A proton carrying `1 MeV` kinetic energy is moving in a circular path of radius `R` in uniform magnetic field. What should be the energy of an `alpha-` particle to describe a circle of the same radius in the same field?

Options

A. `2 MeV`
B. `1 MeV`
C. `0.5 MeV`
D. `4 MeV`

Explanation

To solve the problem, we need to find the kinetic energy of an alpha particle that moves in a circular path of the same radius \( R \) as a proton moving in a uniform magnetic field. The proton has a kinetic energy of \( 1 \, \text{MeV} \).

### Step-by-step Solution:

1. **Understanding the Motion in a Magnetic Field**:
- A charged particle moving in a magnetic field experiences a magnetic force that acts as the centripetal force required for circular motion. The magnetic force \( F_B \) on a charged particle is given by:
\[ F_B = QvB \]
- The centripetal force \( F_c \) required for circular motion is given by:
\[ F_c = \frac{mv^2}{R} \]

2. **Equating Forces**:
- For a particle moving in a circular path, we set the magnetic force equal to the centripetal force:
\[ QvB = \frac{mv^2}{R} \]

3. **Solving for Velocity**:
- Rearranging the equation gives:
\[ v = \frac{QRB}{m} \]

4. **Kinetic Energy Expression**:
- The kinetic energy \( KE \) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
- Substituting the expression for \( v \):
\[ KE = \frac{1}{2} m \left(\frac{QRB}{m}\right)^2 = \frac{Q^2 R^2 B^2}{2m} \]

5. **Applying to Proton**:
- For the proton, let \( KE_P = 1 \, \text{MeV} \), \( Q_P = e \) (charge of proton), and \( m_P \) (mass of proton):
\[ KE_P = \frac{e^2 R^2 B^2}{2m_P} \]

6. **Applying to Alpha Particle**:
- For the alpha particle, \( Q_\alpha = 2e \) (charge of alpha particle), \( m_\alpha = 4m_P \) (mass of alpha particle):
\[ KE_\alpha = \frac{(2e)^2 R^2 B^2}{2(4m_P)} = \frac{4e^2 R^2 B^2}{8m_P} = \frac{e^2 R^2 B^2}{2m_P} \]

7. **Relating the Energies**:
- Notice that the expression for \( KE_\alpha \) simplifies to:
\[ KE_\alpha = KE_P \]
- Since \( KE_P = 1 \, \text{MeV} \), we conclude that:
\[ KE_\alpha = 1 \, \text{MeV} \]

### Conclusion:
The energy of the alpha particle to describe a circle of the same radius \( R \) in the same magnetic field is also \( 1 \, \text{MeV} \).