IB Physics HL Topic 5.2: Quantum physics (HL) | Notes & Flashcards

IB Physics HL — Quantum physics (HL)

Topic 5.2 of IB Physics covers Quantum physics (HL), which is part of Unit 5: Nuclear and quantum physics. Students explore key concepts including Wave–particle duality, De Broglie wavelength and diffraction. A strong understanding of quantum physics (hl) is essential for IB Physics HL exams and builds the foundation for connected topics across the syllabus.

Higher Level students should use this topic hub as a map: start with the shared sub-topics, then follow the HL-only extensions and exam-skill links where this topic asks for deeper analysis.

Key concepts in Quantum physics (HL)

Key Idea: Light and matter are both wave AND particle.

Photoelectric effect

Eₘₐₓ = hf − Φ
Threshold f₀ = Φ/h
Intensity: number, not energy

Compton scattering

Photon bounces off an electron
Longer wavelength afterwards
Δλ = (h/mₑc)(1 − cos θ)

Matter waves

λ = h/p
Electron diffraction rings
First minimum: sin θ ≈ λ/b

Light comes in photons: the photoelectric effect and Compton scattering prove it. Particles like electrons have a wavelength: electron diffraction proves it. This wave–particle duality is the heart of quantum physics. It is HL only (E.2).

The formulas you're given

\begin{aligned} E = hf \\ E_{\max} = hf - \Phi \\ \lambda = \frac{h}{p} \\ \Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta) \end{aligned}

Given in the data booklet (E.2).

$\Phi$: work function (J)
$p$: momentum (kg m s⁻¹)
$\theta$: photon scattering angle

Not in the booklet: build them from the ones above.

\begin{aligned} \lambda = \frac{h}{\sqrt{2m_e eV}} \\ eV_s = E_{\max} \end{aligned}

An electron accelerated through V (from λ = h/p and eV = p²/2m), and the stopping voltage.

$V$: accelerating pd (V)
$V_s$: stopping voltage (V)

IB-style worked examples

IB-style questionDetermine[2 marks]

Light of frequency 8.0 × 10¹⁴ Hz hits a metal of work function 3.0 × 10⁻¹⁹ J. Determine the maximum kinetic energy of an ejected electron.

Model answer plan

See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.

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IB-style questionDetermine[2 marks]

An electron (mass 9.11 × 10⁻³¹ kg) moves at 2.0 × 10⁶ m s⁻¹. Determine its de Broglie wavelength.

Model answer plan

See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.

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IB-style questionDetermine[2 marks]

An X-ray photon of wavelength 6.00 × 10⁻¹¹ m is Compton-scattered through 60°. Determine its new wavelength. (h/mₑc = 2.43 × 10⁻¹² m)

Model answer plan

See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.

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Important: 1. In the photoelectric effect, brightness changes the NUMBER of electrons; frequency changes their ENERGY. Below the threshold frequency, nothing happens however bright the light. 2. λ = h/p is an INVERSE relation — more momentum means a SHORTER wavelength. 3. Always find p = mv before using λ = h/p. 4. A Compton-scattered photon has a longer wavelength (it lost energy to the electron), never a shorter one.

Tap each card to reveal the answer.

What is a photon? A packet (quantum) of light energy, E = hf.

Which effect proves light is particle-like? The photoelectric effect (a sharp threshold frequency, instant emission).

Which effect proves particles are wave-like? Electron diffraction — electrons make diffraction patterns off crystals.

Increase the light's intensity (same frequency) — effect? More electrons per second, but the same maximum kinetic energy.

Why don't everyday objects show wave behaviour? Their momentum is huge, so λ = h/p is far too small to ever notice.

What does Compton scattering show that the photoelectric effect does not? That photons carry momentum (p = h/λ): they collide with electrons like particles, coming off with a longer wavelength.

Exam tips

Photoelectric: intensity ↔ number of electrons, frequency ↔ their energy.
Threshold frequency f₀ = Φ/h (set Eₘₐₓ = 0).
Find momentum p = mv before using λ = h/p.
Bigger momentum → shorter de Broglie wavelength (inverse).
Compton: Δλ = (h/mₑc)(1 − cos θ) is largest (2h/mₑc) when the photon bounces straight back.

IB Physics HL — Quantum physics (HL)

Higher Level students should use this topic hub as a map: start with the shared sub-topics, then follow the HL-only extensions and exam-skill links where this topic asks for deeper analysis.

Key concepts in Quantum physics (HL)

Key Idea: Light and matter are both wave AND particle.

Photoelectric effect

Eₘₐₓ = hf − Φ
Threshold f₀ = Φ/h
Intensity: number, not energy

Compton scattering

Photon bounces off an electron
Longer wavelength afterwards
Δλ = (h/mₑc)(1 − cos θ)

Matter waves

λ = h/p
Electron diffraction rings
First minimum: sin θ ≈ λ/b

The formulas you're given

\begin{aligned} E = hf \\ E_{\max} = hf - \Phi \\ \lambda = \frac{h}{p} \\ \Delta\lambda = \frac{h}{m_e c}(1 - \cos\theta) \end{aligned}

Given in the data booklet (E.2).

$\Phi$: work function (J)
$p$: momentum (kg m s⁻¹)
$\theta$: photon scattering angle

Not in the booklet: build them from the ones above.

\begin{aligned} \lambda = \frac{h}{\sqrt{2m_e eV}} \\ eV_s = E_{\max} \end{aligned}

An electron accelerated through V (from λ = h/p and eV = p²/2m), and the stopping voltage.

$V$: accelerating pd (V)
$V_s$: stopping voltage (V)

IB-style worked examples

IB-style questionDetermine[2 marks]

Light of frequency 8.0 × 10¹⁴ Hz hits a metal of work function 3.0 × 10⁻¹⁹ J. Determine the maximum kinetic energy of an ejected electron.

Model answer plan

See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.

Claim your free topic →

IB-style questionDetermine[2 marks]

An electron (mass 9.11 × 10⁻³¹ kg) moves at 2.0 × 10⁶ m s⁻¹. Determine its de Broglie wavelength.

Model answer plan

See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.

Claim your free topic →

IB-style questionDetermine[2 marks]

An X-ray photon of wavelength 6.00 × 10⁻¹¹ m is Compton-scattered through 60°. Determine its new wavelength. (h/mₑc = 2.43 × 10⁻¹² m)

Model answer plan

See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.

Claim your free topic →

Important: 1. In the photoelectric effect, brightness changes the NUMBER of electrons; frequency changes their ENERGY. Below the threshold frequency, nothing happens however bright the light. 2. λ = h/p is an INVERSE relation — more momentum means a SHORTER wavelength. 3. Always find p = mv before using λ = h/p. 4. A Compton-scattered photon has a longer wavelength (it lost energy to the electron), never a shorter one.

Tap each card to reveal the answer.

What is a photon? A packet (quantum) of light energy, E = hf.

Which effect proves light is particle-like? The photoelectric effect (a sharp threshold frequency, instant emission).

Which effect proves particles are wave-like? Electron diffraction — electrons make diffraction patterns off crystals.

Increase the light's intensity (same frequency) — effect? More electrons per second, but the same maximum kinetic energy.

Why don't everyday objects show wave behaviour? Their momentum is huge, so λ = h/p is far too small to ever notice.

Exam tips

Photoelectric: intensity ↔ number of electrons, frequency ↔ their energy.
Threshold frequency f₀ = Φ/h (set Eₘₐₓ = 0).
Find momentum p = mv before using λ = h/p.
Bigger momentum → shorter de Broglie wavelength (inverse).
Compton: Δλ = (h/mₑc)(1 − cos θ) is largest (2h/mₑc) when the photon bounces straight back.

IB Physics HL — Quantum physics (HL)

Key concepts in Quantum physics (HL)

The formulas you're given

IB-style worked examples

Exam tips

What you'll learn in Topic 5.2

Study resources — 5.2 Quantum physics (HL)

Wave–particle duality

De Broglie wavelength and diffraction

Ready to study Quantum physics (HL)?

Ready to practice?

IB Physics HL — Quantum physics (HL)

Key concepts in Quantum physics (HL)

The formulas you're given

IB-style worked examples

Exam tips

What you'll learn in Topic 5.2

Study resources — 5.2 Quantum physics (HL)

Wave–particle duality

De Broglie wavelength and diffraction

Ready to study Quantum physics (HL)?

Ready to practice?