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.
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
01
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.
02
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.
03
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.
04
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.
05
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.
02
Change one variable at a time
Make the relationship visible.
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
03
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.
04
Worked examples
State the rule, substitute, then check units.
EasyFind the energy of one photon of 500 nm green light, in joules and electronvolts.
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.
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.Open subsectionHide 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.
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.
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
electrons carry negative charge
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
a set of concentric bright rings
centred on the point where the undeviated beam meets the screen
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
eV = ½mv² and λ = h/mv, so λ ∝ 1/√V
V is 4 times greater, so λ is halved: factor 0.50 (7.8 × 10⁻¹¹ m falls to 3.9 × 10⁻¹¹ m)
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
λ = h/√(2meV), so the proton wavelength is smaller than the electron wavelength by √(mp/me) ≈ 43
λ ≈ 1.8 × 10⁻¹² m for the protons
this is far smaller than the spacing of the atoms in the crystal, so the diffraction angles are too small to be seen
(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
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.
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
Conventional current flows from the plate through the microammeter towards the negative terminal of the supply, opposite to the electron flow in that wire
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
Photon energy = 1240/400 = 3.10 eV
Maximum kinetic energy = 3.10 − 2.30 = 0.80 eV
Stopping potential difference = 0.80 V, with the collector made negative
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
First curve rises from zero at a negative potential difference and levels off at a constant saturation current once the collector is positive
The arrangement drawn, with the collector held positive, lies on the flat saturated part of that curve
Second curve saturates at the same current, because the number of photons arriving each second is unchanged
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
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
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
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
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
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
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.
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
2.1 eV, the work function labelled on the block
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
Photon energy = 1240/420 = 2.95 eV
= 2.95 × 1.60 × 10⁻¹⁹ = 4.72 × 10⁻¹⁹ J
Rate = 1.5 × 10⁻³/(4.72 × 10⁻¹⁹)
= 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
Maximum kinetic energy = 2.95 − 2.1 = 0.85 eV
Only the velocity component normal to the surface is retarded, so the energy to be removed is 0.85 sin²60° = 0.64 eV
Distance = energy in eV divided by the field in V m⁻¹, d = 0.64/500
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
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
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
Only one arrow is drawn, whereas photoelectrons leave in all directions above the surface
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