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A Level Physics 9702 · A Level · Topic 22 · ~5 min read

Quantum physics

Photons with E = hf and momentum p = E/c, the electronvolt, the photoelectric effect and Einstein’s equation, wave–particle duality and the de Broglie wavelength, and energy levels with emission and absorption line spectra.

Photon energy E = hfPhoton momentumElectronvoltPhotoelectric effectWork function & threshold frequencyhf = Φ + ½mv²maxElectron diffractionde Broglie λ = h/p
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Paper tag = where it is examinedDATA & FORMULAE given in the examC1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Light comes in photons: E = hf = hc/λ, with momentum p = E/c. Energies are often given in electronvolts: 1 eV = 1.60 × 10⁻¹⁹ J.
  • Photoelectric effect: one photon gives all its energy to one electron. hf = Φ + ½mv²max. Below the threshold frequency f₀ = Φ/h, no electrons are emitted, however intense the light.
  • Particles have waves: λ = h/p (de Broglie). Electron diffraction by graphite proves electrons behave as waves.
  • Energy levels: an electron falling between levels emits one photon with hf = E₁ − E₂, giving a line spectrum.

By the end you can

  • Understand photons, and use E = hf and p = E/c; use the electronvolt.
  • Explain the photoelectric effect and why it supports a particle model of light; use hf = Φ + ½mv²max.
  • Describe electron diffraction as evidence for wave behaviour, and use λ = h/p.
  • Explain emission and absorption line spectra using discrete energy levels, and use hf = E₁ − E₂.

1Photons

Energy and momentum

E = hf = hc/λ ; p = E/c ; 1 eV = 1.60 × 10⁻¹⁹ J
h = 6.63 × 10⁻³⁴ J s

Electronvolt — the energy gained by an electron moving through a potential difference of 1 V.

Worked example

Find the energy of a 450 nm photon in joules and in eV.

  1. E = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸/450 × 10⁻⁹ = 4.4 × 10⁻¹⁹ J C1 A1
  2. ÷ 1.60 × 10⁻¹⁹ = 2.8 eV A1

2The photoelectric effect

Evidence for photons

f / 10¹⁴ HzEkf₀ (threshold)intercept = −Φgradient = hEk max = hf − Φbelow f₀: no electrons,however bright the light
Maximum kinetic energy rises linearly with frequency: gradient h, intercept f₀.
ObservationExplanation with photons
No emission below f₀, however intenseone photon must have at least Φ; energy can't build up
Emission is instantaneousa single photon–electron interaction
Max KE depends on frequency, not intensityeach photon has E = hf; intensity is just the number of photons
Photocurrent ∝ intensity (above f₀)more photons per second release more electrons per second

Work function Φ — the minimum energy needed to remove an electron from the surface of a metal.

hf = Φ + ½mv²max
Einstein's photoelectric equation
Worked example

Light of wavelength 400 nm falls on a metal of work function 2.3 eV. Find the maximum kinetic energy of the electrons and the threshold wavelength.

  1. E = hc/λ = 3.11 eV C1
  2. KEmax = 3.11 − 2.3 = 0.81 eV = 1.3 × 10⁻¹⁹ J A1
  3. λ₀ = hc/Φ = 6.63 × 10⁻³⁴ × 3.0 × 10⁸/(2.3 × 1.6 × 10⁻¹⁹) = 540 nm A1
Common mistake"More intense light gives faster electrons". Intensity changes the number of electrons per second, not their maximum energy. This is a perennial Paper 4 error.

3Wave–particle duality

de Broglie

λ = h/p = h/(mv)
for an electron accelerated from rest through V: p = √(2meV)
e⁻ gungraphiterings: electrons diffract like wavesλ = h/p
Electrons through thin graphite form rings: a diffraction pattern, so they behave as waves.
Worked example

Find the de Broglie wavelength of an electron accelerated through 150 V.

  1. p = √(2meV) = √(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ × 150) = 6.6 × 10⁻²⁴ N s C1
  2. λ = h/p = 1.0 × 10⁻¹⁰ m A1 (similar to atomic spacing, so diffraction is visible)

4Energy levels

Line spectra

n = 1-13.60 eVn = 2-3.40 eVn = 3-1.51 eVn = 4-0.85 eVn = 5-0.54 eV0 (ionised)10.2 eV: UV1.89 eV: red, 656 nmEmission: a falling electron emits one photon with E = hf = ΔE. (Not to scale.)
A falling electron emits one photon whose energy equals the gap between the levels.
hf = E₁ − E₂
emission when an electron falls; absorption when a photon of exactly that energy lifts it
emissionabsorptionSame element, same wavelengths: bright lines from a hot gas,dark lines when white light passes through a cooler gas.
Emission spectrum: bright lines on dark. Absorption spectrum: dark lines at the same wavelengths.
Worked example

An electron in hydrogen falls from −3.40 eV to −13.6 eV. Find the wavelength of the photon.

  1. ΔE = 10.2 eV = 1.63 × 10⁻¹⁸ J C1
  2. λ = hc/ΔE = 1.22 × 10⁻⁷ m (ultraviolet) A1
  • Absorption lines: white light through a cool gas loses exactly those photons whose energies match level gaps. The re-emitted photons go in all directions, so those wavelengths look dark.
Linking questionWaves and diffraction are in Superposition; X-ray photons in Medical physics; and spectra of stars in Astronomy and cosmology.

Test yourself — cover the answers first

Quick check

  1. Convert 5.0 eV to joules.
  2. What does the gradient of a KE_max–f graph equal?
  3. What evidence shows electrons behave as waves?
  4. Does brighter light below the threshold frequency release electrons?
  5. State the de Broglie equation.

Exam-style

  1. Find the momentum of a photon of wavelength 600 nm.
  2. The stopping potential for a metal under 250 nm light is 2.1 V. Find the work function in eV.
  3. A proton and an electron have the same de Broglie wavelength. Compare their speeds.
Show answers
  1. 8.0 × 10⁻¹⁹ J
  2. the Planck constant h
  3. electron diffraction
  4. no
  5. λ = h/p
  6. p = h/λ = 1.1 × 10⁻²⁷ N s
  7. E = 4.97 eV; Φ = 4.97 − 2.1 = 2.9 eV
  8. same momentum, so v ∝ 1/m: the electron is about 1840 times faster
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.