AQA A-Level Physics: Turning points in physics (7408)

This study guide teaches section 3.12 Turning points in physics of the AQA AS and A-level Physics specification (7407/7408, version 1.4, AS and A-level exams June 2016 onwards), sub-sections 3.12.1 to 3.12.3. Turning points is one of the options (sections 3.9 to 3.13): you study one option, and it is assessed in Paper 3 Section B (35 of the paper’s 80 marks). Section A of Paper 3 assesses practical skills and data analysis. All of section 3.12 is A-level only.

See the course hub, printable checklist, free diagnostics, revision notes and practice questions.

Core background: Particles and radiation (specific charge, photoelectric effect), Waves (Young’s slits) and Fields and their consequences (charges in fields).

What this topic covers (A-level only, option)

Spec ref	Content
3.12.1.1–3.12.1.4	Cathode rays, thermionic emission, e/mₑ and its significance, Millikan and quantised charge
3.12.2.1–3.12.2.6	Newton vs Huygens, Young, EM waves (Maxwell, Hertz, Fizeau), black body and photoelectricity, de Broglie, TEM and STM
3.12.3.1–3.12.3.5	Michelson-Morley, postulates, time dilation, length contraction, mass and energy, Bertozzi

3.12.1 The discovery of the electron

Cathode rays

A discharge tube holds gas at low pressure with a high pd between cathode and anode. Positive ions are accelerated into the cathode and release electrons, which accelerate towards the anode. They ionise some gas atoms and excite others; excited atoms de-excite and emit light, so the gas glows. The beams from the cathode, called cathode rays, are electrons.

Thermionic emission

A heated metal filament gives some conduction electrons enough kinetic energy to escape the surface: thermionic emission. In an evacuated tube an anode at pd V accelerates them into a beam. The work done on each electron is eV, so

½mv² = eV

Worked example 1. An electron gun has an anode pd of 2500 V. Find the speed of the electrons.

Ek = eV = 1.60 × 10⁻¹⁹ × 2500 = 4.00 × 10⁻¹⁶ J
v = √(2Ek/m) = √(2 × 4.00 × 10⁻¹⁶ / 9.11 × 10⁻³¹) = 2.96 × 10⁷ m s⁻¹

That is about 0.1c, so the non-relativistic formula is acceptable.

Specific charge e/mₑ

You need one method. Fine-beam tube: the beam (anode pd V) enters a uniform magnetic field B at right angles, and the magnetic force provides the centripetal force:

Bev = mv²/r   so   v = Ber/m
½mv² = eV     so   v² = 2eV/m
Combine:  e/m = 2V/(B²r²)

Worked example 2. V = 250 V, B = 0.90 mT, r = 5.9 cm.

e/m = 2 × 250 / ((0.90 × 10⁻³)² × 0.059²) = 1.77 × 10¹¹ C kg⁻¹

(Thomson used crossed fields: no deflection when eE = Bev, so v = E/B.)

Significance. Thomson got the same e/m whatever gas or cathode he used, so the particles are in all matter. The electron’s specific charge (about 1.76 × 10¹¹ C kg⁻¹) is roughly 1800 times the specific charge of the hydrogen ion (about 9.6 × 10⁷ C kg⁻¹). For equal charge, the electron’s mass is far smaller than the lightest ion’s: it is sub-atomic.

Millikan’s oil drop experiment

Charged oil drops fall between horizontal plates a distance d apart.

• Field off: at terminal speed v, weight = drag (ignoring upthrust): mg = 6πηrv, with m = (4/3)πr³ρ, so r = √(9ηv/(2ρg)), then the mass.
• Field on, drop stationary: QV/d = mg, so Q = mgd/V.
• Field on, drop rising at terminal speed v: QV/d = mg + 6πηrv.

Worked example 3. d = 6.0 mm, ρ = 880 kg m⁻³, η = 1.8 × 10⁻⁵ N s m⁻². Field off, the drop falls at 1.2 × 10⁻⁴ m s⁻¹; it is held stationary at 540 V.

r = √(9 × 1.8 × 10⁻⁵ × 1.2 × 10⁻⁴ / (2 × 880 × 9.81)) = 1.06 × 10⁻⁶ m
m = (4/3)π(1.06 × 10⁻⁶)³ × 880 = 4.40 × 10⁻¹⁵ kg
Q = mgd/V = 4.40 × 10⁻¹⁵ × 9.81 × 6.0 × 10⁻³ / 540 = 4.80 × 10⁻¹⁹ C = 3e

Significance. Every charge was a whole-number multiple of about 1.6 × 10⁻¹⁹ C: charge is quantised, with e the smallest unit. With e/m, this gave the electron’s mass.

3.12.2 Wave-particle duality

Newton and Huygens

Newton’s corpuscular theory: light is a stream of tiny particles, reflecting like bouncing balls and refracting because they are attracted into the denser medium, so light would be faster in water. Huygens’ wave theory: each point on a wavefront is a source of secondary wavelets; light is slower in the denser medium.

Newton’s theory was preferred because of his great reputation, because it explained reflection, refraction and sharp shadows simply, and because the speed of light in water could not yet be measured to decide.

Young’s double slits

Light from one source passes through two close slits. Where waves meet in phase they reinforce (bright fringes); in antiphase they cancel (dark fringes). Corpuscles cannot explain interference. Acceptance of the wave theory was still delayed, largely because of Newton’s authority.

Electromagnetic waves

An EM wave is electric and magnetic fields oscillating in phase, at right angles to each other and to the direction of travel; it needs no medium. Maxwell predicted its speed in a vacuum:

c = 1/√(μ₀ε₀)

ε₀ (permittivity of free space) relates to the electric field strength due to a charged object; μ₀ (permeability of free space) to the magnetic flux density due to a current-carrying wire.

Worked example 4. μ₀ = 4π × 10⁻⁷ H m⁻¹, ε₀ = 8.85 × 10⁻¹² F m⁻¹ gives c = 1/√(4π × 10⁻⁷ × 8.85 × 10⁻¹²) = 3.00 × 10⁸ m s⁻¹.

Hertz produced radio waves with a spark gap and detected them with a wire loop with a small gap, where sparks appeared. Reflection from a metal sheet made stationary waves with nodes half a wavelength apart; with the oscillator frequency known, c = fλ gave the speed of radio waves, equal to the speed of light.

Fizeau sent light through a gap in a rotating toothed wheel to a mirror a distance d away. At rotation frequency f the returning light was blocked by the next tooth. With N teeth, gap to next tooth takes t = 1/(2Nf), so c = 2d/t.

Worked example 5. N = 500, d = 6.0 km, first blocked at 25 rev s⁻¹.

t = 1/(2 × 500 × 25) = 4.0 × 10⁻⁵ s
c = 2 × 6000 / 4.0 × 10⁻⁵ = 3.0 × 10⁸ m s⁻¹

Implication: the measured speed of light matched Maxwell’s predicted speed for EM waves, so light is an electromagnetic wave.

The discovery of photoelectricity

A black body absorbs all radiation falling on it. Its measured emission peaks at one wavelength and falls to zero at short wavelengths, but classical wave theory predicted intensity rising without limit as wavelength falls: the ultraviolet catastrophe. Planck fixed this by assuming energy is emitted in quanta, E = hf.

Wave theory predicts that any frequency would emit electrons if bright enough, that weak light would cause a delay while energy built up, and that brighter light would give faster electrons. Observed: a threshold frequency; instant emission; maximum kinetic energy depends on frequency only, intensity changing only the number of electrons.

Einstein: light is photons of energy hf, and one electron absorbs one photon (hf = ϕ + Ek(max)). So EM radiation has a particle nature as well as a wave nature.

Wave-particle duality

de Broglie: particles have a wavelength, p = h/λ. For an electron accelerated from rest through V, p = √(2meV), so λ = h/√(2meV).

Low-energy electron diffraction: a beam through a thin crystal or graphite film gives rings. Higher speed means larger p and shorter λ, so less diffraction and smaller rings.

Electron microscopes

To resolve atom-sized detail (about 10⁻¹⁰ m), λ must be of that order.

Worked example 6. Estimate the anode pd for λ = 5.0 × 10⁻¹¹ m.

V = h²/(2meλ²) = (6.63 × 10⁻³⁴)² / (2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ × (5.0 × 10⁻¹¹)²)
  = 603 V ≈ 600 V

TEM: electrons from a heated filament are accelerated by the anode. Magnetic lenses focus them: a condenser lens makes a parallel beam through a very thin sample; objective and projector lenses form a magnified image on a fluorescent screen. Detail is limited by wavelength, sample thickness (which slows electrons and spreads their wavelengths) and lens aberrations.

STM: a fine conducting tip sits about 1 nm above a conducting surface, with a small pd. Electrons cross the gap by quantum tunnelling, and the current depends very strongly on the gap. In constant-current mode the tip height is adjusted to keep the current fixed, so the tip follows the surface; in constant-height mode current changes map it.

3.12.3 Special relativity

The Michelson-Morley experiment

The interferometer splits light at a semi-silvered plate into two perpendicular beams, which reflect from mirrors and recombine to give fringes. If the Earth moved through an “ether” (absolute rest frame), the times along the two arms would differ, and rotating the apparatus through 90° should shift the fringes. No shift was detected. Absolute motion cannot be detected, and the speed of light is the same whatever the motion of the apparatus: the invariance of the speed of light.

Einstein’s postulates

An inertial frame is not accelerating; Newton’s first law holds in it. The postulates:

1. Physical laws have the same form in all inertial frames.
2. The speed of light in free space is invariant.

Time dilation

Proper time t₀ is measured by an observer at rest relative to the clock. An observer moving at v relative to it measures a longer time: t = t₀/√(1 − v²/c²).

Worked example 7. Muons at 0.995c pass two detectors 2.0 km apart vertically. Proper half-life 1.5 μs.

Earth-frame time = 2000/(0.995 × 3.00 × 10⁸) = 6.70 μs
γ = 1/√(1 − 0.995²) = 10.0, so half-life = 15.0 μs
surviving: 0.5^(6.70/15.0) = 0.73 (not 0.5^(6.70/1.5) = 0.045)

Far more muons arrive than predicted without dilation: evidence for time dilation.

Length contraction

An object of proper length l₀ moving at v has length l = l₀√(1 − v²/c²) along its motion.

Worked example 8. A 120 m spacecraft at 0.80c: l = 120 × √(1 − 0.64) = 72 m.

Mass and energy

E = mc², with m = m₀/√(1 − v²/c²), so E = m₀c²/√(1 − v²/c²). Kinetic energy is Ek = mc² − m₀c².

Graphs: mass starts at m₀ and rises towards infinity as v → c. Kinetic energy follows ½m₀v² at low speed, then rises without limit as v → c.

Worked example 9. Find the speed of an electron with Ek = 1.5 MeV.

m₀c² = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 8.20 × 10⁻¹⁴ J (0.512 MeV)
E = 1.5 + 0.512 = 2.01 MeV;  γ = 2.01/0.512 = 3.93
v = c√(1 − 1/γ²) = 0.967c

Newtonian ½mv² gives 7.3 × 10⁸ m s⁻¹, faster than light: impossible.

Bertozzi’s experiment accelerated electrons to several known energies and timed their flight over a known distance; kinetic energy was checked from the heating of the target they struck. Speed approached but never exceeded c, matching the relativistic graph: direct evidence for the variation of kinetic energy with speed.

Common errors

• Using ½mv² = eV for relativistic electrons; check v/c first.
• Finding the Millikan radius from the field-on equation; r comes from the field-off terminal speed.
• Swapping t and t₀: proper time is the shorter, measured in the clock’s own frame.
• Describing the STM as using electron diffraction; it uses tunnelling.

For Paper 3 planning see exam preparation.

Official syllabus

AQA AS and A-level Physics specification (7407/7408), version 1.4, July 2026, AS and A-level exams June 2016 onwards: section 3.12 Turning points in physics (A-level only). Published by AQA.