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AQA A-level Physics revision notes

Turning points in physics (A-level only)

Section 3.12
Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.
15 specification points
Optional · choose 1 of 5

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 7408 section 3.12

Checked against AQA 7408 section 3.12. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

How this checking works

In the exam: Data and formulae booklet provided · calculator allowed in every paper

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3.12.1.1

Cathode rays

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Cathode rays are produced in a discharge tube containing gas at low pressure and a large potential difference between electrodes. The field ionises gas atoms.
  • Positive ions accelerate towards the negative cathode and release electrons on impact; these electrons accelerate away from the cathode towards the anode.
  • A hole in the anode can form a narrow beam, which may produce fluorescence when it reaches the glass or a screen.
  • Electric deflection towards a positive plate identifies negative charge, while magnetic deflection confirms moving charged particles.
  • The independence of the ray's behaviour from gas and electrode material supported the conclusion that electrons are constituents of all atoms.
Electrons released at the cathode pass through the anode aperture to form a cathode ray.
Worked example

Explain how a cathode ray is produced when a large potential difference is applied across a low-pressure discharge tube.

  1. 1.The strong electric field ionises atoms of the low-pressure gas.
  2. 2.Positive ions accelerate to the negative cathode and release electrons when they strike it.
  3. 3.The released electrons accelerate from the cathode towards the anode and form the cathode ray.

Answer: Gas ionisation supplies positive ions whose cathode impacts release the electrons accelerated into the ray.

Common mistakes

  • Don't describe the cathode ray as electromagnetic radiation rather than a stream of electrons.
  • Don't send electrons from the positive anode towards the negative cathode.
  • Don't omit the low gas pressure and ionisation stage when explaining how the ray is produced.

Exam tip

For a production question, give the causal sequence gas ionisation, positive-ion impact, electron release and electron acceleration.

Tier 1 · Easy

ORIGINAL

State the direction in which a cathode ray travels inside a discharge tube and identify the particles in the ray.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A large potential difference is connected across a discharge tube containing low-pressure gas. Describe the sequence that creates the electron beam.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A beam from the cathode bends towards a positive plate. Reversing a transverse magnetic field reverses its magnetic deflection, and the results are unchanged when the gas and cathode metal are replaced. Explain the conclusions drawn from these observations and their importance for the atomic model.

[5 marks]

Total for this question: 5

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3.12.1.2

Thermionic emission of electrons

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Thermionic emission occurs when heating a metal gives some conduction electrons enough energy to escape from its surface. A heated cathode therefore supplies electrons continuously, and a positive anode can accelerate them into a beam.
  • When an electron starts from rest and moves through potential difference VV, electrical work eVeV becomes kinetic energy: eV=12mev2eV=\frac12m_{\mathrm e}v^2.
  • This relation assumes negligible energy loss and non-relativistic speed.
  • Potential difference must be in volts, ee in coulombs and mem_{\mathrm e} in kilograms.
  • The examiner may ask for the emission principle, an energy-transfer explanation, or a speed calculated by rearranging the energy equation.
Worked example

Electrons emitted from a heated cathode are accelerated from rest through 1.80kV1.80\,\text{kV}. Calculate their speed using e=1.60×1019Ce=1.60\times10^{-19}\,\text{C} and me=9.11×1031kgm_{\mathrm e}=9.11\times10^{-31}\,\text{kg}.

  1. 1.Convert the potential difference: V=1.80×103VV=1.80\times10^3\,\text{V}.
  2. 2.Use eV=12mev2eV=\frac12m_{\mathrm e}v^2, so v=2eV/mev=\sqrt{2eV/m_{\mathrm e}}.
  3. 3.v=2(1.60×1019)(1.80×103)/(9.11×1031)v=\sqrt{2(1.60\times10^{-19})(1.80\times10^3)/(9.11\times10^{-31})}.

Answer: v=2.51×107m s1v=2.51\times10^7\,\text{m s}^{-1}.

Common mistakes

  • Don't say heating ejects positive ions from the cathode rather than allowing conduction electrons to escape.
  • Don't substitute kilovolts directly into eVeV without converting to volts.
  • Don't use eV=mev2eV=m_{\mathrm e}v^2 and omit the factor of one half.

Exam tip

A calculation should state the energy transfer eV=12mev2eV=\frac12m_{\mathrm e}v^2 before numerical substitution.

Tier 1 · Easy

ORIGINAL

Explain why heating a metal cathode can cause electrons to leave its surface.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Electrons emitted from a heated cathode start from rest and are accelerated through 2.50kV2.50\,\text{kV}. Calculate their speed using a non-relativistic model.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An electron gun accelerates a steady current of 3.00mA3.00\,\text{mA} through 7.50kV7.50\,\text{kV}. Calculate the electron speed, the number of electrons emitted each second, and the power transferred to the beam. Use a non-relativistic model.

[5 marks]

Total for this question: 5

3.12.1.3

Specific charge of the electron

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Specific charge is charge divided by mass; the electron's magnitude is e/mee/m_{\mathrm e} in C kg1\text{C kg}^{-1}. In one determination, crossed electric and magnetic fields are adjusted so the beam is undeflected.
  • Balancing eE=evBeE=evB gives the selected speed v=E/Bv=E/B.
  • The electrons then enter a perpendicular magnetic field and follow a circular path because evB=mev2/revB=m_{\mathrm e}v^2/r, giving e/me=v/(Br)e/m_{\mathrm e}=v/(Br).
  • Thomson found a magnitude far greater than the hydrogen-ion specific charge.
  • Since charge magnitudes are comparable, this showed that the electron has far smaller mass and is a subatomic constituent rather than a whole atom.
Crossed fields select the electron speed before magnetic curvature determines e/mee/m_{\mathrm e}.
Worked example

An undeflected beam passes through E=1.8×104V m1E=1.8\times10^4\,\text{V m}^{-1} and B=6.0×103TB=6.0\times10^{-3}\,\text{T}. It then curves with radius 0.085m0.085\,\text{m} in B=2.0×104TB=2.0\times10^{-4}\,\text{T}. Determine e/mee/m_{\mathrm e}.

  1. 1.The selector speed is v=E/B=(1.8×104)/(6.0×103)=3.0×106m s1v=E/B=(1.8\times10^4)/(6.0\times10^{-3})=3.0\times10^6\,\text{m s}^{-1}.
  2. 2.Use magnetic circular motion: e/me=v/(Br)e/m_{\mathrm e}=v/(Br).
  3. 3.e/me=(3.0×106)/[(2.0×104)(0.085)]e/m_{\mathrm e}=(3.0\times10^6)/[(2.0\times10^{-4})(0.085)].

Answer: e/me=1.8×1011C kg1e/m_{\mathrm e}=1.8\times10^{11}\,\text{C kg}^{-1}.

Common mistakes

  • Don't use the selector magnetic field again in the circular-path calculation when the two fields have different values.
  • Don't invert the circular-motion result and write e/me=Br/ve/m_{\mathrm e}=Br/v.
  • Don't compare electron and hydrogen-ion specific charges without linking the much larger electron value to its much smaller mass.

Exam tip

Keep the speed-selection and magnetic-deflection stages on separate lines, with each magnetic field clearly labelled.

Tier 1 · Easy

ORIGINAL

An electron beam passes undeflected through crossed fields of strength E=2.40×104V m1E=2.40\times10^4\,\text{V m}^{-1} and B=8.00×103TB=8.00\times10^{-3}\,\text{T}. Calculate the electron speed.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Electrons selected at 3.00×106m s13.00\times10^6\,\text{m s}^{-1} enter a perpendicular magnetic field of 1.80×104T1.80\times10^{-4}\,\text{T} and follow a circular path of radius 9.50cm9.50\,\text{cm}. Determine the magnitude of their specific charge.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

In a Thomson-style experiment, electrons pass between plates separated by 15.0mm15.0\,\text{mm} with 360V360\,\text{V} across them. A crossed field of 8.00mT8.00\,\text{mT} makes the beam undeflected. The selected beam then follows a circular path of radius 10.7mm10.7\,\text{mm} in a separate 1.60mT1.60\,\text{mT} field. Determine e/mee/m_{\mathrm e} and compare it with the hydrogen-ion specific charge 9.58×107C kg19.58\times10^7\,\text{C kg}^{-1}.

[5 marks]

Total for this question: 5

3.12.1.4

Principle of Millikan's determination of the electronic charge, e

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • In Millikan's method, a charged oil droplet lies between oppositely charged parallel plates.
  • When it is stationary, upward electric force balances weight: QE=mgQE=mg, with E=V/dE=V/d, so QV/d=mgQV/d=mg.
  • With the field removed, a falling droplet reaches terminal speed when Stokes' viscous force 6πηrv6\pi\eta rv balances its effective weight; together with the droplet density, this determines radius and mass.
  • Motion may also be observed with the field applied to find the charge.
  • Repeating the measurement produced charges that were integer multiples of a smallest value, ee, demonstrating that electric charge is quantised rather than continuous.
For a stationary negatively charged droplet, electric force balances its weight.
Worked example

A droplet of mass 4.9×1015kg4.9\times10^{-15}\,\text{kg} is stationary between plates 4.0mm4.0\,\text{mm} apart with 750V750\,\text{V} across them. Calculate its charge magnitude.

  1. 1.The electric field is E=V/d=750/(4.0×103)=1.875×105V m1E=V/d=750/(4.0\times10^{-3})=1.875\times10^5\,\text{V m}^{-1}.
  2. 2.For equilibrium, QE=mgQE=mg, so Q=mg/EQ=mg/E.
  3. 3.Q=(4.9×1015)(9.81)/(1.875×105)=2.56×1019CQ=(4.9\times10^{-15})(9.81)/(1.875\times10^5)=2.56\times10^{-19}\,\text{C}.

Answer: Q=2.6×1019CQ=2.6\times10^{-19}\,\text{C} to two significant figures.

Common mistakes

  • Don't use Q=mg/VQ=mg/V and omit the plate separation in E=V/dE=V/d.
  • Don't apply 6πηrv6\pi\eta rv before the falling droplet has reached terminal speed.
  • Don't claim one measured charge alone proves quantisation instead of comparing many charges for integer multiples of ee.

Exam tip

A force-balance answer should show the field direction and charge sign before equating force magnitudes.

Tier 1 · Easy

ORIGINAL

A negatively charged oil droplet is held stationary between horizontal plates. State the relationship between the magnitudes of the electric force and the droplet's weight.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

An oil droplet of mass 6.52×1015kg6.52\times10^{-15}\,\text{kg} is held stationary between plates 3.00mm3.00\,\text{mm} apart with 600V600\,\text{V} across them. Calculate the magnitude of its charge and express the result as a multiple of ee.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

With the electric field off, an oil droplet of density 850kg m3850\,\text{kg m}^{-3} falls through air at terminal speed 8.00×105m s18.00\times10^{-5}\,\text{m s}^{-1}. The air viscosity is 1.80×105Pa s1.80\times10^{-5}\,\text{Pa s}. Neglect air buoyancy. The droplet is then held stationary by a field of 7.50×104V m17.50\times10^4\,\text{V m}^{-1}. Use Stokes' law to determine the droplet charge and show how the result supports charge quantisation.

[5 marks]

Total for this question: 5

3.12.2.1

Newton's corpuscular theory of light

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Newton's corpuscular theory treated light as tiny particles emitted by sources and travelling in straight lines. This readily explained sharp shadows, and forces acting on corpuscles at boundaries were invoked to account for reflection and refraction.
  • Huygens instead treated every point on a wavefront as a source of secondary wavelets.
  • Newton's theory was preferred because his mechanical ideas were highly successful and authoritative, ray-like propagation was familiar, and diffraction was not obvious in everyday conditions.
  • The models made different predictions: Newton's corpuscles should speed up in a denser optical medium, whereas Huygens' waves should slow.
  • Later speed measurements and interference evidence supported the wave account.
Worked example

Compare Newton's and Huygens' predictions for light entering glass and explain why a measured lower speed in glass mattered.

  1. 1.Newton's boundary-force account predicted that corpuscles speed up in glass.
  2. 2.Huygens' wave construction predicted that light slows in glass.
  3. 3.The measured lower speed agreed with the wave prediction and contradicted the corpuscular prediction.

Answer: The speed measurement discriminated between the theories and supported Huygens' wave model.

Common mistakes

  • Don't say Newton's corpuscular theory predicted that light slows in a denser medium.
  • Don't explain historical preference only by saying Newton was correct, without identifying authority or apparent explanatory success.
  • Don't treat Huygens' theory as a modern photon theory rather than a wavefront model.

Exam tip

A compare question needs both models' distinct predictions and the observation that selects between them.

Tier 1 · Easy

ORIGINAL

State one observation that made Newton's corpuscular model of light appear plausible.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Newton modelled light as corpuscles, whereas Huygens used wavefronts. Compare their speed predictions for light entering glass from air and state why the later measurement mattered.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Discuss why Newton's particle account of light was accepted for so long and why later optical evidence forced physicists to replace it with a wave account.

[5 marks]

Total for this question: 5

3.12.2.2

Significance of Young's double slits experiment

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Young's double-slit experiment uses one monochromatic source to illuminate two narrow slits, producing waves with a stable phase relationship. Diffraction at each slit allows the two waves to overlap on a screen.
  • Where they arrive in phase, constructive superposition forms a bright fringe; where they arrive in antiphase, destructive superposition forms a dark fringe.
  • Only a qualitative account is required here, not fringe-spacing calculations.
  • The repeated bright and dark pattern was powerful evidence for wave behaviour because independent classical corpuscles could add intensity but could not explain cancellation to darkness.
  • Acceptance of Huygens' wave theory was nevertheless delayed by Newton's authority and the established corpuscular view.
Waves from both slits overlap at the screen to produce bright and dark fringes.
Worked example

Explain why a dark fringe in Young's experiment supported a wave theory of light.

  1. 1.Both slits send diffracted waves to the dark-fringe position.
  2. 2.The waves arrive in antiphase, so their displacements cancel by destructive superposition.
  3. 3.Independent classical corpuscles could not explain the cancellation of arriving light to zero intensity.

Answer: The dark fringe is evidence of destructive interference, a characteristic wave effect.

Common mistakes

  • Don't say dark fringes form because no light travels from either slit to those positions.
  • Don't use two unrelated sources and ignore the need for a stable phase relationship.
  • Don't perform a fringe-spacing calculation even though this turning-points specification requires only a general explanation.

Exam tip

For significance, identify destructive superposition and state why classical corpuscles could not produce it.

Tier 1 · Easy

ORIGINAL

State the feature of Young's double-slit pattern that provides evidence for interference.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Explain qualitatively how bright and dark fringes are formed when monochromatic light passes through Young's two slits.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A screen behind two illuminated narrow slits shows many regularly spaced dark bands between bright bands rather than two bright images. Explain why this result was a turning point in the debate between Newton's and Huygens' models of light.

[5 marks]

Total for this question: 5

3.12.2.3

Electromagnetic waves

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • An electromagnetic wave consists of oscillating electric and magnetic fields that are perpendicular to each other and to the propagation direction. Maxwell predicted the vacuum speed c=1/μ0ε0c=1/\sqrt{\mu_0\varepsilon_0}.
  • The permittivity ε0\varepsilon_0 relates to electric-field strength produced by charge in free space, while permeability μ0\mu_0 relates to magnetic flux density produced by current.
  • Agreement between Maxwell's value and the measured speed of light identified light as electromagnetic.
  • Hertz generated and detected radio waves, measured a speed close to cc and observed wave behaviour, supporting Maxwell.
  • Fizeau's toothed-wheel experiment independently measured the finite speed of light on Earth, strengthening the connection.
The electric and magnetic fields of an electromagnetic wave are transverse and mutually perpendicular.
Worked example

Calculate Maxwell's predicted wave speed using μ0=1.26×106H m1\mu_0=1.26\times10^{-6}\,\text{H m}^{-1} and ε0=8.85×1012F m1\varepsilon_0=8.85\times10^{-12}\,\text{F m}^{-1}.

  1. 1.Use c=1/μ0ε0c=1/\sqrt{\mu_0\varepsilon_0}.
  2. 2.Substitute: c=1/(1.26×106)(8.85×1012)c=1/\sqrt{(1.26\times10^{-6})(8.85\times10^{-12})}.
  3. 3.Evaluate and compare with the measured speed of light.

Answer: c=3.00×108m s1c=3.00\times10^8\,\text{m s}^{-1}, agreeing with the speed of light.

Common mistakes

  • Don't draw the electric or magnetic field oscillation parallel to the direction of travel.
  • Don't interchange ε0\varepsilon_0 and μ0\mu_0 when relating them to electric and magnetic fields.
  • Don't state that Hertz first measured visible-light speed rather than producing and detecting radio waves.

Exam tip

A significance answer should compare Maxwell's calculated value with measured light and radio-wave speeds.

Tier 1 · Easy

ORIGINAL

Describe the relative directions of the electric field, magnetic field and travel direction in a plane electromagnetic wave.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Use μ0=4π×107H m1\mu_0=4\pi\times10^{-7}\,\text{H m}^{-1} and ε0=8.85×1012F m1\varepsilon_0=8.85\times10^{-12}\,\text{F m}^{-1} to calculate the speed predicted by Maxwell for an electromagnetic wave in a vacuum.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

In a Fizeau-type experiment, light travels to a mirror 8.63km8.63\,\text{km} away and back through a wheel with 720720 teeth. The first extinction occurs at 12.1Hz12.1\,\text{Hz} because the wheel turns from a gap to the adjacent tooth during the round trip. Calculate the speed of light and explain how Fizeau's and Hertz's results supported Maxwell's theory.

[5 marks]

Total for this question: 5

3.12.2.4

The discovery of photoelectricity

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Classical theory predicted the ultraviolet catastrophe: black-body intensity should rise without limit at high frequency, contrary to observed spectra. Planck resolved this by proposing that energy is exchanged in quanta E=hfE=hf, making high-frequency quanta less likely at a fixed temperature.
  • Classical wave theory also failed for photoelectricity: it could not explain threshold frequency, immediate emission, or maximum electron kinetic energy depending on frequency rather than intensity.
  • Einstein treated each quantum as a photon absorbed by one electron, giving hf=ϕ+Kmaxhf=\phi+K_{\max}.
  • Above threshold, greater intensity supplies more photons and raises the emission rate, but at fixed frequency it does not raise KmaxK_{\max}.
  • This established a particle aspect of electromagnetic radiation.
One photon transfers energy hfhf to one electron, which escapes with maximum energy hfϕhf-\phi.
Worked example

A metal has work function 2.4eV2.4\,\text{eV} and is illuminated by 4.1eV4.1\,\text{eV} photons. Calculate the maximum photoelectron kinetic energy.

  1. 1.Use Einstein's equation hf=ϕ+Kmaxhf=\phi+K_{\max}.
  2. 2.Kmax=4.12.4=1.7eVK_{\max}=4.1-2.4=1.7\,\text{eV}.
  3. 3.Convert if required: 1.7(1.60×1019)=2.72×1019J1.7(1.60\times10^{-19})=2.72\times10^{-19}\,\text{J}.

Answer: Kmax=1.7eV=2.7×1019JK_{\max}=1.7\,\text{eV}=2.7\times10^{-19}\,\text{J}.

Common mistakes

  • Don't say increasing intensity at fixed frequency increases the maximum electron kinetic energy.
  • Don't explain threshold frequency by sharing one photon's energy among several electrons.
  • Don't state that classical theory predicted the observed fall in black-body intensity at ultraviolet frequencies.

Exam tip

For an explain question, link each observation to the one-photon–one-electron energy transfer.

Tier 1 · Easy

ORIGINAL

State one photoelectric observation that classical wave theory could not explain.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Explain how Planck's quantum hypothesis avoided the ultraviolet catastrophe in the black-body spectrum.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Light of irradiance 2.50×106W m22.50\times10^{-6}\,\text{W m}^{-2} falls on a metal whose work function is 2.30eV2.30\,\text{eV}. In a classical continuous-wave model, suppose one electron accumulates all the energy incident on a surface patch of area 1.80×1020m21.80\times10^{-20}\,\text{m}^2. Calculate the shortest predicted delay before that electron can escape, and contrast this with immediate emission under the photon model.

[5 marks]

Total for this question: 5

3.12.2.5

Wave-particle duality

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • de Broglie's hypothesis assigns wavelength λ=h/p\lambda=h/p to a particle of momentum pp. For a non-relativistic electron accelerated from rest through potential difference VV, eV=p2/(2me)eV=p^2/(2m_{\mathrm e}), so λ=h/2meeV\lambda=h/\sqrt{2m_{\mathrm e}eV}.
  • Low-energy electrons diffracted by a crystal produce rings or maxima, demonstrating wave superposition, while localised detection retains particle behaviour.
  • Increasing electron speed or accelerating voltage raises momentum and reduces wavelength.
  • With fixed crystal spacing, a smaller wavelength produces diffraction maxima at smaller angles.
  • Calculations require potential difference in volts and energy eVeV in joules; the voltage itself is not an energy.
Worked example

An electron is accelerated from rest through 250V250\,\text{V}. Calculate its de Broglie wavelength.

  1. 1.Use λ=h/2meeV\lambda=h/\sqrt{2m_{\mathrm e}eV}.
  2. 2.Substitute h=6.63×1034J sh=6.63\times10^{-34}\,\text{J s}, me=9.11×1031kgm_{\mathrm e}=9.11\times10^{-31}\,\text{kg} and e=1.60×1019Ce=1.60\times10^{-19}\,\text{C}.
  3. 3.λ=(6.63×1034)/2(9.11×1031)(1.60×1019)(250)\lambda=(6.63\times10^{-34})/\sqrt{2(9.11\times10^{-31})(1.60\times10^{-19})(250)}.

Answer: λ=7.76×1011m\lambda=7.76\times10^{-11}\,\text{m}.

Common mistakes

  • Don't use λ=h/v\lambda=h/v and omit the electron mass from momentum.
  • Don't treat potential difference VV as kinetic energy instead of using eVeV.
  • Don't claim increasing electron speed makes the diffraction angle larger despite the reduced wavelength.

Exam tip

For a qualitative pattern change, write the chain higher speed, greater momentum, shorter wavelength, smaller diffraction angle.

Tier 1 · Easy

ORIGINAL

The momentum of electrons incident on a fixed crystal is increased by a factor of 33. State the factor by which their de Broglie wavelength changes and describe the resulting change in the spacing of the diffraction maxima.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

The surface rows of atoms in a crystal act as a diffraction grating of spacing d=0.215nmd=0.215\,\text{nm}. The first-order electron-diffraction maximum is observed at 24.624.6^\circ to the straight-through direction, so that nλ=dsinθn\lambda=d\sin\theta with n=1n=1. Determine the electron wavelength and momentum.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Electrons accelerated through 100V100\,\text{V} produce a first diffraction maximum from a crystal plane at angle 37.837.8^\circ. The accelerating voltage is raised to 400V400\,\text{V}. Calculate the new de Broglie wavelength and, using dsinθ=λd\sin\theta=\lambda, the new angle. Explain the turning-point significance of electron diffraction.

[5 marks]

Total for this question: 5

3.12.2.6

Electron microscopes

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Electron microscopes exploit electron de Broglie wavelengths that can be comparable with atomic dimensions. For non-relativistic electrons accelerated through anode voltage VV, λ=h/2meeV\lambda=h/\sqrt{2m_{\mathrm e}eV}, so the voltage required for an atomic-scale wavelength can be estimated.
  • In a transmission electron microscope, electrons are accelerated, pass through a thin specimen, and electromagnetic lenses form an image from transmitted and scattered electrons.
  • A scanning tunnelling microscope instead moves a sharp conducting tip very close to a conducting surface.
  • Quantum tunnelling current changes steeply with tip–surface separation, allowing surface height or electronic structure to be mapped.
  • The STM's atomic sensitivity is not produced by a TEM-style imaging beam.
A simplified TEM sends electrons through electromagnetic lenses and a thin specimen to form an image.
Worked example

Estimate the anode voltage needed for electrons of wavelength 8.0×1011m8.0\times10^{-11}\,\text{m}.

  1. 1.Rearrange λ=h/2meeV\lambda=h/\sqrt{2m_{\mathrm e}eV} to V=h2/(2meeλ2)V=h^2/(2m_{\mathrm e}e\lambda^2).
  2. 2.Substitute h=6.63×1034h=6.63\times10^{-34}, me=9.11×1031m_{\mathrm e}=9.11\times10^{-31}, e=1.60×1019e=1.60\times10^{-19} and λ=8.0×1011\lambda=8.0\times10^{-11} in SI units.
  3. 3.Evaluate the denominator before dividing by h2h^2.

Answer: V=2.36×102VV=2.36\times10^2\,\text{V}, or about 240V240\,\text{V}.

Common mistakes

  • Don't say a TEM uses glass lenses rather than magnetic fields in electromagnetic lenses.
  • Don't describe an STM as transmitting electrons through a thin specimen.
  • Don't attribute STM atomic resolution to a short imaging wavelength instead of the separation-sensitive tunnelling current.

Exam tip

A compare question should separate TEM transmission and lens imaging from STM surface scanning and tunnelling.

Tier 1 · Easy

ORIGINAL

Explain why increasing the anode voltage of a transmission electron microscope can improve its resolution.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Estimate the anode voltage required to give non-relativistic electrons a wavelength of 0.100nm0.100\,\text{nm}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A TEM is designed for an electron wavelength of 5.00pm5.00\,\text{pm}. Estimate the non-relativistic anode voltage and compare how a TEM and an STM obtain atomic-scale information.

[5 marks]

Total for this question: 5

3.12.3.1

The Michelson-Morley experiment

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A Michelson interferometer sends coherent light to a half-silvered beam splitter, producing two beams that travel along perpendicular arms. Mirrors return the beams to the splitter, where they recombine and form interference fringes.
  • If Earth moved through a stationary ether, an 'ether wind' would make the round-trip light times depend on arm orientation.
  • Rotating the apparatus should then change the phase difference and shift the fringes.
  • Michelson and Morley found no significant periodic shift.
  • This null result did not prove Earth stationary; it failed to detect absolute motion, undermined the stationary-ether model and supported the invariance of the speed of light.
A Michelson interferometer compares round trips along two perpendicular optical arms.
Worked example

Explain why rotating a Michelson interferometer through 9090^\circ was expected to reveal motion through a stationary ether.

  1. 1.The two perpendicular arms were expected to have different round-trip times relative to an ether wind.
  2. 2.A 9090^\circ rotation exchanges the arm parallel to the proposed motion with the perpendicular arm.
  3. 3.The changed travel-time difference should change phase difference and shift the interference fringes.

Answer: The absence of the predicted rotation-dependent fringe shift meant absolute ether motion was not detected.

Common mistakes

  • Don't claim the null result proved that Earth was stationary.
  • Don't describe the apparatus as comparing two unrelated light sources rather than split coherent beams.
  • Don't omit rotation, because this does not explain how the predicted orientation-dependent phase change was tested.

Exam tip

For significance, state both what was predicted—an orientation-dependent fringe shift—and what the null result ruled against.

Tier 1 · Easy

ORIGINAL

State the effect that Michelson and Morley expected to observe if Earth moved through a stationary ether.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Explain why a Michelson interferometer was rotated through 9090^\circ and state the significance of the null result.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

For an interferometer with equal arm length L=11.0mL=11.0\,\text{m} moving at v=3.00×104m s1v=3.00\times10^4\,\text{m s}^{-1} through a proposed ether, use ΔtLv2/c3\Delta t\approx Lv^2/c^3 for the initial difference in round-trip times. Calculate the fringe shift predicted on rotating the apparatus through 9090^\circ for light of wavelength 550nm550\,\text{nm}, using n=2cΔt/λn=2c\Delta t/\lambda. Explain why observing no such shift was decisive.

[5 marks]

Total for this question: 5

3.12.3.2

Einstein's theory of special relativity

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • An inertial frame is non-accelerating: an object with no resultant force moves at constant velocity within it.
  • Einstein's first postulate states that physical laws have the same form in all inertial frames, so no inertial frame is privileged and no experiment within one reveals absolute uniform motion.
  • The second postulate states that the speed of light in free space is invariant: every inertial observer measures cc, independent of the source's or observer's motion.
  • Keeping both postulates requires measured time intervals, lengths and simultaneity to depend on reference frame.
  • Galilean addition cannot be applied to make a measured light speed greater or smaller than cc.
Worked example

A spacecraft moves at constant velocity and emits a forward light pulse. State the light speed measured inside the craft and by an inertial observer outside.

  1. 1.Both the spacecraft frame and the outside observer's frame are inertial.
  2. 2.The laws of physics have the same form in both frames.
  3. 3.The second postulate requires each observer to measure the pulse speed as cc.

Answer: Both observers measure 3.00×108m s13.00\times10^8\,\text{m s}^{-1}, not cc plus the spacecraft speed.

Common mistakes

  • Don't call a rotating or accelerating frame inertial.
  • Don't add the source velocity to cc using Galilean velocity addition.
  • Don't state that all observers measure the same light frequency or wavelength rather than the same free-space speed.

Exam tip

When asked to state the postulates, use the exact ideas same form of physical laws and invariant free-space light speed.

Tier 1 · Easy

ORIGINAL

Define an inertial frame of reference.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

State Einstein's two postulates of special relativity and explain how they account for the Michelson-Morley null result.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Two flashes occur simultaneously at the front and rear of a platform according to an observer at its midpoint. A train moves towards the front flash, and a passenger is at the train's midpoint as the flashes occur. Explain why the passenger does not judge the flashes to be simultaneous and why this follows from Einstein's postulates rather than from light travelling faster from one end.

[5 marks]

Total for this question: 5

3.12.3.3

Time dilation

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Proper time t0t_0 is measured by one clock present at both events, so the events occur at the same position in that clock's frame.
  • An observer for whom the clock moves measures the dilated interval t=t0/1v2/c2=γt0t=t_0/\sqrt{1-v^2/c^2}=\gamma t_0, which is longer than t0t_0.
  • The effect is a consequence of special relativity, not a mechanical fault in the clock.
  • Atmospheric muons provide evidence: their proper mean lifetime is short, but in Earth's frame their moving lifetime is dilated, allowing many more to reach the surface than classical timing predicts.
  • Calculations must first identify which interval is proper and which belongs to the laboratory frame.
Time dilation in Earth's frame allows fast atmospheric muons to travel farther before decaying.
Worked example

A particle has proper lifetime 1.50μs1.50\,\mu\text{s} and travels at 0.900c0.900c. Calculate its mean lifetime in the laboratory.

  1. 1.γ=1/10.9002=2.294\gamma=1/\sqrt{1-0.900^2}=2.294.
  2. 2.The proper lifetime belongs to the particle rest frame.
  3. 3.t=γt0=(2.294)(1.50μs)=3.44μst=\gamma t_0=(2.294)(1.50\,\mu\text{s})=3.44\,\mu\text{s}.

Answer: The laboratory mean lifetime is 3.44μs3.44\,\mu\text{s}.

Common mistakes

  • Don't use the laboratory interval as proper time even though the two events occur at different laboratory positions.
  • Don't divide the proper lifetime by γ\gamma and predict a shorter laboratory lifetime.
  • Don't explain sea-level muons by saying their proper lifetime changes in their own rest frame.

Exam tip

Identify the one-clock frame explicitly before choosing t0t_0 in a time-dilation calculation.

Tier 1 · Easy

ORIGINAL

Define proper time.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A muon has proper mean lifetime 2.20μs2.20\,\mu\text{s} and moves through a laboratory at 0.800c0.800c. Calculate its mean lifetime in the laboratory and the mean distance it travels there.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Muons are created 10.0km10.0\,\text{km} above sea level with speed 0.995c0.995c and proper mean lifetime 2.20μs2.20\,\mu\text{s}. Assuming exponential decay, calculate the fraction that survive to sea level. Compare this with the prediction if time dilation were ignored.

[5 marks]

Total for this question: 5

3.12.3.4

Length contraction

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Proper length l0l_0 is measured in the object's rest frame, where its endpoints have fixed positions. An observer who sees the object moving parallel to its length measures l=l01v2/c2=l0/γl=l_0\sqrt{1-v^2/c^2}=l_0/\gamma, so the moving length is shorter.
  • Only the component parallel to relative motion contracts; transverse dimensions do not.
  • Measuring a moving length requires recording both endpoint positions simultaneously in the observer's frame, linking contraction to the relativity of simultaneity.
  • A laboratory track is therefore its own proper length in the laboratory frame, while a passing particle describes that track as contracted.
  • Examiners expect correct identification of the rest frame before substitution.
The length parallel to motion is shorter than the proper length measured in the object's rest frame.
Worked example

A spacecraft has proper length 80m80\,\text{m} and moves parallel to its length at 0.750c0.750c. Calculate the observed length.

  1. 1.The spacecraft rest frame supplies l0=80ml_0=80\,\text{m}.
  2. 2.Use l=l01v2/c2l=l_0\sqrt{1-v^2/c^2}.
  3. 3.l=8010.7502=52.9ml=80\sqrt{1-0.750^2}=52.9\,\text{m}.

Answer: The observed spacecraft length is 53m53\,\text{m} to two significant figures.

Common mistakes

  • Don't call the moving length ll the proper length.
  • Don't apply contraction to a dimension perpendicular to the relative motion.
  • Don't contract a laboratory-fixed distance in the laboratory frame rather than in the moving object's frame.

Exam tip

Write the rest frame beside l0l_0 before using the contraction formula.

Tier 1 · Easy

ORIGINAL

Define the proper length of an object.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A spacecraft has proper length 120m120\,\text{m} and passes an observer parallel to its length at 0.600c0.600c. Calculate the length measured by the observer.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A particle moves at 0.980c0.980c along a straight accelerator of proper length 3.00km3.00\,\text{km} in the laboratory. Calculate the accelerator length and the transit time in the particle's frame. Show that the result is consistent with the laboratory transit time and time dilation.

[5 marks]

Total for this question: 5

3.12.3.5

Mass and energy

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Mass and energy are equivalent: E=mc2E=mc^2.
  • In the specification's relativistic-mass convention, $m=m_0/\sqrt{1-v2/c2}=\gamma m_0$, so total energy is E=γm0c2E=\gamma m_0c^2 and kinetic energy is K=(γ1)m0c2K=(\gamma-1)m_0c^2.
  • At low speed this approaches 12m0v2\frac12m_0v^2.
  • Graphs of relativistic mass and kinetic energy against speed rise increasingly steeply and tend towards infinity as vv approaches cc, so a massive particle cannot reach cc.
  • Bertozzi independently measured electron kinetic energy and speed: added energy produced ever smaller speed increases near cc, directly supporting the relativistic relation and rejecting the classical prediction of speeds above cc.
Relativistic mass and kinetic energy rise increasingly steeply as speed approaches cc.
Worked example

A particle of rest mass 2.0×1027kg2.0\times10^{-27}\,\text{kg} moves at 0.800c0.800c. Calculate its total energy and kinetic energy.

  1. 1.γ=1/10.8002=1.667\gamma=1/\sqrt{1-0.800^2}=1.667.
  2. 2.E=γm0c2=(1.667)(2.0×1027)(3.00×108)2=3.00×1010JE=\gamma m_0c^2=(1.667)(2.0\times10^{-27})(3.00\times10^8)^2=3.00\times10^{-10}\,\text{J}.
  3. 3.K=(γ1)m0c2=(0.667)(2.0×1027)(3.00×108)2=1.20×1010JK=(\gamma-1)m_0c^2=(0.667)(2.0\times10^{-27})(3.00\times10^8)^2=1.20\times10^{-10}\,\text{J}.

Answer: E=3.0×1010JE=3.0\times10^{-10}\,\text{J} and K=1.2×1010JK=1.2\times10^{-10}\,\text{J}.

Common mistakes

  • Don't use E=m0c2E=m_0c^2 for total energy at non-zero relativistic speed.
  • Don't call total energy γm0c2\gamma m_0c^2 the kinetic energy without subtracting rest energy.
  • Don't draw mass or kinetic energy reaching a finite maximum at v=cv=c.

Exam tip

On a graph question, show both the increasingly steep rise and the asymptotic behaviour as vv approaches cc.

Tier 1 · Easy

ORIGINAL

A reaction reduces the total rest mass of a system by 3.00×1011kg3.00\times10^{-11}\,\text{kg}. Calculate the energy released.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

An electron moves at 0.800c0.800c. Calculate its total energy and kinetic energy. Use m0=9.11×1031kgm_0=9.11\times10^{-31}\,\text{kg}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

In a Bertozzi-type accelerator experiment, an electron has kinetic energy 4.00MeV4.00\,\text{MeV}. Its rest energy is 0.511MeV0.511\,\text{MeV}. Calculate its relativistic speed and the speed predicted by K=12m0v2K=\frac{1}{2}m_0v^2. Explain why the comparison supports special relativity.

[5 marks]

Total for this question: 5

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