A proton of mass m and charge `+e` is moving in a circular orbit in a magnetic field with energy `1 MeV`. What should be the energy of alpha-particle (mass=`4m` and charge=`+2e`), so that it can revolve in the path of same radius?

Options

A. 1 Me V
B. 4 Me V
C. 2 Me V
D. 0.5 Me V

Answer

To solve the problem, we need to find the energy of an alpha particle that allows it to revolve in the same circular path as a proton in a 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,
- \( q \) is the charge of the particle,
- \( B \) is the magnetic field strength.

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

3. **Substituting Velocity into the Radius Formula**:
Substituting the expression for \( v \) into the radius formula, we get:
\[ r = \frac{m \sqrt{\frac{2K}{m}}}{qB} = \frac{\sqrt{2Km}}{qB} \]
Rearranging gives:
\[ K = \frac{q^2 B^2 r^2}{2m} \]

4. **Setting Up the Proportionality**:
For the proton:
- Mass \( m_p = m \)
- Charge \( q_p = e \)
- Energy \( K_p = 1 \text{ MeV} \)

For the alpha particle:
- Mass \( m_{\alpha} = 4m \)
- Charge \( q_{\alpha} = 2e \)

5. **Finding the Energy of the Alpha Particle**:
The energy of the alpha particle \( K_{\alpha} \) can be expressed in terms of the proton's energy:
\[ K_{\alpha} = \frac{(q_{\alpha})^2 B^2 r^2}{2m_{\alpha}} = \frac{(2e)^2 B^2 r^2}{2(4m)} = \frac{4e^2 B^2 r^2}{8m} = \frac{e^2 B^2 r^2}{2m} \]
Notice that this is the same expression as for the proton:
\[ K_p = \frac{(e)^2 B^2 r^2}{2m} \]

6. **Conclusion**:
Since both expressions for \( K_{\alpha} \) and \( K_p \) are equal, we conclude:
\[ K_{\alpha} = K_p = 1 \text{ MeV} \]

Thus, the energy of the alpha particle should also be **1 MeV** to revolve in the same path as the proton.

### Final Answer:
The energy of the alpha particle should be **1 MeV**.