To solve the problem of finding the frequency and energy of an electron moving in a magnetic field, we can follow these steps:
### Step 1: Understand the motion of the electron in the magnetic field
When a charged particle, such as an electron, enters a magnetic field perpendicularly, it moves in a circular path due to the magnetic force acting as the centripetal force.
### Step 2: Use the formula for the frequency of revolution
The frequency of revolution (ν) of a charged particle in a magnetic field is given by the formula:
\[
\nu = \frac{Bq}{2\pi m}
\]
where:
- \(B\) is the magnetic field strength (in Tesla),
- \(q\) is the charge of the electron (in Coulombs),
- \(m\) is the mass of the electron (in kg).
### Step 3: Substitute the known values into the frequency formula
Given:
- \(B = 6 \times 10^{-4} \, \text{T}\)
- Charge of the electron, \(q = 1.6 \times 10^{-19} \, \text{C}\)
- Mass of the electron, \(m = 9.1 \times 10^{-31} \, \text{kg}\)
Substituting these values into the formula:
\[
\nu = \frac{(6 \times 10^{-4}) \times (1.6 \times 10^{-19})}{2\pi (9.1 \times 10^{-31})}
\]
### Step 4: Calculate the frequency
Calculating the numerator:
\[
6 \times 10^{-4} \times 1.6 \times 10^{-19} = 9.6 \times 10^{-23}
\]
Calculating the denominator:
\[
2\pi \times 9.1 \times 10^{-31} \approx 5.73 \times 10^{-30}
\]
Now, calculating the frequency:
\[
\nu = \frac{9.6 \times 10^{-23}}{5.73 \times 10^{-30}} \approx 1.67 \times 10^{7} \, \text{Hz} \approx 16.7 \, \text{MHz}
\]
### Step 5: Calculate the kinetic energy of the electron
The kinetic energy (KE) of the electron is given by the formula:
\[
KE = \frac{1}{2} mv^2
\]
Substituting the known values:
\[
KE = \frac{1}{2} \times (9.1 \times 10^{-31}) \times (3 \times 10^{7})^2
\]
Calculating \(v^2\):
\[
(3 \times 10^{7})^2 = 9 \times 10^{14}
\]
Now substituting back:
\[
KE = \frac{1}{2} \times (9.1 \times 10^{-31}) \times (9 \times 10^{14}) = 4.095 \times 10^{-16} \, \text{J}
\]
### Step 6: Convert the kinetic energy to electron volts
To convert Joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[
KE \text{ (in eV)} = \frac{4.095 \times 10^{-16}}{1.6 \times 10^{-19}} \approx 2.56 \times 10^{3} \, \text{eV} \approx 2.56 \, \text{keV}
\]
### Final Answers
- Frequency: Approximately \(20 \, \text{MHz}\)
- Energy: Approximately \(2.5 \, \text{keV}\)
