Section 10.3: Photoelectric Effect
The photoelectric effect occurs when light of sufficient frequency shines on a metal surface and electrons are emitted. Classical wave theory could not explain why the effect depends on frequency, not intensity.
- Einstein’s Equation: \[ K_\text{max} = h \nu - \phi \] where \( K_\text{max} \) = maximum kinetic energy of ejected electrons, \( h \) = Planck’s constant, \( \nu \) = frequency of light, \( \phi \) = work function of the metal.
- Threshold Frequency: Minimum frequency of light needed to emit electrons: \[ \nu_\text{threshold} = \frac{\phi}{h} \]
- Increasing light intensity increases the number of emitted electrons, but not their maximum kinetic energy.
Example: Photoelectric Energy
Light of frequency \( 8.0 \times 10^{14} \, \text{Hz} \) shines on a metal with work function \( 3.0 \times 10^{-19} \, \text{J} \). Find the maximum kinetic energy of emitted electrons.
\( K_\text{max} = h\nu - \phi = (6.626 \times 10^{-34} \cdot 8.0 \times 10^{14}) - 3.0 \times 10^{-19} \approx 2.3 \times 10^{-19} \, \text{J} \)
Practice Problems
- Calculate the threshold frequency for a metal with work function \( 4.5 \times 10^{-19} \, \text{J} \).
- Light with wavelength 400 nm shines on a metal. Find the maximum kinetic energy of electrons if the work function is \( 2.0 \times 10^{-19} \, \text{J} \).
- Explain why increasing light intensity does not change the kinetic energy of photoelectrons.
- A metal requires a minimum frequency of \( 5.0 \times 10^{14} \, \text{Hz} \) for electron emission. What is its work function?
- Discuss one experimental observation that supports the particle nature of light.