H2 Physics Quantum Physics Notes | A-Level 9478

Study guideUpdated 21 Aug 2026
Q: What do these H2 Physics quantum notes cover?
A: They cover photoelectric effect, de Broglie wavelength, wavefunctions, Heisenberg uncertainty, particle in a box, and spectra for A-Level 9478.
TL;DR
These A-Level quantum physics notes cover the photoelectric effect, de Broglie matter waves, wavefunctions, uncertainty, particle-in-a-box energy ladders, and line spectra for H2 Physics 9478. Master wave-particle duality, ψ \psi -algebra, uncertainty maths, and spectra links to move faster in Paper 1 MCQ, Paper 2 explanations, and Modern Physics revision.

Concrete example: how to use this page

If a photoelectric question changes intensity, think number of photons. If it changes frequency, think photon energy. Keeping those two levers separate makes the explanation much shorter.

Revisit the Modern Physics sequence (photoelectric effect → quantum → nuclear) via the H2 Physics notes hub so each derivation here links straight to the nuclear/particle follow-ups. For the full topic map and paper weightings, see our H2 Physics Syllabus 2026-27 overview.

A-Level quantum physics notes: start here, then route

Use this page as the Topic 19 owner for a level quantum physics, quantum physics a level, quantum physics a level notes, and a level physics quantum queries. Do not start with the formula sheet alone. The SEAB 9478 topic tests model selection: photon energy, de Broglie wavelength, wavefunction probability, uncertainty, particle-in-a-box energy, and spectra.

Search clueFirst ownerNext route
a level quantum physics or quantum physics a levelThis pageUse the route-selection map below before choosing an equation.
quantum physics a level notes or quantum physics notesThis pageRead the top route map, then jump to the subtopic that matches the question stem.
quantum physics a level notes pdfThis page and its PDFThen use the H2 Physics notes hub for the full 20-topic sequence.
h2 physics formula sheet while doing quantumH2 Physics Data and Formulae 2026Return here to decide whether the formula is photon, matter-wave, or box-model.
Late-JC2 quantum weaknessA-Level Physics tuitionUse tuition for feedback on explanation structure, spectra, and timed Paper 2 or 3 work.

When Quantum Physics needs marked feedback

Use the notes first if the mistake is recall: the photon equations, de Broglie route, wavefunction language, uncertainty relation, or particle-in-a-box energy formula. Move to A-Level Physics tuition only when the error repeats after self-study.

Repeated script signalWhat feedback should targetNext route
You can quote E=hfE = hf, but the explanation mixes intensity and frequency.Separate photon count from photon energy in timed sentences.Attempt two photoelectric explanations, then get marked feedback if the wording still drifts.
You know λ=h/p\lambda = h/p, but choose the wrong momentum route.Decide whether the particle is a photon, electron with speed, or electron with kinetic energy.

Quantum route-selection map

Use this map before substituting constants. Quantum questions often share the same constants, so the main challenge is deciding which physical model is being tested.

What the question changes or showsFirst moveMain routeMisconception check
Light intensity or frequency in the photoelectric effectSeparate photon count from photon energy.Use E=hfE = hf, then compare with the work function and stopping potential if needed.Higher intensity below threshold still gives no photoelectrons.
Massive particle with momentum or speedCheck whether the particle is massive or a photon.Use de Broglie wavelength λ=h/p\lambda = h/p

1 Particle nature of light

1.1 Photoelectric effect

  • Threshold frequency: No electrons emerge when incident light has a frequency below a critical threshold f0 f_0 ; intensity alone cannot compensate. This contradicts classical wave theory and signals discrete packets of energy.
  • Photon energy: Each packet carries E=hf E = h f . Memorise h=6.63×1034 Js h = 6.63 \times 10^{-34} \space \pu{J.s}

p=Ec=hλ. p = \frac{E}{c} = \frac{h}{\lambda}.

Photoelectric lever checkpoint

When a question changes the light or the metal, name the lever before writing an equation.

Change in the questionWhat changes physicallyWhat to calculateCommon trap
Increase intensity, same frequency above thresholdMore photons arrive each second.Emission rate or photocurrent increases.Saying each electron leaves with more kinetic energy.
Increase frequency, same intensityEach photon carries more energy.Kmax=hfϕK_{\max} = hf - \phi

Worked check: if hf=5.0 eVhf = \pu{5.0 eV} and ϕ=2.0 eV\phi = \pu{2.0 eV}, then Kmax=3.0 eVK_{\max} = \pu{3.0 eV}

Misconception check: one photon interacts with one electron in the basic H2 model. Intensity changes the number of attempts per second; frequency changes the energy per attempt.

Mini-drill

Calculate the momentum of a 500 nm 500 \space \pu{nm} photon.

p=hλ=6.63×10345.00×107=1.33×1027kgms1. p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{5.00 \times 10^{-7}} = 1.33 \times 10^{-27} \pu{kg.m.s-1}.


2 Wave nature of matter

2.1 Electron diffraction

  • The Davisson-Germer nickel crystal experiment produced concentric rings identical to X-ray patterns, confirming electron wavelengths.

2.2 Firing single particles at a double-slit

  • Fire electrons one by one: the screen still builds an interference fringe, proving every particle's wavefunction passes through both slits until detected.

2.3 de Broglie formula

λ=hp=hmv.(1) \lambda = \frac{h}{p} = \frac{h}{m v}. \tag{1}

Use Eq. (1)\text{(1)} to explain why soccer balls never show diffraction - their λ \lambda is <1034 m < 10^{-34} \space \pu{m} .

de Broglie route checkpoint

Before substituting into λ=h/p\lambda = h/p, decide how the question gives momentum. Most errors come from using a photon formula for an electron, or from forgetting to convert energy into joules before finding momentum.

Given informationFirst momentum moveThen useCommon trap
Massive particle speed is givenUse p=mvp = mv.λ=h/(mv)\lambda = h/(mv).Using p=E/cp = E/c

Worked check: an electron accelerated through 150 V\pu{150 V} gains kinetic energy K=eV=150eVK = eV = 150\,\text{eV}. Convert this to joules before using p=2mKp = \sqrt{2mK}

Misconception check: de Broglie wavelength belongs to matter waves, but the route to pp depends on the data given. The formula λ=h/p\lambda = h/p is shared; the momentum step is not.


3 Wavefunctions & probability

  • A particle's state is ψ(x,t) \psi(x, t) ; ψ2 \lvert \psi \rvert^2 yields a probability density that integrates to 1 after normalisation.
  • Superposition lets us add legitimate ψ \psi functions. Sum two slits and you recover the interference pattern in §2.2, or clamp endpoints to get standing waves in a box.

Quick normalisation hack
For ψ=Asin(nπxL) \psi = A \sin \left( \dfrac{n \pi x}{L} \right) in [0,L] [0, L] :

0Lψ2dx=A2L2=1A=2L. \int_0^L \lvert \psi \rvert^2 \mathrm{d}x = A^2 \frac{L}{2} = 1 \quad \Rightarrow \quad A = \sqrt{\frac{2}{L}}.


4 Heisenberg uncertainty

Localising a particle into Δx \Delta x demands a spread of momenta Δp \Delta p . Their product obeys

Δx Δph. \Delta x \space \Delta p \gtrsim h.

Tight boxes (Δx) (\downarrow \Delta x) force high kinetic energy (Δp) (\uparrow \Delta p) .

Exam cue: the SEAB syllabus uses the order-of-hh form above; when you need the full textbook constant, swap in ΔxΔph4π=2 \Delta x \Delta p \ge \tfrac{h}{4 \pi} = \tfrac{\hbar}{2}


5 Particle in a box

Solving the 1-D Schrödinger equation with ψ(0)=ψ(L)=0 \psi(0) = \psi(L) = 0 yields

En=h28mL2n2(n=1,2,3,).(2) E_n = \frac{h^2}{8 m L^2} n^2 \quad (n = 1, 2, 3, \dots). \tag{2}

  • Zero-point energy: even at n=1 n = 1 , the electron cannot be at rest.
  • Energy gaps widen with smaller L L - reason organic dyes with shorter conjugation lengths absorb bluer light (box model for π \pi electrons).

Particle-in-a-box energy checkpoint

Before substituting into En=h2n28mL2E_n = \dfrac{h^2 n^2}{8mL^2}

Question cueFirst moveWhat changesCommon trap
"Find the energy of level nn"Substitute the stated nn into n2n^2.Energy is proportional to n2n^2

Worked check: if a particle drops from n=3n=3 to n=1n=1, the emitted photon has energy E3E1=(91)h28mL2=8h28mL2E_3-E_1 = (9-1)\dfrac{h^2}{8mL^2} = 8\dfrac{h^2}{8mL^2}

Misconception check: the quantum number labels the standing-wave mode. It is not the energy itself, so changing from n=1n=1 to n=2n=2 makes the energy four times larger, not twice as large.


6 Quantised atoms & spectra

  • Electrons in hydrogen occupy discrete orbits. Transitions release/absorb photons that match energy differences, giving line spectra.
  • Emission lines appear when excited electrons drop to lower levels; absorption lines appear when ground-state electrons jump up, leaving dark gaps in a continuous spectrum. James Webb's spectrographs use the same principle to fingerprint exoplanet atmospheres.

Problem type

A 656 nm 656 \space \pu{nm} photon (Balmer H-α \alpha ) is emitted. Find the energy gap.
E=hcλ=3.03×1019J E = \tfrac{h c}{\lambda} = 3.03 \times 10^{-19} \pu{J}

7 WA timing rules (modern-paper edition)

  1. List givens under every quantum problem before manipulating equations - avoids dropping h h or c c .
  2. Box problems: write Eq. (2)\text{(2)} once, then plug numbers; do not derive under exam pressure.
  3. Convert wavelengths to energy first; the rest is simple bookkeeping.

Need structured practice on Quantum Physics? Our H2 Physics tuition programme covers this topic with weekly problem sets and Paper 4 practical drills.


Comprehensive revision pack

9478 Section VI, Topic 19 Syllabus outcomes

Candidates should be able to:

  • (a) show an understanding that the existence of a threshold frequency in the photoelectric effect provides evidence that supports the particulate nature of electromagnetic radiation while phenomena such as interference and diffraction provide evidence that supports its wave nature.
  • (b) state that a photon is a quantum of electromagnetic radiation, and recall and use the equation E=hf E = hf for the energy of a photon to solve problems, where h h is the Planck constant.
  • (c) show an understanding that while a photon is massless, it has a momentum given by p=Ec p = \dfrac{E}{c}

Concept map (in words)

Photons deliver energy E=hf E = h f to liberate electrons above threshold frequency. Matter exhibits wave behaviour with λ=hp \lambda = \dfrac{h}{p} .

Wavefunctions ψ \psi give probability distributions when ψ2 \lvert \psi \rvert^2 is normalised.

Confining particles quantises energy via En=n2h28mL2 E_n = \dfrac{n^2 h^2}{8 m L^2}

Spectral lines correspond to transitions obeying ΔE=hf=hcλ \Delta E = h f = \dfrac{h c}{\lambda} .

Key relations

ConceptExpression / reminder
Photon energyE=hf=hcλ E = h f = \dfrac{h c}{\lambda}
Photoelectric equationhf=ϕ+12mvmax2 h f = \phi + \tfrac{1}{2} m v_{\max}^2

Derivations & reasoning to master

  1. Photoelectric graph: derive the linear relation between stopping potential and frequency; the slope gives he \dfrac{h}{e} .
  2. de Broglie: show consistency with diffraction experiments (Davisson-Germer) using λ=hp \lambda = \dfrac{h}{p}

Worked example 1 - stopping potential

Ultraviolet light of frequency 8.0×1014 Hz 8.0 \times 10^{14} \space \pu{Hz} shines on a metal with work function 2.6 eV \pu{2.6 eV} . Determine maximum kinetic energy and stopping potential for emitted electrons.

Solution: Kmax=hfϕ K_{\max} = h f - \phi . Convert to joules with 1 eV=1.60×1019J 1 \space \pu{eV} = 1.60 \times 10^{-19} \pu{J}

hf=(6.63×1034)(8.0×1014)=5.30×1019J. hf = (6.63\times 10^{-34})(8.0\times 10^{14}) = 5.30\times 10^{-19}\,\pu{J}.

ϕ=2.6eV=2.6(1.60×1019)=4.16×1019J. \phi = 2.6\,\text{eV} = 2.6(1.60\times 10^{-19}) = 4.16\times 10^{-19}\,\pu{J}.

Kmax=hfϕ=1.14×1019JVs=Kmaxe0.71 V. K_{\max} = hf-\phi = 1.14\times 10^{-19}\,\pu{J} \Rightarrow V_s = \dfrac{K_{\max}}{e} \approx \pu{0.71 V}.

Worked example 2 - particle in a box transition

An electron confined to a one-dimensional box of length 0.35 nm 0.35 \space \pu{nm} makes a transition from n=3 n = 3 to n=2 n = 2 . Calculate the photon wavelength emitted.

Method: Use En=n2h28mL2 E_n = \dfrac{n^2 h^2}{8 m L^2}

ΔE=(3222)h28mL2=5h28mL2. \Delta E = (3^2-2^2)\dfrac{h^2}{8mL^2}=5\dfrac{h^2}{8mL^2}.

With L=0.35nm=3.5×1010mL=0.35\,\text{nm}=3.5\times 10^{-10}\,\pu{m}, h=6.63×1034Jsh=6.63\times 10^{-34}\,\pu{J.s}

ΔE2.46×1018J,λ=hcΔE8.1108 m81 nm. \Delta E \approx 2.46\times 10^{-18}\,\pu{J},\qquad \lambda = \dfrac{hc}{\Delta E} \approx \pu{8.1e-8 m} \approx \pu{81 nm}.

So the photon is in the ultraviolet (visible light is roughly 400 to 700 nm).

Practical & data tasks

  • Analyse photoelectric experiment data (Vs V_s vs f f ) to extract Planck's constant.
  • Simulate particle-in-box wavefunctions with Desmos/Python; verify nodal structure for each n n .
  • Examine hydrogen spectra using a school spectroscope; identify Balmer lines.

Common misconceptions & exam traps

  • Believing intensity affects photoelectron kinetic energy (it affects number only).
  • Forgetting to convert eV \pu{eV} to J \pu{J} or vice versa.
  • Mixing de Broglie wavelength with photon wavelength when comparing massive vs massless particles.
  • Ignoring normalisation when interpreting probability densities.

Quick self-check quiz

  1. What happens to photoelectron emission if light frequency drops below threshold? - No electrons emitted regardless of intensity.
  2. State de Broglie's relation. - λ=hp \lambda = \dfrac{h}{p} .
  3. Why can't an electron in a box have zero energy? - Boundary conditions enforce non-zero momentum, so E1>0 E_1 > 0

Revision workflow

  1. Re-derive the photoelectric equation and practise slope/intercept interpretation weekly.
  2. Solve mixed problems on de Broglie wavelengths for electrons, neutrons, and atoms.
  3. Build flashcards linking spectral series (Lyman, Balmer, Paschen) with wavelength ranges.
  4. Attempt uncertainty principle estimation questions to build physical intuition.

8 Further reading


Last updated 3 Jun 2026. Review after SEAB's 2027 draft release.


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