PhysicsQuantum phenomena › The photoelectric effect

The photoelectric effect

Shine light on a metal and electrons can leap out, but the details do not match the predictions of a wave model. One experiment forced physics to accept photons, and its equation reads as a three-term energy budget you can balance in a single line.

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.
  • Read the Ek(max) against frequency graph, naming its gradient and both intercepts.
  • Estimate the Planck constant from the pd at which an LED first lights, using eV = hc/λ.

COMMON MISCONCEPTION

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 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 is a statement about ordinary light, and it deserves its condition stated. At lamp and laser-pointer intensities the chance of a second photon reaching the same electron while it still holds the first is negligible, so the one-to-one rule holds throughout this topic. Focus a pulsed laser hard enough and it stops being true. Multiphoton photoemission is real, and electrons then do come out under light below the threshold frequency. A-level assumes the low-intensity regime 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)one photon in, at most one electron outdim lightbrighter, same colourbelow thresholdmetal surfaceelectrons ejected:none, ever
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)ON THE AQA DATA SHEET

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.

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, 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. The photoelectric argument sits in that one comparison.

Reading the graph

Rearranged as Ek(max) = hf − φ, the equation plots as a straight line against frequency, gradient h, crossing the frequency axis at the threshold.

Maximum kinetic energy against frequency: a straight line of gradient h starting at the threshold frequencyfEₖ (max)f₀below f₀:no photoelectrons, 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.

Every metal draws the same gradient, because h belongs to the light and not to the surface. Change the metal and φ changes, sliding the line sideways without tilting it. There is one more measurable worth naming. 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), and an awkward kinetic energy becomes an easy voltage reading.

GUIDED PRACTICE

Reading the straight line

A metal's photoelectric graph of maximum kinetic energy against frequency cuts the frequency axis at 5.0 × 1014 Hz. Find the maximum kinetic energy at 9.0 × 1014 Hz, working from the graph's meaning.

Show the working

The intercept is the threshold frequency, and the line's gradient is h, so Ek(max) = h(f − f0).

Ek(max) = 6.63 × 10−34 × 4.0 × 1014 = 2.7 × 10−19 J, about 1.7 eV. Notice that you never needed to know which metal it was. Only the intercept moves from metal to metal.

INDEPENDENT PRACTICE

Photons by the trillion

A 1.0 mW laser pointer emits 650 nm light. Find the energy of one photon and the number emitted each second.

Show the working

E = hc/λ = (6.63 × 10−34 × 3.00 × 108)/(6.5 × 10−7) = 3.1 × 10−19 J.

Rate = power/energy per photon = 1.0 × 10−3/(3.1 × 10−19) = 3.3 × 1015 per second. Millions of billions arrive every second, which is how something relentlessly grainy manages to feel continuous.

Measuring h with LEDs

Reading h off that gradient is a genuine measurement of the constant, and it takes a photocell in a vacuum tube to make. A cheaper route to the same number runs the photon story backwards. A light emitting diode emits instead of absorbing. Forward bias pushes electrons across its junction, each electron drops through the energy step built into the junction, and the drops that emit light each put out one photon of the diode's own colour (some drops hand their energy to the crystal instead, which is why the argument is a threshold estimate rather than a bookkeeping of every electron).

So turn the pd across an LED up from zero. Nothing shows until each electron carries enough energy to make a photon of that colour, and at that threshold pd V the electron delivers eV while the photon carries hc/λ. Setting them equal, an idealisation that treats every joule of the electron's energy as reaching the photon, gives eV=hcλeV = \frac{hc}{λ}, and rearranging gives h=eVλch = \frac{eVλ}{c}. One voltmeter reading and the wavelength printed on the packet are the whole measurement.

In practice the LED runs from a potentiometer with a protective resistor in series, the voltmeter goes across the LED itself rather than across the pair, and the room is darkened so that the first faint glow can be seen at all. Take the pd at which that glow appears, then repeat with LEDs of several different colours. Plotting the threshold pd against 1/λ gives a straight line of gradient hc/e, one that passes close to the origin if the thresholds have been judged well, though the method's approximations can leave a small intercept, and h comes out as gradient × e/c from all the readings together instead of resting on one.

WORKED EXAMPLE

The constant from a red LED

A red LED of quoted wavelength 630 nm first glows at 1.9 V. Estimate the Planck constant.

h = eVλ/c = (1.60 × 10−19 × 1.9 × 630 × 10−9)/(3.00 × 108) = 6.4 × 10−34 J s.

Against the accepted 6.63 × 10−34 J s that is about four per cent low, which is roughly what this method delivers.

The word to use for it is estimate. Judging the exact pd at which a glow begins is an eye's decision taken in a dark room, an LED emits a band of wavelengths around the one on the packet, and a little of each electron's energy is spent inside the junction rather than reaching the photon. All three are reasons to quote the result as an estimate rather than a measurement.

GUIDED PRACTICE

The same measurement, taken from a gradient

Threshold pds for LEDs of five colours are plotted against 1/λ, giving a straight line of gradient 1.24 × 10−6 V m. Find the Planck constant, and say why the gradient is more reliable than any single LED's reading.

Show the working

The gradient is hc/e, so h = gradient × e/c = (1.24 × 10−6 × 1.60 × 10−19)/(3.00 × 108) = 6.6 × 10−34 J s.

Every threshold has been judged by the same eye in the same way, so much of that judgement error is likely to be shared between the points. Error the points share moves the line bodily and lands mainly in the intercept, while the gradient rests on the differences between the LEDs. That is a plausibility argument rather than a guarantee, and it is why the gradient is quoted rather than any single reading.

ASSESSMENT FOCUS

  • Three observations rule out the wave model, and each is worth marks. 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. The word minimum carries the mark on its own.
  • 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.
  • The Ek(max) against f graph gets asked three ways, so learn all three. Gradient h. Intercept on the f axis at f0. Intercept on the energy axis at −φ.
  • Work functions come quoted in eV while h works in joules. Multiply by 1.60 × 10−19 before substituting, never after.
  • For the LED method, say what the threshold pd means before using it: an electron crossing the junction can hand its energy to one photon, so eV = hc/λ at the pd where light first appears, as an idealised threshold. Quote the answer as an estimate and name a reason, judging the first glow by eye being the readiest one.
  • 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

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the photoelectric effect questions page.

9 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

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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.
  • Read the Ek(max) against frequency graph, naming its gradient and both intercepts.
  • Estimate the Planck constant from the pd at which an LED first lights, using eV = hc/λ.

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