The Photoelectric Effect: Grade 12 Physical Sciences
Shine a bright enough light on a metal and nothing happens. Shine a dim ultraviolet lamp on the same metal and electrons come flying off. That result broke the wave theory of light and it won Einstein his Nobel Prize.
This page covers the photoelectric effect for Grade 12: the demonstration, Planck's constant and E = hf, work function and threshold frequency, the equation you get marked on, and the distinction the exam tests every single year, which is frequency decides whether, intensity decides how many.
What the photoelectric effect is
The photoelectric effect is the emission of electrons from a metal surface when light of a suitable frequency shines on it.
The electrons that come off are called photoelectrons. They are ordinary electrons; the name only says where they came from.
Hertz stumbled on it in 1887 while working on something else entirely. It sat unexplained for eighteen years because the wave model of light says it should not happen that way, and nobody could see past that.
The demonstration, and it is the whole argument in four steps
A clean zinc plate sits on the cap of a gold leaf electroscope. Charge the electroscope negatively, so the leaf stands out, then try four different things.
| What you shine on the zinc | What the leaf does | What it means |
|---|---|---|
| Ultraviolet lamp, about 6 W | Falls. The electroscope discharges | Electrons are being knocked out of the zinc |
| 40 W light bulb | Nothing | Visible light does not do it |
| 200 W light bulb, much brighter | Still nothing | Brighter does not help. This is the important one |
| Ultraviolet lamp through a sheet of window glass | Nothing | Glass blocks ultraviolet. Take the glass away and it works again |
Sit with the third row for a moment. A 200 W bulb pours more than thirty times the energy onto that plate than the 6 W ultraviolet lamp does, and it cannot shift a single electron. The dim lamp can.
That is not how waves behave.
Why the wave model cannot survive this
If light were purely a wave, its energy would arrive spread out and continuously. Turn up the brightness and you pour in more energy per second, so electrons should eventually absorb enough to escape. It might take longer with a dim source, but it should still work.
| The wave model predicts | What actually happens |
|---|---|
| Bright enough light always works, eventually | Below a certain frequency it never works, at any brightness, for any length of time |
| Brighter light gives faster electrons | Brighter light gives more electrons, all at the same maximum speed |
| There should be a delay while energy builds up | Emission is immediate |
Three predictions, three failures. The problem was not the arithmetic. It was the picture.
Planck's constant and E = hf
In 1901 Max Planck suggested that energy is not radiated continuously but in fixed packets called quanta, and that the energy of each packet depends only on frequency.
E = hf
| Symbol | Means | Value or unit |
|---|---|---|
| E | Energy of one photon | J |
| h | Planck's constant | 6,63 × 10-34 J·s |
| f | Frequency of the light | Hz |
Use 6,63 × 10-34 J·s. That is the value printed on the CAPS data sheet, and it is the one to use in an exam. You will see 6,6 × 10-34 in some books and 6,626 × 10-34 in others. They are the same constant rounded differently, and the difference will not change a Grade 12 answer, but use the data sheet value and you can never be marked down for it.
When the light is described by its wavelength instead of its frequency, use c = fλ to swap between them:
E = hf = hc ÷ λ, with c = 3 × 108 m·s-1
Worked example: energy from frequency
Calculate the energy of one photon of red light of frequency 4,5 × 1014 Hz.
- E = hf = (6,63 × 10-34)(4,5 × 1014)
- E = 2,98 × 10-19 J
Worked example: energy from wavelength
Calculate the energy of one photon of blue light of wavelength 450 nm.
- Convert to metres first. 450 nm = 450 × 10-9 m = 4,5 × 10-7 m
- E = hc ÷ λ = (6,63 × 10-34)(3 × 108) ÷ (4,5 × 10-7)
- E = 4,42 × 10-19 J
Blue light carries more energy per photon than red light, because it has the higher frequency. That single fact is the answer to most of this chapter.
Work function and threshold frequency
An electron sitting in a metal is held there. It takes a certain minimum amount of energy to get one out, and that amount depends on the metal.
| Term | Symbol | Definition |
|---|---|---|
| Work function | W0 | The minimum energy needed to eject an electron from that particular metal |
| Threshold frequency, also called the cut-off frequency | f0 | The minimum frequency of incident light that will eject an electron from that metal |
They are the same requirement stated two ways, and the bridge between them is E = hf:
W0 = hf0
Both are properties of the metal, not of the light. Every metal has its own pair, which is why one lamp will free electrons from caesium and do nothing at all to zinc.
One photon, one electron, and that is the key to the whole thing
A photon gives all of its energy to a single electron. It cannot split the energy between two, and two photons cannot team up on one electron.
So the question is never how much light there is. It is whether one photon carries enough energy on its own.
If it does not, nothing happens, and it does not matter how many of them arrive. A million photons that are each too weak are still each too weak.
That is why the 200 W bulb fails. It sends out an enormous number of photons and every one of them is below zinc's threshold.
The equation you get marked on
E = W0 + Ek(max)
In words: the energy the photon brought in is spent on freeing the electron, and whatever is left over becomes the electron's kinetic energy.
Rearranged the way you will usually use it:
Ek(max) = hf − W0, and since Ek(max) = ½mv2max, ½mv2max = hf − hf0
The mass of an electron is 9,11 × 10-31 kg and it is on the data sheet.
The subscript "max" is not decoration. Electrons deeper in the metal need more than the minimum to escape, so they come out slower. W0 is the easiest case, so hf − W0 is the fastest electron, not the average one.
Worked example: the full question
A metal has a work function of 3,68 × 10-19 J. Light of frequency 8,0 × 1014 Hz shines on it.
Does the photoelectric effect occur?
- E = hf = (6,63 × 10-34)(8,0 × 1014) = 5,30 × 10-19 J
- 5,30 × 10-19 J is greater than 3,68 × 10-19 J
- Yes. Electrons are emitted.
Find the threshold frequency of the metal.
- f0 = W0 ÷ h = (3,68 × 10-19) ÷ (6,63 × 10-34)
- f0 = 5,55 × 1014 Hz
Find the maximum kinetic energy of a photoelectron.
- Ek(max) = hf − W0 = 5,30 × 10-19 − 3,68 × 10-19
- Ek(max) = 1,62 × 10-19 J
Find the maximum speed of a photoelectron.
- ½mv2 = 1,62 × 10-19
- v2 = (2 × 1,62 × 10-19) ÷ (9,11 × 10-31)
- v = 5,97 × 105 m·s-1
Check the order of magnitude. Photoelectron speeds come out around 105 to 106 m·s-1. An answer of 60 m·s-1 or 1012 m·s-1 is an arithmetic error, and the second one is faster than light.
Frequency decides whether. Intensity decides how many.
This one sentence is worth more marks in this chapter than anything else on this page.
| If you increase the | Then | And |
|---|---|---|
| Frequency of the light | Each photon carries more energy | The photoelectrons come out faster. Their maximum kinetic energy rises |
| Intensity of the light, at the same frequency | More photons arrive per second | More photoelectrons come out per second. Their maximum speed does not change at all |
| Intensity, while below the threshold frequency | More photons arrive, all still too weak | Nothing happens. Ever. |
Exam papers set the two questions side by side on purpose. "What determines whether photoelectrons are emitted?" and "What determines how many photoelectrons are emitted per second?" look almost identical and have completely different answers.
The first is frequency. The second is intensity.
Where this fits in the curriculum
| Subject | Physical Sciences |
|---|---|
| Grade | 12 |
| Term | 3 |
| Topic | Matter and materials, optical phenomena |
| Status | Examinable theory. No formal assessment attached |
| Paper | Physics, Paper 1 |
There is no prescribed practical here, and the demonstration is often skipped because the apparatus is not in the cupboard. The chapter still turns up in Paper 1 as definitions, one or two calculations, and almost always a question that separates frequency from intensity.
The dual nature of light
Is light a wave or a particle? The honest answer is that it is neither, and the question is the wrong shape.
| Behaviour | Explained by |
|---|---|
| Diffraction, interference, refraction | The wave model |
| The photoelectric effect | The particle model |
Light is a form of energy that shows wave properties in some experiments and particle properties in others. Neither model explains everything, and no experiment has ever caught it behaving as both at once.
That is not physics dodging the question. It is physics admitting that a thing can be genuinely unlike anything at human scale.
Where it is used
| Application | How |
|---|---|
| Solar cells and solar-powered calculators | Light frees electrons in a semiconductor and the movement of those electrons is a current |
| Photodiodes and light meters | A cathode coated in a low-threshold material emits electrons; the current is proportional to the light falling on it |
| Automatic lights and dusk sensors | The same photodiode, wired to a switch |
| Image sensors in cameras and phones | Each pixel counts the electrons freed by the light landing on it |
The calculator on your desk is running on it. That is a better answer to "when will I ever use this" than most chapters can offer.
Emission and absorption spectra
The second half of this chapter is about what light comes out of atoms rather than what it knocks out of metals, and it rests on the same idea: energy comes in fixed amounts.
An electron in an atom can only sit at certain energy levels. To move up it must absorb exactly the right amount of energy, and when it drops back down it emits exactly that amount as a photon of a particular frequency.
| Spectrum | What you see | How it is made |
|---|---|---|
| Continuous emission spectrum | An unbroken band of all colours | Light from a hot filament, spread out by a prism or a diffraction grating |
| Line emission spectrum | Bright coloured lines on a dark background | An excited gas, in a discharge tube or a flame, viewed through a grating |
| Absorption spectrum | Dark lines on a continuous coloured band | White light passed through a cool gas, which removes exactly the frequencies it would have emitted |
The lines are a fingerprint. Hydrogen always gives the same pattern, sodium always gives its two close yellow lines, and no two elements share a set. That is how the composition of a star is worked out from its light.
And an element's absorption lines sit at exactly the same frequencies as its emission lines, because the same energy gaps do both jobs. If you can explain why, you have understood energy levels.
This is the same physics behind red shift: astronomers know where a hydrogen line should be, so a shifted line tells them how the source is moving.
The mistakes that cost the marks
| The mistake | What to do instead |
|---|---|
| Saying brighter light gives faster electrons | It gives more electrons at the same maximum speed. Frequency controls speed |
| Answering "intensity" when the question asks what determines whether emission happens | That is frequency. Read which of the two questions is being asked |
| Using 6,6 or 6,626 for h | Use the data sheet value, 6,63 × 10-34 J·s |
| Forgetting to convert nm to m | Divide by 109. A wavelength in nanometres put straight into hc ÷ λ is out by a factor of a billion |
| Treating the work function as a property of the light | It belongs to the metal. Change the lamp and W0 does not move |
| Leaving out "maximum" in Ek(max) | Only the most loosely held electrons come out at that speed |
| Saying light is both a wave and a particle at the same time | It shows wave behaviour in some situations and particle behaviour in others. Never both at once |
If the demonstration does not work
| What happens | Why |
|---|---|
| The leaf does not fall under the ultraviolet lamp | The zinc is oxidised. A layer of zinc oxide raises the effective work function. Clean the plate with sandpaper or steel wool immediately before the lesson |
| The electroscope will not hold a charge at all | Humidity. Charge leaks away through damp air. A dry day, or a hair dryer over the apparatus first |
| The leaf falls even with no lamp on | Same thing. Charge is leaking, so time how long it takes without the lamp and compare |
| The leaf falls under the 200 W bulb too | The bulb is warming the apparatus and the air. Keep it well back, and use it only briefly |
| You cannot charge it negatively | Polythene rubbed with wool gives a negative charge. Perspex rubbed with silk gives a positive one, which is the wrong sign and will not discharge |
| Nothing you do works | The zinc plate has to be in electrical contact with the cap. Paint, lacquer or a sticker underneath will stop the whole thing |
The charge must be negative. A positively charged electroscope will not discharge under ultraviolet light, because the plate is short of electrons already and any that are ejected are pulled straight back. That is a good extension question and a common exam question.
How the 40 marks are made up
| Section | Marks |
|---|---|
| Definitions and terminology | 8 |
| Photon energy calculations | 8 |
| Work function, threshold frequency and kinetic energy | 10 |
| Multiple choice | 10 |
| The demonstration and its interpretation | 4 |
No mark allocation is prescribed for this one. The split above is ours, weighted towards the calculations because that is what a Paper 1 question does with this topic.
Free worksheet and marking memo
Both free, no sign up, straight to the PDF.
- Learner worksheet, 40 marks, with multiple choice, the definitions, photon energy from both frequency and wavelength, a full work function question on zinc, and the demonstration
- Marking memorandum, with full working, the mark breakdown line by line and a note on the seven places learners drop marks
Related pages
- The Doppler effect, Grade 12. Red shift runs on the same spectral lines
- Electrostatic charge investigation. Where the electroscope comes from, and worth doing first
- Snell's Law and refraction. The wave side of light, Grade 11
Apparatus
An honest note: most South African schools cannot run this demonstration, and we are currently no help.
It needs a gold leaf electroscope, a clean zinc plate and an ultraviolet lamp. Our electroscopes are out of stock, and we do not list a laboratory ultraviolet lamp or a zinc plate at all. Diffraction gratings and spectroscopes for the spectra half of the chapter are in the same position.
We would rather say that than sell you a page pretending otherwise. The light and optics range is where these items will appear when they are back.
In the meantime the chapter teaches perfectly well from the four-row table near the top of this page, because the argument is what matters and the argument is short: the dim lamp works, the bright bulb does not, and no wave model survives that.