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The photoelectric effect and the photon

Light ejects electrons from a metal surface only above a threshold frequency, an observation no wave model of light accounts for. Einstein's photon explanation gives the photoelectric equation hf = φ + Ek(max), where the work function φ is the least energy an electron needs to escape and the stopping potential measures Ek(max).

The photoelectric effect

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Year 12AQA 3.2.2.1CIE 22.1.1, 22.1.2, 22.1.3, 22.1.4, 22.1.5, 22.2.1, 22.2.2, 22.2.3, 22.2.4, 22.2.5OCR A 4.5.1(a), 4.5.1(b), 4.5.1(c), 4.5.1(d), 4.5.2(a)(i), 4.5.2(b), 4.5.2(c), 4.5.2(d), 4.5.2(e), 4.5.2(f)EDEXCEL 91, 92, 93, 94, 95

The photoelectric effect, part 1 of 2. Part 2 is The photoelectric graph and measuring h.

IN THIS TOPIC

  • Describe the photoelectric observations that a wave model of light cannot explain.
  • Explain threshold frequency using photons, and define work function and stopping potential.
  • Apply the photoelectric equation hf = φ + Ek(max), converting a work function from eV first.

COMMON MISCONCEPTION

Brighter light gives the electrons more energy.

Brightness sets only how many photons arrive, not the energy of each. One electron absorbs one photon of energy E = hf, so below the threshold frequency no electrons escape however bright the light.

What actually happens

Shine light on a clean metal surface and, under the right conditions, electrons are ejected: the photoelectric effect. The observations are what a classical wave model cannot explain. Below a certain threshold frequency f0, no electrons leave, however bright the light and however long you wait. Above it, they leave immediately, even in the dimmest glow, and making the light brighter ejects more electrons without making any of them faster.

A wave should behave differently on every count. Waves deliver energy continuously, so a bright red lamp should eventually shake electrons loose, and a brighter beam of any colour should eject faster electrons after a pause while energy accumulates. None of that happens. The frequency decides whether emission happens at all and sets the maximum kinetic energy, the intensity sets only the number ejected, and the emission is effectively instantaneous.

The photon explanation

Einstein's resolution was to let light arrive in photons, packets of energy E = hf, with each ejected electron absorbing one photon, whole or not at all. Escaping the metal requires a fixed minimum energy, the work function φ, the energy needed to remove an electron from the surface.

One photon per electron holds at ordinary intensities, where the chance of a second photon reaching the same electron while it still holds the first is negligible. Under an intense pulsed laser it does not: multiphoton photoemission occurs, and electrons are then emitted by light below the threshold frequency. A-level assumes the low-intensity condition throughout.

A photon carries momentum as well as energy, p = E/c, despite having no mass. The kick is tiny, so sunlight does not knock you over, but it is real enough to push comet tails and steer a spacecraft under solar sail. CIE asks for that relation by name; AQA does not.

The photoelectric effect, event by event (animated figure)
Describe the drawing

Light arrives as photons, and each one above the threshold frees exactly one electron, its speed set by hf minus the work function. First beat: dim light, five photons, five electrons. Second: three times brighter, same colour, so fifteen photons and fifteen electrons at exactly the same speed as before. Third: plenty of photons but each below the threshold, and nothing leaves, ever, which is the observation the wave picture cannot survive.

FIG. 1Light arrives as photons, and each absorbed photon can eject at most one electron, its speed set by hf minus the work function. First beat: dim light, five photons absorbed, five electrons. Second: three times brighter, same colour, so fifteen absorbed and fifteen ejected, the fastest of them no faster than before. (Real surfaces also reflect photons or absorb some without an escape; the animation follows the absorbed ones that succeed.) Third: plenty of photons but each below the threshold, and nothing leaves however long you wait, the observation a wave model cannot explain.

The budget balances in one line, the photoelectric equation.

hf=φ+Ek(max)hf = \phi + E_{k}(max)IN THE AQA DATA BOOKLET

Ek(max) is the maximum kinetic energy, carried by the electrons that started right at the surface. Electrons from deeper down lose extra energy on the way out and emerge slower. Every observation now follows. A photon with hf below φ carries too little energy, so below f0 = φ/h nothing leaves at any ordinary brightness, because turning up the intensity sends more photons and not bigger ones. Above the threshold, one absorbed photon ejects one electron at once, with no waiting about while energy accumulates.

Two identical metal plates in dim and bright blue light. The bright plate ejects more electrons, but every electron arrow is the same length: intensity sets the number, frequency sets the energy.
FIG. 2Dim and bright light of the same colour: the bright beam ejects more electrons, but the fastest are no faster; the maximum energy has not moved.

WORKED EXAMPLE

A threshold frequency from a work function

Zinc's work function is 4.3 eV. Find its threshold frequency, and decide whether visible light can eject electrons from zinc.

Convert first, so φ = 4.3 × 1.60 × 10−19 = 6.9 × 10−19 J. At the threshold a photon supplies exactly the work function and nothing more.

f0 = φ/h = 6.9 × 10−19/(6.63 × 10−34) = 1.0 × 1015 Hz.

Visible light tops out near 7.5 × 1014 Hz, comfortably below that threshold. No visible photon can free an electron from zinc, however bright the lamp, and zinc needs ultraviolet.

One more quantity makes Ek(max) easy to measure. The stopping potential Vs is the pd (the energy each coulomb of charge transfers, met properly in the Electricity unit) needed to bring even the fastest photoelectrons to rest, so eVs = Ek(max).

Reading the graph has moved to the next part of this lesson.

Measuring h with LEDs has moved to the next part of this lesson.

ASSESSMENT FOCUS

  • Three observations rule out the wave model, and a complete answer names all three. A sharp threshold frequency. Intensity changing the number ejected and not their energy. Emission with no measurable delay.
  • Define the work function as the minimum energy needed to remove an electron from the metal's surface. Without the word minimum, the definition is satisfied by any larger energy as well.
  • One photon is absorbed by one electron, and at the intensities this topic deals with no electron collects two. That single sentence explains the instant emission and the flat maximum energy, and it is the line a complete answer needs. Multiphoton absorption does happen under intense pulsed lasers, but nothing on the specification goes near that regime.
  • Ek(max) belongs to the surface electrons. Asked why the emitted electrons come out with a range of energies, say that the deeper ones lose extra energy on the way out.
  • Work functions come quoted in eV while h works in joules. Multiply by 1.60 × 10−19 before substituting, never after.
  • AQA asks for the stopping potential as an idea only, so eVs = Ek(max) is what you need. Measuring one in the lab is outside the specification.

CHECK YOURSELF

Sodium has a work function of 2.3 eV. Light of frequency 7.0 × 1014 Hz falls on it. Find the maximum kinetic energy of the photoelectrons, in eV and in joules. (h = 6.63 × 10−34 J s.)

Show a hint

Find the photon energy in joules first, then run the budget.

Show the answer

Photon energy first. E=hfE = hf = 6.63 × 10−34 × 7.0 × 1014 = 4.64 × 10−19 J, which is 2.9 eV.

Now the budget. Ek(max)=hf-φE_{k}(max) = hf - \phi = 2.9 − 2.3 = 0.6 eV.

In joules, 0.6 × 1.60 × 10−19 = 9.6 × 10−20 J. Each photon either supplies at least the 2.3 eV work function or no electron is emitted. Brightness never enters the calculation.

Frequency sets whether electrons leave, and how fast.

Brightness sets only how many.

WORKBOOK

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  • Describe the photoelectric observations that a wave model of light cannot explain.
  • Explain threshold frequency using photons, and define work function and stopping potential.
  • Apply the photoelectric equation hf = φ + Ek(max), converting a work function from eV first.

Open the full revision checklist to track your progress across the whole unit.