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. the radius `r_(d)` of the deutron path to the radius `r_(p)` of the proton path is `sqrt2`
B. the radius `r_(alpha)` of the alpha particle path to `r_(p)` is 1
C. the radius `r_(d)` of the deutron path to the radius `r_(p)` of the proton path is 2
D. the radius `r_(alpha)` of the alpha particle path to `r_(p)` is `sqrt2`

Answer

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

### Step 1: Define the Mass and Charge of Each Particle
- **Proton (p)**: Mass = \( m \), Charge = \( e \)
- **Deuteron (d)**: Mass = \( 2m \), Charge = \( e \)
- **Alpha Particle (α)**: Mass = \( 4m \), Charge = \( 2e \)

### Step 2: Write the Expression for Kinetic Energy
Since the kinetic energies of all three particles are the same, we can denote this common kinetic energy as \( K \).

The kinetic energy for each particle can be expressed as:
- For Proton: \( K = \frac{1}{2} mv_p^2 \)
- For Deuteron: \( K = \frac{1}{2} (2m) v_d^2 \)
- For Alpha Particle: \( K = \frac{1}{2} (4m) v_\alpha^2 \)

### Step 3: Solve for Velocity of Each Particle
From the kinetic energy equation, we can express the velocities of each particle:
- For Proton:
\[ v_p = \sqrt{\frac{2K}{m}} \]
- For Deuteron:
\[ v_d = \sqrt{\frac{2K}{2m}} = \sqrt{\frac{K}{m}} \]
- For Alpha Particle:
\[ v_\alpha = \sqrt{\frac{2K}{4m}} = \sqrt{\frac{K}{2m}} \]

### Step 4: Write the Expression for Radius in a Magnetic Field
The radius \( r \) of the circular path of a charged particle moving 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

### Step 5: Calculate the Radius for Each Particle
- For Proton:
\[ r_p = \frac{m v_p}{eB} = \frac{m \sqrt{\frac{2K}{m}}}{eB} = \frac{\sqrt{2Km}}{eB} \]

- For Deuteron:
\[ r_d = \frac{(2m) v_d}{eB} = \frac{(2m) \sqrt{\frac{K}{m}}}{eB} = \frac{2\sqrt{Km}}{eB} \]

- For Alpha Particle:
\[ r_\alpha = \frac{(4m) v_\alpha}{2eB} = \frac{(4m) \sqrt{\frac{K}{2m}}}{2eB} = \frac{2\sqrt{2Km}}{eB} \]

### Step 6: Compare the Radii
Now we can write the ratios of the radii:
\[ \frac{r_p}{r_d} = \frac{\frac{\sqrt{2Km}}{eB}}{\frac{2\sqrt{Km}}{eB}} = \frac{1}{2} \]
\[ \frac{r_p}{r_\alpha} = \frac{\frac{\sqrt{2Km}}{eB}}{\frac{2\sqrt{2Km}}{eB}} = \frac{1}{2} \]
\[ \frac{r_d}{r_\alpha} = \frac{\frac{2\sqrt{Km}}{eB}}{\frac{2\sqrt{2Km}}{eB}} = \frac{1}{\sqrt{2}} \]

### Final Ratios
Thus, the final ratios of the radii are:
- \( r_p : r_d : r_\alpha = 1 : \sqrt{2} : 1 \)

### Conclusion
The comparison of the radii of the circular paths is:
- The radius of the proton path to the radius of the deuteron path is \( \sqrt{2} \).
- The radius of the alpha particle path to the radius of the proton path is \( 1 \).