Quantum · 25.1

The Photoelectric Effect

Shine ultraviolet light on zinc and electrons fly out instantly — even when the light is absurdly dim. Shine red light on the same plate and nothing happens, no matter how bright. Classical wave theory cannot explain either fact. Einstein could, by making light grainy.

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Build the model

Connect the measurement to the mechanism.

A beam of light of frequency f delivers its energy in indivisible packets — photons — each carrying E = hf. An electron in a metal absorbs one photon or none; it cannot save up several. Escaping the metal costs a fixed entry fee called the work function φ, so the fastest electrons leave with E(max) = hf − φ.

Below the threshold frequency f₀ = φ/h a photon simply cannot pay the fee, so no electrons appear at any intensity. Intensity only sets how many photons arrive per second — the current, never the electron energy. Reversing the collecting voltage until the current dies measures E(max) directly: eV(s) = E(max).

Simple definition: Light of frequency f is a stream of photons of energy E = hf; one photon can eject one electron from a metal if hf exceeds the work function φ, and the fastest electrons carry away E(max) = hf − φ.
Example: A photographer’s darkroom uses a red safelight because red photons (about 1.9 eV each) cannot trigger the chemistry that blue photons (about 2.8 eV) trigger easily — brightness is irrelevant, the packet size is everything.

Photon energyE = hf = hc/λ

Light of one frequency comes in identical energy packets. Doubling the brightness doubles the number of packets per second — it never makes any single packet bigger.

h = 6.63 × 10⁻³⁴ J s

Work function & thresholdφ = hf₀

The work function is the minimum energy to pull one electron out of the metal surface. It fixes a threshold frequency f₀ below which no photon can free an electron.

φ in eV varies metal to metal

Einstein’s equationE(max) = hf − φ

One photon hands hf to one electron; φ is spent escaping; the remainder is kinetic energy. Electrons from deeper in the metal lose more, which is why E(max) is a maximum.

Energy bookkeeping for one photon, one electron

Stopping potentialeV(s) = E(max)

Make the collector negative and it repels photoelectrons. The voltage that stops even the fastest ones converts straight into their kinetic energy in electronvolts.

The reverse voltage that just kills the current

Photon momentump = E/c = h/λ

A photon has no mass and still carries momentum: divide its energy by c, or divide h by its wavelength. Short wavelength, bigger kick.

Massless, but never momentum-less

Compton shiftΔλ = (h/mₑc)(1 − cos θ)

The wavelength a photon gains scattering off a free electron. It depends on the viewing angle and nothing else — not the starting wavelength, not the target, not the intensity.

h/mₑc = 2.43 pm; largest shift 4.86 pm, straight back

What the wave picture predicted — and got wrong

A wave delivers energy continuously, so any frequency should eject electrons if you wait long enough or turn the intensity up, and brighter light should make faster electrons. Experiment says the opposite: below f₀ nothing happens at any brightness, above f₀ emission is instantaneous even in the dimmest light, and only frequency changes the electron energy.

What each dial does

Frequency sets the energy of each photon, so it sets E(max) and the stopping potential. Intensity sets the number of photons per second, so it sets the photocurrent — the count of ejected electrons — and nothing else. The two roles never mix, which is the clearest fingerprint of the photon model.

Millikan’s reluctant confirmation

Robert Millikan spent a decade trying to disprove Einstein’s equation. Plotting stopping energy against frequency for different metals, he found perfectly straight lines, every one with slope h — the same constant Planck had extracted from glowing objects. The intercepts gave each metal’s work function. Einstein received the 1921 Nobel Prize for this equation, not for relativity.

Light pushes

p = mv is a rule for slow matter, not for light, so a massless photon is not a momentum-less one: it carries p = E/c = h/λ, and a surface that absorbs it feels the recoil. Absorb a beam of power P for one second and it delivers P/c of momentum; reflect it instead and the momentum change doubles, which is why a mirrored solar sail feels twice the push of a black one. The numbers are small — full sunlight on a square metre of mirror pushes with about 9 μN — but they act continuously and for free, which is enough to sail on.

Compton: billiards with light

Fire X-rays into graphite and some emerge with a longer wavelength than they went in with, by an amount that depends only on the angle you look from. A wave cannot do that: an electron shaken by a wave re-radiates at the frequency that shook it. A particle can. Treat the photon as a projectile carrying energy hf and momentum h/λ, collide it elastically with a free electron, and conserving both gives λf − λi = (h/mₑc)(1 − cos θ). The photon hands over some energy, so its wavelength grows; the electron recoils with the rest. The shift is independent of the incoming wavelength and of the target material, which is exactly why it settled the argument.

Change one variable at a time

Make the relationship visible.

Light frequency f1.00 × 10¹⁵ HzWork function φ (choose the metal)2.3 eVIntensity50%

Caesium φ ≈ 1.9, sodium 2.3, zinc 4.3, platinum 5.5 eV. Slide the intensity and watch what does — and does not — change: only the current responds, never the electron energy or the threshold.

Photon energy hf4.14 eV

E(max) = hf − φ1.84 eV

Stopping potential V(s)1.84 V

Photocurrent (∝ intensity)50% of maximum

Catch the common trap

Explain before calculating.

Zinc has φ = 4.3 eV. Which beam ejects photoelectrons from it?

Choose an answer to test the model.

Worked examples

State the rule, substitute, then check units.

EasyFind the energy of one photon of 500 nm green light, in joules and electronvolts.

E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (500 × 10⁻⁹).
E = 3.98 × 10⁻¹⁹ J.
Divide by 1.6 × 10⁻¹⁹ J/eV: E ≈ 2.5 eV.

Answer3.98 × 10⁻¹⁹ J ≈ 2.5 eV

MediumA photosensitive metal has a work function of 5.5 eV. Find the minimum (threshold) frequency of light that can free an electron.

At threshold the photon exactly pays the work function: hf₀ = φ.
φ = 5.5 × 1.6 × 10⁻¹⁹ = 8.8 × 10⁻¹⁹ J.
f₀ = 8.8 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ ≈ 1.3 × 10¹⁵ Hz — ultraviolet.

Answerf₀ ≈ 1.3 × 10¹⁵ Hz

HardLight of frequency 2.5 × 10¹⁵ Hz strikes the same φ = 5.5 eV metal. Find E(max) and the stopping potential.

Photon energy: hf = 6.63 × 10⁻³⁴ × 2.5 × 10¹⁵ = 1.66 × 10⁻¹⁸ J.
E(max) = hf − φ = 1.66 × 10⁻¹⁸ − 8.8 × 10⁻¹⁹ = 7.8 × 10⁻¹⁹ J ≈ 4.9 eV.
eV(s) = E(max), so V(s) ≈ 4.9 V — the reverse voltage that just zeroes the current.

AnswerE(max) ≈ 7.8 × 10⁻¹⁹ J (4.9 eV); V(s) ≈ 4.9 V

Challenging540 nm light ejects electrons of maximum kinetic energy 1.9 eV from an unknown metal. Find its work function, and describe the E(max)–f graph for this metal.

Photon energy: E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (540 × 10⁻⁹) = 3.7 × 10⁻¹⁹ J ≈ 2.3 eV.
φ = hf − E(max) = 2.3 − 1.9 = 0.4 eV — an unusually easy metal to strip, like caesium coatings in photocells.
The graph of E(max) against f is a straight line: slope h (identical for every metal), x-intercept f₀ = φ/h, y-intercept −φ. Only the intercepts move between metals.

Answerφ ≈ 0.4 eV; straight line of universal slope h with intercept −φ

Exam diagrams for this topic3 figures to inspect and practiseQuestions, hints and marking points in one compact subsection.

See it. Read it. Work it.

These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.

01
InspectRead the figure comment.
02
TraceFollow labels, arrows and axes.
03
AnswerWork one part at a time.
04
CheckReveal hints and marking points.

A Level

01Fig. 11.1Quantum physics · Nuclear physicsA Level

Figure comment

Fig. 11.1An evacuated tube is drawn as a long horizontal rectangle, closed at its right-hand end by a bar labelled fluorescent screen. Near the left-hand end is a short vertical cathode, and beyond it a vertical anode plate with a gap at its centre; leads from both pass out of the tube to a cell labelled 250 V, whose positive terminal is joined to the anode. An arrow along the axis shows a beam of electrons passing through the gap in the anode and travelling to a thin crystal mounted upright across the middle of the tube. The screen is drawn blank.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

22 Quantum physics 23 Nuclear physics Open A Level question · A Level extension Q11

Guided questions 5 parts

Reading cue. The 250 V acts only between cathode and anode; beyond the anode gap there is no field, so the electrons meet the crystal at the speed they had on leaving it.

aExplain Explain why the cell in Fig. 11.1 is connected with its positive terminal to the anode.
recall2 marks
Check answer 2 marks
1. electrons carry negative charge
2. they are repelled from the negative cathode and attracted to the positive anode, so this connection accelerates them along the tube towards the screen
bDescribe Describe what is seen on the blank fluorescent screen once the electron beam has passed through the thin crystal.
routine3 marks
Check answer 3 marks
1. a set of concentric bright rings
2. centred on the point where the undeviated beam meets the screen
3. the pattern is brightest at the centre, with the rings becoming fainter further out
cDetermine The accelerating p.d. is increased from 250 V to 1000 V. Determine the factor by which the de Broglie wavelength of the electrons changes, and describe the effect on the pattern on the screen.
demanding3 marks
Check answer 3 marks
1. eV = ½mv² and λ = h/mv, so λ ∝ 1/√V
2. V is 4 times greater, so λ is halved: factor 0.50 (7.8 × 10⁻¹¹ m falls to 3.9 × 10⁻¹¹ m)
3. the diffraction angles are smaller, so the rings close in towards the centre of the screen
dSuggest Protons are accelerated through the same 250 V and directed at the same crystal. Suggest why no ring pattern appears on the screen.
top of the paper4 marks
Check answer 4 marks
1. λ = h/√(2meV), so the proton wavelength is smaller than the electron wavelength by √(mp/me) ≈ 43
2. λ ≈ 1.8 × 10⁻¹² m for the protons
3. this is far smaller than the spacing of the atoms in the crystal, so the diffraction angles are too small to be seen
4. (the much heavier protons would also be absorbed within the crystal)

Transfer challenge

Neutrons in a reactor are slowed until their kinetic energy is 0.025 eV. Calculate their de Broglie wavelength, and suggest why such neutrons are useful for investigating crystal structure. (mass of a neutron = 1.67 × 10⁻²⁷ kg)

Check answer 3 marks

E = 0.025 × 1.60 × 10⁻¹⁹ = 4.0 × 10⁻²¹ J, and p = √(2mE) = 3.7 × 10⁻²⁴ N s
λ = h/p = 6.63 × 10⁻³⁴ / 3.7 × 10⁻²⁴ = 1.8 × 10⁻¹⁰ m
this is comparable with the spacing of atoms in a crystal, so the neutrons are strongly diffracted and the pattern reveals that spacing

IB

02Fig. 6.1Quantum physicsIB

Figure comment

Fig. 6.1A photoelectric cell drawn as an evacuated tube. Inside it a flat metal plate stands on the left and a smaller collecting electrode on the right. A beam of monochromatic light enters through the top of the tube and falls on the face of the plate that faces the collector, and an arrow across the vacuum shows photoelectrons travelling from the plate to the collector. Outside the tube the plate is connected through a microammeter and along a wire to a d.c. supply, the positive terminal of which is the one joined back to the collector.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

E.2 Quantum physics Open IB question · Nuclear and quantum physics Q6

Guided questions 5 parts

Reading cue. Follow the supply plus terminal: it reaches the collector, so electrons are being pulled across. As drawn the meter reads a collected current, not anything about a stopping potential.

aState State the direction of the conventional current in the wire joining the plate to the microammeter, and state why the tube must be evacuated.
recall2 marks
Check answer 2 marks
1. Conventional current flows from the plate through the microammeter towards the negative terminal of the supply, opposite to the electron flow in that wire
2. Evacuated so that photoelectrons are not scattered or absorbed by gas molecules before reaching the collector
bDetermine The incident light has a wavelength of 400 nm and the plate has a work function of 2.30 eV. Determine the reverse potential difference that would have to be applied to the collector to bring the microammeter reading to zero, and determine the reading in μA when 1.2 × 10¹² electrons leave the plate each second and all are collected. Use hc = 1240 eV nm.
routine4 marks
Check answer 4 marks
1. Photon energy = 1240/400 = 3.10 eV
2. Maximum kinetic energy = 3.10 − 2.30 = 0.80 eV
3. Stopping potential difference = 0.80 V, with the collector made negative
4. Current = 1.2 × 10¹² × 1.60 × 10⁻¹⁹ = 1.9 × 10⁻⁷ A, that is 0.19 μA
cSketch Sketch the variation of the microammeter reading with the potential difference applied to the collector, from a reverse value, through zero, to a forward value large enough for the reading to stop rising. On the same axes sketch the result of replacing the light with light of shorter wavelength delivering the same number of photons per second.
demanding4 marks
Check answer 4 marks
1. First curve rises from zero at a negative potential difference and levels off at a constant saturation current once the collector is positive
2. The arrangement drawn, with the collector held positive, lies on the flat saturated part of that curve
3. Second curve saturates at the same current, because the number of photons arriving each second is unchanged
4. Second curve meets the potential-difference axis at a more negative value, because the photoelectrons now leave with greater maximum kinetic energy
dDiscuss The collecting electrode is drawn small and clear of the light beam. Discuss what the microammeter would record if it were enlarged so that it faced the whole plate and was itself illuminated.
top of the paper5 marks
Check answer 5 marks
1. Light reaching the enlarged collector ejects photoelectrons from the collector as well as from the plate, provided its work function is small enough for the wavelength used
2. With the collector held positive, as drawn, the field between the electrodes returns those electrons to the collector, so the saturation reading is barely altered by them
3. An electrode large enough to face the whole plate does stand in the path of the beam, so less light reaches the plate and the saturation current falls for that reason instead
4. Once the potential difference is reversed to look for the stopping potential, the field then drives the collector's own photoelectrons across to the plate, giving a current in the opposite sense
5. The reading therefore does not fall to zero at the true stopping potential difference, so any maximum kinetic energy obtained from it would be wrong

Transfer challenge

In a separate experiment the stopping potential difference is measured for light of several frequencies falling on one metal, and the graph of stopping potential difference against frequency is a straight line of gradient 4.1 × 10⁻¹⁵ V s. Determine the value of Planck's constant this gives, and state what the intercept on the stopping-potential axis represents.

Check answer 4 marks

eV = hf − φ, so the plotted gradient is h/e
h = 1.60 × 10⁻¹⁹ × 4.1 × 10⁻¹⁵
h = 6.6 × 10⁻³⁴ J s
Intercept on the stopping-potential axis is −φ/e, the work function of the metal expressed in volts, with sign reversed

03Fig. 10.1Quantum physicsIB

Figure comment

Fig. 10.1Three parallel rays of light of wavelength 420 nm slant down to the right and meet the flat upper face of a block labelled as a caesium surface with a work function of 2.1 eV. From a point on that same face, further to the right of where the light lands, a single arrow slants up and to the right, marking one photoelectron leaving the metal.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

E.2 Quantum physics Open IB question · Nuclear and quantum physics Q10

Guided questions 5 parts

Reading cue. Only the 420 nm label fixes the photon energy, the slant of the rays does not. The single drawn arrow is one electron among many, not necessarily the fastest one to leave.

aState State the minimum energy that must be given to a single electron for it to leave the surface drawn, and state how that energy reaches the electron.
recall2 marks
Check answer 2 marks
1. 2.1 eV, the work function labelled on the block
2. From one photon of the incident light, absorbed whole in a single one-to-one interaction
bDetermine The rays drawn represent a beam delivering 1.5 mW to the surface. Determine the number of photons striking the surface each second. Use hc = 1240 eV nm.
routine4 marks
Check answer 4 marks
1. Photon energy = 1240/420 = 2.95 eV
2. = 2.95 × 1.60 × 10⁻¹⁹ = 4.72 × 10⁻¹⁹ J
3. Rate = 1.5 × 10⁻³/(4.72 × 10⁻¹⁹)
4. = 3.2 × 10¹⁵ photons per second
cDetermine The photoelectron drawn leaves at 60° to the surface, carrying the maximum possible kinetic energy. A uniform retarding field of 500 V m⁻¹ is now applied at right angles to the surface. Determine how far from the surface this electron travels before it stops moving away from it.
demanding4 marks
Show a hint
Resolve the velocity into components along and normal to the surface; the field acts on only one of them.
Check answer 4 marks
1. Maximum kinetic energy = 2.95 − 2.1 = 0.85 eV
2. Only the velocity component normal to the surface is retarded, so the energy to be removed is 0.85 sin²60° = 0.64 eV
3. Distance = energy in eV divided by the field in V m⁻¹, d = 0.64/500
4. d = 1.3 × 10⁻³ m (1.3 mm), the electron still moving parallel to the surface at that moment
dDiscuss The single electron is drawn leaving the surface at a point some distance along from where the rays land. Discuss how faithful this drawing is to the photon model of the effect.
top of the paper4 marks
Check answer 4 marks
1. In the photon model one photon is absorbed by one electron and emission follows with no measurable delay, so electrons leave from the illuminated region itself
2. The drawn separation implies the energy travels along the surface before emission, which the model does not allow and which would introduce a delay that is not observed
3. Only one arrow is drawn, whereas photoelectrons leave in all directions above the surface
4. The drawn electron need not carry the maximum kinetic energy either: electrons freed below the surface lose energy on the way out, giving a spread of energies from zero up to 0.85 eV

Transfer challenge

In an X-ray tube, electrons are accelerated from rest through 25 kV and stopped abruptly in a metal target. Determine the shortest wavelength present in the X-rays produced, and explain why no shorter wavelength appears however long the tube is left running.

Check answer 4 marks

Each electron arrives at the target with 25 keV of kinetic energy
Shortest wavelength arises when one electron gives all of that energy to a single photon: λ = 1240/25000 nm
λ = 5.0 × 10⁻² nm (4.96 × 10⁻¹¹ m)
A shorter wavelength would require a photon of more than 25 keV, which no single electron can supply, so the continuous spectrum ends sharply at this wavelength

Cambridge 9702 · focused teaching and practice

Complete the objective

Open a topic for its explanation, worked example and original practice. The AS and additional A-Level requirements remain separate. Practical preparation does not replace supervised laboratory work.

Additional A-Level · 22.2.2

Threshold frequency and maximum threshold wavelength

Teaching, example & practice

General browsing: select Cambridge AS or A2 and your examination year for a recommended route.

A surface's threshold frequency f₀ is the minimum light frequency that can eject photoelectrons in the ordinary single-photon photoelectric model. At threshold a photon supplies just the work function Φ, leaving zero maximum kinetic energy.

Because c = fλ, the threshold wavelength λ₀ is the longest wavelength that can cause emission, not the shortest. Emission is possible when f ≥ f₀ or equivalently λ ≤ λ₀. Light with a larger wavelength has smaller photon energy and cannot overcome the work function, even if its intensity is increased within this model.

hf₀ = Φ

λ₀ = c/f₀ = hc/Φ

Emission: f ≥ f₀ ⇔ λ ≤ λ₀

Worked example

A metal has f₀ = 5.0 × 10¹⁴ Hz. Find λ₀ and decide whether 500 nm and 700 nm light can emit photoelectrons. Use c = 3.0 × 10⁸ m s⁻¹.

λ₀ = 3.0 × 10⁸/(5.0 × 10¹⁴) = 6.0 × 10⁻⁷ m = 600 nm.
500 nm is shorter than 600 nm, so its photons exceed the threshold energy.
700 nm is longer than 600 nm, so no photoemission is predicted by the single-photon model.

λ₀ = 600 nm; 500 nm can cause emission, 700 nm cannot.

Try the original practice · 6 marks →Objective and coverage record

Scope and sources

Original GioPhysics teaching and practice, checked against the syllabus. This section does not certify unrelated videos, downloads or the rest of a shared lesson.

Cambridge syllabus · printed page 35 — 2025–2027: verified numbered demand on the printed page; teaching and questions are newly authored.
Cambridge syllabus · printed page 35 — 2028–2030: checked directly against the downloaded official PDF; the targeted demand is unchanged.

Additional A-Level · 22.3.1

What light's wave and particle evidence each explains

Teaching, example & practice

General browsing: select Cambridge AS or A2 and your examination year for a recommended route.

Interference and diffraction test how light propagates. In a double-slit arrangement, coherent contributions overlap to make alternating bright and dark fringes. The dark regions are evidence of cancellation through superposition, and changing wavelength changes the fringe spacing. A slit or obstacle also produces spreading and a diffraction pattern; these observations support a wave description.

The photoelectric effect tests how light transfers energy to matter. Electrons are emitted only when the frequency reaches a material's threshold, and their maximum kinetic energy increases with frequency rather than intensity. A photon of energy hf transfers energy in one interaction; it must supply the work function before any energy is available as electron kinetic energy.

Wave–particle duality does not mean choosing one experiment and declaring the other false. Wave behaviour describes the interference/diffraction probabilities, while energy is exchanged in discrete photon events. For this objective, compare evidence from light itself; electron diffraction is separate evidence about matter waves.

Observation with light	Key feature	What it supports
Two-slit interference	Bright/dark pattern from reinforcement/cancellation	Wave superposition
Single-slit diffraction	Spreading and a structured intensity pattern	Wave propagation
Photoelectric emission	Frequency threshold and frequency-dependent maximum electron energy	Discrete photon energy transfer

Worked example

One experiment produces dark fringes where two beams overlap. Another finds that intense red light emits no electrons from a surface but weak ultraviolet does. Explain which model addresses each.

Dark overlap regions are explained by destructive interference of waves.
Ultraviolet photons have larger hf and can exceed the work function; red photons may each fall below it.
Intensity changes the number of photons arriving per second, not each photon's energy at fixed frequency.

Interference supports wave behaviour; the photoelectric threshold supports photon energy transfer.

Try the original practice · 6 marks →Objective and coverage record

Scope and sources

Original GioPhysics teaching and practice, checked against the syllabus. This section does not certify unrelated videos, downloads or the rest of a shared lesson.

Cambridge syllabus · printed page 35 — 2025–2027: verified numbered demand on the printed page; teaching and questions are newly authored.
Cambridge syllabus · printed page 35 — 2028–2030: checked directly against the downloaded official PDF; the targeted demand is unchanged.

Extra challenging questions

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