An electron is moving in a circular path under the influence of a transverse magnetic field of `3.57xx10^(-2)` T.If the value of e/m is `1.76xx10^11` c/kg, the frequency of revolution of the electron is

Options

A. 6.82 MHz
B. 1 GHz
C. 100 MHz `
D. 62.8 MHz

Explanation

To find the frequency of revolution of an electron moving in a circular path under the influence of a magnetic field, we can follow these steps:

### Step 1: Understand the relationship between force and motion
The magnetic force acting on the electron provides the centripetal force necessary for circular motion. The magnetic force is given by:
\[ F = qvB \]
where:
- \( F \) is the magnetic force,
- \( q \) is the charge of the electron,
- \( v \) is the velocity of the electron,
- \( B \) is the magnetic field strength.

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

### Step 2: Set the forces equal
Since the magnetic force provides the centripetal force, we can set the two equations equal to each other:
\[ qvB = \frac{mv^2}{r} \]

### Step 3: Solve for velocity \( v \)
Rearranging the equation to solve for \( v \):
\[ qvB = \frac{mv^2}{r} \implies v = \frac{qBr}{m} \]

### Step 4: Relate frequency to velocity
The frequency \( f \) of revolution is defined as the number of revolutions per second. The relationship between frequency and the time period \( T \) is:
\[ f = \frac{1}{T} \]
The time period \( T \) for one complete revolution is given by the distance traveled in one revolution divided by the velocity:
\[ T = \frac{2\pi r}{v} \]

### Step 5: Substitute for \( v \)
Substituting \( v \) from Step 3 into the equation for \( T \):
\[ T = \frac{2\pi r}{\frac{qBr}{m}} = \frac{2\pi m}{qB} \]

### Step 6: Find frequency \( f \)
Now substituting \( T \) into the frequency equation:
\[ f = \frac{1}{T} = \frac{qB}{2\pi m} \]

### Step 7: Substitute known values
We know:
- \( q = e = 1.6 \times 10^{-19} \, \text{C} \) (charge of the electron),
- \( B = 3.57 \times 10^{-2} \, \text{T} \),
- \( \frac{e}{m} = 1.76 \times 10^{11} \, \text{C/kg} \).

From the relationship \( \frac{q}{m} = \frac{e}{m} \), we can substitute \( q \) as \( e \):
\[ f = \frac{(1.6 \times 10^{-19}) (3.57 \times 10^{-2})}{2\pi (1.76 \times 10^{11})} \]

### Step 8: Calculate the frequency
Calculating the values:
\[ f = \frac{(1.6 \times 10^{-19}) (3.57 \times 10^{-2})}{2\pi (1.76 \times 10^{11})} \]

Calculating the numerator:
\[ 1.6 \times 10^{-19} \times 3.57 \times 10^{-2} = 5.712 \times 10^{-21} \]

Calculating the denominator:
\[ 2\pi \times 1.76 \times 10^{11} \approx 1.107 \times 10^{12} \]

Now, substituting these values into the frequency equation:
\[ f \approx \frac{5.712 \times 10^{-21}}{1.107 \times 10^{12}} \approx 5.16 \times 10^{-33} \, \text{Hz} \]

### Final Answer
Thus, the frequency of revolution of the electron is approximately:
\[ f \approx 5.16 \times 10^{-33} \, \text{Hz} \]