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. 1 MeV
B. 0.5 MeV
C. 4 MeV
D. 2 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 as a proton moving with a kinetic energy of 1 MeV in a uniform magnetic field.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The radius \( R \) of a charged particle moving in a magnetic field is given by the formula:
\[ R = \frac{mv}{qB} \]
where:
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle,
- \( q \) is the charge of the particle,
- \( B \) is the magnetic field strength.

2. **Kinetic Energy Relation**:
The kinetic energy \( K \) of a particle is given by:
\[ K = \frac{1}{2}mv^2 \]
From this, we can express velocity \( v \) in terms of kinetic energy:
\[ v = \sqrt{\frac{2K}{m}} \]

3. **Substituting Velocity in Radius Formula**:
We can substitute \( v \) into the radius formula:
\[ R = \frac{m \sqrt{\frac{2K}{m}}}{qB} = \frac{\sqrt{2Km}}{qB} \]

4. **Setting Up the Equation for Proton and Alpha Particle**:
Let:
- \( K_1 = 1 \, \text{MeV} \) (kinetic energy of the proton),
- \( m_1 \) = mass of the proton,
- \( q_1 \) = charge of the proton,
- \( K_2 \) = kinetic energy of the alpha particle,
- \( m_2 = 4m_1 \) (mass of the alpha particle),
- \( q_2 = 2q_1 \) (charge of the alpha particle).

Since the radii are equal, we have:
\[ R_1 = R_2 \implies \frac{\sqrt{2K_1 m_1}}{q_1 B} = \frac{\sqrt{2K_2 m_2}}{q_2 B} \]

5. **Canceling Common Terms**:
Canceling \( B \) from both sides and squaring both sides gives:
\[ \frac{2K_1 m_1}{q_1^2} = \frac{2K_2 m_2}{q_2^2} \]
Simplifying further:
\[ \frac{K_1 m_1}{q_1^2} = \frac{K_2 (4m_1)}{(2q_1)^2} \]

6. **Substituting Known Values**:
Substituting \( m_2 = 4m_1 \) and \( q_2 = 2q_1 \):
\[ \frac{K_1}{q_1^2} = \frac{K_2 \cdot 4m_1}{4q_1^2} \]
This simplifies to:
\[ K_1 = K_2 \]

7. **Final Calculation**:
Since \( K_1 = 1 \, \text{MeV} \):
\[ K_2 = 1 \, \text{MeV} \]

### Conclusion:
The energy of the alpha particle required to describe a circle of the same radius in the same magnetic field is **1 MeV**.