Physics › ⚛️ Modern & Quantum

The Photoelectric Effect

Why a dim blue light can eject electrons that a blinding red light never touches.

AP Physics 2 levelUni Year 1
The Photoelectric Effect — illustration
Illustrative hero image.
💡
The big idea: Light doesn't deliver energy smoothly like a wave — it arrives in discrete packets called photons, each carrying energy E = hf. Whether a photon can knock an electron out of a metal depends entirely on its frequency, not on how many photons (how bright the light is) arrive per second. This one experiment convinced physicists that light itself is quantized.
🎯 By the end, you'll be able to
  • explain why the wave model of light fails to predict the photoelectric effect, while the photon model succeeds
  • calculate a photon's energy from its frequency or wavelength using E = hf
  • find a metal's work function and threshold frequency from experimental data
  • apply Einstein's equation KEmax = hf - W to find the maximum kinetic energy of ejected electrons
📎 You should already know
  • Wave properties of light (frequency, wavelength, speed)
  • Kinetic energy basics
  • Working with units: Joules vs. electron-volts

A Puzzle Classical Physics Couldn't Solve

In the late 1800s, physicists shone light on metal surfaces and watched electrons pop out — the photoelectric effect. It seemed easy to explain if light is a wave: brighter light carries more energy, so brighter light should knock electrons out with more energy. But that's not what happened.

Turn up the intensity of dim red light as much as you like, and nothing comes out. Switch to a faint blue light, and electrons fly out immediately. The color (frequency) of the light mattered — not how bright it was. Something was deeply wrong with treating light as a smooth, continuous wave.

🔑 Einstein's Insight: Light Comes in Packets

In 1905, Einstein proposed that light isn't continuous when it interacts with matter — it arrives in discrete packets of energy called photons. Each photon carries a fixed amount of energy that depends only on the light's frequency, never its intensity. A single photon either has enough energy to knock an electron free, or it doesn't. Turning up the intensity just sends more photons per second; it doesn't make any one photon more energetic.

\[ E = hf \]
The energy of a single photon depends only on its frequency \(f\); \(h\) is Planck's constant (\(6.63\times10^{-34}\,\text{J·s}\), or \(4.14\times10^{-15}\,\text{eV·s}\)).
\[ f_0 = \dfrac{W}{h} \quad\text{or}\quad \lambda_0 = \dfrac{hc}{W} \]
Every metal has a work function \(W\) — the minimum energy needed to pull an electron free. Below the threshold frequency \(f_0\) (equivalently, above the threshold wavelength \(\lambda_0\)), no electrons are ejected, no matter the intensity.
\[ KE_{max} = hf - W \]
Einstein's photoelectric equation: the maximum kinetic energy of an ejected electron equals the photon's energy minus the energy spent escaping the metal.
🎮 Interactive: The Photoelectric Effect LIVE
Adjust the light's frequency and intensity and watch how many electrons are ejected and how fast they fly off. Notice: intensity changes the current, not the electrons' speed.

Reading the Stopping Voltage

Experimentally, physicists measure \(KE_{max}\) by applying a reverse voltage just strong enough to stop the fastest electrons from reaching a collector — the stopping potential \(V_0\), where \(eV_0 = KE_{max}\). Plot \(KE_{max}\) against frequency for a given metal and you get a straight line. Its slope is exactly \(h\) — the same constant no matter which metal you test — and its x-intercept is the threshold frequency \(f_0\). That graph, built from real data, is some of the cleanest evidence we have that light is quantized.

📝 Worked example: Light of wavelength 400 nm strikes a sodium surface, whose work function is 2.28 eV. Find the maximum kinetic energy of the ejected photoelectrons.
  1. Find the photon energy using \(E = hf = \dfrac{hc}{\lambda}\). Using \(hc = 1240\,\text{eV·nm}\): \(E = \dfrac{1240}{400} = 3.10\,\text{eV}\).
  2. Apply Einstein's equation: \(KE_{max} = hf - W = 3.10\,\text{eV} - 2.28\,\text{eV}\).
  3. \(KE_{max} = 0.82\,\text{eV}\).
✓ \(KE_{max} \approx 0.82\,\text{eV}\) (about \(1.31\times10^{-19}\,\text{J}\)).
📝 Worked example: A metal has a threshold wavelength of 580 nm. Find its work function (in eV) and its threshold frequency.
  1. Relate work function to threshold wavelength: \(W = \dfrac{hc}{\lambda_0}\). Using \(hc = 1240\,\text{eV·nm}\): \(W = \dfrac{1240}{580} \approx 2.14\,\text{eV}\).
  2. Convert to joules: \(W = 2.14\,\text{eV} \times 1.60\times10^{-19}\,\text{J/eV} \approx 3.43\times10^{-19}\,\text{J}\).
  3. Find the threshold frequency from \(f_0 = W/h\): \(f_0 = \dfrac{3.43\times10^{-19}}{6.63\times10^{-34}} \approx 5.17\times10^{14}\,\text{Hz}\).
✓ \(W \approx 2.14\,\text{eV}\); \(f_0 \approx 5.17\times10^{14}\,\text{Hz}\).
⚠️ Common Trap: Intensity ≠ Energy Per Electron

It's tempting to think brighter light means faster electrons. It doesn't. Intensity controls how many photons arrive per second, so it changes the number of electrons ejected (the photocurrent) — not their maximum speed. Only a higher frequency (bluer light) increases \(KE_{max}\), because that's what increases the energy of each individual photon. Keep intensity and frequency firmly separated in your mind, and this topic gets much less confusing.

Check your understanding

1. According to Einstein's photon model, what property of light determines whether electrons are ejected from a metal at all?
Each photon's energy depends only on frequency (E = hf). A photon below the threshold frequency never carries enough energy to free an electron, no matter how many arrive — that's why intensity alone can't trigger emission.
2. A metal has a work function of 2.30 eV. Light of wavelength 300 nm shines on it. What is the maximum kinetic energy of the ejected photoelectrons?
Photon energy = hc/λ = 1240/300 ≈ 4.13 eV. Subtracting the work function: 4.13 eV − 2.30 eV = 1.83 eV = KEmax.
3. Light below a metal's threshold frequency is made much more intense. What happens?
Below the threshold frequency, every individual photon carries less energy than the work function, so no single photon can free an electron. Adding more of them (higher intensity) just means more photons that still can't do the job — electrons don't accumulate energy from multiple photons.
4. In a plot of KEmax versus frequency for a given metal, what does the slope of the line represent?
Rearranging KEmax = hf − W into the form y = mx + b shows a line with slope h and y-intercept −W. The slope is Planck's constant — identical for every metal, since h is a fundamental constant, not a property of the material.
✅ Key takeaways
  • Light delivers energy in discrete packets called photons, each carrying E = hf — determined by frequency alone, never by intensity.
  • Every metal has a work function W: the minimum energy needed to free an electron from its surface.
  • Only light above the threshold frequency f0 = W/h (or below the threshold wavelength λ0 = hc/W) can eject electrons, no matter how bright it is.
  • Einstein's equation KEmax = hf − W predicts the maximum kinetic energy of ejected electrons; intensity only changes how many electrons are ejected, not how fast.