An electron having mass 'm' and kinetic energy E enter in uniform magnetic field B perpendicularly, then its frequency will be : -

Options

A. `(eE)/(qVB)`
B. `(2pim)/(eB)`
C. `(eB)/(2pim)`
D. `(2m)/(eBE)`

Answer

To solve the problem, we need to determine the frequency of an electron moving in a magnetic field when it enters perpendicularly. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Motion in a Magnetic Field:**
When a charged particle, like an electron, enters a magnetic field perpendicularly, it experiences a magnetic force that causes it to move in a circular path. The magnetic force acts as the centripetal force required for circular motion.

2. **Magnetic Force and Centripetal Force:**
The magnetic force \( F \) on the electron is given by:
\[ F = q v B \]
where \( q \) is the charge of the electron, \( v \) is its velocity, and \( B \) is the magnetic field strength.

The centripetal force required to keep the electron in circular motion is given by:
\[ F = \frac{m v^2}{r} \]
where \( m \) is the mass of the electron and \( r \) is the radius of the circular path.

3. **Equating the Forces:**
Setting the magnetic force equal to the centripetal force gives:
\[ q v B = \frac{m v^2}{r} \]

4. **Finding the Radius of Circular Path:**
Rearranging the equation to solve for the radius \( r \):
\[ r = \frac{m v}{q B} \]

5. **Relating Velocity to Kinetic Energy:**
The kinetic energy \( E \) of the electron is given by:
\[ E = \frac{1}{2} m v^2 \]
From this, we can express \( v \) in terms of \( E \):
\[ v = \sqrt{\frac{2E}{m}} \]

6. **Substituting Velocity into Radius:**
Substitute \( v \) back into the equation for \( r \):
\[ r = \frac{m \sqrt{\frac{2E}{m}}}{q B} = \frac{\sqrt{2Em}}{q B} \]

7. **Finding the Time Period of Revolution:**
The time period \( T \) of the electron's revolution is given by:
\[ T = \frac{2 \pi r}{v} \]
Substituting for \( r \) and \( v \):
\[ T = \frac{2 \pi \left(\frac{\sqrt{2Em}}{q B}\right)}{\sqrt{\frac{2E}{m}}} \]
Simplifying this gives:
\[ T = \frac{2 \pi m}{q B} \]

8. **Finding the Frequency:**
The frequency \( f \) is the reciprocal of the time period:
\[ f = \frac{1}{T} = \frac{q B}{2 \pi m} \]

9. **Substituting Charge of Electron:**
For an electron, the charge \( q \) is \( e \):
\[ f = \frac{e B}{2 \pi m} \]

### Final Answer:
The frequency of the electron moving in the magnetic field is:
\[ f = \frac{e B}{2 \pi m} \]