IB DP Physics: Photoelectric effect, matter waves and Compton scattering (DP Physics)

This guide covers sub-topic E.2, Quantum physics, from Topic E (Nuclear and quantum physics) of the IB Diploma Programme Physics guide, first assessment 2025. E.2 is entirely additional higher level (HL only): the guide lists no standard level content, so it is for HL candidates only. Use it with the revision notes and the practice set. The course hub is at /boards/ib/ib-dp/physics/ and the printable checklist is at /checklists/ib/ib-dp/physics/. For the bigger picture see the DP Physics syllabus guide and exam preparation. To find your weak spots first, try the free diagnostics.

The guide’s two guiding questions set the theme: how can light create an electric current, and what is meant by wave-particle duality?

What E.2 asks you to do

Guide section	What you must be able to do	Tier
E.2	Explain the photoelectric effect as evidence for the particle nature of light	HL only
E.2	Explain the threshold frequency	HL only
E.2	Use Einstein’s explanation, Emax = hf - Φ, with the work function Φ	HL only
E.2	Discuss which features of the effect classical wave theory cannot explain	HL only
E.2	Explain particle diffraction as evidence for the wave nature of matter, and wave-particle duality	HL only
E.2	Use the de Broglie wavelength λ = h/p; describe a scattering experiment and locate the minimum intensity	HL only
E.2	Explain Compton scattering and use Δλ = (h/mec)(1 - cos θ) (no derivation needed)	HL only

1. The photoelectric effect

When light of suitable frequency falls on a clean metal surface, electrons are emitted. They are called photoelectrons. If the emitted electrons are collected, a current flows in an external circuit. This is the sense in which light can be used to create a current.

Three observations matter.

1. Below a certain frequency, called the threshold frequency f₀, no electrons are emitted. This holds however intense the light is and however long you wait.
2. Above f₀, the maximum kinetic energy of the electrons depends on the frequency of the light. It does not depend on the intensity.
3. Emission starts with no measurable delay once the light arrives, even for very dim light.

Increasing the intensity above the threshold increases the number of electrons released each second, so the photocurrent rises. It does not change their maximum energy.

Why classical wave theory fails

A classical wave spreads its energy over the whole wavefront. A brighter beam carries more energy per second, and a dim beam should deliver energy to an electron gradually until the electron escapes. The guide asks you to discuss the features that this picture cannot explain:

• Threshold frequency. Waves of any frequency should free electrons if the intensity is high enough or the exposure long enough. In practice, low-frequency light never does.
• Emax independent of intensity. A more intense wave should give electrons more energy. In practice only the frequency changes Emax.
• No time delay. A dim wave should need time to build up enough energy. In practice emission is immediate.

Einstein’s photon explanation

Light arrives in packets, photons, each with energy E = hf (h is the Planck constant). One photon gives all of its energy to one electron in a single event.

The work function Φ is the least energy that must be supplied to free an electron at the metal’s surface. An electron that uses the least energy, and loses nothing on the way out, leaves with the maximum kinetic energy:

Emax = hf - Φ

At threshold, Emax = 0, so hf₀ = Φ. Photons below f₀ each carry less than Φ, and since one photon acts on one electron, none can eject an electron. More intense light means more photons per second and so more electrons per second, with unchanged Emax. A photon is absorbed all at once, which explains the lack of delay. The same three observations that defeat the wave model therefore follow from the photon model.

Stopping potential and the graph

If a retarding potential difference Vs just stops the fastest electrons, Emax = eVs. Then eVs = hf - Φ, so a graph of Vs against f is a straight line:

• gradient = h/e
• intercept on the frequency axis = f₀
• intercept on the Vs axis = -Φ/e

Worked example 1: one metal, three beams

A metal has Φ = 2.90 eV. Take h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C and electron mass 9.11 × 10⁻³¹ kg.

(a) Threshold frequency: Φ = 2.90 × 1.60 × 10⁻¹⁹ = 4.64 × 10⁻¹⁹ J, so f₀ = Φ/h = 4.64 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 7.00 × 10¹⁴ Hz. The matching wavelength is c/f₀ = 429 nm.

(b) Light of wavelength 310 nm: photon energy = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 310 × 10⁻⁹ = 6.42 × 10⁻¹⁹ J = 4.01 eV. Then Emax = 4.01 - 2.90 = 1.11 eV (1.78 × 10⁻¹⁹ J). The stopping potential is 1.11 V. The maximum speed comes from ½mv² = 1.78 × 10⁻¹⁹ J: v = √(2 × 1.78 × 10⁻¹⁹ / 9.11 × 10⁻³¹) = 6.24 × 10⁵ m s⁻¹.

(c) Light of wavelength 450 nm has photon energy 2.76 eV, which is less than 2.90 eV. No electrons are emitted, however bright the beam.

Worked example 2: photons per second and current

A metal has Φ = 2.10 eV. A 0.40 mW beam of wavelength 350 nm falls on it. On average 1 photon in 1000 releases an electron.

• Photon energy = hc/λ = 5.68 × 10⁻¹⁹ J = 3.55 eV.
• Photons per second = 0.40 × 10⁻³ / 5.68 × 10⁻¹⁹ = 7.04 × 10¹⁴ s⁻¹.
• Electrons per second = 7.04 × 10¹¹, so current = 7.04 × 10¹¹ × 1.60 × 10⁻¹⁹ = 1.13 × 10⁻⁷ A.
• Emax = 3.55 - 2.10 = 1.45 eV.

If the power is doubled, the current doubles to 2.25 × 10⁻⁷ A but Emax stays at 1.45 eV.

Worked example 3: finding h and Φ from data

A student measures the stopping potential Vs for a metal at six frequencies. Two of the points are (6.00 × 10¹⁴ Hz, 0.09 V) and (11.0 × 10¹⁴ Hz, 2.16 V), both lying on the best-fit line.

Gradient = (2.16 - 0.09) / (5.0 × 10¹⁴) = 4.14 × 10⁻¹⁵ V Hz⁻¹. Since gradient = h/e, h = 4.14 × 10⁻¹⁵ × 1.60 × 10⁻¹⁹ = 6.62 × 10⁻³⁴ J s.

The line meets the frequency axis where Vs = 0. Going back from the first point: f₀ = 6.00 × 10¹⁴ - 0.09 / 4.14 × 10⁻¹⁵ = 5.78 × 10¹⁴ Hz. Then Φ = hf₀ = 6.62 × 10⁻³⁴ × 5.78 × 10¹⁴ = 3.83 × 10⁻¹⁹ J = 2.39 eV.

2. Matter waves and de Broglie

Diffraction is a wave property, so if particles diffract, they must have a wave aspect. Beams of electrons (and of neutrons and atoms) do produce diffraction patterns when they pass through gaps or crystals of the right size. This is the guide’s evidence for the wave nature of matter. Combined with photons, it gives wave-particle duality: matter and radiation both show wave-like and particle-like behaviour, depending on the experiment.

The de Broglie wavelength of a particle with momentum p is

λ = h/p

For a non-relativistic particle of mass m accelerated through a potential difference V, p = √(2meV) for an electron.

Worked example 4: an electron and a ball

An electron is accelerated from rest through 85.0 V. Its kinetic energy is eV = 1.36 × 10⁻¹⁷ J, so p = √(2 × 9.11 × 10⁻³¹ × 1.36 × 10⁻¹⁷) = 4.98 × 10⁻²⁴ kg m s⁻¹ and λ = 1.33 × 10⁻¹⁰ m. Its speed is p/m = 5.46 × 10⁶ m s⁻¹.

A 0.058 kg ball moving at 40 m s⁻¹ has p = 2.32 kg m s⁻¹, so λ = 2.86 × 10⁻³⁴ m. That is far smaller than any gap you could build, so its wave behaviour never shows.

The scattering experiment and the minimum

You must be able to describe an experiment in which a beam of particles is diffracted and say where the minimum intensity is.

1. A source produces a beam of electrons of a known kinetic energy (set by an accelerating potential difference), in a vacuum.
2. The beam passes through a very narrow gap, or a thin crystalline target, whose spacing is comparable to λ.
3. A detector, such as a fluorescent screen, records intensity against angle.
4. A pattern of maxima and minima appears, as for waves.

For a single gap of width b, the first minimum on either side of the central maximum lies at an angle θ where sin θ = λ/b, with λ = h/p. Faster electrons have larger p, smaller λ and so a narrower pattern.

Worked example 5: where is the minimum?

The electrons of example 4 (λ = 1.33 × 10⁻¹⁰ m) pass through a gap of width 1.00 nm. sin θ = 1.33 × 10⁻¹⁰ / 1.00 × 10⁻⁹ = 0.133, so θ = 7.64°. If the accelerating voltage were raised, λ would fall and the minimum would move closer to the centre.

3. Compton scattering

In Compton scattering a photon scatters from an electron. The scattered photon has a longer wavelength than the incident one: λf > λi. Treating the photon as a particle with energy hc/λ, it hands some energy to the electron, which recoils. A wave picture that scatters light without changing its frequency cannot account for the shift.

The change in wavelength is

Δλ = λf - λi = (h/mec)(1 - cos θ)

where θ is the angle through which the photon is deflected and me is the electron mass. You do not need the derivation. The quantity h/mec = 2.43 × 10⁻¹² m. The shift is zero at θ = 0 and greatest at θ = 180° (backscatter), where it is twice this value, 4.86 × 10⁻¹² m. It does not depend on the incident wavelength.

The guide’s linking question asks how this resembles and differs from two solid balls colliding. Both exchange energy and momentum in a single event. The difference is that the “ball” here is a photon, which has no rest mass and always travels at c.

Worked example 6

A photon of wavelength 0.0500 nm scatters through 90° from an electron.

• Δλ = 2.43 × 10⁻¹² × (1 - cos 90°) = 2.43 × 10⁻¹² m.
• λf = 5.00 × 10⁻¹¹ + 0.243 × 10⁻¹¹ = 5.24 × 10⁻¹¹ m, an increase of 4.86%.
• Taking the electron as free and initially at rest, the energy it gains is hc/λi - hc/λf. These are 24.9 keV and 23.7 keV, so the electron gains about 1.15 keV.

To find an angle from a measured shift of 1.00 × 10⁻¹² m: cos θ = 1 - 1.00 × 10⁻¹² / 2.43 × 10⁻¹² = 0.588, so θ = 54.0°.

Why Compton is stronger evidence

The photoelectric effect can be explained with a quantum of energy alone. Compton scattering shows the photon also behaves like a particle in a collision, with a definite shift linked to the deflection angle. That is why the guide’s linking question calls it more convincing evidence.

Common errors

• Using the intensity to change Emax. Only frequency (or wavelength) does.
• Forgetting to convert eV to J (or the reverse) before using h.
• Using the work function as the energy of the emitted electron. Emax = hf - Φ.
• Writing λ = h/mv for the photon. For a photon, p = h/λ.
• Giving a Compton shift in the wrong direction: the scattered wavelength is longer.
• Mixing θ in a Compton calculation with the diffraction angle of a gap.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. This page covers sub-topic E.2 (HL only).