University Physics IV · Limits of Classical Physics · 1.6

The photoelectric effect and the light quantum

One lamp dimmed and one lamp bright, same colour: the bright one frees more electrons, never faster ones. That single stubborn fact is what forces light to hand over its energy in lumps of hν — and it is why a stopping-potential graph gives you Planck's constant honestly and a work function only if you read it right.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Classical electromagnetism makes three checkable predictions about light falling on a metal, and measurement refuses all three. A wave delivers energy continuously and spreads it over the whole surface, so a weak beam should need an accumulation time — minutes, for microwatt irradiance on an atom-sized catchment — yet emission is seen inside a nanosecond. A wave of any frequency should eventually do the job given power and patience; instead there is a threshold ν₀ below which no current flows at any ordinary intensity.

And a stronger field should shake electrons out faster; instead intensity changes only how many leave per second, never how fast. Einstein's move is one sentence: light exchanges energy with matter in quanta of hν, one quantum absorbed by one electron, and the electron pays the surface's work function φ to escape, leaving Kₘₐₓ = hν − φ. Everything measurable follows — the threshold at φ/h, a maximum energy that climbs linearly with frequency at a slope every metal shares, a saturation current that alone tracks intensity.

Two assumptions ride inside the equation, single-photon absorption and a cold clean emitter, and one experimental trap sits outside it: a photocell's voltmeter reads the stopping potential against the collector's work function, not the emitter's, so the slope of the fit measures h while the intercept never measures the cathode's φ.

Simple definition: The photoelectric effect is the emission of electrons from an illuminated surface in which one light quantum of energy hν is absorbed by one electron, which then pays the work function φ to escape: Kₘₐₓ = hν − φ.
Example: Sodium has φ = 2.28 eV, so 400 nm light (hc/λ = 1240/400 = 3.10 eV) gives Kₘₐₓ = 0.82 eV; at 600 nm the photon carries only 2.07 eV, below φ, and no current flows however bright the lamp.

Einstein's photoelectric equationKₘₐₓ = hν − φ

One line holds the whole effect: a slope that is the same for every metal, and an offset that belongs to one surface.

Kₘₐₓ in eV or J, ν in Hz, φ the emitter's work function; h = 4.136 × 10⁻¹⁵ eV s = 6.626 × 10⁻³⁴ J s.

Threshold frequency and wavelengthν₀ = φ/h, λ₀ = hc/φ = (1240 eV nm)/φ[eV]

Decides whether there is any current at all. No lamp power, exposure time or beam area shifts it by a nanometre.

Caesium φ = 2.10 eV → λ₀ = 590 nm; copper φ = 4.70 eV → λ₀ = 264 nm.

Photon energy in laboratory unitsE = hν = hc/λ = (1240 eV nm)/λ[nm]

Turns a filter wavelength into an energy in one step: 365 nm → 3.40 eV, 405 nm → 3.06 eV, 546 nm → 2.27 eV.

λ in nm returns E in eV directly; 1 eV = 1.602 × 10⁻¹⁹ J.

What the stopping potential measureseVₛ = hν − φ_A

Slope h/e = 4.136 × 10⁻¹⁵ V s gives h to a fraction of a percent; the intercept is −φ_A/e, and no amount of cleaning the cathode changes it.

Vₛ is the retarding p.d. that kills the current; φ_A is the collector's work function, not the emitter's.

Saturation current and quantum efficiencyIₛₐₜ = η e P / (hν)

The plateau alone carries the intensity: double P and Iₛₐₜ doubles, while Vₛ does not shift by a microvolt.

P is absorbed optical power in W, η electrons per photon (10⁻⁴ to 10⁻² for a clean metal), Iₛₐₜ in A.

Classical accumulation timet ≈ φ / (I · Aₐₜₒₘ), Aₐₜₒₘ ≈ 3 × 10⁻²⁰ m²

Puts a number on 'instantaneous': the wave picture asks for twenty minutes where experiment bounds the delay below a nanosecond.

I is irradiance in W m⁻². At 1.0 μW cm⁻² = 10⁻² W m⁻² with φ = 2.1 eV, t ≈ 1.1 × 10³ s.

What the wave picture predicts, and how badly

Treat light as a classical wave and three predictions follow. First, energy arrives continuously and spreads over the whole surface, so a weak beam must be gathered before any electron can leave: at an irradiance of 1.0 μW cm⁻² = 1.0 × 10⁻² W m⁻², with an atom-sized catchment of about 3 × 10⁻²⁰ m², the 2.1 eV = 3.4 × 10⁻¹⁹ J needed to free a caesium electron takes t ≈ 3.4 × 10⁻¹⁹ / (3 × 10⁻²²) ≈ 1.1 × 10³ s, close to twenty minutes. Second, the field amplitude sets the force on an electron, so red light should work as well as blue given enough power and time — there can be no threshold. Third, a brighter beam pushes harder, so Kₘₐₓ should climb with intensity. Exactly one classical prediction survives contact with the data: that the number of electrons per second is proportional to intensity. It survives because it is a statement about counting, not about energy.

Four measurements, and which prediction each one kills

The onset of photocurrent has been bounded below 10⁻⁹ s at irradiances where the wave estimate demands minutes — prediction one dead by twelve orders of magnitude. Every clean surface shows a threshold: caesium stops responding beyond about 590 nm however bright the lamp, and the threshold does not move when lamp power is raised by four decades. That kills prediction two. Plot maximum kinetic energy against frequency and you get a straight line whose slope is identical for caesium, sodium, zinc and platinum while the intercept differs, and the line stays put when the intensity is changed. That kills prediction three, and the slope is also the clue, because a slope shared by every metal cannot be a property of any metal. The fourth measurement is the classical survivor: at fixed frequency the saturation current is proportional to intensity. Frequency governs the energy per event, intensity the number of events, and no continuous-wave model separates the two knobs that cleanly.

One quantum, one electron, and the escape fee φ

Einstein's 1905 proposal is a rule about exchange: light of frequency ν gives up energy only in units of hν, and one such unit is absorbed by one electron. The electron must then pay to leave. The work function φ is the minimum energy needed to take an electron from the top of the metal's filled levels — the Fermi level — to rest just outside the surface, so Kₘₐₓ = hν − φ. Two things are built into the word maximum. Electrons starting below the Fermi level need more than φ, so the emitted electrons run over a continuous spread from zero up to Kₘₐₓ, and only the upper edge carries the physics. And φ is a property of the surface, not the bulk: two crystal faces of tungsten differ by about 0.7 eV, an oxide layer raises it, and a monolayer of caesium on tungsten drops it from roughly 4.5 eV to below 2 eV. Numbers follow at once. Potassium, φ = 2.30 eV, under the 365 nm mercury line: E = 1240/365 = 3.40 eV, so Kₘₐₓ = 1.10 eV and the threshold sits at 1240/2.30 = 539 nm.

Reading the I–V curve: saturation, cut-off, and the tail

The apparatus is a photocell: an illuminated cathode, a collecting anode, a variable bias and a picoammeter. Sweep the bias forward and the current rises to a plateau — the saturation current, where every emitted electron is collected. Its height is proportional to intensity and says nothing about energy. Sweep the bias backward and the current falls as the retarding field turns back progressively faster electrons, reaching zero at the stopping potential Vₛ. That single voltage is the energy measurement, because it selects the fastest electron in the distribution. Raise the intensity tenfold and the plateau rises tenfold while Vₛ does not move: the cleanest single statement of the whole effect. Real curves are messier. Thermal spread at the emitter (kT = 0.026 eV at 300 K) rounds the cut-off, and stray light on the anode drives a small reverse photocurrent, so the measured curve crosses zero slightly early. Both are why a cold, clean surface is part of the model, and both are systematics you must bound before quoting a value of h.

Why the intercept belongs to the collector, not the emitter

Connect the two electrodes through the external circuit and, at zero applied voltage, their Fermi levels align, so the difference in work functions appears as a contact potential difference across the gap — and it is already inside every reading the voltmeter gives, because a voltmeter compares Fermi levels, not vacuum levels. Do the bookkeeping from the cathode's Fermi level. The fastest emitted electron leaves the cathode with total energy φ_C + Kₘₐₓ = hν above it. A retarding potential Vₛ lifts the anode's Fermi level to eVₛ above that reference, and the anode's vacuum level to eVₛ + φ_A. The electron just fails to arrive when hν = eVₛ + φ_A, so eVₛ = hν − φ_A: φ_C has cancelled exactly, and the fit returns the collector's work function. The slope is untouched, because a constant offset shifts a line without tilting it — which is why Millikan, who measured the contact potential difference separately and corrected for it, could still extract h to about half a percent from the slope alone. When φ_A > φ_C there is a band of frequencies above the cathode's threshold where Vₛ comes out negative: current flows, and a small accelerating bias is needed to collect it.

Two assumptions inside the equation, and where each fails

Kₘₐₓ = hν − φ assumes that one electron absorbs one photon, and that the emitter is cold and clean. The first holds at laboratory irradiance because a photoexcited electron shares its energy with the rest of the metal within about 10 fs, and a second photon almost never lands on the same electron inside that window: at 1 W m⁻² of 3 eV light the photon flux is about 2 × 10¹⁸ m⁻² s⁻¹, so an atom-sized area of 3 × 10⁻²⁰ m² receives roughly 0.06 photons per second, and the chance of a second one within 10 fs is of order 10⁻¹⁵. Focus a femtosecond pulse to 10¹⁵ W m⁻² and that chance becomes of order one: two-photon photoemission then produces current below the threshold, scaling as the square of the intensity, and the threshold is gone. The second assumption fails more gently. At finite temperature the Fermi edge is smeared over a few kT, so a few electrons start above E_F and escape at photon energies slightly below φ; the wall becomes a rounded knee about kT wide, 0.026 eV at room temperature, and Fowler's 1931 analysis makes the near-threshold yield T² times a universal function of (hν − φ)/kT that rises as (hν − φ)² once well above threshold. Clean means atomically clean: adsorbed oxygen or water shifts φ by up to an electronvolt, which is why Millikan shaved his alkali surfaces with a knife inside the evacuated tube before each run. Once both assumptions hold, the fit establishes that energy is exchanged in units of hν, and no more; whether the field itself is quantised is a separate question, taken up later in the unit.

Change one variable at a time

Make the relationship visible.

Interactive model

Photon frequency ν8.2 ×10¹⁴ Hz

Cathode work function φ_C2.1 eV

Anode work function φ_A2.6 eV

Drag φ_A alone: the line slides up and down but never tilts, so the intercept moves while the slope h/e does not. Drag φ_C alone: only the threshold marker and the line's left end move. Set φ_A above φ_C and hunt for the band where Vₛ is negative.

PHOTON ENERGY hν3.39 eV

K max AT CATHODE1.29 eV

STOPPING POTENTIAL0.79 V

CONTACT P.D. (φ_C−φ_A)/e-0.50 V

Live interpretationPHOTON ENERGY hν: 3.39 eV. K max AT CATHODE: 1.29 eV. STOPPING POTENTIAL: 0.79 V. CONTACT P.D. (φ_C−φ_A)/e: −0.50 V

Catch the common trap

Explain before calculating.

Misconception check

The error is reading the emitting surface's work function off the stopping-potential graph. Students fit Vₛ against ν, take the magnitude of the intercept as φ/e, and quote it as the cathode's — then are puzzled when a caesium photocathode returns 2.6 eV instead of 2.10 eV, or when the same cathode gives different answers in two different tubes. The voltmeter measures the difference between the electrodes' Fermi levels, not their vacuum levels, so the contact potential difference (φ_C − φ_A)/e rides on every reading and cancels φ_C out of the algebra: eVₛ = hν − φ_A. Because that offset is the same at every frequency it leaves the slope, and so h, untouched. To get the cathode's φ, find the threshold frequency at which photocurrent first appears, or measure the contact potential difference separately with a Kelvin probe.

A photocell has a caesium cathode (φ_C = 2.10 eV) and a copper anode (φ_A = 4.70 eV). The stopping potential Vₛ is measured at several frequencies, plotted against ν, and a straight line is fitted. What do the slope and the frequency-axis intercept of that line give?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe 365 nm line of a mercury lamp falls on a potassium surface, φ = 2.30 eV. Find the photon energy, the maximum kinetic energy of the emitted electrons, their maximum speed, and the longest wavelength that produces any current at all. Take hc = 1240 eV nm.

Photon energy in laboratory units: E = hc/λ = 1240 eV nm / 365 nm = 3.40 eV.
One quantum to one electron, and the electron pays φ to leave: Kₘₐₓ = 3.40 − 2.30 = 1.10 eV.
Convert and use K = ½mv²: 1.10 eV = 1.10 × 1.602 × 10⁻¹⁹ = 1.76 × 10⁻¹⁹ J, so vₘₐₓ = √(2 × 1.76 × 10⁻¹⁹ / 9.11 × 10⁻³¹) = √(3.86 × 10¹¹) = 6.2 × 10⁵ m s⁻¹ — only 0.21% of c, so the non-relativistic formula is safe.
Threshold: current stops when hc/λ falls to φ, at λ₀ = 1240/2.30 = 539 nm. The 546 nm green line is already too red, however bright the lamp and however long you wait.

AnswerE = 3.40 eV, Kₘₐₓ = 1.10 eV, vₘₐₓ = 6.2 × 10⁵ m s⁻¹, λ₀ = 539 nm.

MediumIn a photocell with a caesium cathode (φ_C = 2.10 eV) the stopping potential is 0.90 V at 365.0 nm and 2.39 V at 253.7 nm. Take c = 2.998 × 10⁸ m s⁻¹ and e = 1.602 × 10⁻¹⁹ C. Find Planck's constant from the slope, find the work function the intercept returns, and say which electrode that work function belongs to.

Convert to frequency: ν₁ = c/λ₁ = 2.998 × 10⁸ / 365.0 × 10⁻⁹ = 8.214 × 10¹⁴ Hz, and ν₂ = 2.998 × 10⁸ / 253.7 × 10⁻⁹ = 1.1817 × 10¹⁵ Hz.
Slope of Vₛ against ν: (2.39 − 0.90) V ÷ (1.1817 × 10¹⁵ − 8.214 × 10¹⁴) Hz = 1.49 ÷ 3.603 × 10¹⁴ = 4.135 × 10⁻¹⁵ V s.
That slope is h/e, so h = 4.135 × 10⁻¹⁵ × 1.602 × 10⁻¹⁹ = 6.624 × 10⁻³⁴ J s — 0.03% below the accepted 6.626 × 10⁻³⁴ J s, a shortfall that rounding each Vₛ to 0.01 V fully accounts for.
Intercept from either point: φ/e = (h/e)ν₁ − Vₛ₁ = 4.135 × 10⁻¹⁵ × 8.214 × 10¹⁴ − 0.90 = 3.40 − 0.90 = 2.50 V. The second point gives 4.89 − 2.39 = 2.50 V, so the two agree: the fit returns 2.50 eV.
That 2.50 eV is the anode's work function, because eVₛ = hν − φ_A. The caesium cathode's 2.10 eV never enters the fit; it appears only as the frequency at which current first flows, ν₀ = 2.10 / (4.136 × 10⁻¹⁵) = 5.08 × 10¹⁴ Hz, that is 590 nm.
Consequence worth checking: at the 546.1 nm green line hν = 1240/546.1 = 2.27 eV. Electrons do escape, since 2.27 > 2.10, but Vₛ = 2.27 − 2.50 = −0.23 V — you must apply 0.23 V of accelerating bias to collect them at all.

Answerh = 6.624 × 10⁻³⁴ J s from the slope; the intercept gives 2.50 eV, which is the anode's work function, not the caesium cathode's 2.10 eV.

HardA 405 nm laser diode delivers 1.50 mW onto a caesium photocathode (φ_C = 2.10 eV) in a tube whose anode has φ_A = 2.60 eV. The quantum efficiency is 0.30% (electrons out per photon in) and the beam is focused to a 1.0 mm diameter spot. Find the photon arrival rate, the saturation current, the electrons' maximum kinetic energy, the stopping potential the voltmeter will actually show, and the delay a classical wave model would predict.

Photon energy: E = 1240/405 = 3.06 eV = 3.06 × 1.602 × 10⁻¹⁹ = 4.90 × 10⁻¹⁹ J.
Photon rate: N = P/E = 1.50 × 10⁻³ ÷ 4.90 × 10⁻¹⁹ = 3.06 × 10¹⁵ photons per second.
Saturation current: Iₛₐₜ = ηeN = 0.0030 × 1.602 × 10⁻¹⁹ × 3.06 × 10¹⁵ = 1.47 × 10⁻⁶ A = 1.47 μA. It is fixed entirely by the photon rate; the colour was already spent in step 1.
Electron energy at the emitting surface: Kₘₐₓ = hν − φ_C = 3.06 − 2.10 = 0.96 eV.
What the voltmeter reads is not that. eVₛ = hν − φ_A = 3.06 − 2.60, so Vₛ = 0.46 V. The 0.50 V shortfall is the contact potential difference (φ_C − φ_A)/e = −0.50 V, and it sits in every reading at every frequency.
Classical check: irradiance = 1.50 × 10⁻³ ÷ π(0.50 × 10⁻³)² = 1.50 × 10⁻³ ÷ 7.85 × 10⁻⁷ = 1.9 × 10³ W m⁻², about twice noon sunlight. With a 3 × 10⁻²⁰ m² catchment one electron absorbs 5.7 × 10⁻¹⁷ W, so gathering φ_C = 3.36 × 10⁻¹⁹ J takes 5.9 × 10⁻³ s — six milliseconds. The measured onset is under a nanosecond.

AnswerN = 3.06 × 10¹⁵ s⁻¹, Iₛₐₜ = 1.47 μA, Kₘₐₓ = 0.96 eV, but the measured stopping potential is only 0.46 V because the intercept carries φ_A = 2.60 eV. The classical delay would be ≈ 6 ms against an observed onset below 1 ns.