DSE Physics | Photoelectric Effect and the Photon Model: Tutor Lesson Plan

2026-11-03

Tutor lesson-prep guides · Secondary Physics

Students recite that brighter light means a bigger current, then freeze when asked about frequency. This lesson does three things: convert wavelength to electron volts, treat the work function as an entry fee, and turn the leftover energy into a stopping voltage.

Objectives — Students can: (1) use 1240 divided by wavelength in nm, getting 3.10 eV at 400 nm; (2) apply KEmax = hf minus phi to get 1.10 eV for a 2.0 eV metal; (3) state that the stopping voltage is 1.10 V and does not depend on intensity; (4) compute 4.13 eV at 300 nm and a threshold of 620 nm, and explain why brighter red light still gives nothing.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Wavelength and frequency	Roughly what frequency is 400 nm light?	About 7.5 x 10^14 Hz
Electron volt	How many joules is 1 eV?	1.6 x 10^-19 J
Conservation of energy	Can energy vanish?	No, it only changes form
Current and voltage	Does a larger stopping voltage mean more or less kinetic energy?	More

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Opening conflict	Ask whether a red laser ejects electrons from zinc	Answer with enough energy, not enough brightness
10-22 min	Photon energy	Show 1240 divided by 400	Compute 300 nm and 620 nm themselves
22-34 min	Einstein equation	Draw the energy budget: photon in, work function out, kinetic left	Write KEmax = hf minus phi
34-44 min	Stopping voltage	Convert 1.10 eV into 1.10 V	Explain why a reverse voltage stops them
44-52 min	Four question types	Work from threshold to intensity comparisons	Name the equation before calculating
52-60 min	Independent practice	Three questions in eight minutes	Check answers and label the error

03 Teaching script (read it out)

Scene 1: Intensity sets the count, frequency sets the possibility

Intensity is how many photons arrive each second; frequency is how much energy each one carries. A red photon below the 2.0 eV entry fee gives nothing no matter how many arrive, because the fee is charged per particle, not on the total. Break this first, otherwise every later calculation is aimed the wrong way. Write 100 photons at 1.8 eV on the board and show the product still failing.

Ask the student whether doubling the intensity changes the maximum kinetic energy, and why.

Scene 2: Explain where 1240 comes from

The number 1240 is hc expressed in electron volts and nanometres, so energy equals 1240 divided by wavelength in nm. That gives 3.10 eV at 400 nm and 2.00 eV at 620 nm, which equals the work function, so 620 nm is the threshold. Students only need to remember that the wavelength must be in nm; feeding in metres is the usual disaster. Deliberately substitute 400 m and let the absurd answer surface.

Ask the student to compute 1240 divided by 500 and state the unit.

Scene 3: The stopping voltage is kinetic energy spoken in volts

Leaving 1.10 eV of kinetic energy requires 1.10 V of reverse voltage, because an electron loses 1 eV crossing 1 V. The number is the same, only the unit changes from eV to V. Many students multiply 1.10 by 1.6 x 10^-19 and call it a voltage, which mixes energy with potential difference. Make them add the sentence that an electron stops after crossing 1.1 V.

Ask what the stopping voltage becomes when KEmax rises from 1.10 to 2.13 eV.

04 Question types and worked steps

Type	Example	Steps	Answer
Photon energy	Photon energy at 400 nm	Divide 1240 by 400, label eV, convert to joules if needed	3.10 eV
Maximum kinetic energy	Work function 2.0 eV, incident light 400 nm	Find photon energy, subtract the work function, negative means no emission	1.10 eV
Stopping voltage	A cell with KEmax of 1.10 eV	Copy the number, change the unit to V, state independence from intensity	1.10 V
Threshold wavelength	A metal with a 2.0 eV work function	Divide 1240 by the work function, compare with the incident wavelength	620 nm

Sanity check (teach this): Order check: visible photons always land between 1.8 and 3.1 eV, so an answer in the thousands means metres were used. KEmax cannot be negative; a negative value means no emission, not negative energy. The threshold must be longer than the incident wavelength for emission to happen, and 620 nm is longer than 400 nm, so the case works.

05 Independent practice (3 questions)

Question	Answer	What to watch
Photon energy at 350 nm for a 2.5 eV metal	3.54 eV	Did you divide 1240 by 350
Maximum kinetic energy in that case	1.04 eV	Is the answer below the photon energy
Threshold wavelength of that metal	496 nm	Is it longer than 350 nm

06 Common mistakes and what to say

Mistake	Student writes	Say this
Using intensity to explain emission	The light is not bright enough, so no electrons come out	The fee is charged per photon; doubling intensity only adds more photons that are still too weak.
Wavelength in metres	1240 divided by 4 x 10^-7 gives 3.1 x 10^9 eV	The 1240 shortcut already assumes nm; using metres claims a single photon carries billions of electron volts.
Mixing eV and V	The stopping voltage is 1.10 x 1.6 x 10^-19 V	1.10 eV is stopped by exactly 1.10 V; the number stays, the unit changes, and multiplying invents a voltage that does not exist.
Reporting negative kinetic energy	KEmax equals 3.10 minus 2.50, so minus 1.04 eV	A negative value means no photoelectrons; say so instead of quoting a negative energy.

07 Homework (paper, section and question numbers)

Physics chapter index — Revise radioactivity and half-life, then questions 1 and 2
Radioactivity and half-life — Question 3 also deals with particle energy
Electricity and circuits — Revise the link between eV and voltage

08 Exit check (last 5 minutes)

Question	Answer	Decision
Energy of a 400 nm photon	3.10 eV	Error: wavelength in metres
Maximum kinetic energy for a 2.0 eV metal	1.10 eV	Error: forgot the work function
Corresponding stopping voltage	1.10 V	Error: multiplied by 1.6 x 10^-19
Threshold wavelength of that metal	620 nm	Error: quoted the incident wavelength

09 Tutor FAQ

Why does intensity not change the maximum kinetic energy?

One photon interacts with one electron, so energy depends on frequency; intensity only changes how many electrons are released.

Why use 1240 instead of Planck constant?

Both are correct; 1240 is hc in eV and nm, so it is faster and avoids powers of ten.

What if the metal surface is oxidised?

The layer absorbs energy and lowers the measured stopping voltage; questions usually assume a clean surface.

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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.