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IB DP Physics · Theme E Nuclear and quantum physics · E.2 (HL only) · ~6 min read

E.2 Quantum physics

The photon model of light, the photoelectric effect and Einstein's equation, threshold frequency and work function, de Broglie wavelength and electron diffraction, photon momentum and Compton scattering.

PhotonsPhotoelectric effectWork functionStopping potentialde Broglie wavelengthElectron diffractionCompton scattering
RSby Riyaz Sir
mentorriyaz.in
No tag = SL and HLHL ONLY HL students onlyDATA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • E.2 is HL only. Light comes in photons of energy E = hf. In the photoelectric effect one photon frees one electron: Ek max = hf − Φ.
  • Below the threshold frequency f0 = Φ/h no electrons are emitted, however bright the light: brightness changes the number of photons, not their energy.
  • Matter behaves like a wave too: λ = h/p (de Broglie). Electron diffraction proves it.
  • Photons carry momentum p = h/λ. Compton scattering of X-rays by electrons (λ′ − λ = h(1 − cos θ)/mec) is the clearest evidence.

By the end you can

  • Explain the photoelectric effect and why the wave model fails to explain it. HL
  • Use Ek max = hf − Φ, threshold frequency and stopping potential. HL
  • Interpret the graph of Ek max against frequency. HL
  • Calculate de Broglie wavelengths and describe electron diffraction as evidence for matter waves. HL
  • Use photon momentum and the Compton scattering equation. HL

1The photoelectric effect HL

  • Observations: (1) no electrons below a threshold frequency, whatever the intensity; (2) above it, electrons come off instantly, even in dim light; (3) the maximum kinetic energy depends on frequency, not intensity; (4) the number of electrons per second is proportional to the intensity.
  • Why waves fail: a wave's energy depends on its amplitude, so bright enough light of any frequency should eventually free electrons, after a delay. Neither prediction is seen.
E = hf DATA BOOKLET
energy of one photon
Ek max = hf − Φ DATA BOOKLET
Φ = work function, the minimum energy needed to free an electron from the surface
f / 10¹⁴ HzEkf₀ (threshold)intercept = −Φgradient = hEk max = hf − Φbelow f₀: no electrons,however bright the light
One graph, three facts: gradient h (the same for every metal), x-intercept f0, y-intercept −Φ.
Worked example

Light of 400 nm falls on sodium (Φ = 2.3 eV). Find the maximum kinetic energy of the electrons, the stopping potential and the threshold frequency.

  1. photon energy = hc/λ = 4.97 × 10−19 J = 3.11 eV M1
  2. Ek max = 3.11 − 2.3 = 0.81 eV A1
  3. stopping potential Vs = Ek max/e = 0.81 V A1
  4. f0 = Φ/h = 2.3 × 1.6 × 10−19/6.63 × 10−34 = 5.6 × 1014 Hz A1

Stopping potential Vs — the reverse potential difference that just stops the fastest photoelectrons: eVs = Ek max.

Common mistakeSaying brighter light gives faster electrons. Brighter light means more photons, so more electrons, but each photon still has the same energy hf.

2Matter waves HL

λ = hp DATA BOOKLET
de Broglie wavelength of any particle with momentum p
e⁻ gungraphiterings: electrons diffract like wavesλ = h/p
Electrons passing through a thin graphite film form rings, exactly like X-ray diffraction. Particles are behaving as waves.
Worked example

Find the de Broglie wavelength of an electron accelerated from rest through 100 V.

  1. Ek = 100 eV, p = √(2meEk) = √(2 × 9.11 × 10−31 × 1.6 × 10−17) = 5.4 × 10−24 kg m s−1 M1
  2. λ = h/p = 1.2 × 10−10 m, similar to atomic spacings, so it diffracts in crystals A1
  • Faster electrons (higher accelerating voltage) have a shorter wavelength, so their diffraction rings get smaller.
  • Wave–particle duality: light shows wave behaviour (interference, C.3) and particle behaviour (photoelectric effect); electrons show both too. Which one you see depends on the experiment.

3Photon momentum and Compton scattering HL

p = hλ
momentum of a photon (the same de Broglie equation)
photon λe⁻ at restscattered photon λ′ > λrecoiling electronθ
A photon collides with an electron like a billiard ball. It loses energy, so its wavelength increases.
Δλ = hmec(1 − cos θ) DATA BOOKLET
Compton shift; h/mec = 2.43 × 10−12 m
Worked example

X-rays scatter from electrons at 90°. Find the change in wavelength.

  1. Δλ = (h/mec)(1 − cos 90°) = 6.63 × 10−34/(9.11 × 10−31 × 3.00 × 108) M1
  2. = 2.4 × 10−12 m A1
  • Why it proves photons: treating light as particles with momentum and applying conservation of energy and momentum to each collision predicts Δλ exactly. A wave model gives no shift.
Exam tipUse eV for energies until the last step. hf in eV: E (eV) = 1240/λ (nm) is a quick check (not in the booklet, so show the full calculation).
Linking questionQuantised energy levels in atoms (E.1) and photon energies in greenhouse absorption (B.2) all rest on E = hf. Electron diffraction uses the grating equation from C.3.
Idea for your IA / explorationUse LEDs of different colours to estimate Planck's constant: measure each LED's turn-on voltage V and use eV ≈ hc/λ, plotting V against 1/λ.

Test yourself — cover the answers first

Quick check

  1. What does the work function represent?
  2. Why are no electrons emitted below f0?
  3. What does the gradient of an Ek max–f graph equal?
  4. What happens to the photocurrent if the intensity doubles (above f0)?
  5. What evidence shows electrons behave as waves?
  6. Does a scattered Compton photon have a longer or shorter wavelength?

Exam-style

  1. The stopping potential for a metal is 1.2 V with 300 nm light. Find the work function in eV.
  2. Find the de Broglie wavelength of a 0.15 kg ball at 40 m s−1 and explain why it does not diffract.
Show answers
  1. the minimum energy to free an electron from the metal surface
  2. one photon has too little energy to free an electron, and electrons cannot add photon energies
  3. Planck's constant h
  4. it doubles
  5. electron diffraction
  6. longer
  7. photon energy = hc/λ = 6.63 × 10−19 J = 4.14 eV
    Φ = 4.14 − 1.2
    = 2.9 eV
  8. λ = h/mv = 6.63 × 10−34/(0.15 × 40)
    = 1.1 × 10−34 m
    far smaller than any gap or object, so diffraction is undetectable
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.