A proton, a deuteron and an `alpha`- particle having the same kinetic energy are moving in circular trajectors in a constant magnetic field. If `r_p, r_d` and `r_(alpha)` denote respectively the radii of the trajectories of these particles then

Options

A. `r_(alpha)=r_(p)ltr_(d)`
B. `r_(alpha)gtr_(d)gtr_(p)`
C. `r_(alpha)=r_(d)gtr_(p)`
D. `r_(p)=r_(d)=r_(alpha)`

Answer

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

### Step 1: Understand the motion of charged particles in a magnetic field
When a charged particle moves in a magnetic field, it experiences a magnetic force that acts as the centripetal force, causing it to move in a circular path. The magnetic force \( F_B \) is given by:
\[ F_B = qvB \]
where \( q \) is the charge of the particle, \( v \) is its velocity, and \( B \) is the magnetic field strength.

### Step 2: Set up the centripetal force equation
The centripetal force required to keep a particle moving in a circular path is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the particle, \( v \) is its velocity, and \( r \) is the radius of the circular path.

### Step 3: Equate the magnetic force and centripetal force
Since the magnetic force provides the centripetal force, we can set these two equations equal to each other:
\[ qvB = \frac{mv^2}{r} \]

### Step 4: Solve for the radius \( r \)
Rearranging the equation gives us:
\[ r = \frac{mv}{qB} \]

### Step 5: Relate kinetic energy to velocity
The kinetic energy \( K \) of a particle is given by:
\[ K = \frac{1}{2}mv^2 \]
Since the kinetic energy is the same for all three particles, we can express the velocity in terms of kinetic energy:
\[ v = \sqrt{\frac{2K}{m}} \]

### Step 6: Substitute velocity into the radius equation
Substituting \( v \) into the radius equation gives:
\[ r = \frac{m \sqrt{\frac{2K}{m}}}{qB} = \frac{\sqrt{2Km}}{qB} \]

### Step 7: Analyze the radius for each particle
Now we can analyze the radius for each particle:
- For the proton: \( m_p = 1 \, \text{amu}, \, q_p = 1e \)
- For the deuteron: \( m_d = 2 \, \text{amu}, \, q_d = 1e \)
- For the alpha particle: \( m_{\alpha} = 4 \, \text{amu}, \, q_{\alpha} = 2e \)

### Step 8: Write the radius expressions
- For the proton:
\[ r_p = \frac{\sqrt{2K \cdot 1}}{1 \cdot B} \]
- For the deuteron:
\[ r_d = \frac{\sqrt{2K \cdot 2}}{1 \cdot B} \]
- For the alpha particle:
\[ r_{\alpha} = \frac{\sqrt{2K \cdot 4}}{2 \cdot B} \]

### Step 9: Simplify the expressions
- For the proton:
\[ r_p = \frac{\sqrt{2K}}{B} \]
- For the deuteron:
\[ r_d = \frac{\sqrt{4K}}{B} = 2 \cdot \frac{\sqrt{K}}{B} \]
- For the alpha particle:
\[ r_{\alpha} = \frac{2\sqrt{K}}{2B} = \frac{\sqrt{K}}{B} \]

### Step 10: Compare the radii
Now we can compare the radii:
- \( r_p \) (proton) is proportional to \( 1 \)
- \( r_d \) (deuteron) is proportional to \( 2 \)
- \( r_{\alpha} \) (alpha particle) is proportional to \( 1 \)

### Conclusion
Thus, we find that:
\[ r_d > r_p = r_{\alpha} \]
This means the radius of the deuteron is greater than that of the proton and the alpha particle, which have the same radius.