Which of the follwing particles will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field?

Options

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

Answer

To solve the problem of determining which particle will have the minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field, we can follow these steps:

### Step 1: Understand the motion of charged particles in a magnetic field
When a charged particle moves perpendicular to a magnetic field, it experiences a magnetic force that causes it to move in a circular path. The magnetic force acting on the particle is given by:
\[ F = qvB \]
where \( q \) is the charge of the particle, \( v \) is its velocity, and \( B \) is the magnetic field strength.

### Step 2: Relate magnetic force to centripetal force
The magnetic force acts as the centripetal force required for circular motion. Therefore, we can equate the magnetic force to the centripetal force:
\[ qvB = \frac{mv^2}{r} \]
where \( m \) is the mass of the particle and \( r \) is the radius of the circular path.

### Step 3: Solve for the radius of the circular path
Rearranging the equation gives us the radius \( r \):
\[ r = \frac{mv}{qB} \]

### Step 4: Determine the time period of revolution
The time period \( T \) for one complete revolution can be calculated as the circumference of the circular path divided by the speed:
\[ T = \frac{2\pi r}{v} \]
Substituting the expression for \( r \):
\[ T = \frac{2\pi \left(\frac{mv}{qB}\right)}{v} = \frac{2\pi m}{qB} \]

### Step 5: Calculate the frequency of revolution
Frequency \( f \) is the reciprocal of the time period:
\[ f = \frac{1}{T} = \frac{qB}{2\pi m} \]

### Step 6: Analyze the frequency expression
From the frequency expression, we see that:
\[ f \propto \frac{q}{m} \]
This means that the frequency of revolution is directly proportional to the charge-to-mass ratio \( \frac{q}{m} \).

### Step 7: Compare the charge-to-mass ratios of the given particles
To find the particle with the minimum frequency, we need to find the particle with the minimum charge-to-mass ratio. The particles given are:
- Electron: \( q_e \) and \( m_e \)
- Proton: \( q_p \) and \( m_p \)
- Helium ion (He\(^+\)): \( q_{He} = 2q_p \) and \( m_{He} = 4m_p \)
- Lithium ion (Li\(^+\)): \( q_{Li} = q_p \) and \( m_{Li} = 6m_p \)

Calculating the charge-to-mass ratios:
1. For the electron: \( \frac{q_e}{m_e} \)
2. For the proton: \( \frac{q_p}{m_p} \)
3. For the helium ion: \( \frac{2q_p}{4m_p} = \frac{q_p}{2m_p} \)
4. For the lithium ion: \( \frac{q_p}{6m_p} \)

### Step 8: Determine the minimum charge-to-mass ratio
From the ratios:
- \( \frac{q_e}{m_e} \) (highest)
- \( \frac{q_p}{m_p} \) (lower than electron)
- \( \frac{q_p}{2m_p} \) (lower than proton)
- \( \frac{q_p}{6m_p} \) (lowest)

### Conclusion
The lithium ion (Li\(^+\)) has the minimum charge-to-mass ratio, and thus will have the minimum frequency of revolution when projected with the same velocity perpendicular to the magnetic field.

### Final Answer
The particle with the minimum frequency of revolution is the lithium ion (Li\(^+\)).

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