratio**:
After simplifying, we find:
\[ R_p : R_d : R_{\alpha} = 1 : \sqrt{2} : 1 \]

### Conclusion
The final result shows that the ratio of the radii of the circular paths of the proton, deuteron, and alpha particle is:
\[ R_p : R_d : R_{\alpha} = 1 : \sqrt{2} : 1 \]