To solve the problem, we need to follow these steps:
### Step 1: Calculate the Kinetic Energy of the Electrons
When electrons are accelerated from rest by applying a voltage \( V \), the kinetic energy \( KE \) gained by the electrons can be expressed as:
\[
KE = eV
\]
where:
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, C \))
- \( V \) is the voltage applied (500 V)
Substituting the values:
\[
KE = (1.6 \times 10^{-19} \, C) \times (500 \, V) = 8.0 \times 10^{-17} \, J
\]
### Step 2: Relate Kinetic Energy to Velocity
The kinetic energy can also be expressed in terms of the mass \( m \) and velocity \( v \) of the electron:
\[
KE = \frac{1}{2} mv^2
\]
Setting the two expressions for kinetic energy equal gives:
\[
eV = \frac{1}{2} mv^2
\]
Rearranging this to solve for \( v^2 \):
\[
v^2 = \frac{2eV}{m}
\]
### Step 3: Substitute Values to Find Velocity
Substituting the known values:
- \( e = 1.6 \times 10^{-19} \, C \)
- \( V = 500 \, V \)
- \( m = 9.1 \times 10^{-31} \, kg \)
Calculating \( v^2 \):
\[
v^2 = \frac{2 \times (1.6 \times 10^{-19}) \times (500)}{9.1 \times 10^{-31}}
\]
\[
v^2 = \frac{1.6 \times 10^{-16}}{9.1 \times 10^{-31}} \approx 1.75824176 \times 10^{14}
\]
Taking the square root to find \( v \):
\[
v \approx \sqrt{1.75824176 \times 10^{14}} \approx 1.327 \times 10^7 \, m/s
\]
### Step 4: Calculate the Radius of the Path in the Magnetic Field
The radius \( r \) of the circular path of the electron in a magnetic field \( B \) can be calculated using the formula:
\[
r = \frac{mv}{qB}
\]
where:
- \( q \) is the charge of the electron (\( 1.6 \times 10^{-19} \, C \))
- \( B \) is the magnetic field strength (100 mT = \( 100 \times 10^{-3} \, T = 0.1 \, T \))
Substituting the values:
\[
r = \frac{(9.1 \times 10^{-31} \, kg) \times (1.327 \times 10^7 \, m/s)}{(1.6 \times 10^{-19} \, C) \times (0.1 \, T)}
\]
Calculating the denominator:
\[
qB = (1.6 \times 10^{-19}) \times (0.1) = 1.6 \times 10^{-20}
\]
Now substituting back:
\[
r = \frac{(9.1 \times 10^{-31}) \times (1.327 \times 10^7)}{1.6 \times 10^{-20}}
\]
Calculating the numerator:
\[
9.1 \times 10^{-31} \times 1.327 \times 10^7 \approx 1.206 \times 10^{-23}
\]
Now calculating \( r \):
\[
r = \frac{1.206 \times 10^{-23}}{1.6 \times 10^{-20}} \approx 7.54 \times 10^{-4} \, m
\]
Thus, the radius of the path is approximately:
\[
r \approx 7.54 \times 10^{-4} \, m \, \text{or} \, 0.754 \, mm
\]
### Final Answer
The radius of the path of the electrons in the magnetic field is approximately \( 7.54 \times 10^{-4} \, m \).
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