amazon.co.ukA-Level Physics revision: photoelectric effect with Einstein's equation, energy levels and line spectra, wave-particle duality, de Broglie wavelength and particle physics basics (standard model, conservation laws) - AQA 7408 3.2, OCR A H556 4.4/6.4-6.5, Edexcel 9PH0 Topics 5/8.
Shine light on a metal and electrons can be ejected - the photoelectric effect (AQA 7408 3.2.2.1; OCR A H556 4.4.1; Edexcel 9PH0 Topic 5). Three observations killed the wave theory: (1) there is a threshold frequency f_0 below which no electrons are emitted, however intense the light (wave theory: any frequency should work given enough intensity); (2) emission is instantaneous even for dim light (wave theory: energy should take time to accumulate); (3) the maximum kinetic energy of photoelectrons depends only on frequency, not intensity (wave theory: brighter light should mean faster electrons).
Einstein's photon model: light arrives in discrete packets of energy E = hf. One photon is absorbed by one electron in a one-to-one interaction: the electron gains hf, spends the work function φ (minimum energy to escape the surface) and leaves with the remainder: hf = φ + KE_max. Instantaneous, because the energy arrives in one packet. Intensity changes the number of photons per second, so it changes the rate of emission, not the energy per electron. Threshold: KE_max = 0 gives f_0 = φ/h. Electrons deeper in the metal need more than φ, so they leave with less than KE_max - hence a range of energies up to the maximum.
Stopping potential derivation: apply a reverse pd V_s that just stops the fastest electrons. The electrons do work eV_s against the field, losing all their KE: eV_s = KE_max = hf - φ. Plotting V_s against f gives a straight line of gradient h/e and intercept -φ/h - one experimental route to Planck's constant. The gold-leaf electroscope demonstration: a negatively charged zinc plate loses its charge instantly under UV light (electrons ejected) but not under visible light even at high intensity - below threshold for zinc (φ ≈ 4.3 eV, UV ≈ 5+ eV).
Light of wavelength 420 nm is incident on a metal with work function 1.9 eV. Find the maximum kinetic energy of the photoelectrons. (h = 6.63 × 10^-34 J s, c = 3.00 × 10^8 m/s)
Line 1: Photon energy E = hc/λ = (6.63 × 10^-34 × 3.00 × 10^8)/(420 × 10^-9) = 4.74 × 10^-19 J (M1 A1).
Line 2: φ = 1.9 × 1.60 × 10^-19 = 3.04 × 10^-19 J (M1: eV to J conversion).
Line 3: KE_max = hf - φ = 4.74 × 10^-19 - 3.04 × 10^-19 = 1.7 × 10^-19 J (A1) ≈ 1.1 eV.
Electrons in atoms occupy discrete energy levels - negative values because the electron is bound (zero is free, at infinity) (AQA 7408 3.2.2.2). An electron can move between levels only by absorbing or emitting exactly the energy gap as a photon: ΔE = hf = hc/λ. Emission spectra (bright lines on dark) come from excited electrons falling to lower levels; absorption spectra (dark lines on continuous) from cool gas absorbing exactly those photons out of white light passing through. Each element has a unique set of levels, hence unique spectral lines - the fingerprint used to identify elements in stars.
Hydrogen: E_n = -13.6/n^2 eV (n = 1 ground: -13.6 eV; n = 2: -3.40 eV; n = 3: -1.51 eV; n = ∞: 0, ionisation). Series names: transitions ending on n = 1 (Lyman, UV), n = 2 (Balmer, visible), n = 3 (Paschen, IR). Fluorescence: absorb UV, electrons excited, then cascade down emitting visible photons of lower energy each - energy is conserved overall. Excitation by collision: electrons in a fluorescent tube collide with gas atoms, transferring exactly the excitation energies; the fluorescent coating converts UV to visible. Line spectra evidence: discrete lines prove discrete levels - a continuum of possible energies would give a continuous spectrum.
A hydrogen electron falls from n = 3 (E_3 = -1.51 eV) to n = 2 (E_2 = -3.40 eV). Find the photon energy and wavelength.
Line 1: ΔE = E_3 - E_2 = -1.51 - (-3.40) = 1.89 eV = 1.89 × 1.60 × 10^-19 = 3.02 × 10^-19 J (M1 A1: energy difference, sign handled).
Line 2: λ = hc/ΔE = (6.63 × 10^-34 × 3.00 × 10^8)/3.02 × 10^-19 (M1).
Line 3: λ = 6.6 × 10^-7 m = 660 nm - red light, a Balmer line (A1).
Light behaves as a wave (interference, diffraction) and as particles (photoelectric effect, photon momentum) - the photon model with E = hf and p = h/λ (AQA 7408 3.2.2.2). de Broglie's hypothesis extends duality to matter: every moving particle has an associated wavelength λ = h/p. For a non-relativistic particle with kinetic energy E_k, p = √(2mE_k) (from E_k = p^2/2m), so λ = h/√(2mE_k); for an electron accelerated through pd V: E_k = eV, so λ = h/√(2meV).
Evidence: electron diffraction. Accelerate electrons through a pd and fire them at a thin polycrystalline graphite film: the screen shows concentric diffraction rings - the spacing matches the predicted λ = h/p. Faster electrons (higher accelerating voltage) give smaller λ and closer rings - the quantitative test. Neutron diffraction similarly probes crystal structure. Why don't we see cricket-ball diffraction? λ = h/p with a macroscopic p is ~10^-34 m - unobservably small; duality only shows when λ is comparable to the spacing of the obstacles (atomic spacings ~10^-10 m for electrons through ~100 V).
Electron microscopes exploit the small λ: resolving power improves as wavelength shrinks, so electrons at ~100 kV (λ ~ 4 pm) resolve far finer detail than visible light (λ ~ 500 nm).
An electron (m = 9.11 × 10^-31 kg) is accelerated through 150 V. Find its de Broglie wavelength.
Line 1: E_k = eV = 1.60 × 10^-19 × 150 = 2.4 × 10^-17 J (M1 A1).
Line 2: p = √(2mE_k) = √(2 × 9.11 × 10^-31 × 2.4 × 10^-17) = 6.61 × 10^-24 kg m/s (M1 A1).
Line 3: λ = h/p = 6.63 × 10^-34/6.61 × 10^-24 = 1.0 × 10^-10 m = 0.10 nm (A1) - comparable to atomic spacing, hence electron diffraction works.
Quarks (u, d, s, c, b, t) and antiquarks build hadrons - particles feeling the strong interaction (AQA 7408 3.2.1; Edexcel 9PH0 Topic 8; OCR A H556 6.5). Baryons are three quarks (proton uud, neutron udd); mesons are quark-antiquark pairs (pions, kaons). Leptons are fundamental, non-strong: electron, muon, tau and their neutrinos (ν_e, ν_μ, ν_τ). Quark charges: u = +2/3e, d = -1/3e - add them for hadron charges (proton: 2/3 + 2/3 - 1/3 = +1e).
Interactions: the strong force binds quarks (gluons) and holds nuclei together - very short range (~10^-15 m); the weak force mediates beta decay (W±, Z^0 bosons) and can change quark flavour; the electromagnetic force acts on charged particles (photons); gravity is negligible at particle scales. Conservation laws for deciding if a reaction happens: charge, baryon number B (baryon +1, antiquark -1/3 per quark... in practice: baryon +1, meson 0, lepton 0), lepton number L (electron family: electron +1, electron neutrino +1, their antiparticles -1), and strangeness - conserved in strong interactions, changed by ±1 in weak interactions (kaons are strange particles, created in pairs by the strong force).
Beta-minus decay at quark level: n → p + e^- + ν̄_e. At quark level, one d quark changes to a u quark via the weak interaction, emitting a W^- boson which decays into an electron and an antineutrino: d → u + W^-, then W^- → e^- + ν̄_e (Feynman diagram in words: incoming d line bends to u, W^- zigzag leaves the vertex and splits into the electron-antineutrino pair). Check conservation: charge 0 → (+1) + (-1) + 0 ✓; baryon number 1 → 1 ✓; lepton number 0 → 0 + (+1) + (-1) ✓. Pair production (photon → particle-antiparticle pair, needs E ≥ 2mc^2) and annihilation (particle-antiparticle → two photons) are the standard energy-bookkeeping examples - PET scanners use positron-electron annihilation.
Is p + π^- → K^+ + K^0 possible in a strong interaction? (π^-, K^+ and K^0 contain strange/anti-strange quarks; K^0 has strangeness +1.)
Line 1: Charge: +1 - 1 = 0; K^+ + K^0 = +1 + 0 = +1 - not conserved (M1).
Line 2: Charge conservation fails, so the reaction is impossible however the other numbers work (A1). (Check the others for practice: B = 1 + 0 → 0 + 0 fails too.)
The standard practical for quantum phenomena is estimating h from LEDs (OCR PAG5; Edexcel practical work on quantum; AQA covers it under 3.2.2). Each LED emits photons of a known colour/wavelength λ when the pd across it reaches the activation voltage V_0: the electrical energy per electron eV_0 equals the photon energy hf = hc/λ. So eV_0 = hc/λ, giving h = eV_0 λ/c.
Method: connect an LED in series with a resistor and a variable supply; increase the pd slowly in the dark until the LED just begins to emit - record that activation voltage (take several readings, average). Repeat for LEDs of several wavelengths (red, yellow, green, blue). Plot V_0 against 1/λ: a straight line through the origin of gradient hc/e, so h = gradient × e/c. Using the gradient across all LEDs averages random error and tests the model (a straight line confirms eV_0 ∝ 1/λ).
Evaluation points: judging the exact activation point is subjective (the LED glows dimly before it is obvious) - use a consistent, dim criterion in darkness; the pd measured is across the LED plus any voltmeter loading - use a high-resistance voltmeter; some energy is lost as heat in the LED (non-radiative recombination), so the measured h tends to be high; wavelength spreads (LEDs are not perfectly monochromatic) add uncertainty. Quote h with its uncertainty and compare with 6.63 × 10^-34 J s using the percentage difference.
A blue LED (λ = 470 nm) just lights at V_0 = 2.62 V.
Line 1: eV_0 = hc/λ, so h = eV_0 λ/c (M1: energy balance).
Line 2: h = (1.60 × 10^-19 × 2.62 × 470 × 10^-9)/(3.00 × 10^8) (M1: substitution).
Line 3: h = 1.969 × 10^-25/3.00 × 10^8 = 6.6 × 10^-34 J s (A1) - within a few percent of the accepted 6.63 × 10^-34 J s.