A proton, a deutron and an alpha particle, after being accelerated through the same potential difference enter a region of uniform magnetic field `vecB`, in a direction perpendicular to `vecB`. Find the ratio of their kinetic energies. If the radius of proton's circular path is `7cm`, what will be the radii of the paths of deutron and alpha particle?

Options

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

Answer

To solve the problem, we need to find the ratio of the kinetic energies of a proton, a deuteron, and an alpha particle after being accelerated through the same potential difference, and then determine the radii of their paths in a magnetic field.

### Step 1: Determine the Kinetic Energy Ratio

The kinetic energy (KE) of a charged particle accelerated through a potential difference \( V \) is given by the formula:
\[ KE = qV \]
where \( q \) is the charge of the particle.

- For the proton (\( q_p = e \)), the kinetic energy is:
\[ KE_p = eV \]
- For the deuteron (\( q_d = e \)), the kinetic energy is:
\[ KE_d = eV \]
- For the alpha particle (\( q_{\alpha} = 2e \)), the kinetic energy is:
\[ KE_{\alpha} = 2eV \]

Now, we can find the ratio of their kinetic energies:
\[ \text{Ratio of KE} = KE_p : KE_d : KE_{\alpha} = eV : eV : 2eV = 1 : 1 : 2 \]

### Step 2: Calculate the Radius of the Path

The radius \( r \) of the circular path of a charged particle moving in a magnetic field is given by:
\[ r = \frac{mv}{qB} \]
where \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field strength.

From the kinetic energy, we can express the velocity \( v \) as:
\[ KE = \frac{1}{2} mv^2 \implies v = \sqrt{\frac{2KE}{m}} \]

Substituting the expression for \( v \) into the radius formula gives:
\[ r = \frac{m \sqrt{\frac{2KE}{m}}}{qB} = \frac{\sqrt{2mKE}}{qB} \]

Since \( KE \) is proportional to \( q \) (as we derived earlier), we can express the radius in terms of the charge and mass:
\[ r \propto \frac{q \sqrt{m}}{B} \]

### Step 3: Find the Radii of Deuteron and Alpha Particle

Given that the radius of the proton's path is \( r_p = 7 \, \text{cm} \), we can find the radii of the deuteron and alpha particle using their charge and mass ratios.

1. **For the Deuteron**:
- Charge: \( q_d = e \)
- Mass: \( m_d = 2m_p \)
- Ratio of radius:
\[ \frac{r_d}{r_p} = \frac{q_d \sqrt{m_d}}{q_p \sqrt{m_p}} = \frac{e \sqrt{2m_p}}{e \sqrt{m_p}} = \sqrt{2} \]
\[ r_d = r_p \cdot \sqrt{2} = 7 \cdot \sqrt{2} \approx 9.89 \, \text{cm} \]

2. **For the Alpha Particle**:
- Charge: \( q_{\alpha} = 2e \)
- Mass: \( m_{\alpha} = 4m_p \)
- Ratio of radius:
\[ \frac{r_{\alpha}}{r_p} = \frac{q_{\alpha} \sqrt{m_{\alpha}}}{q_p \sqrt{m_p}} = \frac{2e \sqrt{4m_p}}{e \sqrt{m_p}} = 4 \]
\[ r_{\alpha} = r_p \cdot 2 = 7 \cdot 2 = 14 \, \text{cm} \]

### Final Results
- The ratio of kinetic energies is \( 1 : 1 : 2 \).
- The radius of the deuteron's path is approximately \( 9.89 \, \text{cm} \).
- The radius of the alpha particle's path is \( 14 \, \text{cm} \).