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

Explanation

To solve the problem, we need to find the energy of an alpha particle that can revolve in the same circular path as a proton in a magnetic field. Let's break it down step by step.

### Step 1: Understand the relationship between radius, mass, charge, and energy
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.

### Step 2: Express kinetic energy in terms of mass and velocity
The kinetic energy (KE) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
From the radius formula, we can express \( v \) in terms of \( r \), \( m \), \( q \), and \( B \):
\[ v = \frac{qBr}{m} \]
Substituting this expression for \( v \) into the kinetic energy formula gives:
\[ KE = \frac{1}{2} m \left(\frac{qBr}{m}\right)^2 = \frac{q^2B^2r^2}{2m} \]

### Step 3: Calculate the kinetic energy for the proton
For the proton:
- Mass = \( m \)
- Charge = \( +e \)
- Energy = \( 1 \text{ MeV} = 1 \times 10^6 \text{ eV} \)

Thus, the kinetic energy of the proton can be expressed as:
\[ KE_p = \frac{e^2B^2r^2}{2m} \]

### Step 4: Calculate the kinetic energy for the alpha particle
For the alpha particle:
- Mass = \( 4m \)
- Charge = \( +2e \)

The kinetic energy of the alpha particle can be expressed as:
\[ KE_\alpha = \frac{(2e)^2B^2r^2}{2(4m)} = \frac{4e^2B^2r^2}{8m} = \frac{e^2B^2r^2}{2m} \]

### Step 5: Set the kinetic energies equal
Since both particles revolve in the same radius, we can set their kinetic energies equal to each other:
\[ KE_p = KE_\alpha \]
Substituting the expressions we derived:
\[ \frac{e^2B^2r^2}{2m} = \frac{e^2B^2r^2}{2(4m)} \]

### Step 6: Solve for the energy of the alpha particle
From the equality of kinetic energies, we see that:
\[ KE_\alpha = 4 \times KE_p \]
Given that \( KE_p = 1 \text{ MeV} \):
\[ KE_\alpha = 4 \times 1 \text{ MeV} = 4 \text{ MeV} \]

### Final Answer
The energy of the alpha particle should be **4 MeV**.

---