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Turning points in physics (optional)

This is one of the five optional A-Level modules (AQA Paper 3, Section B); a student studies just one option. Turning points traces three revolutions that reshaped physics: the discovery of the electron and the measurement of its charge and mass, the long argument over whether light is a wave or a particle and its resolution in wave-particle duality, and Einstein's special relativity born from the failure of the aether.

4 sections·~12 min reading time·3 competencies·Level Standard 1 · Advanced 3

T·121212 / 13
Exam profile
AO1 · Describe the discovery of the electron, the development of the wave-particle model of light and the postulates of special relativity (optional module)AO2 · Calculate specific charge, charge quantisation, the Lorentz factor, time dilation, length contraction and mass-energyAO3 · Evaluate the Michelson-Morley and muon experiments as evidence for special relativity and the photoelectric effect as evidence for the photon
Operators:describeexplaincalculatedetermineevaluateshow that

basic level

This is an optional A2 module examined in Paper 3 Section B; it is studied only if chosen as the school's option. It requires the discovery of the electron, the wave-particle story and the outline of special relativity.

higher level

The full option demands quantitative treatment of specific charge and Millikan's experiment, the evidence for wave-particle duality, and the Lorentz factor, time dilation, length contraction and mass-energy.

Depth

Reading depth: In depth

Text

Text size: Standard

Contents · 4 sections▾
  1. Turning points in physics (optional)
    • 01The discovery of the electron and Millikan's experiment◐
    • 02The nature of light: wave and particle models●
    • 03The Michelson-Morley experiment and Einstein's postulates●
    • 04Time dilation, length contraction and mass-energy●
§ 01

The discovery of the electron and Millikan's experiment#

●●○StandardLPAQA 7408 3.12.1LPDfE GCE Physics - turning points option

Millikan's oil-drop experiment

Oil-drop apparatusSchematic diagram with 7 elements, + plate, - plate, charged oil drop, weight mg, electric force qE, atomiser, field balances gravity+ plate- platecharged oil dropweight mgelectric forceqEatomiserfield balancesgravity
Fig. 1A charged oil drop is held stationary between two horizontal plates when the upward electric force qEqEqE exactly balances its downward weight mgmgmg.

Key points

The electron was discovered by J. J. Thomson through the study of cathode rays - the glow produced when a high voltage is applied across an evacuated tube. Thomson showed that these rays were deflected by both electric and magnetic fields towards the positive plate, so they were streams of negatively charged particles, far lighter than any atom. By balancing the electric and magnetic deflections he measured their specific charge e/me/me/m, and found it enormous - about 1.76×1011 C kg−11.76\times10^{11}\,\text{C kg}^{-1}1.76×1011C kg−1 - the same whatever the gas or the electrode metal, proving the electron a universal constituent of matter.
Thermionic emission provides electrons for such experiments: a heated metal filament gives its free electrons enough energy to escape its surface. Accelerated through a potential difference VVV, an electron gains kinetic energy eV=12mv2eV = \tfrac{1}{2}mv^2eV=21​mv2, so its speed and momentum can be controlled - the principle of the electron gun in the cathode-ray tube.
Thomson's e/me/me/m gave the ratio but not the charge and mass separately. Millikan supplied the missing piece with his oil-drop experiment: tiny charged oil droplets were held stationary between two horizontal plates, so that the upward electric force qEqEqE exactly balanced the downward weight mgmgmg, giving q=mgEq = \dfrac{mg}{E}q=Emg​. Measuring the charge on many droplets, Millikan found that every value was a whole-number multiple of a single smallest charge.
That smallest charge is the elementary charge e=1.60×10−19 Ce = 1.60\times10^{-19}\,\text{C}e=1.60×10−19C - electric charge is quantised. Combining Millikan's eee with Thomson's e/me/me/m finally gave the electron's tiny mass, 9.11×10−31 kg9.11\times10^{-31}\,\text{kg}9.11×10−31kg. Together the two experiments established the electron as a fundamental particle with a definite charge and mass.
eV=12mv2eV = \tfrac{1}{2}mv^2eV=21​mv2

Accelerated electron

Kinetic energy gained equals the work done by the accelerating pd.

q=mgE=mgdVq = \dfrac{mg}{E} = \dfrac{mgd}{V}q=Emg​=Vmgd​

Balanced oil drop

The electric force balances the weight; E=V/dE = V/dE=V/d.

Worked example

Charge on a balanced oil drop

An oil drop of mass 4.9×10−15 kg4.9\times10^{-15}\,\text{kg}4.9×10−15kg is held stationary between plates 10 mm10\,\text{mm}10mm apart at a potential difference of 1000 V1000\,\text{V}1000V. Find the charge on the drop and the number of electrons it carries. Take g=9.81 m s−2g = 9.81\,\text{m s}^{-2}g=9.81m s−2 and e=1.60×10−19 Ce = 1.60\times10^{-19}\,\text{C}e=1.60×10−19C.

  1. 01Field between the plates

    E=V/d=1000/0.010=1.0×105 V m−1E = V/d = 1000/0.010 = 1.0\times10^{5}\,\text{V m}^{-1}E=V/d=1000/0.010=1.0×105V m−1.

  2. 02Charge from the force balance

    q=mg/E=(4.9×10−15×9.81)/(1.0×105)q = mg/E = (4.9\times10^{-15} \times 9.81)/(1.0\times10^{5})q=mg/E=(4.9×10−15×9.81)/(1.0×105).

    q=4.81×10−141.0×105=4.8×10−19 Cq = \dfrac{4.81\times10^{-14}}{1.0\times10^{5}} = 4.8\times10^{-19}\,\text{C}q=1.0×1054.81×10−14​=4.8×10−19C
  3. 03Number of electrons

    n=q/e=4.8×10−19/1.60×10−19=3.0n = q/e = 4.8\times10^{-19}/1.60\times10^{-19} = 3.0n=q/e=4.8×10−19/1.60×10−19=3.0.

Result: The drop carries 4.8×10−19 C4.8\times10^{-19}\,\text{C}4.8×10−19C, exactly three electron charges - evidence that charge is quantised.

Exam focus

  • Describe Thomson's determination of e/me/me/m from the deflection of cathode rays in electric and magnetic fields.
  • Explain how Millikan's oil-drop experiment shows charge is quantised and find the charge on a balanced droplet from q=mg/Eq = mg/Eq=mg/E.

Typical mistakes

  • Confusing the specific charge e/me/me/m (Thomson) with the charge eee itself (Millikan).
  • Forgetting that the electric field between the plates is E=V/dE = V/dE=V/d when finding the charge on a droplet.

Active revision

An oil drop of mass 6.5×10−15 kg6.5\times10^{-15}\,\text{kg}6.5×10−15kg is held stationary between plates 8.0 mm8.0\,\text{mm}8.0mm apart at a potential difference of 800 V800\,\text{V}800V. Find the charge on the drop and the number of electrons it carries.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Physics 7408 specification (AQA)

§ 02

The nature of light: wave and particle models#

●●●AdvancedLPAQA 7408 3.12.2LPDfE GCE Physics - wave-particle duality

Key points

The nature of light was debated for centuries. Newton favoured a corpuscular (particle) model, picturing light as a stream of tiny particles, which explained reflection and, with some difficulty, refraction. Huygens proposed a wave model, treating each point on a wavefront as a source of secondary wavelets. The wave model correctly predicted that light slows in a denser medium, whereas Newton's particles would have sped up - but Newton's authority kept the particle view dominant for a century.
The wave model triumphed in the nineteenth century. Young's double-slit experiment produced interference fringes that only waves can explain, and Maxwell then showed theoretically that electric and magnetic fields can propagate as electromagnetic waves at the speed of light, unifying light with the whole electromagnetic spectrum. Hertz confirmed this by producing and detecting radio waves, demonstrating they reflect, refract and interfere just like light.
But the wave model could not explain the photoelectric effect - the emission of electrons from a metal by light above a threshold frequency, independent of intensity. Einstein resolved it by reviving the particle idea in quantum form: light comes in photons of energy hfhfhf, each interacting with one electron. Light was neither purely wave nor purely particle but both - wave-particle duality.
De Broglie then extended the symmetry to matter, proposing that particles have a wavelength λ=h/p\lambda = h/pλ=h/p; electron diffraction confirmed it. The long argument was resolved not by one model defeating the other but by the recognition that wave and particle are complementary descriptions, each revealed by a different experiment - one of the deepest lessons in physics.
E=hf=hcλE = hf = \dfrac{hc}{\lambda}E=hf=λhc​

Photon energy

The particle (photon) description of light.

λ=hp\lambda = \dfrac{h}{p}λ=ph​

de Broglie wavelength

The wave description of a particle - the symmetry that completed the duality.

Worked example

Photon energy and emission

Light of wavelength 500 nm500\,\text{nm}500nm strikes a metal of work function 1.9 eV1.9\,\text{eV}1.9eV. Find the photon energy in eV and state whether electrons are emitted. Take h=6.63×10−34 J sh = 6.63\times10^{-34}\,\text{J s}h=6.63×10−34J s, c=3.00×108 m s−1c = 3.00\times10^{8}\,\text{m s}^{-1}c=3.00×108m s−1, 1 eV=1.60×10−19 J1\,\text{eV} = 1.60\times10^{-19}\,\text{J}1eV=1.60×10−19J.

  1. 01Photon energy in joules

    E=hc/λ=(6.63×10−34×3.00×108)/(500×10−9)=3.98×10−19 JE = hc/\lambda = (6.63\times10^{-34} \times 3.00\times10^{8})/(500\times10^{-9}) = 3.98\times10^{-19}\,\text{J}E=hc/λ=(6.63×10−34×3.00×108)/(500×10−9)=3.98×10−19J.

  2. 02Convert to eV

    3.98×10−19/1.60×10−19=2.49 eV3.98\times10^{-19}/1.60\times10^{-19} = 2.49\,\text{eV}3.98×10−19/1.60×10−19=2.49eV.

  3. 03Compare with the work function

    2.49 eV>1.9 eV2.49\,\text{eV} > 1.9\,\text{eV}2.49eV>1.9eV, so emission occurs.

    Ek,max⁡=2.49−1.9=0.6 eVE_{k,\max} = 2.49 - 1.9 = 0.6\,\text{eV}Ek,max​=2.49−1.9=0.6eV

Result: The photon carries 2.49 eV2.49\,\text{eV}2.49eV, above the work function, so electrons are emitted with up to 0.6 eV0.6\,\text{eV}0.6eV of kinetic energy.

Exam focus

  • Contrast Newton's corpuscular and Huygens' wave models and state which predicted the correct change of speed on refraction.
  • Explain how the photoelectric effect required the photon and how electron diffraction confirmed wave-particle duality.

Typical mistakes

  • Saying the particle model predicts light slows in glass - Newton's corpuscular model wrongly predicted it speeds up.
  • Treating wave-particle duality as light being wave and particle at the same instant - it shows one behaviour or the other depending on the experiment.

Active revision

Light of wavelength 450 nm450\,\text{nm}450nm falls on a metal of work function 2.2 eV2.2\,\text{eV}2.2eV. Calculate the photon energy in eV and state whether photoemission occurs.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

§ 03

The Michelson-Morley experiment and Einstein's postulates#

●●●AdvancedLPAQA 7408 3.12.3LPDfE GCE Physics - special relativity

The Michelson-Morley interferometer

Michelson-Morley interferometerSchematic diagram with 9 elements, light source, light, beam splitter, mirror M1, mirror M2, detector (fringes), recombined beams interferelight sourcelightbeam splittermirror M1mirror M2detector(fringes)recombined beamsinterfere
Fig. 2A beam is split, sent along two perpendicular arms to mirrors and recombined; a difference in travel time along the arms would shift the interference fringes at the detector.

Key points

Nineteenth-century physics assumed light waves travelled through an all-pervading medium, the aether, in which the Earth moved. The Michelson-Morley experiment was designed to detect this motion. It split a light beam in two with a half-silvered mirror, sent the halves along two perpendicular arms to mirrors and back, and recombined them to form interference fringes. If the Earth moved through the aether, light travelling along and across the 'aether wind' would take different times, and rotating the apparatus should shift the fringes.
The result was null: no fringe shift was seen, at any orientation or time of year. This was one of the most important negative results in the history of science. It meant either that the Earth dragged the aether with it (ruled out by other observations) or that the aether did not exist and the speed of light was the same in every direction regardless of the observer's motion.
Einstein took the second, radical option and built special relativity on two postulates. First, the laws of physics are the same in all inertial (non-accelerating) frames of reference - no experiment can tell you your absolute velocity. Second, the speed of light in a vacuum is the same for all inertial observers, independent of the motion of the source or observer. The Michelson-Morley null result is exactly what the second postulate demands.
These innocent-looking statements have startling consequences, because holding the speed of light constant forces space and time themselves to be relative. The central quantity is the Lorentz factor γ=11−v2/c2\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}}γ=1−v2/c2​1​, which is essentially 111 at everyday speeds but grows without limit as vvv approaches ccc. It governs how measured times and lengths change between observers, developed in the next section.
γ=11−v2/c2\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}}γ=1−v2/c2​1​

Lorentz factor

Grows without limit as v→cv \to cv→c; essentially 111 at everyday speeds.

Worked example

The Lorentz factor

A particle moves at 0.60c0.60c0.60c. Calculate the Lorentz factor γ\gammaγ.

  1. 01Substitute the speed

    γ=11−(0.60)2=11−0.36\gamma = \dfrac{1}{\sqrt{1 - (0.60)^2}} = \dfrac{1}{\sqrt{1 - 0.36}}γ=1−(0.60)2​1​=1−0.36​1​.

  2. 02Evaluate

    0.64=0.80\sqrt{0.64} = 0.800.64​=0.80, so γ=1/0.80\gamma = 1/0.80γ=1/0.80.

    γ=10.80=1.25\gamma = \dfrac{1}{0.80} = 1.25γ=0.801​=1.25

Result: The Lorentz factor is 1.251.251.25 - relativistic effects are already a noticeable 25%25\%25% at 0.6c0.6c0.6c.

Exam focus

  • Describe the Michelson-Morley experiment and explain the significance of its null result.
  • State Einstein's two postulates and calculate the Lorentz factor γ\gammaγ at a given speed.

Typical mistakes

  • Describing Michelson-Morley as detecting the aether - it famously failed to, and that null result was the point.
  • Thinking the Lorentz factor is significant at everyday speeds - it is essentially 111 until vvv is a large fraction of ccc.

Active revision

Calculate the Lorentz factor for a spacecraft travelling at 0.80c0.80c0.80c, and comment on how measured time aboard compares with that on Earth.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

§ 04

Time dilation, length contraction and mass-energy#

●●●AdvancedLPAQA 7408 3.12.3LPDfE GCE Physics - relativistic effects

The Lorentz factor against speed

Lorentz factor gamma vs v/cGraph of gamma, y-intercept at y = 1, increasing, on the interval x from 0 to 0.950.20.40.60.812345gamma ~ 2 at 0.87cgammaLorentz factor gammaSpeed v / c
Fig. 3The Lorentz factor γ\gammaγ is close to 111 for everyday speeds but rises steeply as vvv approaches the speed of light, diverging to infinity at v=cv = cv=c.

Key points

Because the speed of light is the same for every observer, time itself runs differently for observers in relative motion. A clock moving relative to an observer runs slow: a time interval t0t_0t0​ measured in the clock's own rest frame (the proper time) is measured as the longer interval t=γt0t = \gamma t_0t=γt0​ by an observer who sees the clock move. This is time dilation, and it is real, not an illusion.
The decisive evidence is the muon experiment. Muons are created high in the atmosphere by cosmic rays and travel towards the ground at close to the speed of light, yet their short half-life should let almost none survive the journey. Far more reach the ground than expected, because in our frame their internal clocks run slow (time dilation) - so their half-life is dilated by γ\gammaγ and they live long enough to arrive. In the muons' own frame, the same result appears as length contraction: the atmosphere is thinner.
Length contraction is the companion effect: an object's length is greatest in its own rest frame (the proper length L0L_0L0​), and is measured as the shorter length L=L0γL = \dfrac{L_0}{\gamma}L=γL0​​ by an observer past whom it moves, contracted along the direction of motion. Time dilation and length contraction are two views of the same relativity of space and time.
Relativity also links mass and energy. As a body speeds up, its measured (relativistic) mass increases as m=γm0m = \gamma m_0m=γm0​, so it takes ever more force to accelerate and no massive object can reach the speed of light. Mass and energy are equivalent through E=mc2E = mc^2E=mc2: the total energy of a moving body is γm0c2\gamma m_0 c^2γm0​c2, and its rest energy m0c2m_0 c^2m0​c2 is released when mass is converted to energy - the principle behind nuclear energy met earlier.
t=γt0t = \gamma t_0t=γt0​

Time dilation

A moving clock runs slow; t0t_0t0​ is the proper time.

L=L0γL = \dfrac{L_0}{\gamma}L=γL0​​

Length contraction

A moving object is shortened along its motion; L0L_0L0​ is the proper length.

m=γm0,E=mc2m = \gamma m_0, \qquad E = mc^2m=γm0​,E=mc2

Relativistic mass and energy

Mass increases with speed; mass and energy are equivalent.

Worked example

Muon time dilation

Muons have a mean lifetime of 2.2 μs2.2\,\mu\text{s}2.2μs at rest. Travelling at 0.99c0.99c0.99c, find their mean lifetime measured in the laboratory frame. Take γ=7.09\gamma = 7.09γ=7.09 at this speed.

  1. 01Time dilation

    The lab measures the dilated lifetime t=γt0t = \gamma t_0t=γt0​.

  2. 02Substitute

    t=7.09×2.2 μst = 7.09 \times 2.2\,\mu\text{s}t=7.09×2.2μs.

    t=7.09×2.2=15.6 μst = 7.09 \times 2.2 = 15.6\,\mu\text{s}t=7.09×2.2=15.6μs

Result: The lab measures a mean lifetime of 15.6 μs15.6\,\mu\text{s}15.6μs, seven times longer - which is why so many muons reach the ground.

Exam focus

  • Use t=γt0t = \gamma t_0t=γt0​ and L=L0/γL = L_0/\gammaL=L0​/γ for time dilation and length contraction, and explain the muon experiment.
  • Use relativistic mass m=γm0m = \gamma m_0m=γm0​ and mass-energy E=mc2E = mc^2E=mc2, and explain why ccc is a speed limit.

Typical mistakes

  • Mixing up which frame measures the proper time or proper length - proper time is measured by a clock at rest in its own frame; proper length in the object's rest frame.
  • Applying relativistic formulae at everyday speeds where γ≈1\gamma \approx 1γ≈1 and the effects are negligible.

Active revision

Muons at rest have a mean lifetime of 2.2 μs2.2\,\mu\text{s}2.2μs. Travelling at 0.99c0.99c0.99c (γ=7.1\gamma = 7.1γ=7.1), find their mean lifetime measured from the ground.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA) · GCE AS and A level subject content for the sciences (Department for Education)

Contents

Section -- / 04

    • 01The discovery of the electron and Millikan's experiment◐
    • 02The nature of light: wave and particle models●
    • 03The Michelson-Morley experiment and Einstein's postulates●
    • 04Time dilation, length contraction and mass-energy●

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Department for Education

  • GCE AS and A level subject content for the sciences

AQA

  • AQA A-level Physics 7408 specification

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