Edexcel IAL Physics revision · AS — Quantum physics
Quantum physics
The topic where physics stops being intuitive and starts being examined on precise wording. Three experiments carry all the marks: the photoelectric effect (light as particles), electron diffraction (particles as waves), and line spectra (energy levels).
Dr Desouky Physics Academy · Meet your physics tutor
What the syllabus demands
- —Describe the photon model: E = hf
- —Explain the photoelectric effect and why it contradicts the wave theory of light
- —Define work function and threshold frequency; use hf = Φ + ½mv²max
- —Use the de Broglie wavelength λ = h/p; describe electron diffraction as evidence
- —Explain line spectra using discrete energy levels: hf = E₁ − E₂
Definitions that earn marks
Clear definitions to practise — check your course mark scheme
- Photon
- A discrete packet (quantum) of electromagnetic energy: E = hf.
- Work function (Φ)
- The minimum energy required to remove an electron from the surface of a metal.
- Threshold frequency
- The minimum frequency of electromagnetic radiation that will eject electrons from a given metal surface: f₀ = Φ ÷ h.
- de Broglie wavelength
- The wavelength associated with a moving particle: λ = h ÷ p = h ÷ mv.
The equations
More equations to practise: the Edexcel IAL formula sheet.
Where the marks die
Common mistakes to check
- 01
Explaining the photoelectric effect with intensity. Below the threshold frequency, no electrons are emitted however intense the light — because one electron absorbs one photon, and each photon carries too little energy. Intensity changes the NUMBER of electrons, never their maximum energy.
- 02
Forgetting 'maximum' in ½mv²max. Electrons deeper in the metal lose energy escaping; only surface electrons emerge with the maximum kinetic energy. The word is a mark.
- 03
Energy level diagrams read upside-down: levels are negative, the ground state is the most negative, and a photon is EMITTED when the electron falls to a lower level. Absorption requires the photon energy to match a gap exactly.
- 04
Electronvolt conversions bungled: 1 eV = 1.6 × 10⁻¹⁹ J. Convert before substituting into hf = E₁ − E₂, not after.
One worked example, done properly
Question
Light of wavelength 400 nm falls on a metal of work function 2.0 eV. Find the maximum kinetic energy of the emitted electrons.
Method
- 1.Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸) ÷ (400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J = 3.1 eV.
- 2.½mv²max = hf − Φ = 3.1 − 2.0 = 1.1 eV.
- 3.In joules: 1.1 × 1.6 × 10⁻¹⁹.
KE max ≈ 1.8 × 10⁻¹⁹ J (1.1 eV)