Section 7.2: Photoelectric Effect
The photoelectric effect occurs when light of sufficient frequency shines on a metal surface and ejects electrons. The kinetic energy of emitted electrons depends on the frequency of the light, not its intensity. The maximum kinetic energy is given by: \[ K_\text{max} = h \nu - \phi \] where \( \phi \) is the work function of the metal.
- Light behaves as photons, each with energy \(E = h\nu\).
- Threshold frequency \( \nu_0 \) is the minimum frequency required to eject electrons.
- Increasing intensity increases the number of electrons but not their kinetic energy.
Example: Maximum Kinetic Energy
A metal has work function \( \phi = 2.0 \, \text{eV} \). Light of frequency \( 6.0 \times 10^{14} \, \text{Hz} \) strikes the surface. Calculate the maximum kinetic energy of emitted electrons.
\( E_\text{photon} = h\nu = 6.626 \times 10^{-34} \times 6.0 \times 10^{14} \approx 3.975 \times 10^{-19} \, \text{J} \approx 2.48 \, \text{eV} \)
\( K_\text{max} = E_\text{photon} - \phi = 2.48 - 2.0 = 0.48 \, \text{eV} \)
Practice Problems
- Determine the threshold frequency for a metal with work function \(3.0 \, \text{eV}\).
- Calculate the maximum kinetic energy of electrons if light of wavelength 400 nm strikes a metal with work function \(2.5 \, \text{eV}\).
- Explain why increasing light intensity does not increase electron kinetic energy.
- For a photon of energy 5 eV, calculate the speed of an electron emitted from a metal with work function 2 eV.
- Discuss how the photoelectric effect supports the photon model of light.