# Light as Particles
> Planck quantized the energy *exchanged* between matter and radiation
> ([P.2.1](../c02-radiation-old-quantum/01-blackbody-radiation.md)); Einstein, in 1905, quantized the light itself.
> This lesson follows the evidence that forced the photon on a reluctant physics community: the photoelectric
> effect, X-ray production, Compton scattering, and the creation and annihilation of matter. By the end, light —
> which Young's fringes had settled as a wave — is also, undeniably, a hail of particles. Light, like a cat,
> declines to be only one thing.
## Learning Objectives
After this lesson you will be able to:
1. List the **experimental facts** of the **photoelectric effect** and explain, point by point, why the classical wave picture fails on each.
2. Apply Einstein's **photon** hypothesis $E = h\nu$ to compute **work functions**, $K_{max}$, and **stopping potentials**, and explain what Millikan's slope $h/e$ measures.
3. Derive the **photon momentum** $p = E/c = h/\lambda$ from the relativistic energy–momentum relation.
4. Derive the **Duane–Hunt cutoff** $\lambda_{min} = hc/eV$ and recognize it as the photoelectric effect in reverse.
5. Derive the **Compton shift** $\Delta\lambda = \frac{h}{m_ec}(1-\cos\theta)$ from relativistic energy and momentum conservation.
6. Explain the threshold and kinematics of **pair production** and **annihilation**, including why each requires (or produces) more than one body.
---
## Intuition
Classically, light is a continuous electromagnetic wave: energy flux proportional to the *squared amplitude*
(intensity), spread smoothly over the whole wavefront. Einstein's proposal replaces the smooth flood with rain:
light of frequency $\nu$ arrives in discrete packets — **photons** — each of energy $E = h\nu$. Brightness counts
*how many* packets arrive per second; color sets *how much energy* each carries; one photon is absorbed by one
electron, whole or not at all. Every "impossible" photoelectric fact below becomes obvious in this picture — and
the same bookkeeping, pushed through special relativity, predicts the Compton shift to four significant figures.
---
## Theory
### The photoelectric effect: the facts, and why the wave picture fails on each
Shine light on a clean metal in vacuum; electrons come off and are collected, giving a photocurrent. Four facts,
established between Hertz (1887) and Millikan (1916) — each paired, point by point, with what a classical wave
(energy delivered continuously, at a rate $\propto E_0^2$, spread over the wavefront) predicts:
1. **Threshold frequency.** Each metal has a $\nu_0$ below which *no* electrons are emitted, at any intensity. *Classically*, any frequency should eject electrons if the light is bright enough — there should be no threshold.
2. **No time lag.** Emission starts within $<10^{-9}\,\mathrm{s}$, even in feeble light. *Classically*, an atom of radius $\sim 0.1\,\mathrm{nm}$ intercepts so little of a weak wavefront that accumulating a few eV should take months (E3: 135 days).
3. **Intensity controls the current, not the electron energy.** Doubling intensity doubles the photocurrent; $K_{max}$ is unchanged. *Classically*, a larger amplitude shakes electrons harder, so $K_{max}$ should grow with intensity.
4. **Stopping potential is linear in frequency.** The reverse voltage $V_0$ that just stops the fastest electrons ($eV_0 = K_{max}$) grows linearly with $\nu$, with the *same slope for every metal*. *Classically*, frequency is nearly irrelevant to energy transfer — no universal slope, let alone one equal to $h/e$ with the blackbody $h$ of [P.2.1](../c02-radiation-old-quantum/01-blackbody-radiation.md).
Facts 3 and 4 are read off a single plot. Sweep the collector voltage $V$ and record the photocurrent
$I$. For positive $V$ the current rises and then *saturates* once every emitted electron is collected,
at a level proportional to the light's intensity. Make $V$ negative and fewer electrons can climb the
retarding potential, until the current vanishes at $V = -V_0$, the **stopping potential**. Curves
taken at one frequency but different intensities differ only in their saturation level and all reach
zero at the same $-V_0$; changing the frequency moves $V_0$.
### Einstein's photon and the photoelectric equation
Einstein (1905): light of frequency $\nu$ consists of quanta of energy $E = h\nu$, with
$h = 6.626\times10^{-34}\,\mathrm{J\,s} = 4.136\times10^{-15}\,\mathrm{eV\,s}$. One photon is absorbed by one
electron, which pays the **work function** $\phi$ — its minimum binding energy to the metal, a few eV — and
leaves with at most
$$
K_{max} = h\nu - \phi, \qquad\text{so}\qquad eV_0 = h\nu - \phi
\;\Longrightarrow\; V_0 = \frac{h}{e}\,\nu - \frac{\phi}{e}.
$$
Every fact follows. Threshold: $\nu > \nu_0 = \phi/h$ — below it no single photon suffices (and two-photon
absorption is fantastically unlikely at ordinary intensities). No lag: the energy arrives in one lump. Intensity:
more photons per second means more electrons per second, but each still gets exactly $h\nu$. And $V_0(\nu)$ is a
straight line of **metal-independent slope $h/e$**, intercept $-\phi/e$. Millikan spent a decade trying to refute
this and in 1916 confirmed it, measuring $h = 6.57\times10^{-34}\,\mathrm{J\,s}$ — within 0.5% of the blackbody
value. Independent phenomena, same constant.
Both knobs, side by side, on the sodium cathode of Worked Example 1 below:
```qm-demo photoelectric-stopping-voltage
```
### Photon momentum
Special relativity relates energy, momentum, and mass by $E^2 = (pc)^2 + (mc^2)^2$. A photon travels at $c$,
which no massive particle can do at finite energy, so $m = 0$ and
$$
E = pc \quad\Longrightarrow\quad p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda},
$$
using $c = \lambda\nu$. The momentum is tiny — a green photon carries $1.2\times10^{-27}\,\mathrm{kg\,m/s}$ (E2) —
but real: it reproduces Maxwell's radiation pressure $p = E/c$ in bulk, and Compton verified it photon by photon.
### X-rays: the photoelectric effect in reverse
X-rays are made by accelerating electrons through $V \sim 10$–$100\,\mathrm{kV}$ into a metal anode. Decelerating
in the nuclear electric fields, they radiate a continuous spectrum — **bremsstrahlung** ("braking radiation") —
plus sharp characteristic lines from shell transitions (P.2.2). Classically the continuum should extend to
arbitrarily short wavelength; instead it stops dead at the **Duane–Hunt cutoff**. The photon explains it in one
line: an electron arrives with kinetic energy $eV$, and the most energetic photon takes *all* of it in one quantum,
$$
h\nu_{max} = eV \quad\Longrightarrow\quad \lambda_{min} = \frac{c}{\nu_{max}} = \frac{hc}{eV}.
$$
This is the photoelectric effect run backwards — electron in, photon out ($\phi$, a few eV against tens of keV, is
negligible). The cutoff depends only on $V$, never on the anode material, and historically gave one of the cleanest
measurements of $h$; with $hc = 1240\,\mathrm{eV\,nm}$, $\lambda_{min}[\mathrm{nm}] = 1.240/V[\mathrm{kV}]$, e.g.
$31.0\,\mathrm{pm}$ at $40\,\mathrm{kV}$. Meanwhile the *same* X-rays reflect off crystal planes with constructive
interference at the **Bragg condition** $2d\sin\theta = n\lambda$ — pure wave behavior, and the standard way X-ray
wavelengths are measured. Hold that thought.
### The Compton effect
Compton (1923) scattered $\lambda = 71.1\,\mathrm{pm}$ X-rays off graphite and found scattered radiation at angle
$\theta$ with a *longer* wavelength $\lambda'$, shifted by an amount depending only on $\theta$. A classical wave
cannot do this — an electron driven at $\nu$ reradiates at $\nu$ — but a relativistic photon–electron *collision*
can. Setup: a photon ($E_\gamma = hc/\lambda$, momentum $h/\lambda$ along $x$) hits a free electron at rest; the
photon exits at $\theta$ with wavelength $\lambda'$, the electron at $\varphi$ on the other side with momentum
$p_e$ and energy $E_e = \sqrt{p_e^2c^2 + m_e^2c^4}$. Energy, $x$- and $y$-momentum conservation give three equations:
$$
\frac{hc}{\lambda} + m_ec^2 = \frac{hc}{\lambda'} + \sqrt{p_e^2c^2 + m_e^2c^4}, \qquad
\frac{h}{\lambda} = \frac{h}{\lambda'}\cos\theta + p_e\cos\varphi, \qquad
0 = \frac{h}{\lambda'}\sin\theta - p_e\sin\varphi .
$$
**Step 1 — eliminate $\varphi$.** Isolate the electron terms in the momentum pair, square, add ($\cos^2\varphi + \sin^2\varphi = 1$):
$$
p_e^2 = \Big(\frac{h}{\lambda}\Big)^2 + \Big(\frac{h}{\lambda'}\Big)^2 - \frac{2h^2}{\lambda\lambda'}\cos\theta .
$$
**Step 2 — eliminate $E_e$.** Isolate the square root in the energy equation and square:
$$
p_e^2c^2 = \Big(\frac{hc}{\lambda} - \frac{hc}{\lambda'} + m_ec^2\Big)^2 - m_e^2c^4
= \Big(\frac{hc}{\lambda} - \frac{hc}{\lambda'}\Big)^2 + 2m_ec^2\Big(\frac{hc}{\lambda} - \frac{hc}{\lambda'}\Big).
$$
**Step 3 — equate.** Multiply Step 1 by $c^2$ and set it equal to Step 2; the squared terms
$(hc/\lambda)^2 + (hc/\lambda')^2$ cancel, leaving
$-\tfrac{2(hc)^2}{\lambda\lambda'}\cos\theta = -\tfrac{2(hc)^2}{\lambda\lambda'} + 2m_ec^2\,hc\,\tfrac{\lambda'-\lambda}{\lambda\lambda'}$.
Multiply by $\lambda\lambda'/(2hc\,m_ec^2)$:
$$
\boxed{\;\Delta\lambda = \lambda' - \lambda = \frac{h}{m_ec}\,(1-\cos\theta)\;}
$$
The prefactor is the electron's **Compton wavelength** $\lambda_C = h/m_ec = 2.426\,\mathrm{pm}$, so the shift runs
from $0$ (forward) to $2\lambda_C = 4.85\,\mathrm{pm}$ (backscatter), independent of $\lambda$ and of the material —
exactly what Compton measured, with the recoil electron detected in *coincidence* with the scattered photon. In
energy terms, with $\varepsilon = E_\gamma/m_ec^2$, the scattered photon has $E'_\gamma = E_\gamma/[1 + \varepsilon(1-\cos\theta)]$.
**The unshifted peak.** Compton also saw a component at the original $\lambda$: photons scattering off electrons
so tightly bound that the *whole atom* recoils. Replace $m_e$ by the atomic mass $M \sim 10^4\,m_e$ and the shift
$h/Mc$ shrinks by the same factor — unresolvably small.
> **Caution.** Do not swap one cartoon for another: the photon is *not* a tiny billiard ball. The very same X-ray
> beam Bragg-diffracts off a crystal (wave) and Compton-scatters off its electrons (particle). No classical object
> does both; the resolution is that a *quantum amplitude* propagates like a wave and is detected like a particle —
> the formalism of [P.3.2](02-matter-as-waves.md) and
> [P.4.1](../c04-schrodinger-equation/01-schrodinger-equation-born-rule.md).
### Pair production and annihilation
Above roughly 1 MeV a new channel opens: the photon vanishes and an electron–positron pair appears,
$\gamma \to e^- + e^+$, with threshold set by the rest masses, $E_\gamma \ge 2m_ec^2 = 1.022\,\mathrm{MeV}$. But
this **cannot happen in free space**: each lepton has $E_i = \sqrt{p_i^2c^2 + m_e^2c^4} > p_ic$ strictly (because
$m_e > 0$), so by the triangle inequality
$$
E_- + E_+ > (p_- + p_+)\,c \ge |\vec p_- + \vec p_+|\,c ,
$$
i.e. the pair always has $E > pc$, while the photon has exactly $E = pc$ — energy and momentum cannot both balance.
Pair production therefore happens only near a nucleus, which absorbs the recoil momentum while (being heavy,
$\Delta E = \Delta p^2/2M$) carrying off negligible energy, leaving the threshold at $1.022\,\mathrm{MeV}$. The
time-reverse is **annihilation**: a slowed positron meets an electron essentially at rest, total momentum
$\approx 0$; a *single* photon would carry $p = E/c \ne 0$ — forbidden by the same argument — so annihilation at
rest produces **at least two** photons, back-to-back with equal and opposite momenta,
$E_\gamma = m_ec^2 = 511\,\mathrm{keV}$ each (three-photon decay also occurs, from triplet positronium; two is the
minimum). Each photon's wavelength is $\lambda = hc/m_ec^2 = h/m_ec$, exactly the Compton wavelength
$\lambda_C = 2.426\,\mathrm{pm}$: the Compton wavelength is the wavelength of a photon whose energy
equals the electron's rest energy. These back-to-back photon pairs are the working principle of PET imaging.
---
## Worked Examples
### Example 1 — Photoelectric effect on sodium
Sodium has $\phi = 2.28\,\mathrm{eV}$; shine $\lambda = 400\,\mathrm{nm}$ light on it. Photon energy (with
$hc = 1240\,\mathrm{eV\,nm}$): $E = hc/\lambda = 1240/400 = 3.10\,\mathrm{eV}$, so
$K_{max} = 3.10 - 2.28 = 0.82\,\mathrm{eV}$ and $V_0 = 0.82\,\mathrm{V}$. Threshold:
$\nu_0 = \phi/h = 2.28/(4.136\times10^{-15}) = 5.51\times10^{14}\,\mathrm{Hz}$, i.e.
$\lambda_0 = 1240/2.28 = 544\,\mathrm{nm}$ — green barely works, red does nothing. Doubling the intensity at
400 nm doubles the *current*; $K_{max}$ and $V_0$ do not move.
### Example 2 — Compton scattering at 90°
Compton's beam, $\lambda = 71.1\,\mathrm{pm}$ (Mo K$\alpha$), scattered at $\theta = 90^\circ$:
$\Delta\lambda = \lambda_C(1 - \cos 90^\circ) = 2.43\,\mathrm{pm}$, so $\lambda' = 73.5\,\mathrm{pm}$. Energies:
$E_\gamma = 1240/0.0711 = 17.4\,\mathrm{keV}$, $E'_\gamma = 1240/0.0735 = 16.9\,\mathrm{keV}$; cross-check with
$\varepsilon = 17.4/511 = 0.0341$: $E' = 17.4/1.0341 = 16.9\,\mathrm{keV}$. ✓ The recoil electron carries
$K_e \approx 0.58\,\mathrm{keV}$ — small here, but MeV photons can hand the electron most of their energy (E5).
---
## Hands-on (Python)
Fit synthetic stopping-potential data the way Millikan did — the slope of $V_0(\nu)$ is $h/e$ for *every* metal;
only the intercept $-\phi/e$ changes.
```python
import numpy as np
h_true = 6.62607015e-34 # Planck constant, J s
e = 1.602176634e-19 # elementary charge, C
rng = np.random.default_rng(7)
metals = {"Cs": 2.14, "Na": 2.28, "Zn": 4.33} # work functions phi, eV
for metal, phi in metals.items():
nu0 = phi * e / h_true # threshold frequency, Hz
nu = np.linspace(1.05 * nu0, 2.5 * nu0, 12) # measured frequencies, Hz
V0 = (h_true / e) * nu - phi # ideal line: slope h/e, intercept -phi/e
V0 += rng.normal(0.0, 0.02, V0.size) # 20 mV of measurement noise
slope, intercept = np.polyfit(nu, V0, 1) # least-squares straight line
print(f"{metal}: h = {slope * e:.3e} J s phi = {-intercept:.2f} eV")
# Expected output — h within ~1% of 6.626e-34 for every metal, e.g.:
# Cs: h = 6.645e-34 J s phi = 2.15 eV (Na, Zn similar; varies with the noise)
```
Now the Compton kinematics: the wavelength shift is universal, but the *energy* lost depends strongly on the
incident energy.
```python
import numpy as np
import matplotlib.pyplot as plt
h, c, me = 6.62607015e-34, 2.99792458e8, 9.1093837015e-31
lam_C = h / (me * c) # Compton wavelength = 2.426e-12 m
mec2 = 0.51099895 # electron rest energy, MeV
theta = np.linspace(0.0, np.pi, 500)
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(10, 4))
ax1.plot(np.degrees(theta), lam_C * (1 - np.cos(theta)) * 1e12)
ax1.set(xlabel="θ (deg)", ylabel="Δλ (pm)", title="Compton shift")
# rises from 0 to 2*lam_C = 4.85 pm at 180°, independent of incident energy
for E in (0.1, 0.5, 1.0): # incident photon energies, MeV
ax2.plot(np.degrees(theta), E / (1 + (E / mec2) * (1 - np.cos(theta))), label=f"E = {E} MeV")
ax2.set(xlabel="θ (deg)", ylabel="E' (MeV)", title="scattered photon energy")
ax2.legend()
plt.tight_layout()
plt.show()
# At θ = 180° the 1 MeV curve drops to 1/(1 + 2·1.957) ≈ 0.20 MeV:
# harder photons lose a larger FRACTION of their energy to the electron.
```
---
## Exercises
**E1 (easy).** Cesium has $\phi = 2.14\,\mathrm{eV}$. Can a helium–neon laser ($\lambda = 632.8\,\mathrm{nm}$)
eject photoelectrons from it at *any* intensity? Give the threshold wavelength.
Solution
$E = 1240/632.8 = 1.96\,\mathrm{eV} < 2.14\,\mathrm{eV}$: no single photon suffices, so no emission at any
(ordinary) intensity — intensity only changes the photon *rate*. Threshold:
$\lambda_0 = hc/\phi = 1240/2.14 = 579\,\mathrm{nm}$; anything redder fails.
Solution
$E = 1240/532 = 2.33\,\mathrm{eV} = 3.73\times10^{-19}\,\mathrm{J}$; rate
$N = P/E = 10^{-3}/3.73\times10^{-19} = 2.7\times10^{15}\,\mathrm{s^{-1}}$; momentum
$p = h/\lambda = 1.25\times10^{-27}\,\mathrm{kg\,m/s}$; force (full absorption)
$F = Np = 3.3\times10^{-12}\,\mathrm{N}$ — equal to the classical $P/c$, as it must be.
Solution
Collected power $P = I\pi r^2 = 10^{-6}\pi(10^{-10})^2 = 3.1\times10^{-26}\,\mathrm{W}$; required energy
$\phi = 3.65\times10^{-19}\,\mathrm{J}$; time
$t = 3.65\times10^{-19}/3.1\times10^{-26} = 1.2\times10^{7}\,\mathrm{s} \approx 135$ days. Observed lag: under a nanosecond. The energy is not spread over the wavefront — it arrives in quanta.
Solution
The maximum shift ($\theta = 180^\circ$) is $2\lambda_C = 4.85\,\mathrm{pm}$ regardless of $\lambda$.
(a) $4.85\,\mathrm{pm}/500\,\mathrm{nm} \approx 10^{-5}$ — buried below optical resolution and linewidth.
(b) $4.85/71.1 = 6.8\%$ — easily resolved by Bragg spectrometry. The absolute shift is universal; only short
wavelengths make it *relatively* visible.
Solution
The electron gains most when the photon backscatters, $\theta = 180^\circ$: with $\varepsilon = E_\gamma/m_ec^2$,
$E'_\gamma = E_\gamma/(1+2\varepsilon)$, so $K_{max} = E_\gamma - E'_\gamma = E_\gamma\,\frac{2\varepsilon}{1+2\varepsilon}$.
For $E_\gamma = 511\,\mathrm{keV}$, $\varepsilon = 1$: $K_{max} = 511\times\tfrac23 = 341\,\mathrm{keV}$. This sharp
upper edge in the electron spectrum is a standard feature of gamma-ray detectors — the photon can never give a
free electron *all* its energy (total absorption would violate the same $E$-vs-$pc$ bookkeeping as pair production
in free space).
Answers
1. Threshold frequency, no time lag, intensity-independence of $K_{max}$, linear $V_0(\nu)$. Photons: one quantum
$h\nu$ per electron — below $\phi$ nothing works, energy arrives in a lump, intensity only sets the photon
rate, and $eV_0 = h\nu - \phi$ is linear by construction.
2. The universal slope $h/e$ — nature's constants only; the metal enters solely via the intercept $-\phi/e$.
3. From $E^2 = (pc)^2 + (mc^2)^2$ with $m = 0$: $E = pc$, so $p = E/c = h\nu/c = h/\lambda$. The ingredient:
a particle moving at $c$ must be massless.
4. Some photons scatter off tightly bound electrons — effectively off the whole atom — so the shift $h/Mc$ with
$M \sim 10^4\,m_e$ is unresolvably small.
5. A pair always has $E > pc$ while a photon has $E = pc$, so both conservation laws cannot hold without a nucleus
taking recoil momentum. At rest the total momentum is zero, which one photon ($p = E/c \ne 0$) cannot match —
two back-to-back 511 keV photons can.