Turning points in physics (A-level only)

3.12 Turning points in physics (A-level only) — revision question pack

15 specification points · notes, questions, answers and worked methods

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

Answer all questions in the spaces provided.

3.12.1.1 · Cathode rays

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.The strong electric field ionises atoms of the low-pressure gas.
2.Positive ions accelerate to the negative cathode and release electrons when they strike it.
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

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
State the visible effect produced when a cathode ray reaches a fluorescent screen and explain how the electrons produce it.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
A cathode-ray beam is accelerated through a potential difference and then strikes an earthed metal target. The target becomes warmer. Explain the energy-transfer chain, state what the heating establishes about energy transport, and state one further observation that would be needed to demonstrate momentum transfer.
[3 marks]
Total for this question: 3
A fluorescent screen is placed just beyond the anode aperture of a cathode-ray tube and then moved farther along the tube. A small spot appears on the same line through the aperture at both positions. Explain what these observations show about the ray.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
A metal cross placed in a cathode-ray tube produces a sharp shadow on the glowing glass, and the bright region moves towards a positive plate when an electric field is applied. Explain what these observations establish about cathode rays.
[5 marks]
Total for this question: 5
In a discharge tube, the electric field ionises residual gas atoms; positive ions then accelerate to the cathode and release electrons on impact. Explain why the tube uses gas at low pressure rather than gas at atmospheric pressure or a perfect vacuum.
[4 marks]
Total for this question: 4
A phosphor screen is illuminated first by visible light and then by a cathode-ray beam. Both produce a bright spot. A transverse electric field leaves the light spot fixed but moves the cathode-ray spot towards the positive plate. Evaluate which observation identifies the nature of cathode rays and explain the role of the visible-light trial.
[4 marks]
Total for this question: 4
Two cathode-ray beams pass through the same transverse magnetic field. The beam accelerated through the larger potential difference curves less, although both beams curve in the direction expected for negative charge. Explain how the observations support the particle model of cathode rays.
[4 marks]
Total for this question: 4

3.12.1.2 · Thermionic emission of electrons

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 $V$ , electrical work $eV$ becomes kinetic energy: $eV=\frac12m_{\mathrm e}v^2$ .
This relation assumes negligible energy loss and non-relativistic speed.
Potential difference must be in volts, $e$ in coulombs and $m_{\mathrm e}$ 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 $1.80\,\text{kV}$ . Calculate their speed using $e=1.60\times10^{-19}\,\text{C}$ and $m_{\mathrm e}=9.11\times10^{-31}\,\text{kg}$ .

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

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

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 $eV$ without converting to volts.
Don't use $eV=m_{\mathrm e}v^2$ and omit the factor of one half.

Exam tip

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

Tier 1 · Easy

Explain why heating a metal cathode can cause electrons to leave its surface.
[1 mark]
Total for this question: 1
State the energy equation for an electron accelerated from rest through a potential difference $V$ and state the assumptions it requires.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Electrons emitted from a heated cathode start from rest and are accelerated through $2.50\,\text{kV}$ . Calculate their speed using a non-relativistic model.
[3 marks]
Total for this question: 3
Electrons leave a heated cathode from rest and reach a speed of $3.50\times10^7\,\text{m s}^{-1}$ . Calculate the accelerating potential difference using a non-relativistic model.
[3 marks]
Total for this question: 3
A heated filament continues to glow when the polarity of the anode supply is reversed. Explain why electrons are still emitted but a narrow high-speed beam no longer reaches the anode aperture.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An electron gun accelerates a steady current of $3.00\,\text{mA}$ through $7.50\,\text{kV}$ . 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
An electron emitted from a heated cathode is accelerated from rest through $4.80\,\text{kV}$ . In a collision it loses $30.0\%$ of its kinetic energy. Calculate its speed immediately after the collision and the additional potential difference needed to restore its original kinetic energy. Use a non-relativistic model.
[5 marks]
Total for this question: 5
Electrons start from rest and accelerate uniformly across a $20.0\,\text{mm}$ gap before travelling a further $180\,\text{mm}$ at constant speed. The total flight time is $8.00\,\text{ns}$ . Determine the accelerating potential difference using a non-relativistic model.
[5 marks]
Total for this question: 5
Electrons emitted from a heated cathode are accelerated from rest through $1.84\,\text{kV}$ . The anode potential difference is then raised to $4.21\,\text{kV}$ without changing the filament temperature. Determine the electron momentum at each potential difference and the percentage increase in momentum. Use a non-relativistic model.
[4 marks]
Total for this question: 4
An electron gun has independent supplies for the filament and the accelerating anode. Describe an investigation that distinguishes the effect of increasing filament temperature from the effect of increasing anode potential difference, and predict the observations.
[4 marks]
Total for this question: 4

3.12.1.3 · Specific charge of the electron

Explanation

Specific charge is charge divided by mass; the electron's magnitude is $e/m_{\mathrm e}$ in $\text{C kg}^{-1}$ . In one determination, crossed electric and magnetic fields are adjusted so the beam is undeflected.
Balancing $eE=evB$ gives the selected speed $v=E/B$ .
The electrons then enter a perpendicular magnetic field and follow a circular path because $evB=m_{\mathrm e}v^2/r$ , giving $e/m_{\mathrm e}=v/(Br)$ .
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/m_{\mathrm e}

Worked example

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

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

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

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 $e/m_{\mathrm e}=Br/v$ .
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

An electron beam passes undeflected through crossed fields of strength $E=2.40\times10^4\,\text{V m}^{-1}$ and $B=8.00\times10^{-3}\,\text{T}$ . Calculate the electron speed.
[2 marks]
Total for this question: 2
State the definition and SI unit of specific charge.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Electrons selected at $3.00\times10^6\,\text{m s}^{-1}$ enter a perpendicular magnetic field of $1.80\times10^{-4}\,\text{T}$ and follow a circular path of radius $9.50\,\text{cm}$ . Determine the magnitude of their specific charge.
[3 marks]
Total for this question: 3
An electron beam passes undeflected between plates $6.00\,\text{mm}$ apart with a potential difference of $96.0\,\text{V}$ across them. The electron speed is $3.20\times10^6\,\text{m s}^{-1}$ . Calculate the crossed magnetic flux density.
[3 marks]
Total for this question: 3
In a specific-charge experiment, an electron beam selected by crossed fields has circular-path radius $42.0\,\text{mm}$ in a separate magnetic field. The selector electric field is increased from $18.0$ to $24.0\,\text{kV m}^{-1}$ with its magnetic field unchanged, while the separate deflecting field is increased by a factor of $1.50$ . Determine the new radius.
[3 marks]
Total for this question: 3

Tier 3 · Hard

In a Thomson-style experiment, electrons pass between plates separated by $15.0\,\text{mm}$ with $360\,\text{V}$ across them. A crossed field of $8.00\,\text{mT}$ makes the beam undeflected. The selected beam then follows a circular path of radius $10.7\,\text{mm}$ in a separate $1.60\,\text{mT}$ field. Determine $e/m_{\mathrm e}$ and compare it with the hydrogen-ion specific charge $9.58\times10^7\,\text{C kg}^{-1}$ .
[5 marks]
Total for this question: 5
An electron-beam instrument uses the accepted specific charge $e/m_{\mathrm e}=1.76\times10^{11}\,\text{C kg}^{-1}$ . Electrons accelerated from rest through $240\,\text{V}$ must follow a circular path of radius $18.5\,\text{mm}$ . Calculate the required perpendicular magnetic flux density and determine the new radius if the accelerating potential difference is doubled while this field is unchanged.
[4 marks]
Total for this question: 4
A selector has $182\,\text{V}$ across plates $12.0\,\text{mm}$ apart and magnetic flux density $5.20\,\text{mT}$ . The undeflected electrons then follow a circular path of radius $69.0\,\text{mm}$ in a separate field of $0.240\,\text{mT}$ . Determine the electron specific charge and its absolute uncertainty. The percentage uncertainties in the five measured quantities, in the order given, are $1.0\%$ , $1.0\%$ , $2.0\%$ , $1.5\%$ and $1.5\%$ ; add percentage uncertainties.
[5 marks]
Total for this question: 5
An electron beam takes $120\,\text{ns}$ to cross a field-free distance of $0.375\,\text{m}$ . It then enters a $0.259\,\text{mT}$ magnetic field at right angles and follows a circular path of radius $68.5\,\text{mm}$ . Determine the magnitude of the electron specific charge.
[4 marks]
Total for this question: 4
Inside a $0.394\,\text{mT}$ magnetic region, an electron beam completes $15$ revolutions in $1.36\,\mu\text{s}$ . Show that $e/m_{\mathrm e}=2\pi/(BT)$ , where $T$ is the period of one revolution, and determine the magnitude of the electron specific charge.
[4 marks]
Total for this question: 4

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

Explanation

In Millikan's method, a charged oil droplet lies between oppositely charged parallel plates.
When it is stationary, upward electric force balances weight: $QE=mg$ , with $E=V/d$ , so $QV/d=mg$ .
With the field removed, a falling droplet reaches terminal speed when Stokes' viscous force $6\pi\eta rv$ 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, $e$ , 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 $4.9\times10^{-15}\,\text{kg}$ is stationary between plates $4.0\,\text{mm}$ apart with $750\,\text{V}$ across them. Calculate its charge magnitude.

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

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

Common mistakes

Don't use $Q=mg/V$ and omit the plate separation in $E=V/d$ .
Don't apply $6\pi\eta rv$ 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 $e$ .

Exam tip

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

Tier 1 · Easy

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
Explain why Millikan measured the charges of many oil droplets rather than relying on one droplet.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An oil droplet of mass $6.52\times10^{-15}\,\text{kg}$ is held stationary between plates $3.00\,\text{mm}$ apart with $600\,\text{V}$ across them. Calculate the magnitude of its charge and express the result as a multiple of $e$ .
[3 marks]
Total for this question: 3
An oil droplet of mass $3.75\times10^{-15}\,\text{kg}$ carries charge $3e$ and is stationary between horizontal plates $5.00\,\text{mm}$ apart. Calculate the required potential difference.
[3 marks]
Total for this question: 3
Three droplet charges are measured as $4.80\times10^{-19}\,\text{C}$ , $8.03\times10^{-19}\,\text{C}$ and $11.18\times10^{-19}\,\text{C}$ . An independent estimate places $e$ between $1.5\times10^{-19}\,\text{C}$ and $1.7\times10^{-19}\,\text{C}$ . Determine the best estimate of $e$ from these data.
[3 marks]
Total for this question: 3

Tier 3 · Hard

With the electric field off, an oil droplet of density $850\,\text{kg m}^{-3}$ falls through air at terminal speed $8.00\times10^{-5}\,\text{m s}^{-1}$ . The air viscosity is $1.80\times10^{-5}\,\text{Pa s}$ . Neglect air buoyancy. The droplet is then held stationary by a field of $7.50\times10^4\,\text{V m}^{-1}$ . Use Stokes' law to determine the droplet charge and show how the result supports charge quantisation.
[5 marks]
Total for this question: 5
An oil droplet of radius $0.900\,\mu\text{m}$ falls at terminal speed $8.00\times10^{-5}\,\text{m s}^{-1}$ with no electric field. In an upward electric force it rises at terminal speed $2.00\times10^{-4}\,\text{m s}^{-1}$ . The field strength is $1.37\times10^5\,\text{V m}^{-1}$ and the air viscosity is $1.84\times10^{-5}\,\text{Pa s}$ . Neglect air buoyancy. Determine the droplet charge and express it as a multiple of $e$ .
[5 marks]
Total for this question: 5
A negatively charged oil droplet of mass $7.83\times10^{-15}\,\text{kg}$ is stationary between plates $5.00\,\text{mm}$ apart at $600\,\text{V}$ . After gaining one electron, it is stationary at $480\,\text{V}$ with the field direction unchanged. Determine the elementary charge.
[5 marks]
Total for this question: 5
A spherical oil droplet of radius $1.20\,\mu\text{m}$ and density $870\,\text{kg m}^{-3}$ is held stationary in air of density $1.20\,\text{kg m}^{-3}$ by an electric field of strength $1.29\times10^5\,\text{V m}^{-1}$ . Determine its charge magnitude, express it as a multiple of $e$ , and calculate the percentage by which the charge would be overestimated if air buoyancy were neglected.
[5 marks]
Total for this question: 5
Two oil droplets of the same density fall through the same air at terminal speeds $5.80\times10^{-5}\,\text{m s}^{-1}$ and $1.31\times10^{-4}\,\text{m s}^{-1}$ . A student uses the mass calculated for the slower droplet when applying $Q=mg/E$ to both droplets. Determine the ratio of the actual droplet masses, explain why the student's analysis can hide charge quantisation, and state the correction required.
[4 marks]
Total for this question: 4

3.12.2.1 · Newton's corpuscular theory of light

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.Newton's boundary-force account predicted that corpuscles speed up in glass.
2.Huygens' wave construction predicted that light slows in glass.
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

State one observation that made Newton's corpuscular model of light appear plausible.
[1 mark]
Total for this question: 1
State how Newton's corpuscular theory represented a ray of light.
[1 mark]
Total for this question: 1

Tier 2 · Standard

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
Explain why the formation of sharp shadows initially favoured Newton's corpuscular theory over Huygens' wave theory.
[3 marks]
Total for this question: 3
Describe Huygens' wave model of light in general terms.
[2 marks]
Total for this question: 2

Tier 3 · Hard

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
A narrow light beam reflects from a plane surface with its angle of reflection equal to its angle of incidence. Compare how Newton's corpuscular model and Huygens' wave model account for this observation, and explain why reflection alone could not decide between the models.
[5 marks]
Total for this question: 5

3.12.2.2 · Significance of Young's double slits experiment

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.Both slits send diffracted waves to the dark-fringe position.
2.The waves arrive in antiphase, so their displacements cancel by destructive superposition.
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

State the feature of Young's double-slit pattern that provides evidence for interference.
[1 mark]
Total for this question: 1
State the relationship between the waves from the two slits when both are illuminated by the same monochromatic source.
[1 mark]
Total for this question: 1

Tier 2 · Standard

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
One slit in Young's experiment is covered and the alternating bright and dark fringes disappear. Explain this observation.
[3 marks]
Total for this question: 3
Two separate lasers have the same wavelength but illuminate the two slits independently. Explain why a stable Young fringe pattern is not observed.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Explain why Young's interference evidence did not immediately displace Newton's corpuscular theory despite the appearance of dark fringes.
[5 marks]
Total for this question: 5
With either slit covered, Young observes light at a point that becomes completely dark when both slits are open. Explain what the covered-slit observations control for and state the phase and amplitude conditions required for complete darkness.
[4 marks]
Total for this question: 4
Young's two slits are made progressively wider while their centres and the light source remain unchanged. Explain why the region containing visible two-slit fringes becomes narrower and why this observation supports a wave account of light.
[4 marks]
Total for this question: 4
Young repeats a double-slit observation with blue light after first using red light, leaving the slit geometry and screen position unchanged. The blue bright and dark bands are closer together, but covering either slit removes the alternating sequence for either colour. Explain qualitatively how these comparisons support a wave model of light.
[4 marks]
Total for this question: 4

3.12.2.3 · Electromagnetic waves

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 $c=1/\sqrt{\mu_0\varepsilon_0}$ .
The permittivity $\varepsilon_0$ relates to electric-field strength produced by charge in free space, while permeability $\mu_0$ 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 $c$ 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 $\mu_0=1.26\times10^{-6}\,\text{H m}^{-1}$ and $\varepsilon_0=8.85\times10^{-12}\,\text{F m}^{-1}$ .

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

Answer: $c=3.00\times10^8\,\text{m s}^{-1}$ , 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 $\varepsilon_0$ and $\mu_0$ 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

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
State the physical quantities characterised by $\varepsilon_0$ and $\mu_0$ in free space.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Use $\mu_0=4\pi\times10^{-7}\,\text{H m}^{-1}$ and $\varepsilon_0=8.85\times10^{-12}\,\text{F m}^{-1}$ to calculate the speed predicted by Maxwell for an electromagnetic wave in a vacuum.
[3 marks]
Total for this question: 3
Given $c=3.00\times10^8\,\text{m s}^{-1}$ and $\mu_0=4\pi\times10^{-7}\,\text{H m}^{-1}$ , determine $\varepsilon_0$ from Maxwell's formula.
[3 marks]
Total for this question: 3
Maxwell's constants give an electromagnetic-wave speed of $2.998\times10^8\,\text{m s}^{-1}$ , while an optical experiment gives $3.000\times10^8\,\text{m s}^{-1}$ . Calculate the percentage difference, explain its significance and state why agreement in speed alone was not conclusive.
[3 marks]
Total for this question: 3

Tier 3 · Hard

In a Fizeau-type experiment, light travels to a mirror $8.63\,\text{km}$ away and back through a wheel with $720$ teeth. The first extinction occurs at $12.1\,\text{Hz}$ 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
Hertz produces standing radio waves of frequency $200\,\text{MHz}$ and measures $0.750\,\text{m}$ between adjacent nodes. Calculate the speed of these waves and explain the significance of the result for Maxwell's theory.
[5 marks]
Total for this question: 5
Measurements give $\mu_0=(1.257\times10^{-6}\pm0.2\%)\,\text{H m}^{-1}$ and $\varepsilon_0=(8.85\times10^{-12}\pm0.4\%)\,\text{F m}^{-1}$ . Determine Maxwell's wave speed with its absolute uncertainty and explain whether it agrees with a Hertz measurement of $(3.02\pm0.03)\times10^8\,\text{m s}^{-1}$ . Add percentage uncertainties before applying the power.
[5 marks]
Total for this question: 5
A student claims that the agreement between Maxwell's predicted electromagnetic-wave speed and the measured speed of light was circular because $\mu_0$ and $\varepsilon_0$ had been chosen using optical measurements. Evaluate the claim and explain why Hertz's later work strengthened Maxwell's conclusion.
[4 marks]
Total for this question: 4
In a Hertz experiment, rotating the receiving loop through $90^\circ$ greatly reduces the signal. A metal reflector produces fixed nodes, and the measured node separation and transmitter frequency give a wave speed close to the measured speed of light. Explain how the combined observations support Maxwell's electromagnetic theory.
[5 marks]
Total for this question: 5

3.12.2.4 · The discovery of photoelectricity

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 $E=hf$ , 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 $hf=\phi+K_{\max}$ .
Above threshold, greater intensity supplies more photons and raises the emission rate, but at fixed frequency it does not raise $K_{\max}$ .
This established a particle aspect of electromagnetic radiation.

One photon transfers energy

hf

to one electron, which escapes with maximum energy

hf-\phi

Worked example

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

1.Use Einstein's equation $hf=\phi+K_{\max}$ .
2. $K_{\max}=4.1-2.4=1.7\,\text{eV}$ .
3.Convert if required: $1.7(1.60\times10^{-19})=2.72\times10^{-19}\,\text{J}$ .

Answer: $K_{\max}=1.7\,\text{eV}=2.7\times10^{-19}\,\text{J}$ .

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

State one photoelectric observation that classical wave theory could not explain.
[1 mark]
Total for this question: 1
State the physical endpoint measured by the stopping potential in a photoelectric experiment and relate its magnitude to the maximum photoelectron kinetic energy.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain how Planck's quantum hypothesis avoided the ultraviolet catastrophe in the black-body spectrum.
[3 marks]
Total for this question: 3
A metal has work function $2.35\,\text{eV}$ . It is illuminated separately by monochromatic red, green, violet and ultraviolet beams of frequencies $4.70\times10^{14}$ , $5.45\times10^{14}$ , $7.20\times10^{14}$ and $1.10\times10^{15}\,\text{Hz}$ respectively. Decide which beams can cause photoelectron emission.
[3 marks]
Total for this question: 3
Below a particular frequency, increasing light intensity does not produce photoelectrons. Just above this frequency, emission begins without measurable delay even at low intensity. Explain both observations using photons.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Light of irradiance $2.50\times10^{-6}\,\text{W m}^{-2}$ falls on a metal whose work function is $2.30\,\text{eV}$ . In a classical continuous-wave model, suppose one electron accumulates all the energy incident on a surface patch of area $1.80\times10^{-20}\,\text{m}^2$ . 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
The fastest photoelectrons from a metal move directly against a uniform opposing electric field of magnitude $480\,\text{V m}^{-1}$ and are brought to rest after $7.20\,\text{mm}$ . The metal work function is $2.25\,\text{eV}$ . Determine the maximum photoelectron kinetic energy and the frequency of the incident monochromatic radiation. Ignore collisions.
[5 marks]
Total for this question: 5
The same monochromatic beam of frequency $1.50\times10^{15}\,\text{Hz}$ illuminates potassium and copper surfaces with work functions $2.30\,\text{eV}$ and $4.70\,\text{eV}$ respectively. Determine which surfaces emit photoelectrons and compare the maximum photoelectron momenta. Use a non-relativistic model.
[5 marks]
Total for this question: 5
Light of wavelength $420\,\text{nm}$ gives a stopping potential of $0.720\,\text{V}$ for a metal. Use this result to determine the metal work function. The same surface is then illuminated with light of wavelength $310\,\text{nm}$ . Calculate the new stopping potential.
[5 marks]
Total for this question: 5
A black body is at $1800\,\text{K}$ , for which $kT=2.49\times10^{-20}\,\text{J}$ . Calculate $hf/(kT)$ for frequencies $3.00\times10^{14}\,\text{Hz}$ and $1.20\times10^{15}\,\text{Hz}$ . Use the results to explain how Planck's hypothesis avoids the ultraviolet catastrophe.
[4 marks]
Total for this question: 4

3.12.2.5 · Wave-particle duality

Explanation

de Broglie's hypothesis assigns wavelength $\lambda=h/p$ to a particle of momentum $p$ . For a non-relativistic electron accelerated from rest through potential difference $V$ , $eV=p^2/(2m_{\mathrm e})$ , so $\lambda=h/\sqrt{2m_{\mathrm e}eV}$ .
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 $eV$ in joules; the voltage itself is not an energy.

Worked example

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

1.Use $\lambda=h/\sqrt{2m_{\mathrm e}eV}$ .
2.Substitute $h=6.63\times10^{-34}\,\text{J s}$ , $m_{\mathrm e}=9.11\times10^{-31}\,\text{kg}$ and $e=1.60\times10^{-19}\,\text{C}$ .
3. $\lambda=(6.63\times10^{-34})/\sqrt{2(9.11\times10^{-31})(1.60\times10^{-19})(250)}$ .

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

Common mistakes

Don't use $\lambda=h/v$ and omit the electron mass from momentum.
Don't treat potential difference $V$ as kinetic energy instead of using $eV$ .
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

The momentum of electrons incident on a fixed crystal is increased by a factor of $3$ . 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
State de Broglie's relationship between a particle's momentum and wavelength.
[1 mark]
Total for this question: 1

Tier 2 · Standard

The surface rows of atoms in a crystal act as a diffraction grating of spacing $d=0.215\,\text{nm}$ . The first-order electron-diffraction maximum is observed at $24.6^\circ$ to the straight-through direction, so that $n\lambda=d\sin\theta$ with $n=1$ . Determine the electron wavelength and momentum.
[3 marks]
Total for this question: 3
Electrons accelerated from rest have a de Broglie wavelength of $0.204\,\text{nm}$ . Calculate the accelerating potential difference using a non-relativistic model.
[3 marks]
Total for this question: 3
Alpha particles of mass $6.64\times10^{-27}\,\text{kg}$ travel at $1.20\times10^7\,\text{m s}^{-1}$ . Calculate their de Broglie wavelength and compare it with an atomic spacing of $0.250\,\text{nm}$ to judge whether diffraction would be observable.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Electrons accelerated through $100\,\text{V}$ produce a first diffraction maximum from a crystal plane at angle $37.8^\circ$ . The accelerating voltage is raised to $400\,\text{V}$ . Calculate the new de Broglie wavelength and, using $d\sin\theta=\lambda$ , the new angle. Explain the turning-point significance of electron diffraction.
[5 marks]
Total for this question: 5
At small diffraction angles, electrons accelerated through $200\,\text{V}$ form a ring of diameter $18.0\,\text{mm}$ . A second accelerating voltage produces the corresponding ring with diameter $12.0\,\text{mm}$ . Assume that ring diameter is proportional to electron wavelength. Determine the second voltage.
[4 marks]
Total for this question: 4
An $8.00\,\text{keV}$ X-ray photon and a non-relativistic electron produce a diffraction maximum at the same angle from the same crystal planes and in the same order. Determine the accelerating potential difference of the electron.
[5 marks]
Total for this question: 5
An electron-diffraction experiment is run at such a low beam current that the screen records one isolated impact at a time. The first impacts appear irregularly placed, but after many electrons a stable ring pattern emerges. Explain what each stage shows and why the completed observation is evidence for wave-particle duality.
[4 marks]
Total for this question: 4
An electron beam and a neutron beam are adjusted to have the same momentum before passing separately through the same crystal. Predict how the positions of their diffraction maxima compare and explain what this comparison would test about de Broglie's hypothesis.
[4 marks]
Total for this question: 4

3.12.2.6 · Electron microscopes

Explanation

Electron microscopes exploit electron de Broglie wavelengths that can be comparable with atomic dimensions. For non-relativistic electrons accelerated through anode voltage $V$ , $\lambda=h/\sqrt{2m_{\mathrm e}eV}$ , 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 $8.0\times10^{-11}\,\text{m}$ .

1.Rearrange $\lambda=h/\sqrt{2m_{\mathrm e}eV}$ to $V=h^2/(2m_{\mathrm e}e\lambda^2)$ .
2.Substitute $h=6.63\times10^{-34}$ , $m_{\mathrm e}=9.11\times10^{-31}$ , $e=1.60\times10^{-19}$ and $\lambda=8.0\times10^{-11}$ in SI units.
3.Evaluate the denominator before dividing by $h^2$ .

Answer: $V=2.36\times10^2\,\text{V}$ , or about $240\,\text{V}$ .

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

Explain why increasing the anode voltage of a transmission electron microscope can improve its resolution.
[2 marks]
Total for this question: 2
Explain why a specimen used in a transmission electron microscope must be very thin.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Estimate the anode voltage required to give non-relativistic electrons a wavelength of $0.100\,\text{nm}$ .
[3 marks]
Total for this question: 3
Electrons accelerated from rest through an anode potential difference $V$ have a de Broglie wavelength of $8.00\,\text{pm}$ in an electron microscope. Estimate $V$ using $p=h/\lambda$ , $K=p^2/(2m_{\mathrm e})$ and $V=K/e$ . Then state why this non-relativistic result is only an estimate. Use $h=6.63\times10^{-34}\,\text{J s}$ , $m_{\mathrm e}=9.11\times10^{-31}\,\text{kg}$ and $e=1.60\times10^{-19}\,\text{C}$ .
[3 marks]
Total for this question: 3
As part of the principle of operation of a transmission electron microscope, explain why it uses electromagnetic lenses rather than glass lenses and why the electron column must be evacuated.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A TEM is designed for an electron wavelength of $5.00\,\text{pm}$ . 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
Explain how quantum tunnelling allows a scanning tunnelling microscope to resolve individual surface atoms and why the sample must conduct electricity.
[5 marks]
Total for this question: 5
In constant-height STM operation the tunnelling current changes from $2.00\,\text{nA}$ to $12.0\,\text{nA}$ . Use $I\propto\exp(-2\kappa s)$ with $\kappa=1.00\times10^{10}\,\text{m}^{-1}$ to determine the change in tip–surface separation and explain why such a small change is detectable.
[3 marks]
Total for this question: 3
Two otherwise identical transmission electron microscopes use anode voltages of $18.0\,\text{kV}$ and $72.0\,\text{kV}$ . Determine the ratio of their electron wavelengths and the ideal relative wavelength-limited resolution. Use a non-relativistic model.
[3 marks]
Total for this question: 3
A technician must scan a rough, previously unexamined surface with a scanning tunnelling microscope and can select constant-current or constant-height mode. Justify the choice of constant-current mode for this surface.
[3 marks]
Total for this question: 3

3.12.3.1 · The Michelson-Morley experiment

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 $90^\circ$ was expected to reveal motion through a stationary ether.

1.The two perpendicular arms were expected to have different round-trip times relative to an ether wind.
2.A $90^\circ$ rotation exchanges the arm parallel to the proposed motion with the perpendicular arm.
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

State the effect that Michelson and Morley expected to observe if Earth moved through a stationary ether.
[1 mark]
Total for this question: 1
State the purpose of the half-silvered plate in the Michelson-Morley interferometer.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Explain why a Michelson interferometer was rotated through $90^\circ$ and state the significance of the null result.
[3 marks]
Total for this question: 3
Rotating a Michelson interferometer changes its fringe position by less than $0.010$ fringe when light of wavelength $600\,\text{nm}$ is used. One complete fringe shift corresponds to an optical path change of one wavelength. Calculate the upper limit on the optical path change and compare it with an ether model's prediction of a $0.400$ -fringe shift.
[3 marks]
Total for this question: 3
Explain why the two beams in a Michelson interferometer must be coherent and why the two arm lengths are made nearly equal.
[2 marks]
Total for this question: 2

Tier 3 · Hard

For an interferometer with equal arm length $L=11.0\,\text{m}$ moving at $v=3.00\times10^4\,\text{m s}^{-1}$ through a proposed ether, use $\Delta t\approx Lv^2/c^3$ for the initial difference in round-trip times. Calculate the fringe shift predicted on rotating the apparatus through $90^\circ$ for light of wavelength $550\,\text{nm}$ , using $n=2c\Delta t/\lambda$ . Explain why observing no such shift was decisive.
[5 marks]
Total for this question: 5
A student claims that the Michelson-Morley null result merely showed that Earth happened to be stationary in the ether during the experiment. Explain why this claim is not supported.
[5 marks]
Total for this question: 5
A Michelson-Morley apparatus gives no detectable fringe shift. Before accepting this as a null result, researchers impose a known small displacement on one mirror, enclose the optical paths and repeat rotations in both directions. Explain how these procedures strengthen the conclusion.
[4 marks]
Total for this question: 4
An interferometer has equal arms of length $8.40\,\text{m}$ and uses light of wavelength $520\,\text{nm}$ . Rotation produces no periodic shift greater than $0.012$ fringe. Using $n=2Lv^2/(c^2\lambda)$ for the ether prediction, determine the upper limit on Earth's speed through the proposed ether.
[5 marks]
Total for this question: 5
One $9.20\,\text{m}$ arm of a Michelson interferometer is warmer than the perpendicular arm by $0.015\,\text{K}$ . The arm material has linear expansion coefficient $1.10\times10^{-6}\,\text{K}^{-1}$ and the light wavelength is $610\,\text{nm}$ . Use $\Delta L=\alpha L\Delta T$ . Calculate the resulting fringe shift and evaluate its importance when the predicted ether shift is $0.18$ fringe.
[5 marks]
Total for this question: 5

3.12.3.2 · Einstein's theory of special relativity

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 $c$ , 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 $c$ .

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

Answer: Both observers measure $3.00\times10^8\,\text{m s}^{-1}$ , not $c$ plus the spacecraft speed.

Common mistakes

Don't call a rotating or accelerating frame inertial.
Don't add the source velocity to $c$ 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

Define an inertial frame of reference.
[1 mark]
Total for this question: 1
State Einstein's second postulate of special relativity.
[1 mark]
Total for this question: 1

Tier 2 · Standard

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
A spacecraft travelling at $0.720c$ relative to Earth sends a light pulse forwards. Calculate the pulse speed measured by an observer on Earth and explain why Galilean addition cannot be used.
[3 marks]
Total for this question: 3
Explain why experiments performed entirely inside a windowless spacecraft moving at constant velocity cannot determine the spacecraft's absolute velocity.
[2 marks]
Total for this question: 2

Tier 3 · Hard

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
Spacecraft A coasts past Earth in a straight line at constant velocity. Spacecraft B has the same instantaneous speed but follows a circular orbit. Identify which spacecraft defines an inertial frame and explain why equal speed does not make both frames inertial.
[4 marks]
Total for this question: 4
Two inertial observers measure the same light pulse to have frequencies $6.00\times10^{14}\,\text{Hz}$ and $8.00\times10^{14}\,\text{Hz}$ . Determine both wavelengths and use the results to evaluate the claim that invariant light speed requires frequency or wavelength to be invariant.
[3 marks]
Total for this question: 3
A student argues that Einstein's first postulate alone guarantees that every inertial observer measures the same speed of light. Evaluate the argument and explain the distinct role of each postulate.
[4 marks]
Total for this question: 4
A source moving at $0.650c$ relative to Earth emits light forwards and backwards. Galilean addition predicts Earth-frame speeds of $1.650c$ and $0.350c$ , whereas measurements give $c$ in both directions within experimental uncertainty. Evaluate the two predictions using Einstein's postulates.
[4 marks]
Total for this question: 4

3.12.3.3 · Time dilation

Explanation

Proper time $t_0$ 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 $t=t_0/\sqrt{1-v^2/c^2}=\gamma t_0$ , which is longer than $t_0$ .
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 $1.50\,\mu\text{s}$ and travels at $0.900c$ . Calculate its mean lifetime in the laboratory.

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

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

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 $\gamma$ 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 $t_0$ in a time-dilation calculation.

Tier 1 · Easy

Define proper time.
[1 mark]
Total for this question: 1
A particle is created and later decays while at rest relative to one clock. State which frame measures the proper lifetime.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A muon has proper mean lifetime $2.20\,\mu\text{s}$ and moves through a laboratory at $0.800c$ . Calculate its mean lifetime in the laboratory and the mean distance it travels there.
[3 marks]
Total for this question: 3
An unstable particle has proper lifetime $4.50\,\mu\text{s}$ and laboratory lifetime $6.00\,\mu\text{s}$ . Calculate its speed.
[3 marks]
Total for this question: 3
A clock fixed inside a spacecraft records $18.0\,\text{min}$ between two signals emitted at its position. The spacecraft moves at $0.600c$ relative to Earth. Calculate the interval measured on Earth and identify the proper interval.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Muons are created $10.0\,\text{km}$ above sea level with speed $0.995c$ and proper mean lifetime $2.20\,\mu\text{s}$ . 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
A destination is $9.00$ light-years from Earth in the Earth frame. The journey takes $6.00$ years according to a clock on a spacecraft travelling at constant speed. Determine the spacecraft speed and the journey time measured on Earth.
[4 marks]
Total for this question: 4
A spacecraft travels from Earth to a star $4.00$ light-years away at $0.800c$ and immediately returns at the same speed. Determine the elapsed times for an Earth clock and the spacecraft clock. For each constant-speed leg, identify the proper time used in the time-dilation calculation and justify your choice.
[4 marks]
Total for this question: 4
A spacecraft spends $3.00\,\text{y}$ moving at $0.600c$ relative to Earth and then $2.00\,\text{y}$ moving at $0.800c$ , where both durations are measured on Earth. Ignore the brief change of speed. Determine the elapsed time on the spacecraft clock and compare it with the Earth interval.
[4 marks]
Total for this question: 4
Unstable particles with proper mean lifetime $2.20\,\mu\text{s}$ travel $1.20\,\text{km}$ through a laboratory before reaching a detector. The surviving fraction is $0.250$ . Assuming exponential decay and constant speed, determine the particle speed.
[5 marks]
Total for this question: 5

3.12.3.4 · Length contraction

Explanation

Proper length $l_0$ 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 $l=l_0\sqrt{1-v^2/c^2}=l_0/\gamma$ , 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 $80\,\text{m}$ and moves parallel to its length at $0.750c$ . Calculate the observed length.

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

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

Common mistakes

Don't call the moving length $l$ 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 $l_0$ before using the contraction formula.

Tier 1 · Easy

Define the proper length of an object.
[1 mark]
Total for this question: 1
State which dimension of a moving object undergoes length contraction.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A spacecraft has proper length $120\,\text{m}$ and passes an observer parallel to its length at $0.600c$ . Calculate the length measured by the observer.
[3 marks]
Total for this question: 3
A rod has proper length $75.0\,\text{m}$ . An observer, moving parallel to the rod's length, measures its length as $63.0\,\text{m}$ . Calculate the speed of the observer relative to the rod.
[3 marks]
Total for this question: 3
A student records the front position of a moving rod and, a short time later, records its rear position. Explain why subtracting these positions does not give the rod's length in the student's frame and state the required procedure.
[2 marks]
Total for this question: 2

Tier 3 · Hard

A particle moves at $0.980c$ along a straight accelerator of proper length $3.00\,\text{km}$ 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
A rectangular panel has proper dimensions $8.00\,\text{m}$ by $3.00\,\text{m}$ and moves at $0.640c$ parallel to its longer side relative to a laboratory observer. Determine its dimensions and area measured in the laboratory frame.
[4 marks]
Total for this question: 4
A pole has proper length $12.0\,\text{m}$ and moves at $0.800c$ relative to a barn of proper length $9.00\,\text{m}$ . Calculate the pole length in the barn frame and the barn length in the pole frame. In each calculation identify the proper length.
[4 marks]
Total for this question: 4
In its rest frame, a straight rod has a component $6.40\,\text{m}$ parallel to its direction of motion. It passes a laboratory at $0.750c$ . Determine the parallel component measured in the laboratory frame.
[3 marks]
Total for this question: 3
Markers fixed along a moving survey belt are separated by $2.75\,\text{m}$ in the belt's rest frame. The belt passes a laboratory at $0.860c$ . Determine the marker separation and the number of marker intervals per metre measured simultaneously in the laboratory frame.
[4 marks]
Total for this question: 4

3.12.3.5 · Mass and energy

Explanation

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

Relativistic mass and kinetic energy rise increasingly steeply as speed approaches

c

Worked example

A particle of rest mass $2.0\times10^{-27}\,\text{kg}$ moves at $0.800c$ . Calculate its total energy and kinetic energy.

1. $\gamma=1/\sqrt{1-0.800^2}=1.667$ .
2. $E=\gamma m_0c^2=(1.667)(2.0\times10^{-27})(3.00\times10^8)^2=3.00\times10^{-10}\,\text{J}$ .
3. $K=(\gamma-1)m_0c^2=(0.667)(2.0\times10^{-27})(3.00\times10^8)^2=1.20\times10^{-10}\,\text{J}$ .

Answer: $E=3.0\times10^{-10}\,\text{J}$ and $K=1.2\times10^{-10}\,\text{J}$ .

Common mistakes

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

Exam tip

On a graph question, show both the increasingly steep rise and the asymptotic behaviour as $v$ approaches $c$ .

Tier 1 · Easy

A reaction reduces the total rest mass of a system by $3.00\times10^{-11}\,\text{kg}$ . Calculate the energy released.
[2 marks]
Total for this question: 2
Calculate the mass equivalent of $450\,\text{kJ}$ of energy.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An electron moves at $0.800c$ . Calculate its total energy and kinetic energy. Use $m_0=9.11\times10^{-31}\,\text{kg}$ .
[3 marks]
Total for this question: 3
Sketch the variation of relativistic mass and kinetic energy with speed from rest towards $c$ .
[3 marks]
Total for this question: 3
A closed $12.0\,\text{kg}$ battery stores an additional $7.20\,\text{MJ}$ when charged. Calculate its mass increase and fractional mass increase, then decide whether a balance with fractional resolution $1.0\times10^{-10}$ could detect the change.
[3 marks]
Total for this question: 3

Tier 3 · Hard

In a Bertozzi-type accelerator experiment, an electron has kinetic energy $4.00\,\text{MeV}$ . Its rest energy is $0.511\,\text{MeV}$ . Calculate its relativistic speed and the speed predicted by $K=\frac{1}{2}m_0v^2$ . Explain why the comparison supports special relativity.
[5 marks]
Total for this question: 5
In a Bertozzi-type measurement an electron travels at $0.970c$ . Determine its relativistic kinetic energy, the classical kinetic-energy prediction and their ratio.
[5 marks]
Total for this question: 5
Two identical particles, each of rest mass $1.00\times10^{-27}\,\text{kg}$ and each moving at $0.600c$ in the laboratory frame, approach each other in opposite directions. They combine to form one stationary particle and no energy leaves the system. Determine the rest mass of the new particle and explain its increase over the sum of the original rest masses.
[5 marks]
Total for this question: 5
An accelerator transfers $46.0\,\text{W}$ to a steady electron beam of current $25.0\,\mu\text{A}$ . Determine the kinetic energy per electron, the electron speed and its relativistic mass. Use the electron rest energy $0.511\,\text{MeV}$ and rest mass $9.11\times10^{-31}\,\text{kg}$ .
[5 marks]
Total for this question: 5
Calculate the energy needed to increase an electron's speed from $0.900c$ to $0.990c$ using relativistic kinetic energy. Compare it with the classical prediction for the same speed change and explain the difference. Use the electron rest energy $0.511\,\text{MeV}$ .
[5 marks]
Total for this question: 5

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

3.12.1.1 · Cathode rays

Tier 1 · Easy

Mark scheme for 3.12.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	It travels from the cathode towards the anode and consists of electrons.	The cathode is the negative electrode. The ray leaves this electrode and moves towards the positive anode; its field deflections identify its particles as negatively charged electrons.	2
02.1	The screen glows at the point where the ray strikes it because the incident electrons transfer energy to the fluorescent material, which then emits visible light.	Energetic electrons transfer energy to the fluorescent material. The material then emits visible light, producing a bright spot where the beam arrives.	2

Tier 2 · Standard

Mark scheme for 3.12.1.1 Tier 2 · Standard
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01.1	The field ionises the low-pressure gas; positive ions accelerate to the cathode and release electrons, which accelerate towards the anode as a cathode ray.	The large potential difference creates a strong electric field in the rarefied gas. Collisions ionise gas atoms. Positive ions are accelerated towards the negative cathode and their impacts release electrons. These electrons are accelerated away from the cathode towards the anode, producing the cathode ray.	3
02.1	The electric field does work on the electrons, giving them kinetic energy. Collisions transfer this energy to the target's internal energy, so it warms; the heating therefore shows that the beam carries energy. A separate mechanical observation, such as a measurable force or recoil of the target, would be needed to demonstrate momentum transfer.	Electrical work on the electrons becomes directed electron kinetic energy. Collisions in the target randomise that energy into lattice and electron motion, increasing the target's internal energy and temperature. This heating directly establishes energy transport by the cathode-ray beam. It does not by itself demonstrate momentum transfer; that requires an additional mechanical effect, for example a measurable force on or recoil of the target.	3
03.1	The aperture selects electrons passing through a small region. A small spot at each position shows that the transmitted electrons remain a narrow beam, and the aligned spots show that the beam travels along an approximately straight path.	The anode blocks electrons outside its aperture. Each localised spot on the moved screen shows that the electrons have not spread widely, while the two recorded positions establish a common straight direction beyond the anode. Credit selection, narrowness and the inference from the aligned observations.	3

Tier 3 · Hard

Mark scheme for 3.12.1.1 Tier 3 · Hard
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01.1	Cathode rays are identical negatively charged particles found in every material, so atoms contain smaller constituents and are not indivisible.	Deflection towards the positive plate shows that the beam carries negative charge. Reversal when the magnetic field reverses is consistent with moving charged particles rather than neutral radiation. The same measured behaviour for different gases and cathodes shows that the particles are universal rather than atoms of one material. Thomson identified them as electrons. Because electrons are emitted from atoms of every substance, the evidence overturned the idea that atoms were indivisible.	5
02.1	The rays travel approximately in straight lines, cause fluorescence by striking the glass, and consist of negatively charged particles.	The cross blocks part of the beam, so the sharp shadow shows that the rays propagate approximately in straight lines from the cathode. The glow outside the shadow shows that impacts by the ray transfer energy to the glass and cause fluorescence. Deflection towards a positive plate establishes that the ray carries negative charge. Together with its straight path and localised impacts, this supports a beam of moving electrons rather than neutral light.	5
03.1	Some residual gas is needed for the stated ionisation route, so a perfect vacuum would provide no gas ions. The low density gives ions and electrons long mean free paths, reducing collisions, scattering and energy loss so ions can reach the cathode energetically and electrons can reach the screen as a directed beam. At atmospheric pressure frequent collisions would disrupt this process.	Credit only the pressure trade-off, since the production chain is given. Residual gas is the material required by that chain, whereas a perfect vacuum supplies none. Reducing the gas density increases mean free path, so charged particles undergo fewer energy-losing and scattering collisions. Ions can therefore arrive at the cathode with sufficient energy and the released electrons can retain a directed path to the screen; atmospheric-pressure gas would cause frequent disruptive collisions.	4
04.1	The glow alone shows energy transfer but does not identify a charged particle, because visible light also makes the screen glow. Deflection towards the positive plate shows that the cathode ray carries negative charge and therefore consists of moving electrons. The unchanged light spot is a control showing that the field does not move the screen image itself and that neutral light is not deflected.	Separate the common observation from the discriminating one. Fluorescence occurs after either source transfers energy to the phosphor, so it cannot by itself distinguish particles from electromagnetic radiation. Only the cathode-ray spot responds to the electric field; its direction gives a negative charge sign and hence supports a beam of electrons. Repeating the field trial with visible light controls for movement of the apparatus or screen response and provides the neutral-radiation comparison.	4
05.1	Magnetic deflection shows that each beam contains moving charged particles, and its direction identifies negative charge. The larger accelerating potential gives the electrons greater kinetic energy and momentum. The same magnetic flux density therefore produces a larger path radius, since $r=p/(eB)$ increases with momentum. Neutral light would not show this charge-dependent curvature.	A transverse magnetic field exerts a force on moving charge, so the common deflection direction identifies the negative sign of the beam particles. Electrical work from the accelerating potential becomes electron kinetic energy; the higher-potential beam has greater speed and momentum. A given field then changes its direction less sharply, producing the observed larger path radius. The linked voltage and curvature changes are therefore consistent with moving electrons and not with neutral electromagnetic radiation.	4

3.12.1.2 · Thermionic emission of electrons

Tier 1 · Easy

Mark scheme for 3.12.1.2 Tier 1 · Easy
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01.1	Heating gives some conduction electrons enough energy to overcome the surface barrier and escape.	Thermal energy increases the electrons' energies. Electrons in the high-energy part of the distribution can then overcome the metal's work function and leave the surface.	1
02.1	$eV=\frac12m_{\mathrm e}v^2$ , provided energy losses are negligible and the electron remains non-relativistic.	The electrical work $eV$ becomes the electron's kinetic energy. Starting from rest and neglecting losses gives $eV=\frac12m_{\mathrm e}v^2$ .	2

Tier 2 · Standard

Mark scheme for 3.12.1.2 Tier 2 · Standard
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01.1	$2.96\times10^7\,\text{m s}^{-1}$	Convert the potential difference: $V=2.50\times10^3\,\text{V}$ . Use $eV=\frac{1}{2}m_{\mathrm e}v^2$ , so $v=\sqrt{2eV/m_{\mathrm e}}$ . Hence $v=\sqrt{2(1.60\times10^{-19})(2.50\times10^3)/(9.11\times10^{-31})}=2.96\times10^7\,\text{m s}^{-1}$ .	3
02.1	$3.49\,\text{kV}$ to three significant figures	The electrical work becomes kinetic energy: $eV=\frac12m_{\mathrm e}v^2$ . Hence $V=m_{\mathrm e}v^2/(2e)=(9.11\times10^{-31})(3.50\times10^7)^2/[2(1.60\times10^{-19})]=3.49\times10^3\,\text{V}=3.49\,\text{kV}$ .	3
03.1	Heating still gives conduction electrons enough energy to escape by thermionic emission. The reversed anode is negative, so it repels the emitted electrons instead of accelerating them towards the aperture. The filament supply controls emission whereas the anode supply controls acceleration and beam formation.	Thermionic emission depends on the filament temperature, so reversing a separate anode supply does not stop electrons leaving the metal. A negative anode produces an electric force back towards the cathode. The emitted electrons therefore do not gain directed kinetic energy towards the aperture and no useful high-speed beam is formed.	3

Tier 3 · Hard

Mark scheme for 3.12.1.2 Tier 3 · Hard
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01.1	$v=5.13\times10^7\,\text{m s}^{-1}$ , $1.88\times10^{16}$ electrons per second, and $22.5\,\text{W}$	For one electron, $eV=\frac{1}{2}m_{\mathrm e}v^2$ , so $v=\sqrt{2(1.60\times10^{-19})(7.50\times10^3)/(9.11\times10^{-31})}=5.13\times10^7\,\text{m s}^{-1}$ . The emission rate is $N/t=I/e=(3.00\times10^{-3})/(1.60\times10^{-19})=1.88\times10^{16}\,\text{s}^{-1}$ . The beam power is $P=IV=(3.00\times10^{-3})(7.50\times10^3)=22.5\,\text{W}$ .	5
02.1	$3.44\times10^7\,\text{m s}^{-1}$ and $1.44\,\text{kV}$ , each to three significant figures	Before the collision the kinetic energy is $eV$ . The remaining kinetic energy is $0.700eV=0.700(1.60\times10^{-19})(4.80\times10^3)=5.38\times10^{-16}\,\text{J}$ . Hence $v=\sqrt{2K/m_{\mathrm e}}=\sqrt{2(5.38\times10^{-16})/(9.11\times10^{-31})}=3.44\times10^7\,\text{m s}^{-1}$ . Restoring the lost $0.300eV$ requires an additional potential difference $0.300(4.80\,\text{kV})=1.44\,\text{kV}$ . This is an explicitly non-relativistic estimate: the original speed is about $0.137c$ , so the approximation introduces an error of about $1\%$ .	5
03.1	$2.15\,\text{kV}$ to three significant figures	During uniform acceleration from rest, the average speed is $v/2$ , so the time in the gap is $2d/v$ . The drift time is $L/v$ . Hence $8.00\times10^{-9}=[2(20.0\times10^{-3})+180\times10^{-3}]/v$ , giving $v=2.75\times10^7\,\text{m s}^{-1}$ . Electrical work becomes kinetic energy: $eV=\frac12m_{\mathrm e}v^2$ . Therefore $V=(9.11\times10^{-31})(2.75\times10^7)^2/[2(1.60\times10^{-19})]=2.15\times10^3\,\text{V}$ .	5
04.1	$p_1=2.32\times10^{-23}\,\text{kg m s}^{-1}$ , $p_2=3.50\times10^{-23}\,\text{kg m s}^{-1}$ and a $51.3\%$ increase, each to three significant figures	Electrical work gives $eV=\frac12m_{\mathrm e}v^2=p^2/(2m_{\mathrm e})$ , so $p=\sqrt{2m_{\mathrm e}eV}$ . Hence $p_1=\sqrt{2(9.11\times10^{-31})(1.60\times10^{-19})(1.84\times10^3)}=2.316\times10^{-23}\,\text{kg m s}^{-1}$ and $p_2=\sqrt{2(9.11\times10^{-31})(1.60\times10^{-19})(4.21\times10^3)}=3.503\times10^{-23}\,\text{kg m s}^{-1}$ . The increase is $(p_2/p_1-1)\times100=51.3\%$ .	4
05.1	Vary one supply at a time while keeping the other fixed. Raising filament temperature increases the thermionic emission rate and hence beam current. Raising the positive anode potential increases the energy gained per electron and therefore its speed, with $eV=\frac12m_{\mathrm e}v^2$ ; it need not increase the number emitted if filament temperature is fixed.	Use a beam-current meter and a speed-sensitive measurement while changing only one independent variable. At fixed anode potential, a hotter filament gives more conduction electrons enough energy to escape, so current rises but the electrical energy gained by each transmitted electron is unchanged. At fixed filament temperature, increasing the anode potential makes each electron gain more kinetic energy, so $v\propto\sqrt V$ in the non-relativistic model. Separating the controls prevents a current change from being misidentified as a speed change.	4

3.12.1.3 · Specific charge of the electron

Tier 1 · Easy

Mark scheme for 3.12.1.3 Tier 1 · Easy
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01.1	$3.00\times10^6\,\text{m s}^{-1}$	With no deflection, electric and magnetic forces balance: $eE=evB$ . Therefore $v=E/B=(2.40\times10^4)/(8.00\times10^{-3})=3.00\times10^6\,\text{m s}^{-1}$ .	2
02.1	Specific charge is charge divided by mass and has unit $\text{C kg}^{-1}$ .	For a particle of charge magnitude $Q$ and mass $m$ , the magnitude of its specific charge is $Q/m$ . Dividing coulombs by kilograms gives $\text{C kg}^{-1}$ .	2

Tier 2 · Standard

Mark scheme for 3.12.1.3 Tier 2 · Standard
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01.1	$1.75\times10^{11}\,\text{C kg}^{-1}$	The magnetic force supplies the centripetal force: $evB=m_{\mathrm e}v^2/r$ . Thus $e/m_{\mathrm e}=v/(Br)$ . With $r=9.50\times10^{-2}\,\text{m}$ , $e/m_{\mathrm e}=(3.00\times10^6)/[(1.80\times10^{-4})(9.50\times10^{-2})]=1.75\times10^{11}\,\text{C kg}^{-1}$ .	3
02.1	$5.00\times10^{-3}\,\text{T}$	The plate field is $E=V/d=96.0/(6.00\times10^{-3})=1.60\times10^4\,\text{V m}^{-1}$ . No deflection means $eE=evB$ , so $B=E/v=(1.60\times10^4)/(3.20\times10^6)=5.00\times10^{-3}\,\text{T}$ .	3
03.1	$37.3\,\text{mm}$ to three significant figures	The selector gives $v=E/B$ , so the speed increases by $24.0/18.0=4/3$ . For fixed specific charge, circular motion gives $r=v/[B(e/m_{\mathrm e})]$ . The radius therefore changes by $(4/3)/1.50=8/9$ , giving $r=(42.0)(8/9)=37.3\,\text{mm}$ .	3

Tier 3 · Hard

Mark scheme for 3.12.1.3 Tier 3 · Hard
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01.1	$e/m_{\mathrm e}=1.75\times10^{11}\,\text{C kg}^{-1}$ , about $1.83\times10^3$ times the hydrogen-ion value	The plate field is $E=V/d=360/(15.0\times10^{-3})=2.40\times10^4\,\text{V m}^{-1}$ . Force balance in the selector gives $v=E/B=(2.40\times10^4)/(8.00\times10^{-3})=3.00\times10^6\,\text{m s}^{-1}$ . Magnetic deflection then gives $e/m_{\mathrm e}=v/(Br)=(3.00\times10^6)/[(1.60\times10^{-3})(10.7\times10^{-3})]=1.75\times10^{11}\,\text{C kg}^{-1}$ . The ratio is $(1.75\times10^{11})/(9.58\times10^7)=1.83\times10^3$ . Since the charge magnitudes are equal, this showed that the electron mass is about $1/1830$ of the hydrogen-ion mass.	5
02.1	$B=2.82\,\text{mT}$ and the new radius is $26.2\,\text{mm}$ , each to three significant figures	Combining $eV=\frac12m_{\mathrm e}v^2$ with $evB=m_{\mathrm e}v^2/r$ gives $e/m_{\mathrm e}=2V/(B^2r^2)$ . Hence $B=\sqrt{2V/[(e/m_{\mathrm e})r^2]}=\sqrt{2(240)/[(1.76\times10^{11})(18.5\times10^{-3})^2]}=2.82\times10^{-3}\,\text{T}$ . At fixed $B$ and $e/m_{\mathrm e}$ , $r\propto\sqrt V$ , so doubling $V$ gives $r=(18.5\,\text{mm})\sqrt2=26.2\,\text{mm}$ .	4
03.1	$(1.76\pm0.12)\times10^{11}\,\text{C kg}^{-1}$ to the stated precision	The plate field is $E=V/d=182/(12.0\times10^{-3})=1.52\times10^4\,\text{V m}^{-1}$ . The selected speed is $v=E/B=2.92\times10^6\,\text{m s}^{-1}$ . Circular motion gives $e/m_{\mathrm e}=v/(Br)=(2.92\times10^6)/[(0.240\times10^{-3})(69.0\times10^{-3})]=1.76\times10^{11}\,\text{C kg}^{-1}$ . Since $e/m_{\mathrm e}=V/(dB_{\mathrm s}B_{\mathrm d}r)$ , the percentage uncertainty is $1.0+1.0+2.0+1.5+1.5=7.0\%$ . The absolute uncertainty is $0.070(1.76\times10^{11})=0.12\times10^{11}\,\text{C kg}^{-1}$ .	5
04.1	$1.76\times10^{11}\,\text{C kg}^{-1}$ to three significant figures	The time-of-flight measurement gives $v=L/t=0.375/(120\times10^{-9})=3.125\times10^6\,\text{m s}^{-1}$ . In the magnetic field, $evB=m_{\mathrm e}v^2/r$ , so $e/m_{\mathrm e}=v/(Br)$ . Hence $e/m_{\mathrm e}=(3.125\times10^6)/[(0.259\times10^{-3})(68.5\times10^{-3})]=1.761\times10^{11}\,\text{C kg}^{-1}=1.76\times10^{11}\,\text{C kg}^{-1}$ to three significant figures.	4
05.1	$1.76\times10^{11}\,\text{C kg}^{-1}$ to three significant figures	For circular motion, $evB=m_{\mathrm e}v^2/r$ . With $v=2\pi r/T$ , this becomes $e/m_{\mathrm e}=2\pi/(BT)$ . The period is $T=(1.36\times10^{-6})/15=9.0667\times10^{-8}\,\text{s}$ . Hence $e/m_{\mathrm e}=2\pi/[(0.394\times10^{-3})(9.0667\times10^{-8})]=1.759\times10^{11}\,\text{C kg}^{-1}=1.76\times10^{11}\,\text{C kg}^{-1}$ to three significant figures. The radius and speed cancel, so neither is required.	4

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

Tier 1 · Easy

Mark scheme for 3.12.1.4 Tier 1 · Easy
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01.1	$QE=mg$ , with the electric force upward.	Stationary means zero resultant force. The weight $mg$ acts downward, so the electric force must act upward with equal magnitude: $QE=mg$ .	2
02.1	Many measurements showed that every charge was an integer multiple of the same smallest value, providing evidence that charge is quantised.	One charge value cannot reveal a common step size. Comparing many droplets exposes the repeated smallest increment $e$ and shows that the different measured charges are $ne$ .	2

Tier 2 · Standard

Mark scheme for 3.12.1.4 Tier 2 · Standard
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01.1	$3.20\times10^{-19}\,\text{C}=2.00e$	The field is $E=V/d=600/(3.00\times10^{-3})=2.00\times10^5\,\text{V m}^{-1}$ . At equilibrium, $QE=mg$ , so $Q=(6.52\times10^{-15})(9.81)/(2.00\times10^5)=3.20\times10^{-19}\,\text{C}$ . Dividing by $1.60\times10^{-19}\,\text{C}$ gives $Q/e=2.00$ .	3
02.1	$383\,\text{V}$ to three significant figures	For equilibrium, $QE=mg$ and $E=V/d$ , so $V=mgd/Q$ . With $Q=3(1.60\times10^{-19})=4.80\times10^{-19}\,\text{C}$ , $V=(3.75\times10^{-15})(9.81)(5.00\times10^{-3})/(4.80\times10^{-19})=383\,\text{V}$ .	3
03.1	$e=1.60\times10^{-19}\,\text{C}$ to three significant figures	The allowed range identifies the charges as approximately $3e$ , $5e$ and $7e$ . The three estimates are $4.80/3=1.600$ , $8.03/5=1.606$ and $11.18/7=1.597$ , all in units of $10^{-19}\,\text{C}$ . Their mean is $1.601\times10^{-19}\,\text{C}$ , giving $1.60\times10^{-19}\,\text{C}$ to three significant figures.	3

Tier 3 · Hard

Mark scheme for 3.12.1.4 Tier 3 · Hard
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01.1	$Q=3.19\times10^{-19}\,\text{C}\approx2e$ , supporting quantisation	At terminal speed, $6\pi\eta rv=mg$ and $m=\frac{4}{3}\pi r^3\rho$ . Therefore $6\pi\eta rv=\frac{4}{3}\pi r^3\rho g$ , so $r=\sqrt{9\eta v/(2\rho g)}=\sqrt{9(1.80\times10^{-5})(8.00\times10^{-5})/[2(850)(9.81)]}=8.82\times10^{-7}\,\text{m}$ . The mass is $m=\frac{4}{3}\pi r^3\rho=2.44\times10^{-15}\,\text{kg}$ . When held, $QE=mg$ , giving $Q=(2.44\times10^{-15})(9.81)/(7.50\times10^4)=3.19\times10^{-19}\,\text{C}$ . This is $Q/e=1.99\approx2$ , an integer multiple of the elementary charge within experimental uncertainty.	5
02.1	$6.38\times10^{-19}\,\text{C}$ to three significant figures, equivalent to $3.99e\approx4e$	With buoyancy neglected, the droplet's weight at downward terminal speed is $W=6\pi\eta rv_{\mathrm d}=2.50\times10^{-14}\,\text{N}$ . During upward terminal motion, $QE=W+6\pi\eta rv_{\mathrm u}$ . Hence $Q=6\pi\eta r(v_{\mathrm d}+v_{\mathrm u})/E=6\pi(1.84\times10^{-5})(0.900\times10^{-6})(8.00\times10^{-5}+2.00\times10^{-4})/(1.37\times10^5)=6.38\times10^{-19}\,\text{C}$ . Dividing by $e$ gives $Q/e=3.99\approx4$ .	5
03.1	$1.60\times10^{-19}\,\text{C}$ to three significant figures	Let the initial charge magnitude be $ne$ . For the same mass and plate separation, equilibrium requires $QV$ to be constant, so $n(600)=(n+1)(480)$ . This gives $n=4$ . Initially, $QV/d=mg$ , so $Q=mgd/V=(7.83\times10^{-15})(9.81)(5.00\times10^{-3})/600=6.40\times10^{-19}\,\text{C}$ . Therefore $e=Q/4=1.60\times10^{-19}\,\text{C}$ .	5
04.1	$Q=4.78\times10^{-19}\,\text{C}=2.99e\approx3e$ ; neglecting buoyancy would overestimate the charge by $0.138\%$	The upward electric force balances effective weight, so $QE=\frac43\pi r^3(\rho_{\mathrm{oil}}-\rho_{\mathrm{air}})g$ . Thus $Q=\frac{4\pi(1.20\times10^{-6})^3(870-1.20)(9.81)}{3(1.29\times10^5)}=4.78\times10^{-19}\,\text{C}$ . Dividing by $1.60\times10^{-19}\,\text{C}$ gives $2.99$ , consistent with $3e$ . Neglecting buoyancy replaces $\rho_{\mathrm{oil}}-\rho_{\mathrm{air}}$ by $\rho_{\mathrm{oil}}$ , so the percentage overestimate is $[870/(870-1.20)-1]\times100=0.138\%$ .	5
05.1	The faster droplet has $3.39$ times the mass of the slower droplet. Reusing the smaller mass underestimates the faster droplet's charge by the same factor, so charges that should cluster at integer multiples of $e$ can appear unrelated. The terminal speed of each droplet must be used to determine its own radius and mass before its charge is calculated.	At terminal speed, Stokes' drag and effective weight give $r\propto\sqrt v$ for droplets of the same density in the same air. Since $m\propto r^3$ , it follows that $m\propto v^{3/2}$ . Therefore $m_2/m_1=(1.31\times10^{-4}/5.80\times10^{-5})^{3/2}=3.394=3.39$ . Because the stationary-drop relation is $Q=mg/E$ , inserting $m_1$ for the second droplet gives only $1/3.394$ of its actual charge. Millikan's comparison requires an independent terminal-speed measurement and mass calculation for every droplet.	4

3.12.2.1 · Newton's corpuscular theory of light

Tier 1 · Easy

Mark scheme for 3.12.2.1 Tier 1 · Easy
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01.1	Light appears to travel in straight lines and can form sharp shadows.	Corpuscles moving along straight paths would be blocked geometrically by an obstacle, so the model gave a simple explanation of sharp shadows.	1
02.1	As a stream of small particles emitted by a source.	Newton treated light as corpuscles travelling along particle paths, so a ray traced the direction of their motion.	1

Tier 2 · Standard

Mark scheme for 3.12.2.1 Tier 2 · Standard
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01.1	Newton's theory predicted a speed increase, whereas wave theory predicted a decrease; the measured decrease contradicted the corpuscular account and supported wave theory.	Newton attributed refraction towards the normal to an attractive force increasing a corpuscle's component of velocity in glass, so his model predicted greater speed. Huygens' construction gives bending towards the normal when wave speed falls. Measurement showed that light travels more slowly in glass, selecting the wave prediction and weakening the corpuscular theory.	3
02.1	Straight-moving corpuscles naturally give sharp boundaries, whereas familiar water and sound waves spread around obstacles; diffraction of light was too small to be obvious in ordinary conditions.	Newton's particles were assumed to travel in straight lines, so an opaque object would remove corpuscles from a sharply defined region. Known waves spread into geometrical shadows, which seemed inconsistent with the observed sharpness. Because visible wavelengths are very small, appreciable optical diffraction needs narrow openings and was not apparent in everyday observations.	3
03.1	Every point on a wavefront acts as a source of secondary wavelets. The envelope of those wavelets forms the next wavefront.	Award one mark for points on the existing wavefront acting as sources of secondary wavelets and one for the envelope of those wavelets forming the next wavefront.	2

Tier 3 · Hard

Mark scheme for 3.12.2.1 Tier 3 · Hard
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01.1	It explained ray-like behaviour and carried Newton's authority, but diffraction, interference and slower propagation in dense media matched wave theory and contradicted classical corpuscles.	Straight-line corpuscle paths explained sharp shadows, and boundary forces offered accounts of reflection and refraction. Diffraction was not conspicuous in ordinary conditions, while Newton's success and reputation gave his model great weight. Later, fringes containing dark regions showed that light contributions can cancel by superposition, which independent classical particles cannot explain. Diffraction also showed spreading around obstacles, and measured light speeds in dense media agreed with Huygens rather than Newton. The combined evidence therefore required a wave model despite the earlier model's authority.	5
02.1	In Newton's model a surface impulse reverses the corpuscles' velocity component normal to the surface while leaving the parallel component unchanged, giving equal angles. In Huygens' model secondary wavelets construct a reflected wavefront at the same angle. Because both models predict the observation, reflection is not discriminating evidence; interference or diffraction is needed to distinguish them.	Resolve a corpuscle's velocity into components parallel and perpendicular to the surface. Reversing only the perpendicular component preserves the speed and gives equal incidence and reflection angles. Huygens' construction instead uses secondary wavelets from the reflecting surface to form a reflected wavefront with the same angular result. A shared prediction cannot select a model; dark interference fringes or diffraction provide the distinct wave evidence.	5

3.12.2.2 · Significance of Young's double slits experiment

Tier 1 · Easy

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01.1	Alternating bright and dark fringes.	The repeated maxima and minima show that two contributions sometimes reinforce and sometimes cancel, which is the signature of interference.	1
02.1	They are coherent: they have the same frequency and a constant phase difference.	A common source fixes the phase relationship between the waves emerging from the two slits, which is required for a stable interference pattern.	1

Tier 2 · Standard

Mark scheme for 3.12.2.2 Tier 2 · Standard
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01.1	Diffracted coherent waves overlap; in-phase waves reinforce to form bright fringes and antiphase waves cancel to form dark fringes.	Both slits are illuminated by the same monochromatic source, so the emerging waves have a constant phase relationship. Each slit diffracts the light and the waves overlap. Where the path difference leaves the waves in phase, their amplitudes add and the intensity is high. Where they arrive in antiphase, their amplitudes cancel and a dark fringe forms.	3
02.1	With only one slit there are no two coherent waves to superpose, so path-dependent constructive and destructive interference cannot occur.	The uncovered slit still diffracts light, but each point on the screen receives no second coherent contribution from the covered slit. There is therefore no phase difference between two slit waves to produce repeated reinforcement and cancellation, so the two-slit fringe pattern vanishes.	3
03.1	The lasers do not maintain a constant phase difference. The positions of constructive and destructive interference therefore change rapidly, so time averaging washes out the bright and dark fringes.	Equal wavelength gives equal frequency but does not make independent sources coherent. Their relative phase varies unpredictably. The maxima and minima consequently move during the observation interval, and the detector records an approximately uniform time-averaged intensity instead of stationary fringes.	3

Tier 3 · Hard

Mark scheme for 3.12.2.2 Tier 3 · Hard
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01.1	The pattern required diffraction and destructive superposition, behaviours predicted by wave theory but not by independent classical corpuscles, so it drove acceptance of Huygens' model.	A simple corpuscle model would send particles through either opening and produce two accumulated bright regions; adding more corpuscles cannot create regularly spaced zero-intensity bands. Huygens' model predicts that each slit launches spreading wavelets. Their phase difference varies across the screen, so superposition alternates between reinforcement and cancellation. The dark bands therefore provide discriminating evidence for destructive interference, not merely straight-line propagation. Because this directly matched the wave prediction, it overcame the long-standing preference for Newton's theory and established the wave nature of light.	5
02.1	Newton's corpuscular model already accounted for reflection, refraction and sharp shadows, and Newton's authority made it strongly established. Optical diffraction is inconspicuous at everyday aperture sizes, so the wave alternative initially seemed less compelling. However, independent corpuscles could not naturally produce repeated zero-intensity dark fringes. Replicated interference and diffraction evidence formed a consistent wave account that displaced the corpuscular model.	Newton's model already explained reflection, refraction and sharp shadows in an intuitively mechanical way, and his scientific reputation carried substantial weight. Optical wavelengths are tiny, so the spreading expected for waves is normally inconspicuous. Young's cancellation interpretation was consequently resisted even though independent corpuscles could not naturally create dark bands. Repeated interference and diffraction results accumulated into a consistent wave account and eventually overcame the preference for Newton's theory.	5
03.1	Covering each slit in turn shows that either path alone delivers non-zero light to that point, so the darkness with both open is not caused by one blocked or unilluminated path. With both slits open, the two waves must arrive in antiphase and with equal amplitudes; their displacements then cancel completely.	Credit that each one-slit trial confirms a non-zero contribution at the chosen point, that the two-slit darkness therefore arises from combining the contributions, an antiphase phase relationship and equal amplitudes for complete cancellation. Do not credit historical delayed-acceptance reasons in this item.	4
04.1	Wider slits produce less diffraction, so the waves emerging from the two slits spread through smaller angles. Their overlap region on the screen is therefore narrower, and stable bright and dark fringes occur only within that overlap. The linked changes in diffraction and interference follow from wave superposition rather than independent straight-line corpuscles.	Each slit must diffract the incident light for contributions from both openings to reach the same screen positions. Increasing slit width relative to wavelength reduces the angular spread from each opening. This reduces the common illuminated region in which the two coherent waves can reinforce or cancel. A systematic reduction in fringe extent when diffraction is reduced connects two wave effects and is not explained by simply adding independent particle intensities.	4
05.1	Blue light has a shorter wavelength, so the phase difference between the two paths changes through a full cycle over a smaller change of direction and the bands are closer together. Removing either coherent contribution prevents repeated constructive and destructive superposition, so the alternating sequence disappears. A wavelength-dependent pattern containing dark cancellations is explained by waves and not by independent classical corpuscles travelling from the slits.	Keep the reasoning qualitative, as required for this leaf. Shorter-wavelength blue light reaches successive in-phase and antiphase conditions with smaller angular changes than red light, accounting for the reduced band separation. Covering either slit removes one member of the coherent pair, so there can be no two-path reinforcement or cancellation. The colour comparison and the one-slit control together connect the observation specifically to wavelength and superposition.	4

3.12.2.3 · Electromagnetic waves

Tier 1 · Easy

Mark scheme for 3.12.2.3 Tier 1 · Easy
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01.1	The electric and magnetic fields are mutually perpendicular and both are perpendicular to the direction of travel.	Electromagnetic waves are transverse. The oscillating electric field is at right angles to the oscillating magnetic field, and each field is at right angles to the propagation direction.	2
02.1	$\varepsilon_0$ characterises the electric field strength due to a charged object in free space, while $\mu_0$ characterises the magnetic flux density due to a current-carrying wire in free space.	The permittivity $\varepsilon_0$ enters expressions for electric fields due to charge, while the permeability $\mu_0$ enters expressions for magnetic flux density due to currents.	2

Tier 2 · Standard

Mark scheme for 3.12.2.3 Tier 2 · Standard
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01.1	$3.00\times10^8\,\text{m s}^{-1}$	Use $c=1/\sqrt{\mu_0\varepsilon_0}$ . Thus $c=1/\sqrt{(4\pi\times10^{-7})(8.85\times10^{-12})}=2.999\times10^8\,\text{m s}^{-1}$ , which is $3.00\times10^8\,\text{m s}^{-1}$ to three significant figures.	3
02.1	$8.84\times10^{-12}\,\text{F m}^{-1}$ from the rounded givens $8.85\times10^{-12}\,\text{F m}^{-1}$ using the data-sheet value	From $c=1/\sqrt{\mu_0\varepsilon_0}$ , $\varepsilon_0=1/(\mu_0c^2)$ . Thus $\varepsilon_0=1/[(4\pi\times10^{-7})(3.00\times10^8)^2]=8.84\times10^{-12}\,\text{F m}^{-1}$ .	3
03.1	$0.0667\%$ ; the close agreement supported Maxwell's identification of light as an electromagnetic wave, but further evidence that radio waves share wave properties with light was needed.	The difference is $2.0\times10^5\,\text{m s}^{-1}$ , so the percentage difference is $(2.0\times10^5/3.000\times10^8)\times100=0.0667\%$ . Matching a value predicted from electrical and magnetic constants to the optical speed strongly supported Maxwell's proposal. Hertz's production of radio waves and observation of their wave behaviour provided an independent test of their electromagnetic nature.	3

Tier 3 · Hard

Mark scheme for 3.12.2.3 Tier 3 · Hard
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01.1	$3.01\times10^8\,\text{m s}^{-1}$ ; the agreement of light and radio-wave speeds with Maxwell's prediction supported their common electromagnetic nature	A gap-to-adjacent-tooth turn is $1/(2N)$ of a revolution, so the round-trip time is $t=1/(2Nf)$ . Since the light travels $2D$ , $c=2D/t=4DNf$ . Therefore $c=4(8.63\times10^3)(720)(12.1)=3.01\times10^8\,\text{m s}^{-1}$ . Fizeau showed terrestrially that light has this finite speed. Hertz produced radio waves with wave properties and a speed close to the same value. Their agreement with $1/\sqrt{\mu_0\varepsilon_0}$ supported Maxwell's claim that both light and radio are electromagnetic waves.	5
02.1	$3.00\times10^8\,\text{m s}^{-1}$ ; agreement with the speed of light supports Maxwell's identification of light and radio waves as electromagnetic waves	Adjacent nodes are separated by half a wavelength, so $\lambda=2(0.750)=1.50\,\text{m}$ . The speed is $v=f\lambda=(200\times10^6)(1.50)=3.00\times10^8\,\text{m s}^{-1}$ . Hertz's waves also displayed reflection and interference. Their measured speed matching both the optical value and $1/\sqrt{\mu_0\varepsilon_0}$ supported Maxwell's prediction that radio and visible light share an electromagnetic nature.	5
03.1	$(2.998\pm0.009)\times10^8\,\text{m s}^{-1}$ ; its uncertainty interval overlaps the Hertz interval, so the results agree.	Maxwell's result is $c=1/\sqrt{\mu_0\varepsilon_0}=1/\sqrt{(1.257\times10^{-6})(8.85\times10^{-12})}=2.998\times10^8\,\text{m s}^{-1}$ . The product has percentage uncertainty $0.2+0.4=0.6\%$ , and the power $-1/2$ gives $0.3\%$ in $c$ . The absolute uncertainty is $0.003(2.998\times10^8)=0.009\times10^8\,\text{m s}^{-1}$ . Maxwell's interval, $2.989$ to $3.007$ in units of $10^8\,\text{m s}^{-1}$ , overlaps the Hertz interval $2.99$ to $3.05$ , so the measurements are consistent with a common electromagnetic speed.	5
04.1	The claim is false: $\mu_0$ and $\varepsilon_0$ characterise magnetic and electric behaviour in free space and can be obtained independently of optical speed measurements. Evaluating $1/\sqrt{\mu_0\varepsilon_0}$ with modern constants gives $2.998\times10^8\,\text{m s}^{-1}$ . Hertz then generated and detected radio waves and demonstrated electromagnetic wave behaviour, providing experimental evidence for the new waves rather than only a numerical coincidence.	Distinguish an independently testable prediction from a fitted value. The two constants arise from electrical and magnetic measurements, so their combination was not selected to reproduce optical data. Its agreement with the independently known light speed supported the identification of light as electromagnetic, although matching one speed was not decisive by itself. Hertz supplied a different test by producing radio disturbances and observing properties such as reflection, polarisation or interference expected of Maxwell's transverse electromagnetic waves.	4
05.1	The orientation dependence shows that the detected disturbance has a transverse field direction. Reflection and fixed nodes demonstrate wave superposition and allow its wavelength to be measured. Combining wavelength with frequency gives a speed close to light and to $1/\sqrt{\mu_0\varepsilon_0}$ . A radio disturbance with transverse wave behaviour and Maxwell's predicted speed supports the conclusion that radio waves and light are electromagnetic waves.	The receiver responds to a particular field orientation, so the loss of signal on rotation is evidence for a transverse oscillating field rather than an isotropic disturbance. Incident and reflected waves superpose to create nodes, demonstrating reflection and interference and providing half-wavelength spacings. The independently measured frequency then gives $v=f\lambda$ . Agreement of this speed with both optical measurements and Maxwell's value derived from electrical and magnetic constants links radio and visible radiation as one electromagnetic family.	5

3.12.2.4 · The discovery of photoelectricity

Tier 1 · Easy

Mark scheme for 3.12.2.4 Tier 1 · Easy
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01.1	Below a threshold frequency no electrons are emitted, regardless of intensity. Emission starts without a measurable delay. Maximum electron kinetic energy depends on frequency rather than intensity.	Any one listed observation conflicts with continuous energy delivery by a classical wave: a threshold, immediate one-event transfer, or frequency-controlled electron energy requires quantised photons.	1
02.1	It is the reverse potential at which even the fastest photoelectrons just fail to reach the collector, so the photocurrent falls to zero; $K_{\max}=eV_{\mathrm s}$ for the magnitude $V_{\mathrm s}$ .	The reverse electric field removes kinetic energy from emitted electrons. At the stopping endpoint, its electrical work $eV_{\mathrm s}$ equals the kinetic energy of the fastest emitted electrons, so none reaches the collector and the measured current is zero. The stopping potential therefore measures maximum, not mean, photoelectron kinetic energy.	2

Tier 2 · Standard

Mark scheme for 3.12.2.4 Tier 2 · Standard
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01.1	Energy is exchanged only in packets $E=hf$ , so high-frequency modes require large quanta and are unlikely to be excited at a fixed temperature, suppressing the classical high-frequency divergence.	Classical equipartition allowed every electromagnetic mode to gain energy continuously, producing an unbounded high-frequency intensity. Planck restricted emission and absorption to quanta with energy $hf$ . At high $f$ each quantum is large compared with the available thermal energy, so few such quanta are emitted. The spectrum therefore falls instead of diverging.	3
02.1	The threshold frequency is $5.67\times10^{14}\,\text{Hz}$ . The violet and ultraviolet beams can emit photoelectrons; the red and green beams cannot.	Convert the work function: $\phi=2.35(1.60\times10^{-19})=3.76\times10^{-19}\,\text{J}$ . The threshold is $f_0=\phi/h=(3.76\times10^{-19})/(6.63\times10^{-34})=5.67\times10^{14}\,\text{Hz}$ . A beam can emit electrons only if its frequency is at least $f_0$ , so the two frequencies above the threshold are violet and ultraviolet. Red and green remain below threshold.	3
03.1	One electron absorbs one photon of energy $hf$ . Below threshold, $hf<\phi$ , so no single photon can release an electron however many arrive. Above threshold, one photon can transfer enough energy in one interaction, so emission is immediate.	Photon energy depends on frequency, not intensity. Greater sub-threshold intensity increases only the number of photons, while each remains unable to overcome the work function. Once $hf\geq\phi$ , an electron can receive the required energy in a single absorption event, removing the classical charging delay.	3

Tier 3 · Hard

Mark scheme for 3.12.2.4 Tier 3 · Hard
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01.1	The classical accumulation delay is $8.18\times10^6\,\text{s}$ , or $94.7\,\text{days}$ . Immediate low-irradiance emission is explained instead by one above-threshold photon transferring at least the work-function energy in a single interaction.	Convert the work function without rounding: $\phi=2.30(1.60\times10^{-19})=3.68\times10^{-19}\,\text{J}$ . Even assuming complete transfer to one electron, the classical power available from the patch is only $P=IA=(2.50\times10^{-6})(1.80\times10^{-20})=4.50\times10^{-26}\,\text{W}$ . Hence $t=\phi/P=(3.68\times10^{-19})/(4.50\times10^{-26})=8.18\times10^6\,\text{s}=94.7\,\text{days}$ . The observed absence of such a delay contradicts gradual classical accumulation. In the photon model, one photon with $hf\geq\phi$ supplies the escape energy in one event, so an electron can be emitted as soon as such a photon is absorbed.	5
02.1	$K_{\max}=5.53\times10^{-19}\,\text{J}=3.46\,\text{eV}$ and $f=1.38\times10^{15}\,\text{Hz}$ , each to three significant figures	The opposing field does work $eEs$ while the fastest electron stops, so $K_{\max}=eEs=(1.60\times10^{-19})(480)(7.20\times10^{-3})=5.5296\times10^{-19}\,\text{J}=3.456\,\text{eV}$ . Einstein's equation gives the photon energy as $hf=K_{\max}+\phi=(3.456+2.25)\,\text{eV}=5.706\,\text{eV}$ . Using the unrounded joule energies, $f=[5.5296\times10^{-19}+2.25(1.60\times10^{-19})]/(6.63\times10^{-34})=1.377\times10^{15}\,\text{Hz}$ , which rounds to $1.38\times10^{15}\,\text{Hz}$ .	5
03.1	Both surfaces emit. The maximum momenta are $1.07\times10^{-24}\,\text{kg m s}^{-1}$ for potassium and $6.65\times10^{-25}\,\text{kg m s}^{-1}$ for copper, so $p_{\mathrm K}/p_{\mathrm{Cu}}=1.61$ .	The photon energy is $hf=(6.63\times10^{-34})(1.50\times10^{15})=9.945\times10^{-19}\,\text{J}=6.215625\,\text{eV}$ , which exceeds both work functions. The unrounded maximum kinetic energies are therefore $3.915625\,\text{eV}=6.265\times10^{-19}\,\text{J}$ for potassium and $1.515625\,\text{eV}=2.425\times10^{-19}\,\text{J}$ for copper. With $p=\sqrt{2m_{\mathrm e}K}$ , the momenta are $1.068\times10^{-24}$ and $6.647\times10^{-25}\,\text{kg m s}^{-1}$ respectively. Their ratio is $1.607$ , so potassium's maximum photoelectron momentum is $1.61$ times copper's.	5
04.1	$\phi=2.24\,\text{eV}$ and the new stopping potential is $1.77\,\text{V}$ , each to three significant figures	For $420\,\text{nm}$ , $hf=hc/\lambda=(6.63\times10^{-34})(3.00\times10^8)/(420\times10^{-9})=4.736\times10^{-19}\,\text{J}=2.9598\,\text{eV}$ . Since $K_{\max}=eV_{\mathrm s}=0.720\,\text{eV}$ , $\phi=2.9598-0.720=2.2398\,\text{eV}=2.24\,\text{eV}$ . At $310\,\text{nm}$ the photon energy is $4.0101\,\text{eV}$ , so $V_{\mathrm s}=4.0101-2.2398=1.7703\,\text{V}=1.77\,\text{V}$ .	5
05.1	$hf/(kT)=7.99$ at $3.00\times10^{14}\,\text{Hz}$ and $32.0$ at $1.20\times10^{15}\,\text{Hz}$ . High-frequency modes require quanta far larger than the available thermal energy scale, so they are much less likely to be excited and the spectrum falls instead of diverging.	For the lower frequency, $hf/(kT)=(6.63\times10^{-34})(3.00\times10^{14})/(2.49\times10^{-20})=7.99$ . At the higher frequency the ratio is four times greater, $32.0$ . Planck allowed energy exchange only in packets $hf$ . As frequency rises, one packet becomes increasingly large compared with the thermal energy scale, so occupation of those modes is suppressed. Classical continuous equipartition lacked this restriction and incorrectly predicted unbounded high-frequency intensity.	4

3.12.2.5 · Wave-particle duality

Tier 1 · Easy

Mark scheme for 3.12.2.5 Tier 1 · Easy
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01.1	The wavelength becomes one third of its original value, and the diffraction maxima become more closely spaced because their diffraction angles are smaller.	De Broglie wavelength is inversely proportional to momentum, so multiplying momentum by $3$ divides wavelength by $3$ . The crystal spacing and diffraction order are unchanged; the shorter wavelength therefore satisfies the diffraction condition at smaller angles, bringing the maxima closer together.	2
02.1	$\lambda=h/p$	The de Broglie wavelength equals Planck's constant divided by the particle momentum.	1

Tier 2 · Standard

Mark scheme for 3.12.2.5 Tier 2 · Standard
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01.1	$\lambda=8.95\times10^{-11}\,\text{m}$ and $p=7.41\times10^{-24}\,\text{kg m s}^{-1}$ , each to three significant figures	Use the measured crystal condition first: $\lambda=d\sin\theta=(0.215\times10^{-9})\sin24.6^\circ=8.950037\times10^{-11}\,\text{m}$ . De Broglie's relation then gives $p=h/\lambda=(6.63\times10^{-34})/(8.950037\times10^{-11})=7.40779\times10^{-24}\,\text{kg m s}^{-1}$ . Rounding only at the end gives the stated values.	3
02.1	$36.2\,\text{V}$ to three significant figures	The momentum is $p=h/\lambda=(6.63\times10^{-34})/(0.204\times10^{-9})=3.25\times10^{-24}\,\text{kg m s}^{-1}$ . Since $p^2/(2m_{\mathrm e})=eV$ , $V=p^2/(2m_{\mathrm e}e)=(3.25\times10^{-24})^2/[2(9.11\times10^{-31})(1.60\times10^{-19})]=36.2\,\text{V}$ .	3
03.1	$\lambda=8.32\times10^{-15}\,\text{m}$ . This is about $3.00\times10^4$ times smaller than the $0.250\,\text{nm}$ atomic spacing, so the diffraction angle would be too small for observable diffraction.	The alpha-particle momentum is $p=mv=(6.64\times10^{-27})(1.20\times10^7)=7.968\times10^{-20}\,\text{kg m s}^{-1}$ . Hence $\lambda=h/p=(6.63\times10^{-34})/(7.968\times10^{-20})=8.32078\times10^{-15}\,\text{m}$ , or $8.32\times10^{-15}\,\text{m}$ to three significant figures. The spacing is $0.250\,\text{nm}=2.50\times10^{-10}\,\text{m}$ , so $d/\lambda=3.0045\times10^4$ . Equivalently, first order would require $\sin\theta=\lambda/d=3.33\times10^{-5}$ . The wavelength is therefore orders of magnitude below the atomic spacing and the diffraction angle would be too small to observe.	3

Tier 3 · Hard

Mark scheme for 3.12.2.5 Tier 3 · Hard
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01.1	$6.14\times10^{-11}\,\text{m}$ and $17.8^\circ$	At $100\,\text{V}$ , $\lambda_1=h/\sqrt{2m_{\mathrm e}eV}=1.228\times10^{-10}\,\text{m}$ . Since $\lambda\propto1/\sqrt{V}$ , raising $V$ by a factor of $4$ gives $\lambda_2=\lambda_1/2=6.14\times10^{-11}\,\text{m}$ . The same plane spacing gives $\sin\theta_2/\sin\theta_1=\lambda_2/\lambda_1=1/2$ , so $\theta_2=\sin^{-1}[0.5\sin(37.8^\circ)]=17.8^\circ$ . A diffraction pattern requires coherent wave superposition, so electrons previously treated as particles also have wave behaviour; their localised detection retains the particle aspect.	5
02.1	$450\,\text{V}$ to three significant figures	For the same apparatus and small diffraction angles, ring diameter $D$ is proportional to the electron wavelength. Non-relativistically, $\lambda=h/\sqrt{2m_{\mathrm e}eV}$ , so $D\propto1/\sqrt V$ . Hence $D_1/D_2=\sqrt{V_2/V_1}$ and $V_2=V_1(D_1/D_2)^2=200(18.0/12.0)^2=450\,\text{V}$ .	4
03.1	$62.4\,\text{V}$ to three significant figures	The shared diffraction condition means the photon and electron have the same wavelength. The photon energy is $8.00\times10^3(1.60\times10^{-19})=1.28\times10^{-15}\,\text{J}$ , so $\lambda=hc/E=(6.63\times10^{-34})(3.00\times10^8)/(1.28\times10^{-15})=1.55\times10^{-10}\,\text{m}$ . For the electron, $\lambda=h/\sqrt{2m_{\mathrm e}eV}$ , giving $V=h^2/(2m_{\mathrm e}e\lambda^2)=62.4\,\text{V}$ .	5
04.1	Each isolated impact shows that an electron transfers energy at one localised position, which is particle-like detection. The accumulated rings show a reproducible diffraction probability pattern, which requires wave behaviour associated with the electrons. Since the pattern builds even when electrons arrive separately, it is not caused by collisions between electrons; the same object therefore displays particle-like arrival and wave-like propagation.	Treat the observations at two scales separately. A single screen event is localised rather than a spread-out deposit, which is the particle aspect. The distribution of a large number of those events has diffraction maxima and minima, so their probabilities follow a wave pattern. Low current separates the electrons in time and rules out mutual electron interactions as the source of the rings, leaving a de Broglie wave description for each electron's propagation.	4
05.1	The maxima should occur at the same positions because equal momentum gives both beams the same de Broglie wavelength, $\lambda=h/p$ , and the crystal geometry is unchanged. Agreement would show that matter-wave wavelength depends on momentum rather than on particle charge or identity, extending wave behaviour beyond electrons. Localised detections would still retain the particle aspect.	Apply only de Broglie's relation and a qualitative comparison. The imposed equality of momentum makes $h/p$ identical for the two particles. A fixed crystal selects diffraction directions according to wavelength, so matching wavelengths give matching maxima. Because the neutron is neutral while the electron is charged, a match would also show that the wave relation is not an effect requiring electron charge.	4

3.12.2.6 · Electron microscopes

Tier 1 · Easy

Mark scheme for 3.12.2.6 Tier 1 · Easy
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01.1	The electrons gain momentum, so their de Broglie wavelength decreases and smaller detail can be resolved.	Acceleration through a larger potential difference increases electron momentum. Since $\lambda=h/p$ , the wavelength falls; the diffraction limit is then smaller, allowing finer detail to be distinguished.	2
02.1	Enough electrons must pass through the specimen to form an image, while different regions scatter or absorb different fractions of the beam.	A thick specimen would scatter or stop most electrons. A thin specimen transmits a usable beam whose spatial variations carry information about the internal structure.	2

Tier 2 · Standard

Mark scheme for 3.12.2.6 Tier 2 · Standard
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01.1	$151\,\text{V}$	Rearrange $\lambda=h/\sqrt{2m_{\mathrm e}eV}$ to $V=h^2/(2m_{\mathrm e}e\lambda^2)$ . Thus $V=(6.63\times10^{-34})^2/[2(9.11\times10^{-31})(1.60\times10^{-19})(0.100\times10^{-9})^2]=151\,\text{V}$ .	3
02.1	$V=2.36\times10^4\,\text{V}$ . The classical calculation implies an electron speed of about $0.30c$ , so the non-relativistic relations only estimate $V$ ; using relativistic momentum and energy gives a slightly lower required voltage.	Convert the wavelength: $\lambda=8.00\times10^{-12}\,\text{m}$ . Then $p=h/\lambda=(6.63\times10^{-34})/(8.0\times10^{-12})=8.2875\times10^{-23}\,\text{kg m s}^{-1}$ . The classical kinetic energy is $K=p^2/(2m_{\mathrm e})=(8.2875\times10^{-23})^2/[2(9.11\times10^{-31})]=3.7695\times10^{-15}\,\text{J}$ . Therefore $V=K/e=(3.7695\times10^{-15})/(1.60\times10^{-19})=2.3559\times10^4\,\text{V}$ , giving $2.36\times10^4\,\text{V}$ to three significant figures. The classical momentum implies $v=p/m_{\mathrm e}=9.10\times10^7\,\text{m s}^{-1}\approx0.30c$ . At this fraction of $c$ the classical relations are only approximate: for a fixed momentum the relativistic kinetic energy is slightly smaller than $p^2/(2m)$ , so an exact treatment gives a slightly lower accelerating voltage.	3
03.1	Electrons are charged, so magnetic fields can deflect and focus the beam, whereas glass lenses refract light rather than electron paths. The electron column must be evacuated because air molecules would scatter the electron beam.	A current in an electromagnetic lens produces a magnetic field that exerts a force on moving electrons and brings selected trajectories to a focus. A glass lens has no corresponding refractive-index action on an electron beam. A vacuum prevents air molecules from scattering the electrons before they form the image.	3

Tier 3 · Hard

Mark scheme for 3.12.2.6 Tier 3 · Hard
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01.1	$6.03\times10^4\,\text{V}$	Use $V=h^2/(2m_{\mathrm e}e\lambda^2)$ . With $\lambda=5.00\times10^{-12}\,\text{m}$ , $V=(6.63\times10^{-34})^2/[2(9.11\times10^{-31})(1.60\times10^{-19})(5.00\times10^{-12})^2]=6.03\times10^4\,\text{V}$ . In a TEM the high-voltage electrons pass through a thin specimen and electromagnetic lenses form an image from transmitted and scattered electrons; the short wavelength supports high resolution. An STM instead scans a sharp conducting tip across a conducting surface and measures the strongly distance-dependent tunnelling current. Its atomic sensitivity is therefore a tunnelling effect, not a consequence of TEM-style anode-voltage wavelength reduction.	5
02.1	A potential difference is applied between a sharp conducting tip and the conducting sample. Although the electrons lack the classical energy to cross the gap, their wavefunctions extend through it and produce a tunnelling current. The current changes very rapidly with tip–sample separation. Feedback adjusts the tip height to keep the current constant as the tip scans, and the recorded height changes form an atomic-scale surface map. Both tip and sample must conduct so that a measurable current can flow.	Credit the applied potential difference, quantum tunnelling across the classically forbidden gap, strong current–separation dependence, feedback conversion into a height map and the need for a complete conducting path.	5
03.1	The separation decreases by $8.96\times10^{-11}\,\text{m}=0.0896\,\text{nm}$ . The tunnelling probability, and hence current, depends exponentially on barrier width, so a sub-nanometre change produces a large current ratio.	For the two positions, $I_2/I_1=\exp[-2\kappa(s_2-s_1)]$ . Hence $6.00=\exp(-2\kappa\Delta s)$ and $\Delta s=-\ln6.00/[2(1.00\times10^{10})]=-8.96\times10^{-11}\,\text{m}$ . The negative sign means that the fixed-height tip is closer to the surface. Electron wavefunctions penetrate the classically forbidden gap, and the crossing probability falls exponentially as the gap widens. This strong barrier-width dependence gives atomic-scale sensitivity.	3
04.1	The $72.0\,\text{kV}$ electrons have half the wavelength, giving an ideal twofold improvement in wavelength-limited resolution.	Non-relativistically, $\lambda\propto1/\sqrt V$ , so $\lambda_{72}/\lambda_{18}=\sqrt{18.0/72.0}=0.500$ . A wavelength-limited detail size therefore halves, corresponding to a factor-of-two ideal improvement in resolution at the higher anode voltage.	3
05.1	On a rough surface an unexpectedly high feature would narrow the tip-surface gap; in constant-height mode the tip cannot move away and may strike the feature. In constant-current mode the feedback lifts the tip whenever the current starts to rise, so the separation is maintained and the tip is protected while the surface is followed.	The justification rests on what each mode does when the gap suddenly narrows: constant-height mode records a current change but holds the tip's height fixed, so a tall feature can collide with the tip; constant-current mode responds by retracting the tip to restore the set current, keeping a safe separation over unknown topography. Credit the collision risk, the feedback response, and the conclusion.	3

3.12.3.1 · The Michelson-Morley experiment

Tier 1 · Easy

Mark scheme for 3.12.3.1 Tier 1 · Easy
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01.1	A fringe shift when the interferometer was rotated.	An ether wind was expected to change the relative travel times along the perpendicular arms. Rotation would exchange their orientations and change the phase difference, moving the fringes.	1
02.1	It divides the incident light into beams travelling along two perpendicular arms.	The plate transmits part of the light and reflects the remainder, creating two coherent beams that can be recombined after their round trips.	1

Tier 2 · Standard

Mark scheme for 3.12.3.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Rotation exchanged the arm parallel to the proposed ether wind with the perpendicular arm, so an ether should change the phase difference; no shift meant absolute motion was not detected and supported an invariant light speed.	The beam splitter sends light along two perpendicular paths. In the ether model, their round-trip times differ according to orientation. A $90^\circ$ rotation swaps those orientations, so the predicted time difference changes sign and should move the interference pattern. No significant shift was observed. This removed evidence for a stationary ether or detectable absolute motion and was consistent with light having the same speed in every inertial direction.	3
02.1	The optical path change is less than $6.0\times10^{-9}\,\text{m}$ . The ether prediction corresponds to $2.40\times10^{-7}\,\text{m}$ , 40 times larger, so the null result contradicts that prediction.	Use $\Delta x=n\lambda$ . The measured limit is $(0.010)(600\times10^{-9})=6.0\times10^{-9}\,\text{m}$ . The ether model predicts $(0.400)(600\times10^{-9})=2.40\times10^{-7}\,\text{m}$ , and $(2.40\times10^{-7})/(6.0\times10^{-9})=40$ .	3
03.1	Coherence gives the returning beams a stable phase relationship and therefore stable fringes. The source is not perfectly monochromatic, so fringes are visible only when the path difference is very small (near zero).	Award one mark for coherence giving a stable phase relationship and stationary fringes. Award one for stating that the source is not perfectly monochromatic, so fringes are visible only when the path difference is very small (near zero). The plate's beam-splitting purpose is not a credit point here.	2

Tier 3 · Hard

Mark scheme for 3.12.3.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.400$ fringes	The initial time difference is $\Delta t=Lv^2/c^3=11.0(3.00\times10^4)^2/(3.00\times10^8)^3=3.67\times10^{-16}\,\text{s}$ . Rotation exchanges the arms, so the change is twice the initial path difference and $n=2c\Delta t/\lambda=2(3.00\times10^8)(3.67\times10^{-16})/(550\times10^{-9})=0.400$ . A shift of this scale was the ether model's testable prediction. Its absence meant the expected directional difference in light speed was not present, undermining absolute ether motion and supporting invariant $c$ .	5
02.1	If Earth moved through a stationary ether, rotating the perpendicular arms would change their relative light-travel times and shift the fringes. Earth's orbital velocity also changes direction during the year, so Earth could not remain accidentally at rest relative to the ether for repeated measurements. Persistent null results therefore removed evidence for a privileged ether frame or absolute motion and were consistent with invariant light speed.	The apparatus compared round-trip light times along perpendicular arms. If Earth moved through a stationary ether, the predicted time difference depended on arm orientation, so rotating the instrument should change the fringes. Earth's orbital motion also changes direction through the year, making permanent rest in the ether implausible. No reproducible orientation-dependent shift appeared. The result therefore failed to detect any absolute frame and was consistent with every inertial observer measuring the same free-space light speed.	5
03.1	The known mirror displacement calibrates the response and confirms that the apparatus can resolve a small fringe shift. Enclosing the paths reduces thermal and air-current changes in optical path. Repeating clockwise and anticlockwise rotations tests whether any change is reproducible rather than mechanical backlash or drift. A persistent null after these checks is therefore evidence rather than a failure of sensitivity.	The imposed mirror displacement calibrates the fringe response and proves that a shift of the predicted size would be resolved, rather than hidden by an insensitive detector. Enclosing the paths reduces changes in refractive index and physical length caused by air currents or temperature. Agreement between clockwise and anticlockwise rotations helps separate a reproducible orientation-dependent effect from mechanical backlash or slow drift. A null result that persists after these sensitivity and systematic-error checks is therefore credible.	4
04.1	$v<5.78\times10^3\,\text{m s}^{-1}$ to three significant figures	Rearrange the predicted shift to $v=c\sqrt{n\lambda/(2L)}$ . Using the detection limit gives $v<(3.00\times10^8)\sqrt{(0.012)(520\times10^{-9})/[2(8.40)]}=5.78\times10^3\,\text{m s}^{-1}$ . Any larger ether speed would predict a resolvable periodic shift under the stated model, so the null result places this upper bound rather than proving that Earth is stationary.	5
05.1	The temperature difference produces a $0.498$ -fringe shift, about $2.77$ times the proposed ether shift, so thermal control and rotation-dependent repetition are essential.	The warmer arm lengthens by $\Delta L=\alpha L\Delta T=(1.10\times10^{-6})(9.20)(0.015)=1.518\times10^{-7}\,\text{m}$ . Light traverses the arm twice, so the optical path change is $2\Delta L=3.036\times10^{-7}\,\text{m}$ . The shift is $n=2\Delta L/\lambda=(3.036\times10^{-7})/(610\times10^{-9})=0.498$ fringe. Since $0.498/0.18=2.77$ , this systematic effect could mask or imitate the prediction unless paths are thermally stabilised and only reproducible orientation-dependent changes are accepted.	5

3.12.3.2 · Einstein's theory of special relativity

Tier 1 · Easy

Mark scheme for 3.12.3.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	A non-accelerating frame in which an object with no resultant force moves at constant velocity.	An inertial frame obeys Newton's first law. It may move at a constant velocity relative to another inertial frame, but it must not accelerate or rotate.	1
02.1	The speed of light in free space is invariant for all inertial observers.	Every inertial frame measures the same vacuum light speed $c$ , regardless of the motion of the source or observer.	1

Tier 2 · Standard

Mark scheme for 3.12.3.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Physical laws have the same form in all inertial frames and light in free space has invariant speed $c$ ; therefore perpendicular arms cannot reveal an ether wind through different measured light speeds.	First, no inertial frame is privileged because physical laws have the same form in all of them. Second, every inertial observer measures the vacuum speed of light as $c$ , regardless of source or observer motion. The two interferometer arms therefore do not acquire the classical directional speed difference required by an ether wind, so the absence of a rotation-dependent fringe shift is expected.	3
02.1	$3.00\times10^8\,\text{m s}^{-1}$ . Einstein's second postulate makes $c$ invariant in every inertial frame, so the Galilean result $1.720c$ is not valid for light.	Earth and the constant-velocity spacecraft define inertial frames. Einstein's second postulate requires the Earth observer to measure the free-space pulse speed as $c=3.00\times10^8\,\text{m s}^{-1}$ . Galilean addition would give $1.720c$ , but it does not preserve invariant $c$ and is therefore not valid for light.	3
03.1	The spacecraft is an inertial frame, and the laws of physics have the same form in every inertial frame, so none is a privileged rest frame. Internal experiments therefore cannot reveal absolute uniform motion; velocity can be measured only relative to another frame.	Award one mark for applying the first postulate: all inertial frames are equivalent, with no privileged rest frame. Award one for the conclusion that internal results cannot give an absolute velocity, only velocity relative to another frame.	2

Tier 3 · Hard

Mark scheme for 3.12.3.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The passenger moves towards the front flash and away from the rear flash, so receives the front flash first; because both observers measure each light pulse at $c$ , the passenger concludes the front event occurred earlier, showing simultaneity is frame-dependent.	In the platform frame, equal distances from simultaneous events mean the midpoint observer receives the pulses together. The passenger moves towards the pulse from the front and away from the pulse from the rear, so the front pulse reaches the passenger first. Einstein's second postulate forbids explaining this by assigning different light speeds: the passenger measures both pulses at $c$ . The passenger must instead infer unequal emission times in the train frame. This preserves the same physical laws in both inertial frames and demonstrates that simultaneity, unlike $c$ , is not invariant.	5
02.1	Spacecraft A defines an inertial frame because its velocity is constant in both magnitude and direction. Spacecraft B does not: its velocity direction continually changes, so it has centripetal acceleration even though its speed is constant. Einstein's postulates apply directly to inertial frames, so speed alone is not the criterion.	An inertial frame is non-accelerating. Straight-line motion at constant velocity satisfies this condition for A. In circular motion B's velocity vector changes continuously and the required centripetal acceleration makes its frame non-inertial. Having the same instantaneous speed does not remove that acceleration.	4
03.1	$5.00\times10^{-7}\,\text{m}$ and $3.75\times10^{-7}\,\text{m}$ . The claim is false: both frequency and wavelength may differ between frames, with reciprocal changes that leave each measured product $f\lambda=c$ .	Award one mark for $\lambda_1=c/f_1=5.00\times10^{-7}\,\text{m}$ , one for $\lambda_2=c/f_2=3.75\times10^{-7}\,\text{m}$ , and one for concluding that invariant $c=f\lambda$ does not require either frequency or wavelength separately to be invariant.	3
04.1	The argument is incomplete. The first postulate states that physical laws have the same form in all inertial frames and removes any privileged inertial frame. The second postulate specifically states that the speed of light in free space is invariant for all inertial observers, independent of source or observer motion. Together they require transformations of space and time rather than Galilean addition.	Separate equivalence of frames from the numerical light-speed statement. The first postulate prevents an internal experiment from identifying absolute uniform motion because the same laws apply in every inertial frame. It does not by itself state the measured value of a light speed. The second postulate supplies that additional invariant, $c$ . Preserving both statements forces measured intervals and simultaneity to depend on frame, so classical velocity addition cannot be retained for light.	4
05.1	Earth and the constant-velocity source frame are inertial, so physical laws have the same form in each and neither is a privileged rest frame. The second postulate requires every inertial observer to measure free-space light at $c$ , independent of source motion. The equal measured speeds therefore support Einstein's prediction and reject the Galilean values; source motion may change measured frequency and wavelength, but not their product $f\lambda=c$ .	First identify both constant-velocity frames as inertial. The first postulate prevents Earth from being treated as the unique frame in which the correct laws apply. The second postulate supplies the discriminating numerical prediction: forward and backward light are each measured at $c$ , not $c$ with the source velocity added or subtracted. Agreement with $c$ in both directions therefore contradicts the Galilean model. Frame-dependent Doppler shifts remain possible because frequency and wavelength can change reciprocally while $c$ stays invariant.	4

3.12.3.3 · Time dilation

Tier 1 · Easy

Mark scheme for 3.12.3.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	The time between two events measured by one clock in the frame where the events occur at the same position.	For a proper-time measurement the same clock must be present at both events, so no synchronisation of separated clocks is required.	1
02.1	The particle's rest frame.	Creation and decay occur at the same position in the particle frame, so one co-moving clock can time both events and measures the proper interval.	1

Tier 2 · Standard

Mark scheme for 3.12.3.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.67\,\mu\text{s}$ and $880\,\text{m}$	The Lorentz factor is $\gamma=1/\sqrt{1-0.800^2}=1.667$ . Hence $t=\gamma t_0=(1.667)(2.20\,\mu\text{s})=3.67\,\mu\text{s}$ . The laboratory distance is $x=vt=(0.800)(3.00\times10^8)(3.67\times10^{-6})=8.80\times10^2\,\text{m}$ .	3
02.1	$1.98\times10^8\,\text{m s}^{-1}$	Use the exact lifetime ratio $\gamma=t/t_0=6.00/4.50=4/3$ . Since $\gamma=1/\sqrt{1-v^2/c^2}$ , $v=c\sqrt{1-(3/4)^2}=(3.00\times10^8)\sqrt7/4=1.98\times10^8\,\text{m s}^{-1}$ to three significant figures.	3
03.1	$22.5\,\text{min}$ ; the $18.0\,\text{min}$ spacecraft reading is the proper time.	One spacecraft clock is present at both emission events, so $t_0=18.0\,\text{min}$ . At $0.600c$ , $\gamma=1/\sqrt{1-0.600^2}=1.25$ . Earth measures the dilated interval $t=\gamma t_0=(1.25)(18.0)=22.5\,\text{min}$ .	3

Tier 3 · Hard

Mark scheme for 3.12.3.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.219$ survive with time dilation, compared with $2.44\times10^{-7}$ without it	For $v=0.995c$ , $\gamma=1/\sqrt{1-0.995^2}=10.01$ . The laboratory travel time is $t=10.0\times10^3/[0.995(3.00\times10^8)]=33.50\,\mu\text{s}$ . The proper time experienced by the muons is $t_0=t/\gamma=3.346\,\mu\text{s}$ . Thus the surviving fraction is $N/N_0=e^{-t_0/\tau_0}=e^{-3.346/2.20}=0.219$ . Without time dilation one would use $33.50\,\mu\text{s}$ as the decay time, giving $e^{-33.50/2.20}=2.44\times10^{-7}$ . The observed survival is therefore evidence for time dilation.	5
02.1	$0.832c$ and $10.8$ years, each to three significant figures	The spacecraft time is proper time $t_0=6.00\,\text{y}$ . In the Earth frame, $L=vt=\beta c\gamma t_0$ , so $\beta\gamma=L/(ct_0)=9.00/6.00=1.50$ . Since $(\beta\gamma)^2=\gamma^2-1$ , $\gamma=\sqrt{1+1.50^2}=1.803$ and $\beta=1.50/1.803=0.832$ . Thus $v=0.832c$ . The Earth-frame time is $t=\gamma t_0=(1.803)(6.00)=10.8\,\text{y}$ .	4
03.1	The Earth clock records $10.0\,\text{y}$ and the spacecraft clock records $6.00\,\text{y}$ . On each leg the $3.00\,\text{y}$ interval on the spacecraft clock is the proper time because one co-moving clock is present at both endpoint events, which occur at the same position in that leg's spacecraft frame.	In the Earth frame, each leg takes $4.00/0.800=5.00\,\text{y}$ , so the round trip takes $10.0\,\text{y}$ . At $0.800c$ , $\gamma=1/\sqrt{1-0.800^2}=5/3$ . For either constant-velocity leg, $t_0=t/\gamma=5.00/(5/3)=3.00\,\text{y}$ , giving $6.00\,\text{y}$ over both legs. The departure and arrival events of one leg occur at one location in the co-moving frame and are timed by the same on-board clock, so this leg interval is proper.	4
04.1	The spacecraft clock records $3.60\,\text{y}$ , while Earth records $5.00\,\text{y}$ .	For the first segment, $\gamma_1=1/\sqrt{1-0.600^2}=1.25$ , so the spacecraft proper time is $3.00/1.25=2.40\,\text{y}$ . For the second, $\gamma_2=1/\sqrt{1-0.800^2}=5/3$ , giving $2.00/(5/3)=1.20\,\text{y}$ . The same spacecraft clock times both segments, so it records $2.40+1.20=3.60\,\text{y}$ , compared with the Earth total $3.00+2.00=5.00\,\text{y}$ .	4
05.1	$v=0.795c=2.39\times10^8\,\text{m s}^{-1}$ to three significant figures	The survival law in the particle frame is $N/N_0=\exp(-t_0/\tau_0)$ , so $t_0=-\tau_0\ln(0.250)=3.049\,\mu\text{s}$ . In the laboratory, $L=vt=\beta c\gamma t_0$ , giving $\beta\gamma=L/(ct_0)=1200/[(3.00\times10^8)(3.049\times10^{-6})]=1.3115$ . Since $\beta\gamma=\beta/\sqrt{1-\beta^2}$ , $\beta=(1.3115)/\sqrt{1+(1.3115)^2}=0.795$ . Therefore $v=2.39\times10^8\,\text{m s}^{-1}$ .	5

3.12.3.4 · Length contraction

Tier 1 · Easy

Mark scheme for 3.12.3.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Its length measured in the frame in which it is at rest.	The proper length uses the object's rest frame, where its endpoints have fixed positions and the length is greatest.	1
02.1	Only the component of length parallel to the relative motion contracts.	Special-relativistic contraction affects distances along the direction of motion; transverse dimensions are unchanged.	1

Tier 2 · Standard

Mark scheme for 3.12.3.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$96.0\,\text{m}$	The Lorentz factor is $\gamma=1/\sqrt{1-0.600^2}=1.25$ . Therefore $l=l_0/\gamma=120/1.25=96.0\,\text{m}$ .	3
02.1	$0.543c=1.63\times10^8\,\text{m s}^{-1}$	Use $l/l_0=\sqrt{1-v^2/c^2}$ . Here $l/l_0=63.0/75.0=0.840$ , so $v/c=\sqrt{1-0.840^2}=0.543$ . Hence $v=1.63\times10^8\,\text{m s}^{-1}$ to three significant figures.	3
03.1	The rod moves between the two measurements, so the recorded positions do not belong to the same instant in the student's frame. Both endpoint positions must be measured simultaneously in that frame, and their separation then gives the contracted length.	Length is the spatial separation of an object's endpoints at one time in the measuring frame. Sequential endpoint readings mix a position change caused by motion with the endpoint separation. Synchronised detectors or one simultaneous image must therefore record both ends in the student's frame.	2

Tier 3 · Hard

Mark scheme for 3.12.3.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$597\,\text{m}$ and $2.03\,\mu\text{s}$	The Lorentz factor is $\gamma=1/\sqrt{1-0.980^2}=5.025$ . In the particle frame the moving accelerator has length $l=l_0/\gamma=(3.00\times10^3)/5.025=597\,\text{m}$ . Its transit time there is $t_0=l/v=597/[0.980(3.00\times10^8)]=2.03\,\mu\text{s}$ . In the laboratory, $t=(3.00\times10^3)/[0.980(3.00\times10^8)]=10.2\,\mu\text{s}$ . Since $t/\gamma=10.2/5.025=2.03\,\mu\text{s}$ , the length-contraction and time-dilation descriptions agree.	5
02.1	$6.15\,\text{m}$ by $3.00\,\text{m}$ , with area $18.4\,\text{m}^2$	The Lorentz factor is $\gamma=1/\sqrt{1-0.640^2}=1.301$ . Only the dimension parallel to the motion contracts in the laboratory frame, giving $8.00/1.301=6.15\,\text{m}$ . The transverse side remains $3.00\,\text{m}$ . Therefore the laboratory area is $(6.15)(3.00)=18.4\,\text{m}^2$ to three significant figures.	4
03.1	The barn frame measures the moving pole as $7.20\,\text{m}$ , using pole proper length $12.0\,\text{m}$ . The pole frame measures the moving barn as $5.40\,\text{m}$ , using barn proper length $9.00\,\text{m}$ .	The contraction factor is $\sqrt{1-0.800^2}=0.600$ . In the barn frame the pole moves, so its rest-frame value $l_0=12.0\,\text{m}$ contracts to $l=(12.0)(0.600)=7.20\,\text{m}$ . In the pole frame the barn moves, so the barn's rest-frame value $l_0=9.00\,\text{m}$ contracts to $l=(9.00)(0.600)=5.40\,\text{m}$ . Proper length is the length measured in the object's own rest frame.	4
04.1	$4.23\,\text{m}$ to three significant figures	The stated $6.40\,\text{m}$ is the proper value of the component parallel to the motion. Use $l=l_0\sqrt{1-v^2/c^2}$ : $l=(6.40)\sqrt{1-0.750^2}=(6.40)(0.66144)=4.23\,\text{m}$ .	3
05.1	$1.40\,\text{m}$ and $0.713$ marker intervals per metre, each to three significant figures	The marker spacing is parallel to the motion, so $l=l_0\sqrt{1-v^2/c^2}=(2.75)\sqrt{1-0.860^2}=1.403\,\text{m}=1.40\,\text{m}$ . The laboratory therefore measures $1/l=1/1.403=0.7126\,\text{m}^{-1}=0.713$ marker intervals per metre. The count per metre rises because the simultaneous laboratory spacing is contracted.	4

3.12.3.5 · Mass and energy

Tier 1 · Easy

Mark scheme for 3.12.3.5 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.70\times10^6\,\text{J}$	Use $\Delta E=\Delta mc^2=(3.00\times10^{-11})(3.00\times10^8)^2=2.70\times10^6\,\text{J}$ .	2
02.1	$5.00\times10^{-12}\,\text{kg}$	Use $m=E/c^2=(450\times10^3)/(3.00\times10^8)^2=5.00\times10^{-12}\,\text{kg}$ .	2

Tier 2 · Standard

Mark scheme for 3.12.3.5 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$E=1.37\times10^{-13}\,\text{J}$ and $K=5.47\times10^{-14}\,\text{J}$	The Lorentz factor is $\gamma=1/\sqrt{1-0.800^2}=1.667$ . The rest energy is $m_0c^2=(9.11\times10^{-31})(3.00\times10^8)^2=8.20\times10^{-14}\,\text{J}$ . Hence $E=\gamma m_0c^2=(1.667)(8.20\times10^{-14})=1.37\times10^{-13}\,\text{J}$ . The kinetic energy is $K=E-m_0c^2=(\gamma-1)m_0c^2=5.47\times10^{-14}\,\text{J}$ .	3
02.1	Mass starts at $m_0$ and kinetic energy at zero; both rise increasingly steeply and tend to infinity as $v$ approaches $c$ , without reaching $c$ .	At rest, $\gamma=1$ , so $m=\gamma m_0=m_0$ and $K=(\gamma-1)m_0c^2=0$ . Both curves are initially shallow. Because $\gamma=1/\sqrt{1-v^2/c^2}$ grows without bound, both turn sharply upwards and have a vertical asymptote at $v=c$ .	3
03.1	$8.00\times10^{-11}\,\text{kg}$ and $6.67\times10^{-12}$ ; the balance could not detect the change.	Mass-energy equivalence gives $\Delta m=\Delta E/c^2=(7.20\times10^6)/(3.00\times10^8)^2=8.00\times10^{-11}\,\text{kg}$ . The fractional increase is $\Delta m/m=(8.00\times10^{-11})/12.0=6.67\times10^{-12}$ . This is smaller than $1.0\times10^{-10}$ , so it lies below the balance's stated fractional resolution.	3

Tier 3 · Hard

Mark scheme for 3.12.3.5 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$v_{\mathrm{rel}}=0.9936c=2.98\times10^8\,\text{m s}^{-1}$ ; $v_{\mathrm{class}}=1.19\times10^9\,\text{m s}^{-1}$	Relativistically, $K=(\gamma-1)m_0c^2$ , so $\gamma=1+4.00/0.511=8.83$ . Therefore $v/c=\sqrt{1-1/\gamma^2}=0.9936$ and $v=2.98\times10^8\,\text{m s}^{-1}$ . Classically, $K=4.00\times10^6(1.60\times10^{-19})=6.40\times10^{-13}\,\text{J}$ , so $v=\sqrt{2K/m_0}=\sqrt{2(6.40\times10^{-13})/(9.11\times10^{-31})}=1.19\times10^9\,\text{m s}^{-1}$ , which exceeds $c$ . Bertozzi found that measured speed approached but did not exceed $c$ while kinetic energy continued to rise, matching the relativistic prediction and rejecting the classical one.	5
02.1	$2.55\times10^{-13}\,\text{J}$ , $3.86\times10^{-14}\,\text{J}$ and a ratio of $6.62$ using unrounded energies $2.55\times10^{-13}\,\text{J}$ , $3.86\times10^{-14}\,\text{J}$ and a ratio of $6.61$ using the displayed three-significant-figure energies	At $0.970c$ , $\gamma=1/\sqrt{1-0.970^2}=4.113$ . Relativistically, $K=(\gamma-1)m_0c^2=(3.113)(9.11\times10^{-31})(3.00\times10^8)^2=2.55\times10^{-13}\,\text{J}$ . Classically, $K=\frac12m_0v^2=\frac12(9.11\times10^{-31})[0.970(3.00\times10^8)]^2=3.86\times10^{-14}\,\text{J}$ . Dividing unrounded energies gives $6.618$ , hence $6.62$ ; dividing the displayed three-significant-figure energies gives $6.61$ , which is also acceptable.	5
03.1	$2.50\times10^{-27}\,\text{kg}$ ; the additional $0.50\times10^{-27}\,\text{kg}$ is the mass equivalent of the original particles' kinetic energy.	The opposite momenta cancel, so the product can be stationary. At $0.600c$ , $\gamma=1/\sqrt{1-0.600^2}=1.25$ . The conserved total energy before collision is $2\gamma m_0c^2$ . After collision it is $Mc^2$ , so $M=2\gamma m_0=2(1.25)(1.00\times10^{-27})=2.50\times10^{-27}\,\text{kg}$ . The original rest masses sum to $2.00\times10^{-27}\,\text{kg}$ ; the excess is stored internal energy originating from their kinetic energy.	5
04.1	$K=1.84\,\text{MeV}$ , $v=0.976c=2.93\times10^8\,\text{m s}^{-1}$ and $m=4.19\times10^{-30}\,\text{kg}$ , each to three significant figures	The electron rate is $I/e=(25.0\times10^{-6})/(1.60\times10^{-19})=1.5625\times10^{14}\,\text{s}^{-1}$ . Therefore $K=P/(N/t)=46.0/(1.5625\times10^{14})=2.944\times10^{-13}\,\text{J}=1.84\,\text{MeV}$ . From $K=(\gamma-1)m_0c^2$ , $\gamma=1+1.84/0.511=4.6008$ . Hence $v/c=\sqrt{1-1/\gamma^2}=0.9761$ and $v=2.93\times10^8\,\text{m s}^{-1}$ . The relativistic mass is $m=\gamma m_0=(4.6008)(9.11\times10^{-31})=4.19\times10^{-30}\,\text{kg}$ .	5
05.1	$\Delta K_{\mathrm{rel}}=2.45\,\text{MeV}$ and $\Delta K_{\mathrm{class}}=0.0435\,\text{MeV}$ ; the relativistic requirement is about $56.4$ times larger	The Lorentz factors are $\gamma_{0.900}=1/\sqrt{1-0.900^2}=2.29416$ and $\gamma_{0.990}=1/\sqrt{1-0.990^2}=7.08881$ . Hence $\Delta K=(7.08881-2.29416)(0.511)=2.450\,\text{MeV}$ . Classically, $\Delta K=\frac12m_0c^2(\beta_2^2-\beta_1^2)=\frac12(0.511)(0.990^2-0.900^2)=0.04346\,\text{MeV}$ . The ratio is $2.450/0.04346=56.4$ . Relativistic energy rises increasingly steeply as $v$ approaches $c$ , whereas the classical expression incorrectly permits continued modest energy increments and ultimately speeds beyond $c$ .	5