The path of a charged particle in a uniform magnetic field depends on the angle `theta` between velocity vector and magnetic field, When `theta is 0^(@) or 180^(@), F_(m) = 0` hence path of a charged particle will be linear.
When `theta = 90^(@)`, the magnetic force is perpendicular to velocity at every instant. Hence path is a circle of radius `r = (mv)/(qB)`.
The time period for circular path will be `T = (2pim)/(qB)`
When `theta` is other than `0^(@), 180^(@) and 90^(@)`, velocity can be resolved into two components, one along `vec(B)` and perpendicular to B.
`v_(|/|)=cos theta`
`v_(^)= v sin theta`
The `v_(_|_)` component gives circular path and `v_(|/|)` givestraingt line path. The resultant path is a helical path. The radius of helical path
`r=(mv sin theta)/(qB)`
ich of helix is defined as `P=v_(|/|)T`
`P=(2 i mv cos theta)`
`p=(2 pi mv cos theta)/(qB)`
Which particle will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field.

Options

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

Explanation

To solve the problem regarding the path of a charged particle in a uniform magnetic field and to determine which particle will have the minimum frequency of revolution when projected with the same velocity perpendicular to the magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Magnetic Force**:
The magnetic force \( F_m \) acting on a charged particle is given by:
\[ F_m = q(\vec{v} \times \vec{B}) \]
where \( q \) is the charge of the particle, \( \vec{v} \) is its velocity, and \( \vec{B} \) is the magnetic field.

2. **Analyzing the Angle \( \theta \)**:
- When \( \theta = 0^\circ \) or \( \theta = 180^\circ \), the magnetic force is zero, and the particle moves in a straight line.
- When \( \theta = 90^\circ \), the force is perpendicular to the velocity, resulting in circular motion.

3. **Resolving Velocity into Components**:
For angles other than \( 0^\circ \), \( 180^\circ \), and \( 90^\circ \):
- The velocity can be resolved into two components:
\[ v_{\parallel} = v \cos \theta \quad \text{(along the magnetic field)} \]
\[ v_{\perp} = v \sin \theta \quad \text{(perpendicular to the magnetic field)} \]

4. **Determining the Path**:
- The component \( v_{\perp} \) causes circular motion, while \( v_{\parallel} \) results in linear motion. Thus, the resultant path is helical.

5. **Calculating the Radius of the Helical Path**:
The radius \( r \) of the helical path is given by:
\[ r = \frac{mv \sin \theta}{qB} \]

6. **Finding the Time Period of the Helical Motion**:
The time period \( T \) for one complete revolution in the circular motion is:
\[ T = \frac{2\pi m}{qB} \]

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

8. **Determining Minimum Frequency**:
Since the frequency is inversely proportional to the mass \( m \), a larger mass results in a smaller frequency. Therefore, to find the particle with the minimum frequency, we need to identify the particle with the maximum mass.

9. **Comparing Masses of Particles**:
Among the given options, we compare the masses of the particles. The particle with the largest mass will have the minimum frequency of revolution.

10. **Conclusion**:
If lithium ion \( \text{Li}^+ \) has the maximum mass among the options provided, then it will have the minimum frequency of revolution.

### Final Answer:
The particle with the minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field is **lithium ion \( \text{Li}^+ \)**.