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

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understanding the Relationship Between Radius, Momentum, Charge, and Magnetic Field**:
The radius \( R \) of the circular path 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 its velocity, \( q \) is its charge, and \( B \) is the magnetic field strength.

2. **Expressing Momentum in Terms of Kinetic Energy**:
The kinetic energy \( KE \) of a particle is related to its momentum \( p \) by:
\[ KE = \frac{p^2}{2m} \]
Therefore, we can express momentum as:
\[ p = \sqrt{2m \cdot KE} \]

3. **Substituting Momentum into the Radius Formula**:
We can substitute \( p \) into the radius formula:
\[ R = \frac{\sqrt{2m \cdot KE}}{qB} \]
Rearranging gives us:
\[ R \cdot qB = \sqrt{2m \cdot KE} \]

4. **Finding the Radius for the Proton**:
For the proton, we have:
\[ R = \frac{\sqrt{2m_p \cdot KE_p}}{q_p B} \]
where \( KE_p = 1 \, \text{MeV} = 10^6 \, \text{eV} \).

5. **Finding the Radius for the Alpha Particle**:
For the alpha particle, which has a mass \( m_{\alpha} = 4m_p \) and charge \( q_{\alpha} = 2q_p \), the radius is:
\[ R = \frac{\sqrt{2m_{\alpha} \cdot KE_{\alpha}}}{q_{\alpha} B} \]
Substituting \( m_{\alpha} \) and \( q_{\alpha} \):
\[ R = \frac{\sqrt{2(4m_p) \cdot KE_{\alpha}}}{(2q_p) B} \]

6. **Setting the Radii Equal**:
Since both particles have the same radius \( R \), we can set the two expressions for \( R \) equal to each other:
\[ \frac{\sqrt{2m_p \cdot 10^6}}{q_p B} = \frac{\sqrt{2(4m_p) \cdot KE_{\alpha}}}{(2q_p) B} \]

7. **Simplifying the Equation**:
Cancel \( B \) and \( q_p \) from both sides:
\[ \sqrt{2 \cdot 10^6} = \frac{\sqrt{8m_p \cdot KE_{\alpha}}}{2} \]
Squaring both sides:
\[ 2 \cdot 10^6 = \frac{8m_p \cdot KE_{\alpha}}{4} \]
Simplifying gives:
\[ 2 \cdot 10^6 = 2m_p \cdot KE_{\alpha} \]
Thus:
\[ KE_{\alpha} = 10^6 \, \text{eV} = 1 \, \text{MeV} \]

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