Shine light on a metal and electrons jump out — but only if the light is the right color. The photoelectric effect broke the wave picture of light, and Einstein's fix — the photon — broke physics wide open.
Module 2 · The experiment that won Einstein his Nobel Prize
Beginner Module 02 · Light & Photons ~35 minPrerequisites: Module 1 (Planck's idea that energy comes in packets, \(E = hf\)) and comfort with scientific notation. That's it.
Before we can appreciate how shocking the photoelectric effect was, we need to see why every physicist in 1900 was certain that light is a wave. This wasn't a guess — it was backed by a century of beautiful experiments and one of the greatest theories ever written down.
Thomas Young sent light through two narrow slits and got something remarkable on the screen behind them: not two bright stripes, but a whole series of alternating bright and dark bands. Waves do exactly this — where two ripples meet crest-to-crest they reinforce, and where crest meets trough they cancel. Two thrown pebbles can't cancel each other out; two overlapping ripples can. Interference is the fingerprint of a wave, and light has it. (We'll put this experiment under the microscope in Module 3 — for now, just note that it exists and that it's rock-solid evidence for waves.)
James Clerk Maxwell united electricity and magnetism into four equations — and out of the mathematics popped a traveling wave of electric and magnetic fields, moving at exactly the measured speed of light, \(c = 3.00 \times 10^{8}\) m/s. The conclusion was inescapable: light is an electromagnetic wave. Even better, the theory predicted that such waves should exist at every frequency, not just the visible ones. Heinrich Hertz confirmed it in 1887 by generating radio waves in his lab. Radio, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays — they are all one family, the same wave at different frequencies.
Here's the family portrait. The third column is a spoiler from later in this module: once you know light comes in packets of energy \(E = hf\), each band of the spectrum has a typical photon energy too — and that column will explain everything from sunburn to why your Wi-Fi can't hurt you.
| Band | Typical wavelength \(\lambda\) | Typical frequency \(f\) | Typical photon energy |
|---|---|---|---|
| Radio | ~1 m and longer | ~3 × 108 Hz and below | ~10−6 eV (about a millionth of an eV) |
| Microwave | ~1 cm | ~3 × 1010 Hz | ~10−4 eV |
| Infrared | ~10 μm | ~3 × 1013 Hz | ~0.1 eV |
| Visible | 400–700 nm | 4.3–7.5 × 1014 Hz | 1.8–3.1 eV |
| Ultraviolet | ~10–400 nm | ~1015–1016 Hz | ~3–100 eV |
| X-ray | ~0.01–10 nm | ~1016–1019 Hz | ~100 eV–100 keV |
| Gamma | < 0.01 nm | > 3 × 1019 Hz | > ~100 keV |
The three columns are locked together: wavelength and frequency always satisfy \(c = f\lambda\), and (as we'll see) photon energy is \(E = hf\). Long wavelength → low frequency → feeble photons. Short wavelength → high frequency → energetic photons.
Interference proved light behaves like a wave. Maxwell explained what kind of wave it is. Hertz produced the predicted invisible cousins on demand. By 1900, “light is a wave” wasn't a hypothesis — it was considered one of the most secure facts in all of physics. Which is exactly why what happens next is such a great story.
The setup is almost embarrassingly simple. Take a metal plate, seal it inside a glass tube with the air pumped out (so electrons can fly freely), and place a second, collector plate facing it. Now shine light on the metal. If the light kicks any electrons out of the surface, they drift across the vacuum to the collector, and a sensitive meter in the circuit registers a tiny current. Light in, electrons out — that's the photoelectric effect. It's not exotic: for decades, this exact vacuum-tube gadget (the “phototube”) was the standard electric eye in light sensors and automatic door openers.
Think of a wave delivering energy the way ocean waves lap against a pebble on the beach: continuously, spread over the whole surface, a little at a time. On that picture, an electron sitting in the metal is like the pebble — keep the waves coming and it soaks up energy until, sooner or later, it has enough to be knocked loose. From this, three predictions follow naturally: any frequency of light should eventually eject electrons if you make it bright enough or wait long enough; dim light should work too, just with a delay while the energy accumulates; and brighter light (bigger waves!) should give the escaping electrons more energy. All three predictions are wrong.
| What the wave theory predicts | What actually happens | |
|---|---|---|
| (a) Frequency threshold | Any color of light should work. Below any given frequency, just turn up the brightness — more energy per second must eventually free the electrons. | Below a certain threshold frequency (which depends on the metal), no electrons come out at all — ever — no matter how blindingly bright the light is or how long you wait. |
| (b) Time delay | In dim light, an electron needs time to soak up enough energy — calculations said minutes or even hours before the first electron escapes. | Above the threshold frequency, emission is instantaneous (within nanoseconds), even in extremely dim light. The very first flicker of faint violet light already pops electrons out. |
| (c) What brightness does | Brighter light = bigger waves = each ejected electron should fly out with more energy. | Brightness changes only the number of electrons per second (the current). The maximum energy of each electron depends only on the light's frequency — dim violet light beats floodlight-bright red light every time. |
Picture those gentle ocean waves lapping at the pebble again. The wave theory says: keep them coming and the pebble must eventually wash loose — energy is energy, and it adds up. But the experiment says the equivalent of: a million gentle waves achieve nothing, ever, while one single sharp wave of the right kind knocks the pebble free instantly. Continuous, spread-out energy delivery simply cannot behave like that. Something about how light carries energy had to be fundamentally different from a wave — and remember, we just spent a whole section proving light is a wave. Hold that tension; it's the heart of this course.
In Module 1 we met Max Planck, who in 1900 reluctantly assumed that hot objects emit light energy in discrete lumps of size \(E = hf\) — a mathematical trick he never quite believed in. In 1905, a 26-year-old patent clerk named Albert Einstein did something bolder: he took Planck completely seriously. What if the lumps aren't an accounting trick of how matter emits light? What if light itself travels and is absorbed as indivisible packets of energy — what we now call photons?
A beam of light of frequency \(f\) is a stream of particles, each carrying exactly \(E = hf\) of energy, where \(h = 6.626 \times 10^{-34}\) J·s is Planck's constant. Higher frequency means more energetic photons; brighter light means more photons per second — but each individual packet is the same size.
Now add one rule about absorption: one photon interacts with one electron, all-or-nothing. An electron can't sip half a photon or save up several; it absorbs a single photon whole, or nothing happens. That single rule detonates the whole puzzle.
Electrons don't leave a metal for free — they're bound to it, and escaping costs a minimum amount of energy called the work function, written \(\phi\) (phi). Think of it as the metal's exit fee. Every metal charges a different fee: about 2.28 eV for sodium, 1.95 eV for cesium (a famously cheap exit, which is why cesium was used in phototubes), around 4.7 eV for copper. An electron that absorbs a photon pays the fee on the way out and keeps the change as kinetic energy (speed). The luckiest electrons — the ones right at the surface who pay only the minimum — exit with the maximum possible kinetic energy:
$$KE_{max} = hf - \phi$$That's the photoelectric equation, and it's just energy bookkeeping: energy in (one photon, \(hf\)) minus the exit fee (\(\phi\)) equals the change (kinetic energy). If \(hf < \phi\), the photon can't even cover the fee — the electron stays put, and the photon's energy is shrugged off as a bit of heat.
Imagine a vending machine that charges 2.28 “coins” per snack and has one strange rule: it accepts exactly one coin per transaction and gives no credit. Red-light photons are coins worth 1.77. Feed the machine one of them — nothing. Feed it ten thousand per second — still nothing, because each transaction is one coin, and 1.77 < 2.28 every single time. Coins don't accumulate. Now hand it a single violet coin worth 3.10: instant snack, plus 0.82 in change. That change is the electron's kinetic energy. Brightness is just how many coins per second you feed the machine; frequency is how much each coin is worth. Only the coin's value decides whether anything happens.
Run the three shocking facts back through the photon picture:
Robert Millikan, who frankly disbelieved the photon idea, spent a decade trying to disprove the equation experimentally — and in 1916 ended up confirming it in precise detail, extracting an accurate value of \(h\) in the process. And here's a fact that surprises almost everyone: when Einstein won the 1921 Nobel Prize in Physics, it was not for relativity. The citation reads: “for his services to Theoretical Physics, and especially for his discovery of the law of the photoelectric effect.” The committee considered relativity too controversial; the photon equation was the safe, proven bet.
Joules are clumsy at this scale — a visible photon carries around 0.0000000000000000004 J. So we use the electron-volt: the energy an electron gains crossing a 1-volt battery.
1 eV = 1.602 × 10−19 J
In this currency everything becomes friendly: visible photons carry 1.8–3.1 eV, work functions are 2–5 eV, and a chemical bond is a few eV. Two conversions worth memorizing: to go J → eV, divide by 1.602 × 10−19; and the shortcut \(E \text{ (eV)} \approx 1240 / \lambda \text{ (nm)}\), which follows from \(E = hc/\lambda\). We'll use eV constantly from here on.
Time to plug in numbers. Throughout, we use \(h = 6.626 \times 10^{-34}\) J·s, \(c = 3.00 \times 10^{8}\) m/s, and 1 eV = 1.602 × 10−19 J. Everything is multiplication, division, and scientific notation — nothing more.
Sodium has work function \(\phi = 2.28\) eV. Find its threshold frequency and threshold wavelength — the dividing line between “electrons fly out” and “nothing happens.”
Step 1 — convert the fee to joules.
$$\phi = 2.28 \times (1.602 \times 10^{-19}\ \text{J}) = 3.65 \times 10^{-19}\ \text{J}$$Step 2 — threshold frequency. At threshold the photon exactly covers the fee: \(hf_0 = \phi\), so
$$f_0 = \frac{\phi}{h} = \frac{3.65 \times 10^{-19}\ \text{J}}{6.626 \times 10^{-34}\ \text{J·s}} \approx 5.51 \times 10^{14}\ \text{Hz}$$Step 3 — threshold wavelength. Using \(c = f\lambda\):
$$\lambda_0 = \frac{c}{f_0} = \frac{3.00 \times 10^{8}\ \text{m/s}}{5.51 \times 10^{14}\ \text{Hz}} \approx 5.45 \times 10^{-7}\ \text{m} = 545\ \text{nm}$$Interpretation: 545 nm is green light, right in the middle of the rainbow. Any light with a shorter wavelength (green-blue, blue, violet, UV) ejects electrons from sodium; anything longer does nothing. Check the extremes with the 1240 shortcut: red light at 700 nm carries \(1240/700 \approx 1.77\) eV per photon — below the 2.28 eV fee, so zero electrons no matter how bright. Violet at 400 nm carries \(1240/400 \approx 3.10\) eV — comfortably above, so electrons pop out even in dim violet light.
Shine 400 nm violet light on sodium (\(\phi = 2.28\) eV). What is the maximum kinetic energy of the ejected electrons?
Step 1 — photon energy in joules. Since \(f = c/\lambda\), we can write \(E = hf = hc/\lambda\):
$$E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34}\ \text{J·s})(3.00 \times 10^{8}\ \text{m/s})}{400 \times 10^{-9}\ \text{m}} = \frac{1.988 \times 10^{-25}\ \text{J·m}}{4.00 \times 10^{-7}\ \text{m}} \approx 4.97 \times 10^{-19}\ \text{J}$$Step 2 — convert to eV. Watch the units cancel: joules divided by joules-per-eV leaves eV.
$$E = \frac{4.97 \times 10^{-19}\ \text{J}}{1.602 \times 10^{-19}\ \text{J/eV}} \approx 3.10\ \text{eV}$$Step 3 — pay the exit fee. Because both energies are now in the same currency, the photoelectric equation is one subtraction:
$$KE_{max} = hf - \phi = 3.10\ \text{eV} - 2.28\ \text{eV} = 0.82\ \text{eV}$$Interpretation: each violet photon spends 2.28 eV freeing the electron and hands over the remaining 0.82 eV as motion — that's an electron moving at roughly 540 km/s. And if you double the lamp's brightness? Twice as many electrons per second, each still capped at 0.82 eV.
A green laser pointer emits 532 nm light at 5.0 mW (that is, 5.0 × 10−3 joules of light energy per second). How many photons leave it each second?
Step 1 — energy of one photon.
$$E = \frac{hc}{\lambda} = \frac{1.988 \times 10^{-25}\ \text{J·m}}{532 \times 10^{-9}\ \text{m}} \approx 3.74 \times 10^{-19}\ \text{J} \approx 2.33\ \text{eV}$$Step 2 — divide the power by the packet size. Photons per second = (joules per second) ÷ (joules per photon):
$$N = \frac{5.0 \times 10^{-3}\ \text{J/s}}{3.74 \times 10^{-19}\ \text{J/photon}} \approx 1.3 \times 10^{16}\ \text{photons per second}$$Interpretation: thirteen million billion photons every second, from a keychain toy. This is why nobody noticed the graininess of light for two centuries: with that many packets arriving, light looks perfectly smooth and continuous — just as a sand dune looks smooth from a distance. The photoelectric effect matters because it's an experiment where the packets are caught acting one at a time.
The photoelectric effect isn't museum physics — “one photon in, one electron out” is quietly running the modern world.
A solar panel is the photoelectric idea industrialized. In a silicon solar cell, each absorbed photon (if it carries enough energy — silicon's version of the exit fee is about 1.1 eV) knocks one electron loose inside the material, and the cell's structure sweeps those electrons into a circuit. Sunlight in, current out, one photon-electron transaction at a time, multiplied by trillions per second.
Every photo you've ever taken on a phone is a photon-counting experiment. Each of the millions of pixels in the sensor is a tiny light-sensitive well: arriving photons free electrons, the pixel stores them, and after the exposure the camera literally reads off how many electrons each pixel collected. Bright region = many photons = many electrons = bright pixel. The speckly “noise” in low-light photos is real quantum graininess: when only a handful of photons hit a pixel, the count fluctuates visibly.
The streetlight that switches itself on at dusk uses a photosensor: while daylight lands on it, ejected electrons keep a current flowing that holds the lamp off; when the photon supply fades at sunset, the current stops and the lamp switches on. Older automatic doors, safety light curtains, and film-soundtrack readers all used the same vacuum phototubes we described earlier.
In a photomultiplier tube, one photon ejects one electron from a coated surface; a ladder of charged plates then makes that electron knock loose a few more, and those a few more, until one photon has become a measurable pulse of millions of electrons. These single-photon detectors power night-vision equipment, medical PET scanners, and the giant neutrino observatories buried under mountains — all descendants of the humble photoelectric tube.
Look back at the photon-energy column of the spectrum table. Damaging a molecule in your skin — breaking a chemical bond — costs a few eV, and it's an all-or-nothing transaction, just like the exit fee. A UV photon carries 3–100 eV: one photon, one broken bond — that's sunburn and DNA damage. A radio photon carries ~10−6 eV, about a millionth of the fee, and photons can't pool their money. So a radio tower blasting kilowatts through your body all day breaks exactly zero bonds — the same logic as bright red light ejecting zero electrons from sodium. (Lots of low-energy photons can still gently warm things, which is all a microwave oven does.) Frequency, not brightness, decides what light can do.
Same constants as before: \(h = 6.626 \times 10^{-34}\) J·s, \(c = 3.00 \times 10^{8}\) m/s, 1 eV = 1.602 × 10−19 J — and feel free to use the \(1240/\lambda\) shortcut. Work each one on paper before revealing the solution.
Cesium has the lowest exit fee of the common metals: \(\phi = 1.95\) eV. A red photon with wavelength 650 nm arrives. Does it eject an electron? Show the numbers.
Photon energy: \(E = hc/\lambda = (1.988 \times 10^{-25}\ \text{J·m})/(650 \times 10^{-9}\ \text{m}) \approx 3.06 \times 10^{-19}\ \text{J}\). Converting: \(3.06 \times 10^{-19} / 1.602 \times 10^{-19} \approx 1.91\ \text{eV}\) (shortcut check: \(1240/650 \approx 1.91\) eV). Since 1.91 eV < 1.95 eV, the photon falls short of the work function — by only 0.04 eV, but the transaction is all-or-nothing, so no electron is ejected. “Just barely below threshold” and “far below threshold” give the same result: zero. (Cesium's threshold wavelength works out to about 636 nm, so this 650 nm photon is on the wrong side of the line.)
Now shine 400 nm violet light on the same cesium surface (\(\phi = 1.95\) eV). Compute \(KE_{max}\) of the ejected electrons, in eV.
From Worked Example 2 (same wavelength, so same photon), a 400 nm photon carries \(E = hc/\lambda \approx 4.97 \times 10^{-19}\ \text{J} \approx 3.10\ \text{eV}\). Apply the photoelectric equation: $$KE_{max} = hf - \phi = 3.10\ \text{eV} - 1.95\ \text{eV} = \mathbf{1.15\ \text{eV}}$$ Notice the comparison with sodium: the same violet photon left sodium's electrons with only 0.82 eV, because sodium charges a higher exit fee (2.28 eV vs. 1.95 eV). Same coin, different fee, different change.
A lamp shines below-threshold light on a metal and ejects zero electrons. The experimenter doubles the brightness. Explain, in terms of photons, why the answer is still exactly zero — and say what doubling the brightness would do if the light were above threshold.
Doubling the brightness doubles the number of photons per second — it does nothing to the energy of each photon, which is fixed at \(E = hf\) by the light's frequency. Since each electron can absorb only one photon (all-or-nothing, no saving up), every individual transaction still delivers \(hf < \phi\): the fee isn't covered, so every one of the (now twice as many) photons is refused. Twice zero is zero. Photons below the threshold don't add up; they fail one at a time, no matter how many arrive. If the light were above threshold instead, doubling the brightness would double the number of ejected electrons per second (twice the current) — while \(KE_{max}\) would stay exactly the same, still \(hf - \phi\).
Why the wave picture of light seemed unshakable, exactly how the photoelectric effect shattered it, and how Einstein's photon — energy arriving in indivisible packets of \(E = hf\), spent one packet per electron via \(KE_{max} = hf - \phi\) — explains every detail the wave theory couldn't. You can convert between joules and electron-volts, compute threshold frequencies and kinetic energies, and explain why sunlight can burn you while a radio tower can't.
But wait. Module 2 just proved light is a stream of particles… while the interference experiment from our “Exhibit A” proves, just as solidly, that light is a wave. Both experiments are correct. Both are repeatable. They cannot both be the whole story — and yet they are.
Next up: Module 3 — The Double-Slit Experiment, where we run the interference experiment one photon at a time and confront the full weirdness head-on: light — and, as de Broglie will show us, matter too — behaves as BOTH wave and particle. This is the experiment Feynman called “the only mystery” of quantum mechanics. See you there.