An `alpha` -particle of 1 MeV energy moves in circular path in uniform magnetic field. The kinetic energy (in MeV) of proton in the same magnetic field for circular path of double radius is

Answer

To solve the problem, we need to analyze the motion of an alpha particle and a proton in a magnetic field, focusing on their kinetic energies and the relationship between their radii of circular motion.

### Step-by-Step Solution:

1. **Understanding the alpha particle:**
- An alpha particle has a charge of \( +2e \) and a mass of \( 4m \) (where \( m \) is the mass of a proton).
- The kinetic energy (KE) of the alpha particle is given as \( 1 \, \text{MeV} \).

2. **Kinetic energy of the alpha particle:**
- The kinetic energy of the alpha particle can be expressed as:
\[ KE_{\alpha} = \frac{1}{2} m_{\alpha} v_{\alpha}^2 = 1 \, \text{MeV} \]
- Here, \( m_{\alpha} = 4m \), so we can write:
\[ 1 = \frac{1}{2} (4m) v_{\alpha}^2 \implies v_{\alpha}^2 = \frac{2 \times 1}{4m} = \frac{1}{2m} \]

3. **Radius of the circular path for the alpha particle:**
- The radius \( r \) of the circular path in a magnetic field is given by:
\[ r = \frac{m_{\alpha} v_{\alpha}}{qB} \]
- Substituting \( m_{\alpha} = 4m \) and \( q = 2e \):
\[ r = \frac{(4m) v_{\alpha}}{2eB} = \frac{2mv_{\alpha}}{eB} \]

4. **Kinetic energy of the proton:**
- For a proton, the charge is \( +e \) and the mass is \( m \).
- The radius of the circular path for the proton is given as \( 2r \).
- The kinetic energy of the proton can be expressed as:
\[ KE_{p} = \frac{1}{2} m v_{p}^2 \]

5. **Radius of the proton's circular path:**
- For the proton moving in a magnetic field:
\[ 2r = \frac{mv_{p}}{eB} \]
- Rearranging gives:
\[ v_{p} = \frac{2r eB}{m} \]

6. **Relating the velocities:**
- We know from the previous steps that \( r = \frac{2mv_{\alpha}}{eB} \), so substituting for \( r \):
\[ v_{p} = \frac{2 \left( \frac{2mv_{\alpha}}{eB} \right) eB}{m} = 4v_{\alpha} \]

7. **Calculating the kinetic energy of the proton:**
- Now substituting \( v_{p} = 4v_{\alpha} \) into the kinetic energy equation:
\[ KE_{p} = \frac{1}{2} m (4v_{\alpha})^2 = \frac{1}{2} m (16v_{\alpha}^2) = 8m v_{\alpha}^2 \]
- We already found \( v_{\alpha}^2 = \frac{1}{2m} \):
\[ KE_{p} = 8m \left(\frac{1}{2m}\right) = 4 \, \text{MeV} \]

### Final Answer:
The kinetic energy of the proton in the same magnetic field for a circular path of double radius is **4 MeV**.