F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Quantum MechanicsThe Photoelectric Effect: Evidence That Light Is Made of Particles
F3.1 F2.1 F1.1 F2.4About 70 minutes
Learning goals- I can describe the apparatus and the results of photoelectric effect experiments.
- I can explain stopping voltage and use E(k,max) = eV₀ to find the maximum kinetic energy of photoelectrons.
- I can explain why the wave model of light could not account for the results.
- I can explain the results using Einstein's photon model and the idea of a work function.
Words to know
photoelectric effect
The emission of electrons from a metal surface when light of high enough frequency shines on it.
photoelectron
An electron emitted from a surface by the photoelectric effect.
photocurrent
The current produced by photoelectrons flowing from the emitting surface to a collector.
threshold frequency (f₀)
The minimum frequency of light that can eject electrons from a particular metal.
work function (W)
The minimum energy needed to remove an electron from the surface of a particular metal. W = hf₀.
stopping voltage (V₀)
The reverse voltage that is just large enough to stop even the fastest photoelectrons; E(k,max) = eV₀.
Light that knocks out electrons
In 1887 Heinrich Hertz, while proving that electromagnetic waves exist, noticed something odd: sparks jumped a gap more easily when ultraviolet light shone on the metal. Over the next 15 years, experimenters (notably Philipp Lenard) showed that light can knock electrons out of a metal surface. This is the photoelectric effect.
At first this seemed easy to explain. Light is an electromagnetic wave; its electric field shakes the electrons in the metal; shake hard enough and some fly off. But when the details were measured, almost every one of them contradicted that wave picture. The photoelectric effect became the strongest early evidence for a particle model of light.
The apparatus
Two metal electrodes sit inside an evacuated glass tube (no air, so electrons travel freely). Light shines on one electrode, the emitting surface (cathode). Electrons that escape fly to the collector (anode). An ammeter measures the photocurrent.
A variable power supply lets you change the voltage between the electrodes:
- With the collector positive, it attracts electrons, so almost every emitted electron is collected.
- With the collector negative, it repels electrons. Only electrons with enough kinetic energy can push through. As the reverse voltage grows, fewer and fewer get through. At the stopping voltage V₀, even the fastest electrons are turned back and the current drops to zero.
Measuring kinetic energy with a voltage
An electron moving against a stopping voltage V₀ loses kinetic energy equal to the work done on it, eV₀ (as in the unit Gravitational, Electric and Magnetic Fields). The fastest electrons are just stopped, so
E(k,max) = eV₀
In electron-volts this is very convenient: a stopping voltage of 1.2 V means the fastest electrons had 1.2 eV of kinetic energy.
Worked exampleExample 1: From stopping voltage to speed
Light falling on a metal produces photoelectrons that are stopped by a reverse voltage of 1.2 V. Find (a) the maximum kinetic energy in eV and in joules and (b) the maximum speed of the photoelectrons. (mₑ = 9.11 × 10⁻³¹ kg)
- (a) E(k,max) = eV₀ = 1.2 eV.
- In joules: (1.60 × 10⁻¹⁹ C)(1.2 V) = 1.92 × 10⁻¹⁹ J.
- (b) E(k) = ½mv², so v = √(2E(k)/m) = √(2 × 1.92 × 10⁻¹⁹ ÷ 9.11 × 10⁻³¹).
- v = √(4.22 × 10¹¹) = 6.5 × 10⁵ m/s.
(a) 1.2 eV = 1.9 × 10⁻¹⁹ J (b) 6.5 × 10⁵ m/s. That is fast, but only about 0.2% of the speed of light, so ordinary ½mv² is fine.
(a) Brighter light gives a larger current, but the same stopping voltage. (b) Higher-frequency (blue) light needs a larger stopping voltage than green: its electrons have more kinetic energy. Schematic graphs, not real data.What the experiments showed
Here are the key results, each compared with what the wave model predicted.
- Threshold frequency. For each metal there is a minimum frequency f₀. Below it, no electrons are emitted, no matter how bright the light or how long it shines. Wave model: any frequency should work if the light is bright enough or shines long enough.
- No time delay. Above f₀, electrons come out immediately (within about 10⁻⁹ s), even in very dim light. Wave model: in dim light an electron would need a long time to soak up enough energy (see Example 2).
- Current depends on brightness. Above f₀, brighter light gives more photoelectrons (a larger current). Wave model: agrees on this point.
- Kinetic energy depends on frequency, not brightness. The stopping voltage, and so E(k,max), is the same for dim and bright light of one colour. It increases steadily as the frequency increases. Wave model: brighter light means stronger fields, so electrons should come out faster.
| Observation | Wave model predicts | Experiment shows |
|---|
| Light below some frequency | Works if bright enough | No electrons at all |
| Dim light | Delay while energy builds up | Emission immediate |
| Brighter light | More and faster electrons | More electrons, same E(k,max) |
| Higher frequency | No special effect | Larger E(k,max) |
Worked exampleExample 2: How long would the wave model make us wait?
Dim light of intensity 1.0 × 10⁻² W/m² falls on sodium. Suppose, as the wave model does, that one electron collects all the energy falling on a circle about the size of an atom (radius 1.0 × 10⁻¹⁰ m). How long would it take to collect the 2.0 eV needed to escape?
- Area: A = πr² = π(1.0 × 10⁻¹⁰ m)² = 3.1 × 10⁻²⁰ m².
- Power collected: P = IA = (1.0 × 10⁻²)(3.1 × 10⁻²⁰) = 3.1 × 10⁻²² W.
- Energy needed: 2.0 eV = 3.2 × 10⁻¹⁹ J.
- Time: t = E/P = 3.2 × 10⁻¹⁹ ÷ 3.1 × 10⁻²² ≈ 1.0 × 10³ s.
About 1000 s (17 minutes). In the real experiment electrons appear within about a billionth of a second. The energy is clearly not spread out evenly over the wavefront.
Einstein's explanation (1905)
Einstein said: treat the light as photons, each with energy hf, and assume one photon gives all its energy to one electron.
- Pulling an electron out of the metal costs a minimum energy, called the work function W. It is different for each metal.
- If hf < W, a single photon cannot free an electron, and photons do not team up. No emission, however bright. The threshold is hf₀ = W.
- If hf > W, the electron escapes at once, and the leftover energy becomes kinetic energy. The fastest electrons (those right at the surface) have E(k,max) = hf − W.
- Brighter light means more photons, so more electrons, but each photon still has the same hf, so E(k,max) is unchanged.
Every observation fits. In the next lesson you will use Einstein's equation to solve problems and analyse data.
A change in scientific thought (F1.1)
By 1900 the wave model of light seemed proven beyond doubt: Young's interference experiment, Maxwell's equations and Hertz's radio waves all supported it. The photoelectric effect forced physicists to accept that light also behaves like a stream of particles. Many leading physicists, including Planck, were sceptical of Einstein's photons for years. Robert Millikan set out to test Einstein's equation, expecting it to fail; his careful measurements, published in 1916, confirmed it instead.
Just for funWhy couldn't the dim red light get an electron out of the metal? It kept trying to pay the work function in tiny change, and the metal only accepts whole photons.
Hands-on activityA photoelectric effect simulation (investigation)
You need: a computer or tablet with a browser. Search for the free PhET "Photoelectric Effect" simulation (University of Colorado Boulder). It runs in the browser.
- Choose sodium. Set the intensity to about 50% and the wavelength to 600 nm (orange). Are any electrons emitted?
- Slowly shorten the wavelength until electrons first appear. Record this threshold wavelength. Convert it to a threshold frequency with f₀ = c/λ.
- Pick a wavelength shorter than the threshold wavelength (for example 400 nm). Change the intensity from 20% to 100%. Watch the current and the speed of the electrons. What changes, and what does not?
- At 400 nm, adjust the battery voltage to negative values until the current just reaches zero. Record this stopping voltage. Repeat at 350 nm and 300 nm.
- Write a short conclusion: which results support the photon model, and why?
What to look for: sodium's threshold is near 540 nm (work function about 2.3 eV). Below that frequency nothing happens at any intensity. Above it, intensity changes the number of electrons, not their speed. The stopping voltage grows as the wavelength gets shorter (about 0.8 V at 400 nm, 1.2 V at 350 nm and 1.8 V at 300 nm for a 2.3 eV work function). You will analyse data like these in lesson 5.
Your turn
1A stopping voltage of 0.80 V is measured. What is the maximum kinetic energy of the photoelectrons in eV and in J?
E(k,max) = eV₀ = 0.80 eV = 0.80 × 1.60 × 10⁻¹⁹ = 1.3 × 10⁻¹⁹ J.
2The fastest photoelectrons from a metal have 3.0 eV of kinetic energy. What stopping voltage is needed?
3Find the maximum speed of photoelectrons whose stopping voltage is 2.0 V.
E = 2.0 × 1.60 × 10⁻¹⁹ = 3.2 × 10⁻¹⁹ J. v = √(2 × 3.2 × 10⁻¹⁹ ÷ 9.11 × 10⁻³¹) = 8.4 × 10⁵ m/s.
4Light just above the threshold frequency shines on a metal. The intensity is then tripled. What happens to (a) the photocurrent and (b) the stopping voltage?
(a) It triples (about three times as many photons, so three times as many electrons). (b) It stays the same, because each photon still has the same energy.
5Red light produces no photoelectrons from a certain metal. Will very bright red light work? Explain using photons.
No. Each red photon has energy hf less than the work function, and an electron absorbs only one photon. More photons means more failed attempts, not more energy per electron.
6Which wave-model prediction about the photoelectric effect turned out to be correct?
That brighter light gives more photoelectrons (a larger current). Its predictions about threshold frequency, time delay and kinetic energy were wrong.
7Why must the tube be evacuated?
Air molecules would collide with the photoelectrons, slowing and scattering them, so the current and stopping voltage could not be measured cleanly. (Air could also contaminate the metal surface.)
More practice: warm-up
1What particles are emitted in the photoelectric effect?
2What is the name for the minimum energy needed to free an electron from a metal?
3Does brighter light of the same colour change E(k,max)?
4What happens below the threshold frequency?
No electrons are emitted, at any brightness.
5Write the relationship between maximum kinetic energy and stopping voltage.
More practice: on your own
1Photoelectrons have a maximum speed of 5.0 × 10⁵ m/s. Find the stopping voltage.
E = ½(9.11 × 10⁻³¹)(5.0 × 10⁵)² = 1.14 × 10⁻¹⁹ J = 0.71 eV, so V₀ = 0.71 V.
2Explain, using the wave model, why scientists expected a time delay in dim light, and what was observed instead.
In the wave model the energy is spread over the whole wavefront, so in dim light each electron receives energy slowly and would need seconds to minutes to collect enough to escape. Instead electrons are emitted immediately (within about 10⁻⁹ s), because each photon delivers its energy to one electron all at once.
3Graph (b) above shows blue light needs a larger stopping voltage than green light. What does that tell you?
Electrons ejected by blue light have more kinetic energy. Blue photons have higher frequency and so more energy (hf), leaving more energy after the work function is paid.
4Repeat the Example 2 calculation for light 10 times brighter. Does this rescue the wave model?
Power collected is 10 times larger, so t ≈ 100 s. Still far longer than the observed 10⁻⁹ s, so no.
5Two experimenters use the same metal. One uses dim violet light, the other bright green light (both above threshold). Whose electrons have the larger E(k,max)? Whose current is larger (probably)?
Violet gives the larger E(k,max) (higher frequency). The bright green beam probably gives the larger current, since it likely delivers more photons per second.
Challenge
1A student says: "The stopping voltage measures the kinetic energy of all the photoelectrons." Correct this statement.
The stopping voltage measures only the maximum kinetic energy. Many electrons lose some energy on their way out of the metal (collisions inside it) and come out slower; a smaller reverse voltage already stops those. V₀ is the voltage that stops even the fastest ones.
2In the wave model, the energy of a wave depends on its amplitude. Use this to explain exactly why observation 4 (kinetic energy independent of brightness) contradicted the wave model.
Brighter light = larger amplitude = stronger electric field. A stronger field would push electrons harder and give them more kinetic energy. The experiments showed E(k,max) does not change with brightness at all; it depends only on frequency. That pattern matches energy packets of size hf, not a continuous wave.
3At the same frequency and intensity, metal X gives a stopping voltage of 1.4 V and metal Y gives 0.6 V. Which metal has the larger work function? By how many eV?
Same photon energy hf in both. E(k,max) = hf − W, so the metal with the smaller E(k,max) has the larger W: Y, by 1.4 − 0.6 = 0.8 eV.
Quiz
1. In a photoelectric experiment, the light's intensity is doubled while its frequency stays the same (above threshold). Which change occurs?
2. Which observation could NOT be explained by the wave model of light?
3. The stopping voltage for photoelectrons is 2.5 V. Their maximum kinetic energy is
4. What is the work function of a metal?
5. In Einstein's model, how is a photon's energy given to the metal?
6. Why do photoelectrons appear with no measurable delay, even in dim light?
Big idea (F3.1): the photoelectric effect is the classic evidence for the particle (photon) model of light. Brightness controls how many electrons come out; frequency controls how energetic they are; below a threshold frequency, nothing happens.
How to help: ask your teen to explain the four observations using an analogy, for example a vending machine that only accepts loonies: piling up more dimes (low-frequency photons) never gets you a snack, but each loonie (high-frequency photon) gets one right away. Then ask where the analogy breaks down (leftover energy becomes kinetic energy, like getting change).
Common difficulties:
- Thinking brighter light gives faster electrons. It gives more electrons.
- Mixing up stopping voltage (in V) and kinetic energy (in eV or J): E(k,max) = eV₀.
- Forgetting that the stopping voltage gives the maximum kinetic energy, not every electron's energy.
- Saying the wave model failed completely: it correctly predicted more current with brighter light.