A metal with a work function of 2.3 eV is illuminated with a light of frequency 500 nm. What is the maximum kinetic energy of the electron released?
A metal with a work function of 2.3 eV is illuminated with a light of frequency 500 nm. What is the maximum kinetic energy of the electron released?
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![### Photoelectric Effect Problem
A metal with a work function of 2.3 eV is illuminated with light of frequency 500 nm. What is the maximum kinetic energy of the electron released?
**Explanation:**
- **Work Function (Φ):** The minimum energy needed to remove an electron from the surface of a metal. It is given here as 2.3 eV.
- **Wavelength (λ):** The distance between successive peaks of a wave. The wavelength of the illuminating light is 500 nm (nanometers).
**Solution Approach:**
1. **Convert Wavelength to Frequency:**
- Use the formula \( c = \lambda \nu \) where \( c \) is the speed of light (\( 3 \times 10^8 \) m/s), \( \lambda \) is the wavelength, and \( \nu \) is the frequency.
2. **Calculate Photon Energy (E):**
- Use the formula \( E = h\nu \) where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) Js) and \( \nu \) is the frequency.
3. **Determine Maximum Kinetic Energy (KE_max):**
- \( KE_{max} = E - \Phi \)
In detailed steps:
1. Convert the wavelength (λ) to meters:
\[
\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}
\]
2. Calculate the frequency (\( \nu \)) of the light:
\[
\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \, \text{m/s}}{500 \times 10^{-9} \, \text{m}} = 6.0 \times 10^{14} \, \text{Hz}
\]
3. Calculate the energy of the photon (E):
\[
E = h\nu = 6.626 \times 10^{-34} \, \text{Js} \times 6.0 \times 10^{14} \, \text{Hz} = 3.976 \times 10^{-19} \, \text{J}
\]
Converting this energy to electron](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3fe8677b-d2b4-4cf1-b1ed-08820154fcb5%2F05cb1b3a-7f0a-40a8-8e00-487774966b39%2Fim2mwr_processed.png&w=3840&q=75)
Transcribed Image Text:### Photoelectric Effect Problem
A metal with a work function of 2.3 eV is illuminated with light of frequency 500 nm. What is the maximum kinetic energy of the electron released?
**Explanation:**
- **Work Function (Φ):** The minimum energy needed to remove an electron from the surface of a metal. It is given here as 2.3 eV.
- **Wavelength (λ):** The distance between successive peaks of a wave. The wavelength of the illuminating light is 500 nm (nanometers).
**Solution Approach:**
1. **Convert Wavelength to Frequency:**
- Use the formula \( c = \lambda \nu \) where \( c \) is the speed of light (\( 3 \times 10^8 \) m/s), \( \lambda \) is the wavelength, and \( \nu \) is the frequency.
2. **Calculate Photon Energy (E):**
- Use the formula \( E = h\nu \) where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) Js) and \( \nu \) is the frequency.
3. **Determine Maximum Kinetic Energy (KE_max):**
- \( KE_{max} = E - \Phi \)
In detailed steps:
1. Convert the wavelength (λ) to meters:
\[
\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}
\]
2. Calculate the frequency (\( \nu \)) of the light:
\[
\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \, \text{m/s}}{500 \times 10^{-9} \, \text{m}} = 6.0 \times 10^{14} \, \text{Hz}
\]
3. Calculate the energy of the photon (E):
\[
E = h\nu = 6.626 \times 10^{-34} \, \text{Js} \times 6.0 \times 10^{14} \, \text{Hz} = 3.976 \times 10^{-19} \, \text{J}
\]
Converting this energy to electron
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