A proton and an alpha particle both enters a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is `1 MeV`, the energy acquired by the alpha particles will be :

Options

A. 4 MeV
B. `0.5` MeV
C. `1.5` MeV
D. 1 MeV

Answer

To solve the problem, we need to find the energy acquired by the alpha particle when both a proton and an alpha particle enter a magnetic field with equal radii of their circular paths. The kinetic energy of the proton is given as 1 MeV.

### Step-by-Step Solution:

1. **Identify the Charges and Masses**:
- The charge of a proton (\(Q_p\)) is \(e\).
- The charge of an alpha particle (\(Q_{\alpha}\)) is \(2e\) (since it consists of 2 protons).
- The mass of a proton (\(m_p\)) is \(m\).
- The mass of an alpha particle (\(m_{\alpha}\)) is \(4m\) (since it consists of 2 protons and 2 neutrons).

2. **Use the Formula for Radius of Circular Motion**:
The radius \(R\) of the circular path in a magnetic field is given by:
\[ R = \frac{mv}{qB} \]
where \(m\) is the mass of the particle, \(v\) is its velocity, \(q\) is its charge, and \(B\) is the magnetic field strength.

3. **Set Up the Equations for Proton and Alpha Particle**:
For the proton:
\[ R = \frac{m_p v_p}{Q_p B} \]
For the alpha particle:
\[ R = \frac{m_{\alpha} v_{\alpha}}{Q_{\alpha} B} \]

4. **Equate the Radii**:
Since the radii are equal, we can set the two equations equal to each other:
\[ \frac{m_p v_p}{Q_p B} = \frac{m_{\alpha} v_{\alpha}}{Q_{\alpha} B} \]

5. **Substituting the Values**:
Substitute \(m_{\alpha} = 4m_p\) and \(Q_{\alpha} = 2Q_p\):
\[ \frac{m_p v_p}{Q_p} = \frac{4m_p v_{\alpha}}{2Q_p} \]
Simplifying gives:
\[ v_p = 2v_{\alpha} \]

6. **Relate Kinetic Energies**:
The kinetic energy \(K\) is given by:
\[ K = \frac{1}{2} mv^2 \]
For the proton:
\[ K_p = \frac{1}{2} m_p v_p^2 \]
For the alpha particle:
\[ K_{\alpha} = \frac{1}{2} m_{\alpha} v_{\alpha}^2 = \frac{1}{2} (4m_p) v_{\alpha}^2 \]

7. **Substituting \(v_p\) in Terms of \(v_{\alpha}\)**:
Since \(v_p = 2v_{\alpha}\):
\[ K_p = \frac{1}{2} m_p (2v_{\alpha})^2 = \frac{1}{2} m_p (4v_{\alpha}^2) = 2m_p v_{\alpha}^2 \]

8. **Finding \(K_{\alpha}\)**:
Substitute \(v_{\alpha}\) into \(K_{\alpha}\):
\[ K_{\alpha} = \frac{1}{2} (4m_p) v_{\alpha}^2 = 2m_p v_{\alpha}^2 \]

9. **Relate the Energies**:
Since \(K_p = 1 \text{ MeV}\):
\[ K_{\alpha} = 4K_p \]
Therefore:
\[ K_{\alpha} = 4 \times 1 \text{ MeV} = 4 \text{ MeV} \]

### Final Answer:
The energy acquired by the alpha particle is **4 MeV**.