Iron has a work function of 4.50 eV.
(a) Find the cutoff wavelength and cutoff frequency for the photoelectric effect.
wavelength    	
 nm
frequency    	
 Hz

(b) What is the stopping potential if the incident light has a wavelength of 190 nm?
 V

To solve this problem, we'll start with the given work function and then calculate the cutoff wavelength, cutoff frequency, and stopping potential.

### (a) Cutoff Wavelength and Frequency

1. **Cutoff Wavelength (\( \lambda_0 \))**:

The cutoff wavelength can be calculated using the formula:

\[
\lambda_0 = \frac{hc}{\phi}
\]

where:
- \( h \) (Planck's constant) = \( 4.1357 \times 10^{-15} \, \text{eV s} \)
- \( c \) (speed of light) = \( 3.00 \times 10^8 \, \text{m/s} \)
- \( \phi \) (work function) = 4.50 eV

First, we need to convert \( h \) and \( c \) into compatible units. The energy in Joules can be converted as \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \).

Now, calculate \( hc \):

\[
hc = (4.1357 \times 10^{-15} \, \text{eV s})(3.00 \times 10^8 \, \text{m/s}) = 1.241 \times 10^{-6} \, \text{eV m}
\]

Now plug in the values:

\[
\lambda_0 = \frac{1.241 \times 10^{-6} \, \text{eV m}}{4.50 \, \text{eV}} \approx 2.76 \times 10^{-7} \, \text{m} = 276 \, \text{nm}
\]

2. **Cutoff Frequency (\( f_0 \))**:

The cutoff frequency can be calculated using:

\[
f_0 = \frac{\phi}{h}
\]

Convert the work function to Joules:

\[
\phi = 4.50 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} \approx 7.21 \times 10^{-19} \, \text{J}
\]

Now calculate \( f_0 \):

\[
f_0 = \frac{7.21 \times 10^{-19} \, \text{J}}{4.1357 \times 10^{-15} \, \text{eV s}} \approx 1.74 \times 10^{15} \, \text{Hz}
\]

### Summary of (a):
- **Cutoff Wavelength**: \( 276 \, \text{nm} \)
- **Cutoff Frequency**: \( 1.74 \times 10^{15} \, \text{Hz} \)

### (b) Stopping Potential for Wavelength of 190 nm

The stopping potential (\( V_s \)) can be calculated using the energy of the incident photons minus the work function:

1. **Energy of Incident Photons**:

Using the formula:

\[
E = \frac{hc}{\lambda}
\]

For \( \lambda = 190 \, \text{nm} = 190 \times 10^{-9} \, \text{m} \):

\[
E = \frac{(4.1357 \times 10^{-15} \, \text{eV s})(3.00 \times 10^8 \, \text{m/s})}{190 \times 10^{-9} \, \text{m}} \approx 6.52 \, \text{eV}
\]

2. **Stopping Potential**:

The stopping potential is given by:

\[
V_s = E - \phi
\]

\[
V_s = 6.52 \, \text{eV} - 4.50 \, \text{eV} \approx 2.02 \, \text{V}
\]

### Summary of (b):
- **Stopping Potential**: \( 2.02 \, \text{V} \)

### Final Results:
- **Cutoff Wavelength**: \( 276 \, \text{nm} \)
- **Cutoff Frequency**: \( 1.74 \times 10^{15} \, \text{Hz} \)
- **Stopping Potential**: \( 2.02 \, \text{V} \)

Do you want details on any part of this calculation or have any questions? Here are some related questions you might consider:

1. How does the photoelectric effect depend on the intensity of light?
2. What happens if the wavelength of light is longer than the cutoff wavelength?
3. How does the work function vary for different materials?
4. What are the implications of the photoelectric effect in modern technology?
5. How do photons behave when they have energy less than the work function? 

**Tip**: Remember that the photoelectric effect is a quantum phenomenon, which highlights the particle nature of light.

Iron has a work function of 4.50 eV. (a) Find the cutoff wavelength and cutoff frequency for the photoelectric effect. (b) What is the stopping potential if the incident light has a wavelength of 190 nm?

Explore the calculation of the cutoff wavelength (276 nm) and cutoff frequency (1.74 x 10^15 Hz) for iron with a work function of 4.50 eV. Additionally, find the stopping potential (2.02 V) when light of wavelength 190 nm is used. Suitable for advanced physics students.

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