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The photoelectric effect describes the emission of electrons from a metal surface when light of high enough frequency shines on it. This topic is important because it shows that light transfers energy in discrete packets called photons. A cheat sheet helps students connect frequency, wavelength, energy, work function, and electron kinetic energy in one place. It also supports quick problem solving for AP Physics, honors physics, and introductory modern physics.

Key Facts

  • Photon energy is given by E=hfE = hf, where hh is Planck’s constant and ff is the light frequency.
  • Using wavelength, photon energy can be written as E=hcλE = \frac{hc}{\lambda}, where cc is the speed of light and λ\lambda is wavelength.
  • The work function ϕ\phi is the minimum energy needed to remove one electron from a metal surface.
  • Einstein’s photoelectric equation is Kmax=hfϕK_{\max} = hf - \phi, where KmaxK_{\max} is the maximum kinetic energy of emitted electrons.
  • The threshold frequency is f0=ϕhf_0 = \frac{\phi}{h}, and no electrons are emitted when f<f0f < f_0.
  • The stopping potential satisfies eVs=KmaxeV_s = K_{\max}, so Vs=KmaxeV_s = \frac{K_{\max}}{e}.
  • Increasing light intensity increases the number of emitted electrons if f>f0f > f_0, but it does not increase KmaxK_{\max}.
  • A graph of KmaxK_{\max} versus ff has slope hh and vertical intercept ϕ-\phi.

Vocabulary

Photon
A photon is a discrete packet of electromagnetic energy with energy E=hfE = hf.
Work function
The work function ϕ\phi is the minimum energy required to remove an electron from a metal surface.
Threshold frequency
The threshold frequency f0f_0 is the minimum light frequency needed to eject electrons, given by f0=ϕhf_0 = \frac{\phi}{h}.
Photoelectron
A photoelectron is an electron emitted from a material after absorbing energy from a photon.
Stopping potential
The stopping potential VsV_s is the voltage needed to reduce the maximum photoelectron kinetic energy to 00.
Maximum kinetic energy
The maximum kinetic energy KmaxK_{\max} is the greatest kinetic energy of emitted electrons, equal to hfϕhf - \phi.

Common Mistakes to Avoid

  • Using intensity instead of frequency to decide whether electrons are emitted is wrong because emission requires ff0f \ge f_0, regardless of brightness.
  • Forgetting the work function is wrong because only the leftover photon energy becomes kinetic energy, so Kmax=hfϕK_{\max} = hf - \phi.
  • Mixing joules and electronvolts is wrong because formulas must use consistent units, with 1eV=1.60×1019J1\,\text{eV} = 1.60 \times 10^{-19}\,\text{J}.
  • Assuming longer wavelength means greater photon energy is wrong because E=hcλE = \frac{hc}{\lambda}, so energy decreases as wavelength increases.
  • Using the total number of emitted electrons to find KmaxK_{\max} is wrong because electron count depends mainly on intensity, while KmaxK_{\max} depends on frequency.

Practice Questions

  1. 1 A metal has work function ϕ=2.3eV\phi = 2.3\,\text{eV}. What is the maximum kinetic energy of electrons emitted by light with photon energy E=3.8eVE = 3.8\,\text{eV}?
  2. 2 Find the threshold frequency for a metal with ϕ=4.0×1019J\phi = 4.0 \times 10^{-19}\,\text{J} using h=6.63×1034Jsh = 6.63 \times 10^{-34}\,\text{J}\cdot\text{s}.
  3. 3 Light of wavelength λ=400nm\lambda = 400\,\text{nm} shines on a metal with ϕ=2.0eV\phi = 2.0\,\text{eV}. Find KmaxK_{\max} in eV\text{eV} using hc1240eVnmhc \approx 1240\,\text{eV}\cdot\text{nm}.
  4. 4 If light is below the threshold frequency, explain why increasing the intensity still does not cause photoelectrons to be emitted.