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. `1MeV`
B. `4MeV`
C. `0*5MeV`
D. `1*5MeV`

Explanation

To solve the problem, we need to find the kinetic energy acquired by the alpha particle when both the proton and alpha particle have the same radius of circular motion in a uniform magnetic field.

### Step-by-Step Solution:

1. **Understanding the Motion in a Magnetic Field**:
When a charged particle moves in a magnetic field at right angles, it experiences a magnetic force that causes it to move in a circular path. The radius \( r \) of the circular path is given by the formula:
\[ r = \frac{mv}{Bq} \]
where:
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle,
- \( B \) is the magnetic field strength,
- \( q \) is the charge of the particle.

2. **Finding the Velocity**:
Rearranging the formula for radius, we can express the velocity \( v \):
\[ v = \frac{Bqr}{m} \]

3. **Kinetic Energy Expression**:
The kinetic energy \( KE \) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the expression for \( v \) from step 2 into the kinetic energy formula:
\[ KE = \frac{1}{2} m \left(\frac{Bqr}{m}\right)^2 = \frac{B^2 q^2 r^2}{2m} \]

4. **Comparing Kinetic Energies**:
Since both particles (proton and alpha particle) have the same radius \( r \) and are in the same magnetic field \( B \), we can set up a ratio of their kinetic energies:
\[ \frac{KE_{\text{alpha}}}{KE_{\text{proton}}} = \frac{q_{\text{alpha}}^2 / m_{\text{alpha}}}{q_{\text{proton}}^2 / m_{\text{proton}}} \]

5. **Substituting Values**:
- For a proton, \( q_{\text{proton}} = e \) and \( m_{\text{proton}} = m_p \).
- For an alpha particle, \( q_{\text{alpha}} = 2e \) (since it has 2 protons) and \( m_{\text{alpha}} = 4m_p \) (since it has 2 protons and 2 neutrons).

Thus, we have:
\[ \frac{KE_{\text{alpha}}}{KE_{\text{proton}}} = \frac{(2e)^2 / (4m_p)}{e^2 / m_p} = \frac{4e^2 / 4m_p}{e^2 / m_p} = 1 \]

6. **Calculating Kinetic Energy of Alpha Particle**:
Given that the kinetic energy of the proton is \( KE_{\text{proton}} = 1 \text{ MeV} \), we find:
\[ KE_{\text{alpha}} = KE_{\text{proton}} = 1 \text{ MeV} \]

### Final Answer:
The kinetic energy acquired by the alpha particle is **1 MeV**.