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The photoelectric effect

Shine light on a metal and electrons can leap out, but the details refuse to behave like a wave. The experiment that forced physics to accept photons is first-term A-level content, and its equation is a three-term energy budget.

Year 12AQA 3.2.2.1

IN THIS TOPIC

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

WHAT YOU PROBABLY THINK

Brighter light gives the electrons more energy.

What actually happens

Shine light on a clean metal surface and, under the right conditions, electrons are ejected: the photoelectric effect. The conditions are the scandal. 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 rules everything, the intensity rules only the count, and the emission is effectively instantaneous.

The photon explanation

Einstein's resolution: light arrives in photons, packets of energy E = hf, and each ejected electron has absorbed one photon, whole or not at all. Escaping the metal costs a fixed entry fee, the work function φ, the minimum energy needed to remove an electron from the surface.

One photon, one electron: the photon's energy pays the work function first, and the rest becomes kinetic energymetal surfacephoton: hfelectron outhfφEₖhf = φ + Eₖ: the work function is the exit toll
FIG. 1One photon in, one electron out. The photon's energy hf pays the work function φ, and whatever remains is the electron's kinetic energy.

The budget balances in one line, the photoelectric equation:

hf = φ + Ek(max)ON YOUR DATA SHEET

Ek(max) is the maximum kinetic energy, carried by electrons that started at the surface; those from deeper pay extra on the way out and emerge slower. Every observation now follows. A photon with hf below φ cannot pay the toll, so below f0 = φ/h nothing leaves at any brightness: intensity means more photons, not bigger ones. Above threshold, one absorbed photon ejects one electron at once, no accumulation required.

Brightness changes how many electrons leave, never how fast: only frequency sets the energydim bluebright bluesame speed out of both plates: more photons, not bigger ones
FIG. 2Dim and bright light of the same colour: the bright beam ejects more electrons, and every one leaves with the same maximum energy.

Reading the graph

Plot Ek(max) against frequency and the equation becomes a straight line: Ek(max) = hf − φ, gradient h, meeting the axis at the threshold frequency.

Maximum kinetic energy against frequency: a straight line of gradient h starting at the threshold frequencyfEₖ (max)f₀below f₀:nothing, at any brightnessgradient = h
FIG. 3Nothing below the threshold frequency; above it, maximum kinetic energy climbs along a straight line whose gradient is the Planck constant.

The gradient is the same for every metal, because h belongs to light, not to the surface; changing the metal changes φ and slides the line sideways without tilting it. One more measurable: the stopping potential Vs is the pd needed to bring even the fastest photoelectrons to rest, so eVs = Ek(max). It converts an awkward kinetic energy into an easy voltage reading. AQA asks for the idea, and states that the experimental determination of stopping potential is not required.

THE EXAM BIT

  • The three wave-killing observations, each worth marks: a sharp threshold frequency; intensity changing the number ejected but not their energy; emission without delay.
  • Define the work function as the minimum energy needed to remove an electron from the metal's surface. The word minimum carries the mark.
  • Ek(max) belongs to surface electrons. Asked why emitted electrons have a range of energies, answer that deeper electrons lose extra energy escaping.
  • On the Ek(max) against f graph: gradient h, intercept on the f axis at f0, intercept on the energy axis at −φ. All three get asked.
  • Work functions are usually quoted in eV and h works in joules: convert with 1.60 × 10−19 before substituting, not after.

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: E = hf = 6.63 × 10−34 × 7.0 × 1014 = 4.64 × 10−19 J = 2.9 eV.

Budget: Ek(max) = hf - φ = 2.9 − 2.3 = 0.6 eV.

In joules: 0.6 × 1.60 × 10−19 = 9.6 × 10−20 J. A photon either pays the 2.3 eV toll or nothing happens; brightness never enters the calculation.

Frequency decides if electrons leave, and how fast.

Brightness only decides how many.

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