An electron is moving in a cyclotron at a speed of `3.2 xx 10^(7) m s^(-1)` in a magnetic field of `5 xx 10^(-4) T` perpendicular to it. What is the frequency of this electron? `(q=1.6xx10^(-19)C, m_(c)=9.1xx10^(-31)kg)`

Options

A. `1.4xx10^(5)Hz`
B. `1.4xx10^(7)Hz`
C. `1.4xx10^(6)Hz`
D. `1.4xx10^(9)Hz`

Explanation

To find the frequency of an electron moving in a cyclotron, we can use the relationship between the magnetic force and the centripetal force. Here’s a step-by-step solution:

### Step 1: Write down the formula for frequency
The frequency \( f \) of a charged particle moving in a magnetic field is given by the formula:
\[ f = \frac{qB}{2\pi m} \]
where:
- \( q \) is the charge of the electron,
- \( B \) is the magnetic field strength,
- \( m \) is the mass of the electron.

### Step 2: Substitute the known values
We have the following values:
- Charge of the electron \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Magnetic field \( B = 5 \times 10^{-4} \, \text{T} \)
- Mass of the electron \( m = 9.1 \times 10^{-31} \, \text{kg} \)

Substituting these values into the frequency formula:
\[ f = \frac{(1.6 \times 10^{-19} \, \text{C})(5 \times 10^{-4} \, \text{T})}{2\pi (9.1 \times 10^{-31} \, \text{kg})} \]

### Step 3: Calculate the numerator
Calculating the numerator:
\[ 1.6 \times 10^{-19} \times 5 \times 10^{-4} = 8.0 \times 10^{-23} \]

### Step 4: Calculate the denominator
Calculating the denominator:
\[ 2\pi \times 9.1 \times 10^{-31} \approx 2 \times 3.14 \times 9.1 \times 10^{-31} \approx 5.65 \times 10^{-30} \]

### Step 5: Divide the numerator by the denominator
Now, substituting the calculated values into the frequency formula:
\[ f = \frac{8.0 \times 10^{-23}}{5.65 \times 10^{-30}} \approx 1.42 \times 10^{7} \, \text{Hz} \]

### Step 6: Round off the answer
Rounding off the answer gives:
\[ f \approx 1.4 \times 10^{7} \, \text{Hz} \]

### Final Answer
The frequency of the electron is approximately:
\[ f \approx 1.4 \times 10^{7} \, \text{Hz} \]

---