OCR A-level Physics A (H556) · Module 4: Electrons, Waves and Photons
Mini-Lesson
Electrons, Waves & Photons
Module 4 is the biggest module in H556. Three chapters, one lesson: electricity (charge, resistance, circuits), waves (superposition, interference, refraction, polarisation) and quantum physics (photons, the photoelectric effect, de Broglie).
I = ΔQ/Δt · R = ρL/A · ε = I(R + r) · nλ = d sin θ · hf = φ + KEmax · λ = h/pthe six equations Module 4 is built on
The intellectual climax: the photoelectric effect kills the wave model of light stone dead, and electron diffraction then kills the pure particle model of matter. Both must be waves and particles. Neither picture alone survives contact with the evidence.
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
Module 4 · charge & current
Charge, current and drift velocity
I = ΔQ/Δt · I = AnevA = cross-sectional area · n = number density of charge carriers · e = charge per carrier · v = mean drift velocity
Current is the rate of flow of charge. Conventional current runs from + to −; electrons drift the other way.
The drift velocity in a copper wire is only around 10⁻⁴ m s⁻¹ — walking pace would be a hundred times faster. The lamp lights instantly because the electric field is established at nearly the speed of light and nudges every electron at once.
Number density n is what separates conductors (≈10²⁹ m⁻³), semiconductors and insulators.
Same current through a narrower wire → smaller A → larger drift velocity.
V = W/Q · R = V/I · P = VI = I²R = V²/Rthe volt is one joule per coulomb
Module 4 · resistivity & I–V
Resistivity and I–V characteristics
R = ρL / Aρ = resistivity in Ω m — a property of the MATERIAL, not the sample
Ohmic conductor (metal at constant temperature): I ∝ V, so the I–V graph is a straight line through the origin. That is Ohm's law, and it holds only at constant temperature.
Filament lamp: heating increases lattice vibration → more electron collisions → resistance rises → the graph curves towards the V-axis.
NTC thermistor: heating liberates far more charge carriers → n rises → resistance falls sharply.
LED / diode: conducts in forward bias only, and only above a threshold pd (about 0.6 V for silicon).
Superconductivity: below a critical temperature some materials have exactly zero resistivity — no heat loss at all. Applications: super-strong magnets in MRI scanners and particle accelerators, and lossless power transmission.
Sort it
Series, parallel, or true of both?
Tap a statement, then tap the circuit it describes.
➖ Series circuit
⑂ Parallel circuit
🔁 True of both
Module 4 · circuits
Kirchhoff's laws, internal resistance and potential dividers
Kirchhoff's first law (conservation of charge): the sum of currents into a junction equals the sum out.
Kirchhoff's second law (conservation of energy): around any closed loop, the sum of the EMFs equals the sum of the pds.
ε = I(R + r) → V = ε − IrV = terminal pd · Ir = the LOST VOLTS dissipated inside the source
Plot terminal pd V against current I: the gradient is −r and the y-intercept is ε.
potential divider: Vout = Vin × R₂/(R₁ + R₂)swap R₂ for a thermistor or LDR and you have a sensing circuit
Worked example
ε = 6.0 V, r = 0.80 Ω, external R = 3.4 Ω.
I = ε ÷ (R + r) = 6.0 ÷ 4.2 = 1.43 A
V = ε − Ir = 6.0 − (1.4286 × 0.80) = 6.0 − 1.143 = 4.86 V
Calculate
Your turn — current with internal resistance
1A cell of EMF 6.0 V and internal resistance 0.80 Ω is connected to a 3.4 Ω resistor. Calculate the current. Give your answer in A to 3 significant figures.
2For that same circuit (ε = 6.0 V, r = 0.80 Ω, I = 1.43 A), calculate the terminal potential difference. Give your answer in V to 3 significant figures.
V
Hint: V = ε − Ir = 6.0 − (1.4286 × 0.80) = 6.0 − 1.143.
Quick check
Kirchhoff's first law
?Kirchhoff's first law states that the sum of the currents entering a junction equals the sum leaving. Which conservation principle is this?
Module 4 · waves
Waves, superposition and stationary waves
v = fλ · f = 1/T · I ∝ A²intensity is proportional to the SQUARE of the amplitude
Transverse waves oscillate perpendicular to the direction of energy transfer; longitudinal waves oscillate parallel to it.
Only transverse waves can be polarised — decisive evidence that light is transverse and that sound is not.
Superposition: displacements add. Constructive when the path difference is nλ; destructive when it is (n + ½)λ.
Coherence: same frequency and a constant phase difference — required for a stable interference pattern.
Stationary waves form when two identical waves travel in opposite directions. Nodes (zero amplitude) sit λ/2 apart, with antinodes between them. No net energy is transferred along a stationary wave.
string fixed at both ends: fundamental λ = 2Lso f₀ = v ÷ 2L, and the harmonics are integer multiples of f₀
Calculate
Your turn — stationary wave
3A string of length 1.2 m is fixed at both ends and vibrates in its fundamental mode. Waves travel along it at 240 m s⁻¹. Calculate the fundamental frequency. Give your answer in Hz.
Hz
Hint: In the fundamental mode λ = 2L = 2.4 m. Then f = v/λ = 240 ÷ 2.4.
Calculate
Your turn — diffraction grating
4A grating with 300 lines per mm is illuminated with laser light of wavelength 633 nm. Calculate the angle of the first-order maximum. Give your answer in degrees to 3 significant figures.
°
Hint: d = 1 ÷ (300 × 10³) = 3.333 × 10⁻⁶ m. sin θ = nλ/d = 633 × 10⁻⁹ ÷ 3.333 × 10⁻⁶ = 0.1899. Take the inverse sine.
Module 4 · refraction
Refraction, TIR and polarisation
n = c/v · n₁ sin θ₁ = n₂ sin θ₂ · sin C = 1/n (into air)angles are always measured from the NORMAL
Entering a denser medium, light slows down and bends towards the normal. Its frequency is unchanged; the wavelength shortens.
Total internal reflection requires (i) travel from a more to a less optically dense medium and (ii) an angle of incidence greater than the critical angle C.
Optical fibres: a high-index core in a lower-index cladding traps light by TIR — the basis of all modern high-speed communication.
Polarisation in practice: polarising sunglasses have a vertical transmission axis to cut horizontally polarised glare from wet roads and water. Rotate a polarising filter over an LCD screen and it goes black at 90°.
Quick check
Polarisation
?Light can be polarised; sound cannot. What does this show?
Module 4 · quantum
Photons, the photoelectric effect and the Planck constant
E = hf = hc/λ · hf = φ + KEmaxφ = work function — the minimum energy needed to free an electron from the metal surface
Light arrives in indivisible packets called photons. One photon interacts with one electron, transferring all of its energy. The three observations that demolish the wave model:
Below the threshold frequency f₀ = φ/h, no electrons are emitted, however intense the light or however long you wait.
Emission is instantaneous above f₀, even for feeble intensity.
KEmax depends only on frequency. Greater intensity gives more electrons per second, not faster ones.
Measuring h with an LED: at the threshold pd V at which an LED just lights, eV ≈ hc/λ. Plot V against 1/λ for several colours of LED and the gradient is hc/e — from which h follows. This is a standard H556 practical.
5Calculate the energy of a photon of wavelength 450 nm. Give your answer in eV to 3 significant figures. (h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, 1 eV = 1.60 × 10⁻¹⁹ J)
eV
Hint: E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ 450 × 10⁻⁹ = 4.42 × 10⁻¹⁹ J. Divide by 1.60 × 10⁻¹⁹.
Calculate
Your turn — threshold frequency
6Sodium has a work function of 2.28 eV. Calculate its threshold frequency. Give your answer as a multiple of 10¹⁴ Hz to 3 significant figures. (h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.60 × 10⁻¹⁹ J)
7An electron is accelerated from rest through 100 V. Calculate its de Broglie wavelength. Give your answer in nm to 3 significant figures. (me = 9.11 × 10⁻³¹ kg, e = 1.60 × 10⁻¹⁹ C, h = 6.63 × 10⁻³⁴ J s)
nm
Hint: KE = eV = 1.60 × 10⁻¹⁷ J. p = √(2mE) = √(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁷) = 5.40 × 10⁻²⁴. λ = h/p = 1.23 × 10⁻¹⁰ m. Convert to nm.
Quick check
Below the threshold
?A metal is illuminated with very intense light of a frequency below its threshold frequency. What happens?
Quick check
Potential divider
?A 9.0 V supply is connected across a 2.0 kΩ resistor in series with a 4.0 kΩ resistor. What is the pd across the 4.0 kΩ resistor?
Match it
Match the law to its statement
Tap an item on the left, then its partner on the right.
Law or quantity
Statement
Recap
The big ideas to know
Current: I = ΔQ/Δt and I = Anev; drift velocity is tiny, but the field acts instantly
Resistivity: R = ρL/A; metals get more resistive when hot, NTC thermistors get less
Kirchhoff: 1st law = conservation of charge · 2nd law = conservation of energy
Internal resistance: ε = I(R + r), V = ε − Ir; a V–I graph gives gradient −r and intercept ε
Potential divider: V_out = V_in R₂/(R₁ + R₂)
Waves: v = fλ; only transverse waves polarise; stationary waves have nodes λ/2 apart; grating nλ = d sin θ