Photons and the photoelectric effect

A phototube is a metal cathode facing a collecting anode in vacuum, with an adjustable voltage between them. Light on the cathode drives a current; make the anode negative enough and the current stops. Four results come out of it, and three of them refuse to fit a wave picture. Below a metal's threshold frequency f₀ no current flows at all, however bright the lamp and however long you wait. Above f₀ the stopping voltage depends on frequency alone for a given cathode, and doubling the brightness leaves it exactly where it was. And emission begins immediately: a wave spread over the surface would have to pour energy into roughly one atom's area, about 10⁻²⁰ m², until φ had accumulated, so at an illumination of 1.0 × 10⁻⁶ W m⁻² that is 10⁻²⁶ W paying off a 3.65 × 10⁻¹⁹ J debt, which takes over a year, while Lawrence and Beams bounded the real delay below 3 × 10⁻⁹ s in 1928. The fourth result is the one a wave picture does predict, and it is worth keeping apart from the other three: the saturation current — every emitted electron collected — is proportional to intensity.

Einstein's 1905 proposal was that the energy of light of frequency f is not spread over the wavefront but localized in quanta of hf, and that photoemission is a one-to-one transaction: a single quantum absorbed whole by a single electron. The rest is bookkeeping. Freeing the least tightly bound electron costs the work function φ — 2.28 eV for sodium, 5.65 eV for platinum — so the fastest electron out carries Kₘₐₓ = hf − φ. Electrons starting deeper in the conduction band, or scattering on the way out, arrive with less, which is why Kₘₐₓ is a ceiling rather than a common value. Setting Kₘₐₓ = 0 gives the threshold f₀ = φ/h, or λ₀ = hc/φ, which for sodium is 544 nm: green light works on it, red does not. Notice what is absent from Kₘₐₓ = hf − φ. Intensity appears nowhere; it enters only through how many transactions happen per second.

To measure Kₘₐₓ, make the anode negative. An electron leaving with kinetic energy K climbs the retarding hill only if eV < K, so the current vanishes exactly when eVₛ = Kₘₐₓ — and Vₛ in volts is then Kₘₐₓ in electronvolts, with no arithmetic at all. Substituting Einstein's equation gives Vₛ = (h/e) f − φ/e, a straight line against frequency. Its slope h/e = 4.136 × 10⁻¹⁵ V s is a pure combination of constants, identical for every cathode; the metal chooses only where the line crosses the axis. So the experiment weighs h. Two readings on an unknown cathode, Vₛ = 1.30 V at 365 nm (8.213 × 10¹⁴ Hz) and Vₛ = 0.75 V at 436 nm (6.876 × 10¹⁴ Hz), give a slope of 0.55 V ÷ 1.338 × 10¹⁴ Hz = 4.11 × 10⁻¹⁵ V s, hence h = 6.59 × 10⁻³⁴ J s, low by 0.6%. The intercept follows: φ = 4.11 × 10⁻¹⁵ × 8.213 × 10¹⁴ − 1.30 = 2.08 eV. Millikan spent a decade trying to break this line and ended by publishing the best value of h then available.

Intensity is power per unit area, and power is photons per second times hf, so the arrival rate is R = P/(hf) = Pλ/(hc). The numbers are enormous: 1.0 mW of 500 nm green light is 2.5 × 10¹⁵ photons every second. Double the power and R doubles while every individual transaction stays identical, so the photocurrent doubles and Vₛ does not shift by a microvolt. Not every photon yields a collected electron — most are reflected, or absorbed too deep to escape — so the current is I = η e R, with a quantum efficiency η of a few per cent up to tens of per cent. That is the whole content of intensity effects: rate, never size. It also shows where the rule bends. In a focused femtosecond pulse two photons can reach the same electron within about 10⁻¹⁵ s, so 2hf ≥ φ will do and emission appears below f₀ — two overlapping transactions, not a returning wave, and it needs peak intensities above roughly 10¹² W m⁻² that no lamp approaches.

Photoemission needs the metal, not merely the electron, and conservation laws say why. Work in the frame where a free electron of mass m sits at rest, and let it absorb a photon of energy E. Energy conservation hands the electron a total energy mc² + E; momentum conservation hands it momentum E/c. But any real particle must satisfy Eₜₒₜₐₗ² = (pc)² + (mc²)², and (mc² + E)² = E² + 2mc²E + m²c⁴ overshoots E² + (mc²)² by 2mc²E. The two demands are incompatible unless E = 0, so a free electron absorbing a whole photon is forbidden outright — not improbable, impossible. A bound electron escapes the contradiction because the ion core or the lattice takes the surplus momentum, and being thousands of times heavier it does so at negligible energy cost. The same argument is why Compton's photons scatter instead of vanishing: an electron bound by a few eV looks free to a 24 keV X-ray, and a free electron can only deflect a photon.

Absorption was Einstein's evidence; scattering was Compton's. In 1923 he fired molybdenum X-rays at graphite and found the scattered beam carried a component at longer wavelength, displaced by an amount set by the scattering angle θ alone — not by the incident wavelength, not by the target material. A wave cannot do that, since a classically driven electron re-radiates at the frequency driving it. Treat the event as two particles instead. Conserve energy, hc/λ + m_ec² = hc/λ′ + Eₑ, conserve momentum as vectors with the photon carrying h/λ, then eliminate the electron's energy and angle using Eₑ² = (p_ec)² + (m_ec²)². The algebra collapses to Δλ = (h/m_ec)(1 − cos θ). Because λ_C = 2.426 pm is a constant, the shift is absolute: at 90° it is 2.43 pm whatever you started with. On a 71.1 pm X-ray that is a 3.4% change and plain to see; on 500 nm light the largest possible shift is 4.85 pm, a fraction of 10⁻⁵, which is why the effect waited for X-ray sources.