A proton of energy `8 eV` is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be

Options

A. (a) 4eV
B. (b) 22eV
C. (c) 8eV
D. (d) 6eV

Answer

To solve the problem, we need to analyze the motion of the proton and the alpha particle in a uniform magnetic field. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Motion of Charged Particles in a Magnetic Field
When a charged particle moves in a magnetic field, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path. The centripetal force \( F_c \) can be expressed as:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the particle, \( v \) is its velocity, and \( r \) is the radius of the circular path.

### Step 2: Relate the Centripetal Force to the Magnetic Force
The magnetic force \( F_m \) acting on a charged particle moving with velocity \( v \) in a magnetic field \( B \) is given by:
\[ F_m = qvB \]
where \( q \) is the charge of the particle.

### Step 3: Set the Forces Equal
For a charged particle moving in a circular path, the centripetal force is provided by the magnetic force:
\[ \frac{mv^2}{r} = qvB \]
From this, we can derive the expression for the radius \( r \):
\[ r = \frac{mv}{qB} \]

### Step 4: Write the Radius for Proton and Alpha Particle
Let’s denote:
- For the proton: mass = \( m \), charge = \( e \), velocity = \( v_p \), radius = \( r_p \)
- For the alpha particle: mass = \( 4m \) (since an alpha particle has 2 protons and 2 neutrons), charge = \( 2e \), velocity = \( v_\alpha \), radius = \( r_\alpha \)

Using the radius formula:
\[ r_p = \frac{mv_p}{eB} \]
\[ r_\alpha = \frac{4m v_\alpha}{2eB} = \frac{2mv_\alpha}{eB} \]

### Step 5: Set the Radii Equal
Since both particles are moving in the same circular path:
\[ r_p = r_\alpha \]
This gives us:
\[ \frac{mv_p}{eB} = \frac{2mv_\alpha}{eB} \]
Cancelling out common terms:
\[ v_p = 2v_\alpha \]

### Step 6: Calculate the Kinetic Energies
The kinetic energy \( KE \) of a particle is given by:
\[ KE = \frac{1}{2}mv^2 \]
- For the proton:
\[ KE_p = \frac{1}{2}m v_p^2 \]
- For the alpha particle:
\[ KE_\alpha = \frac{1}{2}(4m)v_\alpha^2 = 2mv_\alpha^2 \]

### Step 7: Substitute \( v_p \) in Terms of \( v_\alpha \)
Substituting \( v_p = 2v_\alpha \) into the kinetic energy of the proton:
\[ KE_p = \frac{1}{2}m(2v_\alpha)^2 = \frac{1}{2}m(4v_\alpha^2) = 2mv_\alpha^2 \]

### Step 8: Compare the Kinetic Energies
Now we can compare the kinetic energies:
\[ KE_p = 2mv_\alpha^2 \]
\[ KE_\alpha = 2mv_\alpha^2 \]
Thus, we find that:
\[ KE_p = KE_\alpha \]

### Conclusion
Since the kinetic energy of the proton is equal to the kinetic energy of the alpha particle, the energy of the alpha particle moving in the same magnetic field and along the same path is also \( 8 \, \text{eV} \).

### Final Answer
The energy of the alpha particle is **8 eV**.

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