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

These notes condense sub-topic E.2, Quantum physics, from the IB Diploma Programme Physics guide, first assessment 2025. E.2 is HL only (additional higher level); the guide has no standard level content here. For full explanations and longer worked examples, read the study guide, then test yourself with the practice set. Hub: /boards/ib/ib-dp/physics/. Checklist: /checklists/ib/ib-dp/physics/. Not sure where to start? Use the free diagnostics. More general advice is in exam preparation.

Values assumed throughout: Planck constant 6.63 × 10⁻³⁴ J s; speed of light 3.00 × 10⁸ m s⁻¹; elementary charge 1.60 × 10⁻¹⁹ C; electron mass 9.11 × 10⁻³¹ kg.

Key results

Idea	Result	Notes
Photon energy	E = hf = hc/λ	One photon, one electron, one event
Photon momentum	p = h/λ	Used in the de Broglie relation
Einstein’s equation	Emax = hf - Φ	HL only
Threshold	hf₀ = Φ	Emax = 0 at f₀
Stopping potential	eVs = Emax	Vs against f is a straight line
de Broglie wavelength	λ = h/p	Any particle, with momentum p
Electron from rest through V	p = √(2meV)	Non-relativistic only
First minimum, single gap	sin θ = λ/b	b is the gap width
Compton shift	Δλ = λf - λi = (h/mec)(1 - cos θ)	Derivation not required
Compton constant	h/mec = 2.43 × 10⁻¹² m	Maximum shift is twice this

Energy conversions: 1 eV = 1.60 × 10⁻¹⁹ J. Convert before substituting into formulas written in joules.

The photoelectric effect

Definitions

• Photoelectron: an electron freed from a surface by radiation falling on it.
• Threshold frequency f₀: the lowest frequency of light that releases photoelectrons from a given metal.
• Work function Φ: the least energy that must be supplied to free an electron at the metal’s surface.

What changes what

You change	Number of electrons per second	Emax
Intensity up (same f above f₀)	Up	No change
Frequency up (same power, above f₀)	Fewer photons per second, each more energetic	Up
Frequency below f₀	Zero	No emission
Metal with larger Φ	Fewer or none	Down

Features classical wave theory cannot explain (the guide requires a discussion):

1. A threshold frequency below which nothing is emitted, whatever the intensity.
2. Emax rising with frequency but not with intensity.
3. Emission with no measurable delay, even for faint light.

Photon explanation, in steps

1. Light consists of photons of energy hf.
2. One photon gives all its energy to one electron.
3. The electron needs at least Φ to leave. The rest becomes kinetic energy.
4. So Emax = hf - Φ. If hf < Φ, nothing happens.

Stopping potential graph. Plot Vs on the vertical axis and f on the horizontal axis. Gradient = h/e. Horizontal intercept = f₀. Vertical intercept = -Φ/e. All metals give parallel lines (same gradient) with different intercepts.

Worked reminder. Light of wavelength 360 nm needs a stopping potential of 0.80 V. Photon energy = hc/λ = 5.52 × 10⁻¹⁹ J = 3.45 eV. Φ = 3.45 - 0.80 = 2.65 eV.

Matter waves

• Electrons, neutrons and atoms diffract. Diffraction is a wave property, so this shows the wave nature of matter.
• Wave-particle duality: matter and light each show wave and particle behaviour. Which one you see depends on the experiment.
• λ = h/p. Larger momentum, shorter wavelength. A heavy, fast object has an immeasurably small wavelength.

Method for a diffraction calculation

1. Find p. For an electron through V, p = √(2meV). Otherwise p = mv.
2. Find λ = h/p.
3. For the first minimum, sin θ = λ/b.
4. If the speed increases, p rises, λ falls and the minimum moves toward the centre.

Description of the experiment (be ready to write this out): an electron beam of known energy in a vacuum, a narrow gap or a thin crystal with spacing comparable to λ, a detector or screen recording intensity against angle, and a pattern of maxima and minima as for waves. The position of the minimum fits the de Broglie wavelength.

Worked reminder. Doubling the accelerating voltage multiplies p by √2, so λ is divided by √2. Quadrupling it halves λ.

Compton scattering

• A photon scatters from an electron and emerges with a longer wavelength, because it transfers energy to the electron.
• Δλ depends on the scattering angle θ only. It is 0 for no deflection, h/mec at 90° and 2h/mec at 180°.
• It is further evidence for the particle nature of light. It is more convincing than the photoelectric effect because the photon behaves like a particle in a collision, with the shift linked to the angle.
• Comparison with two solid balls: energy and momentum are exchanged in one event, but the photon has no rest mass and moves at c.

Method in steps

1. Write Δλ = 2.43 × 10⁻¹² × (1 - cos θ) m.
2. Add Δλ to λi to get λf.
3. For the energy given to the electron, subtract hc/λf from hc/λi (free electron at rest).

Worked reminder. A photon of wavelength 0.0300 nm is scattered through 60°. cos 60° = 0.5, so Δλ = 1.21 × 10⁻¹² m and λf = 3.12 × 10⁻¹¹ m.

Must-know distinctions

Compare	Photoelectric effect	Compton scattering
Photon fate	Absorbed completely	Scattered with lower energy
Wavelength	Not measured after	Increases
Key equation	Emax = hf - Φ	Δλ = (h/mec)(1 - cos θ)
What it shows	Energy comes in quanta	Photons also carry momentum

Do not confuse the work function Φ (a property of the metal) with Emax (a property of the emitted electrons for a given light). Do not confuse the photon wavelength with the de Broglie wavelength of an electron.

Quick self-test

1. A metal has Φ = 2.50 eV. Find the threshold frequency.
2. Light of frequency 9.0 × 10¹⁴ Hz falls on the same metal. Find Emax in eV.
3. Light of wavelength 280 nm falls on a metal with Φ = 4.20 eV. Find Emax in eV.
4. Find the de Broglie wavelength of an electron moving at 1.0 × 10⁶ m s⁻¹.
5. State why raising the intensity of light above the threshold does not change Emax.
6. Electrons with λ = 0.40 nm pass through a gap of width 2.0 nm. Find the angle of the first minimum.
7. A photon of wavelength 0.0300 nm is Compton scattered through 60°. Find its new wavelength.
8. A 2.0 mW beam has wavelength 600 nm. How many photons arrive each second?
9. An electron’s accelerating voltage is raised from 150 V to 600 V. By what factor does λ change?
10. Explain why a wave model predicts a delay before emission.

Answers

1. f₀ = Φ/h = 2.50 × 1.60 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 6.03 × 10¹⁴ Hz.
2. hf = 6.63 × 10⁻³⁴ × 9.0 × 10¹⁴ = 5.97 × 10⁻¹⁹ J = 3.73 eV; Emax = 3.73 - 2.50 = 1.23 eV.
3. hc/λ = 4.44 eV; Emax = 0.24 eV.
4. λ = h/(mv) = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 1.0 × 10⁶) = 7.28 × 10⁻¹⁰ m.
5. More intensity means more photons per second. Each photon still has energy hf, and each acts on one electron, so the maximum energy is unchanged.
6. sin θ = 0.40 / 2.0 = 0.20, so θ = 11.5°.
7. Δλ = 1.21 × 10⁻¹² m; λf = 3.12 × 10⁻¹¹ m.
8. Photon energy = 3.31 × 10⁻¹⁹ J; N = 2.0 × 10⁻³ / 3.31 × 10⁻¹⁹ = 6.03 × 10¹⁵ s⁻¹.
9. The voltage rises by a factor 4, so p rises by 2 and λ falls from 1.00 × 10⁻¹⁰ m to 5.01 × 10⁻¹¹ m: it halves.
10. A classical wave spreads energy continuously, so a dim wave would need time to deliver enough energy to one electron. Emission is observed to be immediate.

Nature-of-science links

The guide pairs this content with questions about how ideas change. Have a sentence ready for each.

• Falsifying a purely wave model of light. The photoelectric observations contradicted a prediction of the wave model, so the model had to be extended. Light needs a particle description as well.
• Compton scattering as stronger evidence. The shift in wavelength is measured and fits a formula in the angle, so a particle-collision picture makes a quantitative prediction that can be tested.
• Particles that diffract. If diffraction is a defining behaviour of waves, an observed pattern of maxima and minima from electrons gives evidence that matter has a wave aspect.
• The Bohr model. The guide asks whether it can help explain the photoelectric effect. Be ready to say that it also treats energy as quantised.

Where marks are usually lost

• Leaving the answer in joules when the question asks for eV, or the other way round.
• Writing Emax = hf instead of hf - Φ.
• Saying that “brighter light gives faster electrons”.
• Saying “the electron needs more energy than the work function” when the correct statement is that the photon energy must be at least Φ.
• Describing the diffraction experiment without stating the vacuum, the gap size compared with λ, or what the detector records.
• Using p = mv for an accelerated electron without first finding its speed from the energy gained.
• Forgetting that Compton scattering increases the wavelength.
• Dropping the “1 -” in the Compton factor, or working in radians when the angle is given in degrees.
• Mixing the photon wavelength and the de Broglie wavelength of the electron in a single line of working.

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).