The Photoelectric Effect
and the Photon Hypothesis
Small packets of energy. Big discoveries.
Explore the threshold, measure the fastest electrons, and use stopping potential to connect light frequency with energy transfer.
Same laws. A bigger universe.
Build the model: light in, electrons out
1Observed facts
In the ordinary, low-intensity photoelectric regime:
- Below a threshold frequency, no electrons escape.
- Maximum kinetic energy depends on frequency, not intensity.
- Emission is prompt, without a long energy build-up delay.
2Einstein’s explanation
A light quantum transfers hν. Part pays the energy cost of escape.
eVs = hν − φ
ν₀ = φ/h
λ₀ = hc/φ
φ: work function; e: positive elementary charge. Vs is the positive magnitude of the ideal retarding cutoff, for emission above threshold.
One quantum. One energy balance.
3I–V curves
Increase power at fixed frequency: more current, the same stopping potential.
V means collector minus emitter. Current ends at V = −Vs. Contact effects are corrected here.
4Why the classical account fails
A simple continuous energy-collection picture predicts a long delay under weak illumination, and no strict frequency threshold.
For φ = 2.3 eV, S = 0.01 W m⁻² and σ = 10⁻¹⁹ m², this naive estimate is about 370 s.
Such slow accumulation is not observed. This estimate illustrates the failed model; it is not a measured response time.
Interactive lab: the photoelectric effect
Change one variable. Predict first, then read the graph.
Live results
- Photon energy hν
- 3.309 eV
- Maximum kinetic energy
- 0.849 eV
- Ideal stopping potential Vs
- 0.849 V
- Threshold frequency ν₀
- 5.948 ×10¹⁴ Hz
- Threshold wavelength λ₀
- 504.0 nm
- Saturation current Isat
- 30.2 μA
Emission allowed. Try changing power: the current changes; the energy cutoff does not.
Model scope: monochromatic light, zero temperature, one-photon absorption, constant efficiency and contact-potential correction. The I–V shape is illustrative, not measured data; saturation is taken at V ≥ 0. Real work functions depend on surface preparation. Presets are representative OpenStax values. At fixed power and efficiency, higher frequency means fewer incident photons per second.
Measuring Planck’s constant
Slope carries h; the intercept needs care.
- Measure cutoff at several frequencies above threshold.
- Fit a straight line, with voltage and frequency uncertainties.
- Multiply the slope in V Hz⁻¹ by e to estimate h.
- A stable contact step shifts the intercept, not the ideal slope.
In the modern SI, h and e are exact defining constants. This experiment tests consistency and calibration; it no longer defines their numerical values.
Try the five-reading fit →Contact potential
Optional apparatus detail · emitter ≠ collector.
Let φE and φC be the two work functions. The contact energy step is φC − φE.
eVr = hν − φC
Vr is signed retardation; applied cutoff V = −Vr. Emission still requires hν > φE. An uncorrected intercept need not give the emitter’s work function.
Explore the contact-potential example
Kmax = 0.354 eV; signed cutoff V = +0.166 V. An accelerating collector voltage is needed before current arrives.
V = collector minus emitter. Uses exact SI h and e; assumes ideal electrodes with a stable contact step.
Scientific interpretation
Separate evidence from the model used to explain it.
- The observations support energy exchange in steps tied to hν.
- Einstein’s light-quantum picture explains the threshold and energy relation.
- Quantised matter driven by a classical field can also reproduce the ordinary photoelectric energy law.
- This experiment alone does not establish every property of a quantised electromagnetic field.
The next topic, Compton scattering, adds evidence involving photon momentum.
The photoelectric effect exposed a limit of fully classical physics. A useful model must also be clear about what its evidence does—and does not—show.
Catch the common trap
The fastest electron is not the average electron.
- Different initial states and scattering produce a range of emitted energies.
- The cutoff measures the upper edge of that range.
- More power at the same frequency increases the arrival rate, not Kmax.
- At fixed power and constant efficiency, raising frequency reduces photon flux.
Practice & worked examples
Preserved university problems, with the assumptions made explicit.
Easy · Threshold wavelengths
A potassium photocathode (φ = 2.29 eV) is lit in turn by three mercury lines, 404.7 nm, 546.1 nm and 577.0 nm, each at the same power. Which lines eject electrons, and what stopping potential does each of those give? Check that the fastest electrons can be treated non-relativistically.
View solution →
- The threshold is λ₀ = hc/φ = (1239.8 eV nm)/(2.29 eV) = 541.4 nm, equivalently ν₀ = φ/h = 5.54 × 10¹⁴ Hz, so only light shorter than 541 nm can eject anything.
- Energy per quantum, hν = hc/λ: 404.7 nm gives 1239.8/404.7 = 3.064 eV; 546.1 nm gives 2.270 eV; 577.0 nm gives 2.149 eV.
- Only 404.7 nm clears 2.29 eV. The 546.1 nm line misses by just 0.020 eV, and raising its power does not help: every absorption still hands one electron 2.270 eV.
- For 404.7 nm, Kₘₐₓ = 3.064 − 2.29 = 0.774 eV, so eVₛ = 0.774 eV and Vₛ = 0.77 V, rounded to the 0.01 eV to which φ is known.
- Check: vₘₐₓ = √(2Kₘₐₓ/mₑ) = √(2 × 0.774 × 1.602 × 10⁻¹⁹ J ÷ 9.109 × 10⁻³¹ kg) = 5.2 × 10⁵ m s⁻¹, only 0.0017c, so K = ½mv² is safe.
Only the 404.7 nm line ejects electrons, with Vₛ = 0.77 V. The 546.1 nm and 577.0 nm lines lie beyond the 541 nm threshold and eject none at any power.
These thresholds assume zero temperature and the stated surface work function. The near-threshold 546.1 nm case can be sensitive to thermal broadening and surface conditions.
Medium · Determine h from a fit
Stopping potentials measured with five mercury lines are 1.38 V at 365.0 nm, 1.08 V at 404.7 nm, 0.83 V at 435.8 nm, 0.28 V at 546.1 nm and 0.14 V at 577.0 nm. Fit Vₛ against ν by least squares, and find h with its standard uncertainty, the work function the intercept implies, and the threshold wavelength.
View solution →
- Convert with ν = c/λ: 8.214, 7.408, 6.879, 5.490 and 5.196 × 10¹⁴ Hz. The means are ν̄ = 6.637 × 10¹⁴ Hz and V̄ = 0.742 V.
- With ν in units of 10¹⁴ Hz, Σ(ν − ν̄)² = 6.532 and Σ(ν − ν̄)(Vₛ − V̄) = 2.685 V. The least-squares slope is their ratio, 2.685/6.532 = 0.4111 V per 10¹⁴ Hz, that is 4.111 × 10⁻¹⁵ V s.
- h = e × slope = (1.602 × 10⁻¹⁹ C)(4.111 × 10⁻¹⁵ V s) = 6.59 × 10⁻³⁴ J s. Dimensions: C V s = J s, as an action should be.
- Residuals from the line are −0.010, +0.021, −0.011, +0.010 and −0.009 V, so s = √(Σr²/(N − 2)) = √(8.65 × 10⁻⁴ V² ÷ 3) = 0.017 V and u(slope) = s/√Σ(ν − ν̄)² = 0.017/√6.532 = 0.0066 V per 10¹⁴ Hz, 1.6%. Hence u(h) = 0.11 × 10⁻³⁴ J s.
- Intercept: V̄ − slope × ν̄ = 0.742 − 0.4111 × 6.637 = −1.99 V, so φ = 1.99 eV. The line crosses Vₛ = 0 at ν₀ = 1.987/0.4111 = 4.83 × 10¹⁴ Hz, so λ₀ = c/ν₀ = 620 nm.
- Check: the accepted 6.626 × 10⁻³⁴ J s lies 0.4 standard uncertainties from the fitted value, so the data agree with it, and a work function near 2 eV is typical of an alkali surface.
h = (6.59 ± 0.11) × 10⁻³⁴ J s. The intercept is −1.99 V, so φ ≈ 1.99 eV, with threshold 4.83 × 10¹⁴ Hz, or 620 nm.
Illustrative exercise readings, not newly collected laboratory data. The quoted uncertainty is the fit’s statistical uncertainty; calibration and contact effects need separate treatment.
Hard · The contact-potential trap
A student's straight-line fit of stopping potential against frequency has slope 4.10 × 10⁻¹⁵ V s and meets the Vₛ axis at −2.38 V. A separate contact-potential measurement shows that the collector's work function exceeds the emitter's by 0.52 eV. Find h, the emitter's work function and its true threshold wavelength. Then decide whether 560 nm light ejects electrons, and what applied voltage first collects them.
View solution →
- h = e × slope = (1.602 × 10⁻¹⁹ C)(4.10 × 10⁻¹⁵ V s) = 6.57 × 10⁻³⁴ J s, 0.9% below the accepted value. The contact potential is the same at every frequency, so it cannot alter the slope.
- Follow the fastest electron. It leaves the emitter with hν − φE, then must climb the applied retarding energy eVᵣ plus the contact step φC − φE. It just fails when hν − φE = eVᵣ + φC − φE, that is when eVᵣ = hν − φC: the emitter's work function cancels.
- So the intercept −2.38 V reports φC = 2.38 eV, and φE = 2.38 − 0.52 = 1.86 eV. The emitter's true threshold is λE = (1239.8 eV nm)/(1.86 eV) = 667 nm, not the 521 nm a naive reading of the intercept gives.
- At 560 nm, hν = 1239.8/560 = 2.214 eV, which exceeds 1.86 eV, so electrons are emitted, with Kₘₐₓ = 2.214 − 1.86 = 0.354 eV.
- Using accepted SI h for this prediction gives eVᵣ = 2.214 − 2.38 = −0.166 eV, a negative stopping voltage. With nothing applied the electrons meet the 0.52 eV contact step with at most 0.354 eV, so none arrive; current starts once an accelerating voltage exceeds 0.52 − 0.354 = 0.166 V.
h = 6.57 × 10⁻³⁴ J s; φE = 1.86 eV, with threshold 667 nm. Light at 560 nm does eject electrons (Kₘₐₓ = 0.35 eV), but they reach the collector only for an accelerating voltage above 0.17 V.
The quoted h is an experimental estimate. The 560 nm prediction and threshold above use accepted SI h for consistency with the displayed contact lab. Using the fitted slope itself gives a cutoff near +0.185 V instead; state which value you use.
Quick check
Five questions · includes the original 400 nm → 300 nm problem · this visit only
Key takeaway & deeper study
Frequency sets energy. Power sets photon arrival rate.
Use hν = φ + Kmax above threshold, then Kmax = eVs for a contact-corrected apparatus. Explain the slope, inspect the intercept, and never assume all electrons leave with the same energy.
Full university study notes · six explanations
Read the current against voltage curve first
Light falls on a clean emitter in vacuum; a collector a few centimetres away is held at a voltage V relative to it, and a meter reads the photocurrent I. Make V positive and every emitted electron is swept across, so I levels off at a saturation value. Make V negative and the collector repels them: only electrons with enough kinetic energy arrive, and at V = −Vₛ the current stops. Energy conservation turns that voltage into an energy, Kₘₐₓ = eVₛ, so a stopping potential of 1.10 V means the fastest electrons left with 1.10 eV = 1.76 × 10⁻¹⁹ J, a speed of 6.2 × 10⁵ m s⁻¹. Three facts come off these curves. Doubling the intensity doubles the saturation current and leaves Vₛ exactly where it was. Changing the frequency moves Vₛ. And below a threshold frequency that depends on the metal, the current is zero at every voltage and every intensity.
What a classical wave predicts, in numbers
A classical wave of intensity S = ½cε₀E₀² drives each electron with a force of amplitude eE₀, so the energy an electron can gather grows with E₀², that is with intensity, and nothing in the account singles out a frequency below which the metal must stay silent. The timing fails worse. Grant the electron the whole cross-section of an atom, σ ≈ 1 × 10⁻¹⁹ m², and illuminate at S = 1.0 × 10⁻² W m⁻², a dim lamp. Power arrives at Sσ = 1 × 10⁻²¹ W, so gathering a work function of 2.3 eV = 3.7 × 10⁻¹⁹ J takes t ≈ φ/(Sσ) ≈ 370 s, six minutes before the first electron could leave. Measurements as early as 1928, by Lawrence and Beams, put any delay below about 3 × 10⁻⁹ s. Brighter light shortens the classical wait but never removes it, and a faint enough source stretches it to days.
One quantum, one electron
Einstein's 1905 proposal replaces the gradual soak with a transaction. Light of frequency ν gives up energy only in lumps hν, and each lump goes whole to one electron. That electron spends at least the work function φ, the energy to lift it from the top of the metal's filled states to rest just outside, and more if it started deeper or collided on the way out; so K ≤ hν − φ, with equality only for the most favoured electrons. Hence Kₘₐₓ = hν − φ, linear in ν and blind to intensity. It gives a threshold ν₀ = φ/h: potassium's φ = 2.29 eV puts ν₀ at 5.54 × 10¹⁴ Hz, λ₀ = 541 nm, so a yellow sodium lamp at 589 nm, 2.10 eV per quantum, ejects nothing however bright. Intensity counts lumps: a beam of power P carries P/(hν) quanta per second, so Iₛₐₜ = ηeP/(hν). And because a single lump can be absorbed at any instant, emission starts at once. The supply is ample: a 1.0 × 10⁻² W m⁻² beam at 400 nm lands 2.0 × 10¹² quanta on each square centimetre every second.
Fit the line, and read h off the slope
Divide the Einstein relation by e and it becomes a straight line in the variables you measure: Vₛ = (h/e)ν − φ/e. Plot Vₛ against ν, not against λ, because only ν makes it straight. The slope is h/e, and multiplying by e, exact since the 2019 SI redefinition, returns h; the intercept on the Vₛ axis is −φ/e and the crossing of the ν axis is the threshold ν₀. The lever arm matters. The mercury lines from 577.0 nm to 365.0 nm span only 3.02 × 10¹⁴ Hz, so errors of 0.02 V pushing the two end points in opposite directions tilt the slope by 0.04 V ÷ 3.02 × 10¹⁴ Hz = 1.3 × 10⁻¹⁶ V s, 3% of h/e. The prize is the cross-check. In the ideal tube assumed so far, changing the metal shifts the line without turning it, and the common slope returns h = 6.63 × 10⁻³⁴ J s, the constant Planck had fitted to blackbody spectra. Millikan, who expected the relation to fail, confirmed it in 1916, with h within about 1% of the blackbody value.
What the intercept hides
The slope is robust; the intercept is not. Emitter and collector are usually different metals; once connected, their Fermi levels line up, leaving a contact step of (φC − φE)/e between their surfaces that adds to whatever voltage you apply. Follow the fastest electron: it pays φE to leave and must then climb eVᵣ + φC − φE, so it is stopped when eVᵣ = hν − φC. The emitter's work function has cancelled. A constant offset leaves the slope at h/e, but the intercept reports the collector unless the contact potential is measured separately. Two further effects blur the cutoff. At 300 K the Fermi edge is smeared over a few kT, with kT = 0.026 eV, so a few electrons leave with slightly more than hν − φ and the current creeps to zero instead of meeting the axis at a definite voltage. And if stray light reaches the collector with hν above φC, the retarding voltage sweeps the electrons it releases back to the emitter; that reverse current makes the total cross zero at a smaller retarding voltage than the true cutoff, by an amount that changes with ν and so can tilt the slope as well.
Quantised absorption is not yet a quantised field
What has been proved? The data demand that the metal takes energy from light of frequency ν only in amounts hν. They do not demand that the light itself exists in lumps. Keep the field a classical wave E₀ cos ωt and make the metal quantum mechanical, with electron states at definite energies. Time-dependent perturbation theory then shows that such a system absorbs only through transitions whose energy matches ħω = hν, at a rate proportional to E₀² and so to intensity, with a probability of emission that grows from t = 0: no lag. Threshold, linear law, rate and promptness are all reproduced, as Lamb and Scully showed in 1969. The minutes-long wait of the classical estimate came from treating the absorber classically, not the light. So the photoelectric effect permits the photon without forcing it. Compton scattering, next, attaches a momentum to the lump; the decisive evidence for a quantised field came from single-photon experiments, such as the anticorrelation at a beam splitter measured by Grangier, Roger and Aspect in 1986.
Prerequisites, sources & model limitations
Prerequisites: wave frequency and wavelength, electronvolts, conservation of energy, electric potential and straight-line fitting. The advanced contact-potential section requires careful sign conventions.
The lab is an ideal teaching model, not a material-specific prediction. It neglects thermal emission, multiphoton absorption, space charge, surface contamination, frequency-dependent efficiency and collector photoemission. The I–V shape is a chosen collection function, not a calculated many-electron spectrum.
- OpenStax · Photoelectric observations, energy balance and representative work functions
- UC Berkeley · Semiclassical radiation and the interpretation of photoelectric evidence
- NIST · SI defining constants
Calculations and lesson presentation reviewed 2 October 2026. These university notes do not assert complete alignment to any secondary syllabus.