PhysicsCore24 min read

Dawn of Modern Physics

The photon, the photoelectric effect and wave-particle duality

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01

The experiment that broke the wave model

Definition

Threshold frequency — The minimum frequency of light that will cause photoelectric emission from a given metal. Below it, no electrons are emitted at any intensity.

By 1900 light was settled: it was a wave. Interference and diffraction had proved it, and a wave model explained everything anyone had measured. Then one experiment refused to fit.

Shine light on a clean metal surface and electrons are emitted. That much a wave model can allow. What it cannot allow is what actually happens.

Below a certain threshold frequency, no electrons come off at all — however bright the light, and however long you wait. Above it, electrons appear immediately, even with a very dim source. Making the light brighter produces more electrons, but never faster ones. Only raising the frequency raises their energy.

Every one of those is wrong on a wave model. A wave delivers energy continuously and spread out, so a dim light should simply take longer to accumulate enough — and brightness, not colour, should control the energy. That is not what nature does.

Drag the frequency below the dashed threshold and the emission stops dead — not "less", but none, at any brightness. Switch metals and the threshold moves, because each has its own work function. That cliff edge is what no wave model can produce.

02

Einstein's photon

Einstein's answer, in 1905, was that light arrives in discrete packets. Each packet — a photon — carries energy E = hf, fixed entirely by the frequency.

One photon is absorbed by one electron, all at once. If that single photon carries enough energy to free the electron from the metal, the electron leaves immediately; if it does not, nothing happens no matter how many such photons arrive.

The energy needed to free an electron is the work function φ, a property of the metal. Anything left over becomes the electron's kinetic energy, which gives Einstein's photoelectric equation.

Everything now fits. The threshold frequency is simply where hf equals φ. Brighter light means more photons, so more electrons — but each photon still carries the same energy, so the electrons are no faster. And emission is instant because it takes one photon, not an accumulation.

Energies here are small, so the electronvolt is used: the energy gained by an electron accelerated through one volt, 1 eV = 1.6 × 10⁻¹⁹ J.

E = h f = h c / λh f = φ + KE_maxKE_max = h f − φh = 6.63 × 10⁻³⁴ J s; below the threshold, hf < φ and nothing is emitted
h
Planck constantJ s
f
frequencyHz
φ
work functionJ or eV
KE_max
maximum kinetic energyJ or eV
Worked example 16 marks

Sodium has a work function of 2.3 eV. Light of wavelength 4.0 × 10⁻⁷ m falls on it. Find the photon energy in eV, the maximum kinetic energy of the emitted electrons, and the threshold wavelength.

  1. E = hc/λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸) / 4.0 × 10⁻⁷.Use hc/λ when given a wavelength rather than a frequency.
  2. E = 4.97 × 10⁻¹⁹ J.
  3. Convert: 4.97 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 3.1 eV.Dividing by the elementary charge converts joules to electronvolts.
  4. KE_max = hf − φ = 3.1 − 2.3.Both in eV, so no further conversion.
  5. KE_max = 0.8 eV.
  6. Threshold: λ₀ = hc/φ = 4.97 × 10⁻¹⁹ × 4.0 × 10⁻⁷ / (2.3 × 1.6 × 10⁻¹⁹) = 5.4 × 10⁻⁷ m.Green light and shorter will work; red will not.

3.1 eV photon, 0.8 eV electrons, threshold 5.4 × 10⁻⁷ m

03

Wave-particle duality

Light now had two faces. Interference and diffraction still demanded a wave; the photoelectric effect demanded particles. Rather than one being wrong, both are needed — light behaves as a wave when it travels and as particles when it exchanges energy with matter.

In 1924 de Broglie asked the obvious opposite question: if a wave can behave as a particle, can a particle behave as a wave? He proposed that any object with momentum has an associated wavelength, λ = h/p.

For anything visible this is far too small to matter. A cricket ball has a de Broglie wavelength around 10⁻³⁴ m — smaller than any length that means anything. But an electron's is comparable to atomic spacing, and that is measurable.

It was measured. Fire electrons at a thin crystal and they produce a diffraction pattern, exactly as X-rays do. Electrons — indisputably particles, with mass and charge — diffract. The electron microscope is built on this: because the electron wavelength is far shorter than light, it resolves detail no optical microscope can reach.

λ = h / p = h / (m v)for an accelerated electron: KE = e Vfaster or heavier means a shorter wavelength — which is why only very light particles diffract measurably
λ
de Broglie wavelengthm
p
momentumkg m s⁻¹
V
accelerating voltageV

Which model applies?

Use the wave model for propagation — interference, diffraction, refraction. Use the particle model for exchanges of energy with matter — the photoelectric effect, emission and absorption spectra. Neither is "the truth"; each is the description that works for what is being asked.

Key points

  1. Below the threshold frequency, no emission occurs at any intensity.
  2. E = hf — a photon's energy depends only on frequency.
  3. hf = φ + KE_max. Brighter light gives more electrons, not faster ones.
  4. Emission is instantaneous because one photon frees one electron.
  5. λ = h/p — everything has a wavelength, but only very light particles have a measurable one.

Practice questions

6 questions · 24 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
State two observations of the photoelectric effect that cannot be explained by a wave model of light.
Model answer

No emission occurs below a threshold frequency, whatever the intensity. And emission is instantaneous even in very dim light, rather than requiring time for energy to build up.

Examiner tip. Any two of: threshold frequency, instant emission, energy depending on frequency not intensity.

SQ2[2 marks]
Explain why increasing the intensity of light above the threshold frequency does not increase the maximum kinetic energy of the emitted electrons.
Model answer

Greater intensity means more photons per second, but each photon still carries the same energy hf. One photon is absorbed by one electron, so the energy per electron is unchanged — only the number emitted rises.

Examiner tip. The one-photon-one-electron idea is the mark. More photons, not more energetic ones.

SQ3[2 marks]
State what is meant by the work function of a metal.
Model answer

The minimum energy needed to remove an electron from the surface of the metal.

Examiner tip. Minimum energy, and from the surface. It is a property of the metal, which is why the threshold differs between metals.

Solved numericals

1 · 5 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[5 marks]
A metal has a work function of 3.2 eV. Light of frequency 1.2 × 10¹⁵ Hz falls on it. Find the photon energy in eV and the maximum kinetic energy of the emitted electrons. Take h = 6.63 × 10⁻³⁴ J s.
Full working
  1. E = hf = 6.63 × 10⁻³⁴ × 1.2 × 10¹⁵[1]
  2. E = 7.96 × 10⁻¹⁹ J[1]
  3. Converts: 7.96 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 4.97 eV[1]
  4. Uses KE_max = hf − φ = 4.97 − 3.2[1]
  5. KE_max = 1.8 eV[1]

photon 4.97 eV, electrons up to 1.8 eV

Examiner tip. Work in electronvolts throughout once you have converted, or in joules throughout. Mixing the two is the usual slip.

Long questions

1 · 8 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[8 marks]
Electrons are accelerated through a potential difference of 2500 V and directed at a thin graphite film.
  1. Calculate the kinetic energy gained by each electron, in joules. [2]
  2. Calculate their speed, taking the electron mass as 9.11 × 10⁻³¹ kg. [3]
  3. Calculate their de Broglie wavelength. [2]
  4. State what is observed on a screen beyond the film, and what it shows. [1]
Mark scheme
  1. Uses KE = eV = 1.6 × 10⁻¹⁹ × 2500[1]
  2. KE = 4.0 × 10⁻¹⁶ J[1]
  3. Uses KE = ½mv²[1]
  4. v² = 2 × 4.0 × 10⁻¹⁶ / 9.11 × 10⁻³¹[1]
  5. v = 3.0 × 10⁷ m s⁻¹[1]
  6. λ = h/mv = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 3.0 × 10⁷)[1]
  7. λ = 2.4 × 10⁻¹¹ mcomparable to atomic spacing, which is why diffraction is observable[1]
  8. A diffraction pattern of rings, showing that electrons behave as waves[1]

4.0 × 10⁻¹⁶ J, 3.0 × 10⁷ m s⁻¹, 2.4 × 10⁻¹¹ m, rings

Examiner tip. The wavelength coming out near atomic spacing is the point of the question — it is why the pattern appears at all.

Exam questions

1 · 5 marks

Multi-part questions with a full mark scheme.

Q1[5 marks]
Light is described as having a dual nature.
  1. State one phenomenon that shows light behaving as a wave, and one that shows it behaving as particles. [2]
  2. Explain why a cricket ball does not show observable wave behaviour. [3]
Mark scheme
  1. Wave: interference or diffraction (for example the double-slit experiment)[1]
  2. Particle: the photoelectric effect, or line spectra[1]
  3. Its de Broglie wavelength is λ = h/mv[1]
  4. Its mass and speed are enormous compared with h, so λ is around 10⁻³⁴ m[1]
  5. That is far smaller than any gap it could pass through, so no diffraction is observable[1]

interference/diffraction for waves; photoelectric effect for particles

Examiner tip. The reason is the size of h. It is so small that only very light, slow objects have a wavelength worth measuring.