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DSE Physics | Photoelectric Effect Lesson-Prep Guide (Tutor Edition) | Work Function and Threshold Wavelength
Students remember that light comes in packets but cannot say why brighter light changes nothing. This lesson uses one equation to settle all three questions: any emission, how much energy, and when it stops.
Objectives — Students can: (1) compute photon energy from wavelength and convert to electron volts; (2) find maximum kinetic energy and stopping voltage; (3) find threshold frequency and wavelength and separate intensity from frequency.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Electron volt | One eV in joules | 1.60 times ten to the minus nineteen |
| Wave equation | Shorter wavelength at fixed speed means what | Higher frequency |
| Energy conservation | Where photon energy goes | Work function plus kinetic energy |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-8 min | Phenomenon | Zinc plate and electroscope, ask why red light fails | State emission versus no emission |
| 8-22 min | Photon energy | E = hc over lambda with the 1240 shortcut at 400 nm | 3.10 eV |
| 22-36 min | Einstein equation | Kinetic energy equals E minus W at 2.14 eV | 0.96 eV |
| 36-46 min | Threshold and stopping | Threshold from hc over W, stopping voltage from energy | 579.4 nm and 0.96 V |
| 46-54 min | Graphs and intensity | Straight line of kinetic energy against frequency | Slope h, intercept minus W |
| 54-60 min | Exit check | Three questions collected and marked | Done independently |
03 Teaching script (read it out)
Scene 1: One shortcut for photon energy
Using hc = 1240 eV nanometre, photon energy is 1240 divided by wavelength in nanometres: 400 nm gives 3.10 eV, and 250 nm gives 4.96 eV. Shorter wavelength means a more energetic single photon, and that single fact drives the whole topic.
Task: convert 3.10 eV into joules as well, so both units feel natural.
Scene 2: Einstein in three sentences
An electron absorbs one photon of energy hf, pays the work function, and keeps the rest as kinetic energy. Caesium at 2.14 eV with 400 nm light gives 0.96 eV and a stopping voltage of 0.96 V. The threshold wavelength is 1240 divided by 2.14, which is 579.4 nm, and the threshold frequency is 5.18 times ten to the fourteen hertz.
Task: decide whether 600 nm red light releases electrons from caesium, comparing the wavelength with 579.4 nm.
Scene 3: Intensity and frequency do different jobs
More intensity means more photons per second, hence more electrons, but each photon carries the same energy, so maximum kinetic energy and stopping voltage stay fixed. Higher frequency raises the energy per photon. Zinc has a work function of 4.30 eV, so its threshold wavelength is 288.4 nm and all visible light fails. On a graph of kinetic energy against frequency the slope is Planck constant, the x intercept is the threshold frequency and the y intercept is minus the work function.
Warn: mixing up brightness and frequency is the most common conceptual error, so answers must name electron number separately from electron energy.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Calculation | Photon energy | 1240 divided by wavelength in nm | 3.10 eV |
| Calculation | Maximum kinetic energy | Energy minus work function | 0.96 eV |
| Calculation | Threshold wavelength | 1240 divided by work function | 579.4 nm |
| Explanation | Graph slope | Slope is h, intercept minus W | Slope independent of metal |
Sanity check (teach this): Kinetic energy cannot be negative: a negative result means no emission, not negative energy. The stopping voltage equals the kinetic energy in eV, so 0.96 V matches 0.96 eV. Emission requires the threshold wavelength to exceed the incident wavelength, and 579.4 exceeds 400.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Kinetic energy for 250 nm on zinc at 4.30 eV | 0.66 eV | 1240 over 250, then subtract W |
| Threshold wavelength of zinc | 288.4 nm | 1240 divided by W |
| Stopping voltage for that case | 0.66 V | Numerically equal to the eV value |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Brighter light for more energy | Predicting larger kinetic energy | “Intensity adds photons, not energy per photon.” |
| Mixing joules and electron volts | Using 3.10 as joules | “1240 belongs to the eV nanometre form.” |
| Threshold frequency reversed | Claiming emission below the threshold | “Emission needs frequency above the threshold.” |
| Reporting negative kinetic energy | Answering minus 1.20 eV | “A negative result means no photoelectrons.” |
07 Homework (paper, section and question numbers)
- Secondary physics chapter index — All chapters
- DSE physics practice paper — Quantum and atomic physics
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Photon energy at 400 nm | 3.10 eV | All three right: move to atomic models |
| Threshold wavelength of caesium at 2.14 eV | 579.4 nm | One wrong: redo the threshold |
| Effect of doubling intensity on maximum kinetic energy | No change, only more electrons | Two wrong: restart act three |
09 Tutor FAQ
Must 1240 be memorised?
Yes, it saves half the time; the precise value is 1239.8.
Why use ultraviolet on zinc?
The work function is 4.30 eV, so visible photons are too weak however bright the source.
Does the photoelectric effect prove light is a wave?
No, it shows particle behaviour; interference shows the wave side.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
