Four charged particles are projected perpendicularly into the magnetic field with equal velocity. Which will have minimum frequency of rotation?

Options

A. Proton
B. Electron
C. `Li^(+)`
D. `He^(+0`

Explanation

To determine which of the four charged particles will have the minimum frequency of rotation when projected perpendicularly into a magnetic field with equal velocity, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion of Charged Particles in a Magnetic Field**:
When a charged particle enters a magnetic field perpendicularly, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path.

2. **Centripetal Force and Magnetic Force Relationship**:
The centripetal force required to keep the particle in circular motion is given by:
\[ F_c = \frac{mv^2}{R} \]
where \( m \) is the mass of the particle, \( v \) is its velocity, and \( R \) is the radius of the circular path.

3. **Magnetic Force on a Charged Particle**:
The magnetic force acting on the charged particle is given by:
\[ F_m = QvB \]
where \( Q \) is the charge of the particle, \( v \) is its velocity, and \( B \) is the magnetic field strength.

4. **Equating Centripetal Force and Magnetic Force**:
Setting the centripetal force equal to the magnetic force, we have:
\[ \frac{mv^2}{R} = QvB \]

5. **Solving for the Radius \( R \)**:
Rearranging the equation gives:
\[ R = \frac{mv}{QB} \]

6. **Finding the Time Period \( T \)**:
The time period \( T \) for one complete revolution is given by:
\[ T = \frac{2\pi R}{v} \]
Substituting the expression for \( R \):
\[ T = \frac{2\pi \left(\frac{mv}{QB}\right)}{v} = \frac{2\pi m}{QB} \]

7. **Calculating the Frequency \( f \)**:
The frequency \( f \) is the reciprocal of the time period:
\[ f = \frac{1}{T} = \frac{QB}{2\pi m} \]

8. **Analyzing the Frequency**:
From the equation \( f = \frac{QB}{2\pi m} \), we can see that the frequency \( f \) is inversely proportional to the mass \( m \) of the particle. This means that as the mass increases, the frequency decreases.

9. **Comparing the Masses of the Particles**:
Given the masses of the particles:
- Proton: \( 1.67 \times 10^{-27} \, \text{kg} \)
- Electron: \( 9.1 \times 10^{-31} \, \text{kg} \)
- Lithium ion (\( \text{Li}^+ \)): \( 1.15 \times 10^{-26} \, \text{kg} \)
- Helium ion (\( \text{He}^+ \)): \( 6.64 \times 10^{-27} \, \text{kg} \)

The lithium ion has the largest mass.

10. **Conclusion**:
Since frequency is inversely proportional to mass, the particle with the largest mass (lithium ion) will have the minimum frequency of rotation.

### Final Answer:
The charged particle with the minimum frequency of rotation is the lithium ion (\( \text{Li}^+ \)).