To solve the problem step by step, we will follow the outlined procedure in the video transcript.
### Step 1: Calculate the Maximum Velocity of the Photoelectron
The formula to find the velocity \( v \) of the photoelectron in a magnetic field is given by:
\[
v = \frac{qBr}{m}
\]
Where:
- \( q \) = charge of the electron = \( 1.6 \times 10^{-19} \, \text{C} \)
- \( B \) = magnetic flux density = \( 5.26 \times 10^{-6} \, \text{T} \)
- \( r \) = radius of the circular path = \( 0.50 \, \text{m} \)
- \( m \) = mass of the electron = \( 9.1 \times 10^{-31} \, \text{kg} \)
Substituting the values into the formula:
\[
v = \frac{(1.6 \times 10^{-19} \, \text{C})(5.26 \times 10^{-6} \, \text{T})(0.50 \, \text{m})}{9.1 \times 10^{-31} \, \text{kg}}
\]
Calculating the numerator:
\[
(1.6 \times 10^{-19})(5.26 \times 10^{-6})(0.50) = 4.208 \times 10^{-25}
\]
Now calculating the velocity:
\[
v = \frac{4.208 \times 10^{-25}}{9.1 \times 10^{-31}} \approx 0.462 \times 10^{6} \, \text{m/s} \approx 0.46 \times 10^{6} \, \text{m/s}
\]
### Step 2: Calculate the Kinetic Energy of the Photoelectron
The kinetic energy \( K \) of the photoelectron can be calculated using the formula:
\[
K = \frac{1}{2} mv^2
\]
Substituting the values:
\[
K = \frac{1}{2} (9.1 \times 10^{-31} \, \text{kg}) (0.46 \times 10^{6} \, \text{m/s})^2
\]
Calculating \( v^2 \):
\[
(0.46 \times 10^{6})^2 = 0.2116 \times 10^{12} \, \text{m}^2/\text{s}^2
\]
Now substituting back into the kinetic energy formula:
\[
K = \frac{1}{2} (9.1 \times 10^{-31}) (0.2116 \times 10^{12}) \approx 9.63 \times 10^{-19} \, \text{J}
\]
To convert this to electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[
K \approx \frac{9.63 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 0.602 \, \text{eV} \approx 0.6 \, \text{eV}
\]
### Step 3: Calculate the Stopping Potential
The stopping potential \( V_0 \) can be calculated using the relationship:
\[
eV_0 = K
\]
Where \( e \) is the charge of the electron. Therefore,
\[
V_0 = \frac{K}{e} = 0.6 \, \text{V}
\]
### Step 4: Calculate the Work Function
The work function \( \phi \) can be calculated using the equation:
\[
\phi = E - K
\]
Where \( E \) is the energy of the incident photons. The energy of the photons can be calculated using:
\[
E = \frac{hc}{\lambda}
\]
Where:
- \( h = 6.63 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^{8} \, \text{m/s} \) (speed of light)
- \( \lambda = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} \)
Calculating \( E \):
\[
E = \frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{4000 \times 10^{-10}} \approx 4.97 \times 10^{-19} \, \text{J}
\]
Converting this to electron volts:
\[
E \approx \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.106 \, \text{eV}
\]
Now calculating the work function:
\[
\phi = E - K \approx 3.106 \, \text{eV} - 0.6 \, \text{eV} \approx 2.506 \, \text{eV} \approx 2.5 \, \text{eV}
\]
### Summary of Results
1. Maximum Velocity of Photoelectron: \( 0.46 \times 10^{6} \, \text{m/s} \)
2. Kinetic Energy: \( 0.6 \, \text{eV} \)
3. Stopping Potential: \( 0.6 \, \text{V} \)
4. Work Function: \( 2.5 \, \text{eV} \)
