S6 Physics: Photoelectric Effect — Tutor Lesson Plan (60 min, 1240 shortcut, four errors)

2026-11-02

Tutor lesson-prep guides · Secondary physics (DSE)

Bright red light does not eject electrons and dim ultraviolet does, and that single fact kills the wave model. This lesson makes students say why before they touch the equation.

Objectives — the student can compute a photon energy in electronvolts from a wavelength, apply Einstein equation to find a maximum kinetic energy, find a threshold wavelength and frequency, read a stopping voltage off a result, and explain why intensity changes the current but not the energy of each electron.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Waves	What is the relation between speed, frequency and wavelength?	c = f times lambda
Energy units	How many joules is one electronvolt?	1.60 times 10 to the minus 19
Rearranging	Make lambda the subject of E = hc over lambda	lambda = hc over E

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Hook	Dim ultraviolet ejects electrons, bright red does not	Explain why the wave model fails
8-18 min	Scene 1	One shortcut: 1240 over wavelength in nanometres	Get 3.10 eV for 400 nm
18-28 min	Scene 2	Einstein equation in one line	Get 0.96 eV for caesium
28-38 min	Scene 3	Threshold wavelength and frequency	Get 579 nm and 5.18 times 10 to the 14
38-52 min	Scene 4	Intensity versus frequency, stopping voltage	Separate current from energy
52-60 min	Exit check	Three items in section 08	Marked at once

03 Teaching script (read it out)

Scene 1: One shortcut worth memorising

Photon energy in electronvolts is 1240 divided by the wavelength in nanometres. So 400 nm gives 3.10 eV and 250 nm gives 4.96 eV. The number comes from hc expressed in electronvolt nanometres, and it removes a factor of 10 to the minus 19 from every line of working, which is where most arithmetic marks are lost. Show the derivation once, then let the class use it.

Ask: “Which is more energetic, 400 nm or 600 nm?” The shorter wavelength wins, and a student who says 600 nm is reading the number instead of the physics.

Scene 2: Einstein equation in three sentences

One photon gives all its energy to one electron. Getting out costs the work function W. What is left is the maximum kinetic energy: hf = W plus KE max. For caesium with W = 2.14 eV and 400 nm light: 3.10 minus 2.14 gives 0.96 eV. The word maximum matters, because electrons deeper in the metal lose more on the way out.

Warn: a negative kinetic energy does not mean a slow electron. It means no emission at all, and the answer is that the photon energy is below the work function.

Scene 3: Intensity moves the current, frequency moves the energy

Brighter light of the same colour carries more photons, so more electrons per second and a larger current, but each photon still has the same energy, so the maximum kinetic energy is unchanged. Because the stopping voltage equals the maximum kinetic energy in electronvolts, 0.96 eV gives 0.96 V, and the threshold is 1240 over 2.14, which is 579.4 nm: 600 nm red light ejects nothing from caesium. For zinc with W = 4.30 eV the threshold is 288 nm, which is why ultraviolet is used in demonstrations.

Ask: “Double the intensity. What doubles?” The current doubles and the stopping voltage does not, and that one answer separates the two graphs in the syllabus.

04 Question types and worked steps

Type	Example	Steps	Answer
Photon energy	Find the photon energy of 400 nm light in electronvolts.	1240 over the wavelength in nanometres	3.10 eV
Kinetic energy	Caesium W = 2.14 eV, light 400 nm. Find the maximum kinetic energy.	Photon energy minus the work function	0.96 eV
Threshold	Find the threshold wavelength of caesium.	1240 over W	579 nm
Stopping voltage	Find the stopping voltage for the caesium case.	Numerically equal to the maximum kinetic energy in electronvolts	0.96 V

Sanity check (teach this): Every answer passes the unit check first: an energy in electronvolts, a wavelength in nanometres, a frequency in hertz. Then the physical check: a kinetic energy must be positive or the answer is “no emission”, a threshold wavelength must be longer than the light that works, and a stopping voltage must never exceed the photon energy divided by the electronic charge.

05 Independent practice (3 questions)

Question	Answer	What to watch
Find the photon energy of 250 nm light.	1240 over 250 = 4.96 eV.	Shorter wavelength, more energy
Zinc W = 4.30 eV is lit with 250 nm light. Find the maximum kinetic energy.	4.96 minus 4.30 = 0.66 eV.	Positive, so emission happens
Find the threshold wavelength of zinc.	1240 over 4.30 = 288 nm.	Ultraviolet only
Will 600 nm light eject electrons from caesium?	No: 600 nm is longer than the 579 nm threshold, so each photon carries too little energy.	Compare wavelengths, not intensities

06 Common mistakes and what to say

Mistake	Student writes	Say this
Using intensity to explain emission	Emission depends on the energy of a single photon, not on how many arrive	Ask what happens with brighter red light
Adding the work function instead of subtracting it	The work function is a cost, so it comes off the photon energy	Ask which number should be smaller
Reporting a negative kinetic energy	It means no emission, which is a valid answer	Ask for the sentence, not the number
Mixing joules and electronvolts in one line	Choose one unit for the whole calculation	Ask for the unit before the arithmetic

07 Homework (paper, section and question numbers)

Secondary physics chapter index — Read this chapter with the quantum physics and the waves chapters
HKDSE Physics Paper 1 practice paper — Photoelectric questions; redo section 05 questions 1 to 4 of this guide

08 Exit check (last 5 minutes)

Question	Answer	Decision
Find the photon energy of 500 nm light.	2.48 eV	All correct: move to the Einstein equation
Caesium W = 2.14 eV, photon energy 3.10 eV. Give the maximum kinetic energy.	0.96 eV	Subtract, do not add
Double the intensity. What happens to the stopping voltage?	Nothing: same photon energy, more electrons per second	Voltage answers energy, not number

09 Tutor FAQ

Must students memorise 1240?

It is worth it, and it is easy to check: 1240 over 400 is about 3, which matches the known value for violet light. Derive it once so it is not magic.

Why use ultraviolet rather than red light in demonstrations?

Most metal work functions sit around 4 eV, which needs light below about 300 nm. Red light is far above that and ejects nothing.

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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.