This mini-lesson covers the AQA 3.12 Turning points in physics option: the discovery of the electron (thermionic emission, Thomson’s specific charge, Millikan’s oil drop), wave–particle duality (Newton vs Huygens, Young, Maxwell, Hertz, the photoelectric effect, electron diffraction), and special relativity (Michelson–Morley, time dilation, length contraction, mass–energy).
Optional topic — examined in Paper 3 Section B.
Optional topic — AQA sections 3.9–3.13 are options: you study exactly ONE of the five (Astrophysics, Medical physics, Engineering physics, Turning points in physics, Electronics). It is examined in Paper 3 Section B. Check with your teacher which option your school teaches before revising this one.
Work through each screen, answer the questions as you go (many are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
3.12.1 · the electron
Thermionic emission and specific charge
Thermionic emission: heat a metal filament and some electrons gain enough kinetic energy to escape the surface. Accelerate them through a pd V in a vacuum and all the electrical work becomes kinetic energy:
eV = ½mv² → v = √(2eV ÷ m)the basis of the cathode ray tube, the electron gun and the X-ray tube
Thomson (1897) deflected cathode rays with electric and magnetic fields. In his velocity selector, the electric and magnetic forces are balanced:
eE = Bev → v = E ÷ Bthen, with the electric field alone, the deflection gives e/m — the specific charge
Thomson found e/m was about 1800× larger than that of a hydrogen ion, and was the same for every cathode material. Conclusion: a universal, sub-atomic particle — the electron.
Millikan (1909) balanced charged oil drops in an electric field (mg = QE, using the terminal velocity under gravity to find the tiny mass). He found every charge was a whole-number multiple of e = 1.6×10−19 C: charge is quantised.
Thomson’s e/m plus Millikan’s e together give the electron mass: m = e ÷ (e/m) = 9.11×10−31 kg.
Worked example
e/m = 1.60×10−19 ÷ 9.11×10−31 = 1.76×1011 C kg−1.
Calculate
Your turn — calculation 1
1Calculate the specific charge of the electron. Give the number in front of ×1011 C kg−1. (e = 1.60×10−19 C, m = 9.11×10−31 kg)
× 10¹¹ C kg⁻¹
Hint: Specific charge = e ÷ m = 1.60×10−19 ÷ 9.11×10−31.
Calculate
Your turn — calculation 2
2An electron is accelerated from rest through a potential difference of 2000 V. Calculate its final speed. Give the number in front of ×107 m s−1. (e = 1.60×10−19 C, m = 9.11×10−31 kg)
× 10⁷ m s⁻¹
Hint: eV = ½mv² → v = √(2eV/m) = √(2 × 1.60×10−19 × 2000 ÷ 9.11×10−31) = √(7.02×1014).
Quick check
Think it through
?In Thomson’s experiment, the electric and magnetic fields are adjusted until the electron beam is undeflected. What does this give?
3.12.1 · Millikan & the CRT
Millikan’s oil-drop experiment and the cathode ray tube
Millikan’s method (1909): spray a fine mist of oil drops, which pick up charge by friction, between two horizontal plates.
With the field off, a drop falls at its terminal velocity: weight = viscous drag (Stokes’ law, F = 6πηrv). Measuring that velocity gives the drop’s radius, and hence its mass.
With the field on, adjust the pd until the drop is stationary: then QE = mg, so Q = mgd ÷ V.
Repeating for thousands of drops, every value of Q was a whole-number multiple of 1.6×10−19 C. Charge is quantised, and that quantum is the charge on the electron.
The cathode ray tube puts the physics to work: thermionic emission from a heated cathode → acceleration by an anode pd (eV = ½mv²) → deflection by electric or magnetic fields → a spot on a fluorescent screen. The same electron gun sits at the heart of the old TV, the oscilloscope and the X-ray tube.
Why it was a turning point: Thomson gave e/m and Millikan gave e. Together they pinned down the electron’s mass as 9.11×10−31 kg — about 1/1836 of a hydrogen atom. The atom was no longer indivisible.
Quick check
Think it through
?In Millikan’s experiment, a charged oil drop is held stationary between two horizontal plates. Which condition applies?
Match it
Match each experiment to the physicist
Tap a card on the left, then its partner on the right.
Statement
Answer
3.12.2 · duality
Wave–particle duality: the long argument
Newton (corpuscles): light is a stream of particles. It explained reflection well, but to explain refraction it needed light to travel faster in glass than in air.
Huygens (waves): light is a wave with secondary wavelets. It also explained refraction — but predicted light travels slower in glass. Newton’s reputation kept the corpuscle theory alive for a century.
Young (1801): double-slit interference. Particles cannot cancel out, so light must be a wave.
Maxwell (1864): predicted electromagnetic waves travelling at a speed calculable from ε0 and µ0 — and it matched the measured speed of light. Hertz then produced and detected radio waves, confirming it.
Foucault (1850) measured the speed of light in water and found it slower than in air — the decisive experimental blow to the corpuscle theory.
The photoelectric effect (1902–05) then showed light also behaves as particles: hf = φ + Ek(max). Below the threshold frequency, no electrons are emitted whatever the intensity; above it, emission is instantaneous.
de Broglie (1924): if waves are particles, particles are waves: λ = h/p. Davisson and Germer then diffracted electrons off a crystal, confirming it. The electron microscope is the direct pay-off.
The final position: light and matter are neither classical waves nor classical particles. Each model works in its own domain — the wave model for propagation, the photon model for emission and absorption.
Sort it
Wave evidence, particle evidence, or relativity?
Tap a statement, then tap the box it belongs in.
🌊 Wave model
⚫ Particle model
🚀 Special relativity
Calculate
Your turn — calculation 3
3Using the speed from earlier (2.65×107 m s−1), calculate the de Broglie wavelength of an electron accelerated through 2000 V. Give the number in front of ×10−11 m. (h = 6.63×10−34 J s, m = 9.11×10−31 kg)
× 10⁻¹¹ m
Hint: p = mv = 9.11×10−31 × 2.65×107 = 2.41×10−23 kg m s−1. Then λ = h ÷ p.
Quick check
Think it through
?Which observation of the photoelectric effect cannot be explained by the wave model of light?
3.12.3 · relativity
Michelson–Morley and special relativity
The problem: if light is a wave, what does it wave in? Physicists proposed a stationary ether. The Michelson–Morley interferometer (1887) split a beam along two perpendicular arms and looked for a fringe shift as the Earth ploughed through the ether.
The result: a null result. No shift, at any time of year. The ether does not exist, and the speed of light is the same for every observer, no matter how they move.
Einstein’s two postulates (1905):
The laws of physics are the same in all inertial frames.
The speed of light in a vacuum is the same for all observers, independent of the motion of the source or observer.
t = t0 ÷ √(1 − v²/c²) · l = l0√(1 − v²/c²)time dilation (moving clocks run slow) · length contraction (moving objects shorten along the direction of motion)
m = m0 ÷ √(1 − v²/c²) · E = mc²mass increases without limit as v → c — which is why nothing with mass can ever reach c
The muon proof: muons created high in the atmosphere have a proper lifetime of only ~2.2 µs and should decay long before reaching the ground. They arrive in large numbers because from our frame their clocks run slow (time dilation) — or equivalently, from the muon’s frame the atmosphere is contracted. Both descriptions give the same answer.
Worked example
A muon travels at 0.80c. √(1 − 0.80²) = √0.36 = 0.60, so γ = 1/0.60 = 1.67.
Its lifetime in our frame: t = 2.20 ÷ 0.60 = 3.67 µs.
Calculate
Your turn — calculation 4
4A muon with a proper lifetime of 2.20 µs travels at 0.80c. Calculate its lifetime as measured in the laboratory frame, in µs.