Why can't the wave model explain the photoelectric effect, and what does it reveal about the nature of light?
Explain the photoelectric effect using the photon model, including threshold frequency, work function and maximum kinetic energy.
Why the photoelectric effect contradicts the wave model, Einstein's photon explanation, and the photoelectric equation linking photon energy, work function and electron kinetic energy.
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What this dot point is asking
You need to explain the photoelectric effect using the photon model and apply the photoelectric equation involving threshold frequency, work function and maximum kinetic energy.
What is observed
When light of high enough frequency hits a metal, electrons (photoelectrons) are emitted. Experiments show:
- There is a threshold frequency : below it, no electrons are emitted no matter how intense or how long the light shines.
- Above , electrons are emitted instantly, even at very low intensity.
- The maximum kinetic energy of the electrons depends on the light's frequency, not its intensity.
- Greater intensity (above threshold) means more electrons per second, but not more energetic ones.
Why the wave model fails
The wave model predicts that energy is spread continuously across the wavefront, so a dim light should take time to build up enough energy in an electron, and any frequency should eventually work given enough intensity. Both predictions are wrong: emission is instant above threshold and impossible below it, regardless of intensity. The wave model cannot explain a threshold frequency or instantaneous emission.
Einstein's photon explanation
To escape the metal, an electron must be given at least the work function - the minimum energy to free an electron from the surface. Any leftover photon energy becomes the electron's kinetic energy.
This explains every observation: below a single photon lacks the energy to free an electron (so no amount of intensity helps); above emission is instant (one photon, one electron); and higher frequency means more leftover energy.
Stopping voltage
The maximum kinetic energy can be measured with a stopping voltage , the reverse voltage that just halts the most energetic electrons: . A graph of (or ) against frequency is a straight line of gradient and frequency-intercept - a classic way to measure Planck's constant.
Exam-style practice questions
Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2025 SACE Stage 23 marksThe work function of a metal target is 1.78 eV. Monochromatic light with a frequency of 5.45 x 10^14 Hz was incident on the target. Calculate the maximum kinetic energy of the electrons emitted from the surface of the target.Show worked answer →
Use Einstein's photoelectric equation: Ek max = h f - W, where W is the work function.
Find the photon energy: h f = (6.63 x 10^-34)(5.45 x 10^14) = 3.61 x 10^-19 J.
Convert the work function to joules: W = 1.78 x 1.60 x 10^-19 = 2.85 x 10^-19 J.
Ek max = 3.61 x 10^-19 - 2.85 x 10^-19 = 7.6 x 10^-20 J.
1 mark for the photon energy, 1 mark for converting the work function to joules, 1 mark for the answer of about 7.6 x 10^-20 J.
2024 SACE Stage 23 marksThe work function of a metal surface is 5.30 x 10^-19 J. Calculate the stopping voltage when light with a frequency of 1.50 x 10^15 Hz is incident on the metal surface.Show worked answer →
First find the maximum kinetic energy from Ek max = h f - W.
h f = (6.63 x 10^-34)(1.50 x 10^15) = 9.945 x 10^-19 J.
Ek max = 9.945 x 10^-19 - 5.30 x 10^-19 = 4.645 x 10^-19 J.
The stopping voltage relates to the maximum kinetic energy by Ek max = e Vs, so Vs = Ek max / e.
Vs = (4.645 x 10^-19) / (1.60 x 10^-19) = 2.90 V.
1 mark for the photon energy, 1 mark for Ek max, 1 mark for Vs = Ek max / e giving about 2.90 V.
2023 SACE Stage 22 marksStudents investigated the photoelectric effect on a caesium surface (accepted work function 2.1 eV). Calculate the maximum kinetic energy of an electron emitted from the caesium surface if light with a frequency of 1.30 x 10^15 Hz was incident upon the surface.Show worked answer →
Use Ek max = h f - W.
Photon energy: h f = (6.63 x 10^-34)(1.30 x 10^15) = 8.62 x 10^-19 J.
Convert the work function: W = 2.1 x 1.60 x 10^-19 = 3.36 x 10^-19 J.
Ek max = 8.62 x 10^-19 - 3.36 x 10^-19 = 5.3 x 10^-19 J.
1 mark for the photon energy, 1 mark for subtracting the work function to get about 5.3 x 10^-19 J.