To solve the problem step by step, we will follow the outlined method in the video transcript while ensuring clarity and precision in each step.
### Step 1: Understand the Problem
We have a solenoid with 20 turns per cm and a current of 2.5 A. Radiation causes electrons to be emitted from a target, and these electrons move in a circular path with a maximum radius of 1 cm. We need to find the wavelength of the radiation, given the work function of the target is 0.5 eV.
### Step 2: Calculate the Magnetic Field (B)
The magnetic field inside a solenoid is given by the formula:
\[ B = \mu_0 n I \]
where:
- \( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space),
- \( n = 20 \, \text{turns/cm} = 2000 \, \text{turns/m} \) (converting to turns per meter),
- \( I = 2.5 \, \text{A} \).
Now substituting the values:
\[
B = (4 \pi \times 10^{-7}) \times (2000) \times (2.5)
\]
Calculating this gives:
\[
B = 4 \pi \times 10^{-7} \times 5000 \approx 6.2832 \times 10^{-3} \, \text{T} \approx 0.0063 \, \text{T}
\]
### Step 3: Calculate the Velocity (V) of the Electrons
The centripetal force acting on the electrons moving in a circular path is provided by the magnetic Lorentz force:
\[
qVB = \frac{mv^2}{r}
\]
Rearranging for \( v \):
\[
v = \frac{qBr}{m}
\]
Where:
- \( q = 1.6 \times 10^{-19} \, \text{C} \) (charge of the electron),
- \( r = 1 \, \text{cm} = 0.01 \, \text{m} \),
- \( m = 9.1 \times 10^{-31} \, \text{kg} \).
Substituting the values:
\[
v = \frac{(1.6 \times 10^{-19}) \times (0.0063) \times (0.01)}{9.1 \times 10^{-31}}
\]
Calculating this gives:
\[
v \approx 1.1 \times 10^{11} \, \text{m/s}
\]
### Step 4: Calculate the Kinetic Energy (KE) of the Electrons
The kinetic energy of the electrons can be calculated using:
\[
KE = \frac{1}{2} mv^2
\]
Substituting for \( v \):
\[
KE = \frac{1}{2} m \left( \frac{qBr}{m} \right)^2 = \frac{q^2 B^2 r^2}{2m}
\]
Substituting the values:
\[
KE = \frac{(1.6 \times 10^{-19})^2 (0.0063)^2 (0.01)^2}{2 \times (9.1 \times 10^{-31})}
\]
Calculating this gives:
\[
KE \approx 2.5 \times 10^{-18} \, \text{J}
\]
### Step 5: Convert Kinetic Energy to Electron Volts
1 eV = \( 1.6 \times 10^{-19} \, \text{J} \), so:
\[
KE \text{ (in eV)} = \frac{2.5 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 15.625 \, \text{eV}
\]
### Step 6: Use the Photoelectric Equation
The energy of the incident radiation is given by:
\[
E = KE + \text{Work Function}
\]
Substituting the values:
\[
E = 15.625 + 0.5 = 16.125 \, \text{eV}
\]
### Step 7: Calculate the Wavelength (\( \lambda \))
Using the formula for energy in terms of wavelength:
\[
E = \frac{hc}{\lambda}
\]
Rearranging gives:
\[
\lambda = \frac{hc}{E}
\]
Substituting:
- \( h = 6.625 \times 10^{-34} \, \text{J s} \),
- \( c = 3 \times 10^8 \, \text{m/s} \),
- \( E = 16.125 \, \text{eV} = 16.125 \times 1.6 \times 10^{-19} \, \text{J} \).
Calculating:
\[
\lambda = \frac{(6.625 \times 10^{-34})(3 \times 10^8)}{16.125 \times 1.6 \times 10^{-19}}
\]
This will yield:
\[
\lambda \approx 1.24 \times 10^{-7} \, \text{m} = 1240 \, \text{nm}
\]
### Final Answer
The wavelength of the radiation is approximately \( 1240 \, \text{nm} \).
