Photons and the photoelectric effect

Treat a beam of frequency f as a stream of quanta each carrying E = hf, and let an electron absorb one whole or none. With hc = 1240 eV nm, a 500 nm photon carries 2.48 eV and a 250 nm photon 4.96 eV. Work functions, the least energy that lifts an electron from the Fermi level to rest outside the surface, run from 2.1 eV for caesium to about 5.6 eV for platinum, so visible light can just free a caesium electron while platinum needs ultraviolet below 1240/5.6 = 221 nm. A classical wave gets three things wrong in kind. Its energy is spread over the wavefront, so an electron's share should grow with intensity; nothing in it makes one frequency a cliff; and the energy has to accumulate. At 1.0 μW m⁻², an atom intercepting 1.0 × 10⁻¹⁹ m² collects 1.0 × 10⁻²⁵ W and needs 3.7 × 10⁶ s, about 43 days, to gather 2.30 eV. Experiment finds electrons within nanoseconds, with energies independent of intensity, and essentially none below a threshold frequency.

One photon, one electron. If the electron starts at the Fermi level, the highest occupied state at T = 0, it spends φ reaching the vacuum level and keeps the rest: Kₘₐₓ = hf − φ. That is the largest kinetic energy, not the typical one. Electrons that start deeper in the band, or lose energy in collisions on the way out, emerge with anything from zero to Kₘₐₓ, so the equation fixes the upper edge of a distribution. Setting Kₘₐₓ = 0 gives the threshold f₀ = φ/h, or λ₀ = hc/φ. For φ = 2.10 eV the threshold is 590 nm, and 420 nm light frees electrons of up to 2.952 − 2.10 = 0.85 eV, moving at √(2Kₘₐₓ/mₑ) = 5.5 × 10⁵ m s⁻¹. Check the mechanics used on the electron: Kₘₐₓ/(mₑc²) = 0.85 eV ÷ 511 keV = 1.7 × 10⁻⁶, and (γ − 1)mₑc² exceeds ½mₑv² by a fraction ¾(v/c)² ≈ 2.5 × 10⁻⁶. Newtonian kinetic energy is safe here and stays safe until hf reaches tens of keV, where Kₘₐₓ/(mₑc²) climbs to a few per cent.

Kₘₐₓ is measured by stopping it. Make the collector negative by V and only electrons with K > eV arrive; the current vanishes at Vₛ, where eVₛ = Kₘₐₓ. Then Vₛ = (h/e)f − φ/e: a line in f whose slope, 4.136 × 10⁻¹⁵ V s, contains nothing about the metal, and whose intercept contains nothing else. Measure Vₛ at several frequencies, fit by least squares, and h = e × slope; the slope's standard error, s/√(Σ(fᵢ − f̄)²) with s the residual scatter on N − 2 degrees of freedom, carries straight into h. Do not force the fit through the origin. The line crosses zero at f₀, and below f₀ there is no current at all, so its negative branch is extrapolation, not data. Read the intercept with care too. Emitter and collector are joined by a wire and share a Fermi level, so the energy an electron needs to land is set by the collector's surface: eVₛ = hf − φ_collector. The slope is untouched, which is why h is what this experiment measures cleanly.

A beam of power P delivers P/(hf) photons per second, and each frees an electron that escapes with probability η, the quantum efficiency: about 10⁻⁴ for bare copper, above 0.2 for the alkali-antimonide cathode of a photomultiplier. So the saturation current is I = ηeP/(hf). Double P and I doubles, while Vₛ does not move by a millivolt, because hf has not changed. Since every photon is an independent trial, the number of photoelectrons in a time t is Poisson-distributed, with mean n̄ = ηPt/(hf) and spread √n̄. That is where the classical picture returns. A 2.6 μA current delivers 1.6 × 10⁷ electrons per microsecond, fluctuating by 0.025%: a smooth current, as a continuous wave would give. Attenuate it by 10⁶ and the same microsecond holds about 16 ± 4 electrons. The grain is plain, and only the mean rate still behaves like a wave's intensity. Correspondence works through large numbers: the photon picture reproduces every classical result that depends on the average energy flux alone.

The sharp threshold assumes a sharp Fermi edge, which is a T = 0 statement. At temperature T the occupation of states near the edge falls off over a few k_B T, with k_B T = 0.0259 eV at 300 K, so a few electrons already sit above it and a photon slightly short of φ can still free them. Fowler's analysis gives the yield per absorbed photon two limits: Y ∝ (hf − φ)² a few k_B T above threshold, and Y ∝ exp(−(φ − hf)/k_B T) well below it, the Boltzmann factor for an electron that far up the tail. Ten k_B T short, 0.26 eV below φ, the yield is down by about e⁻¹⁰ ≈ 5 × 10⁻⁵ on its threshold value: small, yet a photomultiplier counts it. Near 2.5 eV a smear of 4k_B T ≈ 0.10 eV is 1240 × 0.10/2.5² ≈ 20 nm of wavelength, so a threshold read off a room-temperature current plot is soft by that much. The Vₛ line fares better: its foot is rounded over a few k_B T/e, but the slope, fitted well above threshold, is untouched. Read h from the slope, never from where the current seems to vanish.

Einstein's rule is a low-intensity statement. A photon too weak to free an electron alone can still lift it into a short-lived intermediate state; if a second photon reaches that electron within the lifetime τ, of order 10⁻¹⁵ s, it takes both and leaves with Kₘₐₓ = 2hf − φ. The ratio of two-photon to one-photon events is roughly σΦτ, with σ ≈ 10⁻¹⁹ m² an atomic cross-section and Φ = I/(hf) the photon flux. For 2.0 eV photons in sunlight's 10³ W m⁻², Φ = 3 × 10²¹ m⁻² s⁻¹ and σΦτ ≈ 3 × 10⁻¹³: invisible. A focused nanosecond laser at 10¹² W m⁻² gives 3 × 10⁻⁴, negligible beside one-photon emission but the whole signal once hf < φ has closed that channel, and near 10¹⁵ W m⁻² the two compete. The signatures are clean: the n-photon current grows as Iⁿ, so a log–log plot of current against intensity has slope n, and its Vₛ line has slope nh/e. Towards 10²⁰ W m⁻² the laser's field rivals the 5 × 10¹¹ V m⁻¹ that binds an atomic electron, and counting photons stops being the useful picture.