E.5 Fusion and stars: IB Physics HL exam-style questions
Stars are powered by fusion, which needs very high temperatures and densities to overcome the electrostatic repulsion between nuclei. A main-sequence star is stable because radiation pressure balances gravitational collapse, and its later evolution depends on its mass.
Questions use the Hertzsprung–Russell diagram, stellar parallax, Wien's law and the Stefan–Boltzmann law to find the distance, temperature, luminosity and radius of a star, and follow its path to a white dwarf, neutron star or black hole.
55 questions
278 marks
Paper 1A: 30
Paper 1B: 9
Paper 2: 16
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32 practice questions on E.5 Fusion and stars
1E-1A-09
Nuclear fusion·E.5 Fusion and stars
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The core of a contracting protostar is modelled as a fixed mass of ideal gas. As the protostar contracts, the radius of the core halves and the temperature of the core becomes three times larger. The number of particles in the core does not change.
By what factor does the pressure in the core change?
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Step 1PV = NkBT, so P ∝ T/V for a fixed number of particles.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Halving the radius divides the volume by 2³ = 8; tripling T multiplies P by 3.
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Step 3P changes by 3 × 8 = 24: the rise in density and temperature is what makes fusion possible once the core is hot and dense enough.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis multiplies by the volume factor (3 × 1/8) instead of dividing by it — pressure taken as proportional to volume.
BThis takes the volume as proportional to the radius (3 × 2).
CThis takes the volume as proportional to the radius squared, as though it were an area (3 × 4).
DCorrect: V ∝ R³, so P rises by 3 × 2³ = 24.
Syllabus understandingE.5 — the conditions leading to fusion in stars in terms of density and temperature; B.3 — the equations governing the behaviour of ideal gases as given by PV = NkBT (linking question: how can gas laws be used to model stars?) Command term: Deduce
2E-1A-10
Stellar evolution·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
The Hertzsprung–Russell diagram shows the Sun on the main sequence and four possible paths, A, B, C and D, that a star might follow after the hydrogen in its core is used up.
Which path represents the evolution of the Sun?
Hertzsprung–Russell diagram (logarithmic scales; temperature decreases to the right) with four possible paths from the Sun.Show mark scheme
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Step 1A star of about one solar mass expands and cools at its surface to become a red giant: it moves up and to the right.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2It later loses its outer layers as a planetary nebula; the exposed core is a hot, small, faint white dwarf at the lower left.
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Step 3Only path A goes first to the upper right and then to the lower left.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: main sequence → red giant (upper right) → white dwarf (lower left).
BMoving up the main sequence would need a more massive star; when its core hydrogen is used up, a one-solar-mass star swells and its surface cools, so it moves to the right.
CThe red-giant stage is right, but the remnant is a hot white dwarf at the lower left, not a cool faint star at the lower right.
DThis misses the red-giant stage: the star does not go directly from the main sequence to a white dwarf.
Syllabus understandingE.5 — the effect of stellar mass on the evolution of a star; the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions; guidance: the sketching and interpretation of HR diagrams, including the location of main sequence stars, red giants, super giants and white dwarfs Command term: Identify
3E-1A-14
The Hertzsprung–Russell diagram·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A red giant and a white dwarf have the same luminosity. The surface temperature of the white dwarf is five times that of the red giant.
What is the ratio (radius of the red giant) / (radius of the white dwarf)?
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Step 1L = σ(4πR²)T⁴; equal luminosities give Rg²Tg⁴ = Rw²Tw⁴.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Rg/Rw = (Tw/Tg)² = 5².
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Step 3= 25: the cool giant must be much larger to radiate the same power.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is the temperature ratio; radius depends on its square.
BCorrect: R ∝ 1/T² at fixed luminosity.
CThis is the cube of the temperature ratio.
DThis is the fourth power — the ratio of emitted power per unit area, not of radius.
Syllabus understandingE.5 — the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions; how to determine stellar radii; B.1 — the Stefan–Boltzmann law as given by L = σAT⁴ Command term: Determine
4E-1A-21
Wien's law from black-body spectra·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
The graph shows the spectra of the light emitted by two stars, X and Y. Both stars may be treated as black bodies.
What is the ratio (surface temperature of X) / (surface temperature of Y)?
Intensity against wavelength for the light from stars X and Y (the heights are not to the same scale).Show mark scheme
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Step 1Read the peak wavelengths: λmax ≈ 400 nm for X and ≈ 1000 nm for Y.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Wien's law λmaxT = 2.9 × 10−3 m K gives T ∝ 1/λmax.
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Step 3TX/TY = 1000/400 = 2.5 (about 7300 K and 2900 K).
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis takes T ∝ λmax: the ratio is inverted.
BThis takes the square root of 2.5, as though T ∝ 1/√λmax.
CCorrect: T ∝ 1/λmax.
DThis is 2.54, the ratio of the power emitted per unit area (σT⁴), not of the temperatures.
Syllabus understandingE.5 — that the surface temperature of a star can be determined from the stellar spectrum; B.1 — the emission spectrum of a black body and the determination of its temperature using Wien's displacement law λmaxT = 2.9 × 10−3 m K Command term: Determine
5E-1A-26
Apparent brightness·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Star X has luminosity 4L and is at distance 2d from the Earth. Star Y has luminosity L and is at distance d.
What is the ratio (apparent brightness of X)/(apparent brightness of Y)?
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Step 1b = L/4πd².
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2For X: 4L/(4π × 4d²) = L/4πd².
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Step 3The same as Y — ratio 1.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AIgnores the factor 4 in luminosity and uses b ∝ 1/d: 1/2.
BCorrect: the factor 4 in luminosity is cancelled by (2d)² = 4d².
CThis uses 4/2 — inverse-square not applied.
DThis ignores the greater distance of X.
Syllabus understandingB.1 — the concept of apparent brightness b; luminosity L of a body as given by b = L/4πd² (linking question: where do inverse square law relationships appear in other areas of physics?) Command term: Determine
6E-1A-50
Stellar equilibrium·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksExplain
The rate of fusion in the core of a main-sequence star briefly increases.
Which sequence describes how the star returns to equilibrium?
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Step 1More fusion releases energy faster, so the outward radiation (and gas) pressure rises above what is needed to balance gravity.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The core expands; an expanding gas does work and cools, and its density falls.
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Step 3Lower temperature and density reduce the rate of fusion, so the pressure falls back until it balances gravity again (negative feedback).
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThe extra energy release raises the outward pressure; gravity is unchanged, so it does not "win". This sequence would also run away rather than restore equilibrium.
BThe direction of the imbalance is right, but an expanding core cools; a hotter core would make the change grow instead of correcting it.
CThe extra energy release raises the outward pressure rather than lowering it, and a contracting core heats rather than cools.
DCorrect: expansion lowers the core temperature and density, which slows fusion — a self-regulating balance.
Syllabus understandingE.5 — that the stability of stars relies on an equilibrium between outward radiation pressure and inward gravitational forces; the conditions leading to fusion in stars in terms of density and temperature Command term: Explain
7E-1A-51
Stellar parallax·E.5 Fusion and stars
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A star has luminosity L and its apparent brightness at the Earth is b. One parsec is equal to P metres.
What parallax angle, in arc-seconds, would be measured for this star from the Earth?
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Step 1b = L/4πd² gives d = √(L/4πb) metres.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2In parsecs: d = √(L/4πb)/P.
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Step 3p (arc-second) = 1/d(pc) = P√(4πb/L).
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: d from the inverse-square law, converted to parsecs, then inverted.
BThis is the distance in parsecs; it has not been inverted to give the parallax angle.
CThe factor 4π has been omitted from b = L/4πd².
DThe square root has been omitted: this is proportional to 1/d², not 1/d.
Syllabus understandingE.5 — the use of stellar parallax as a method to determine the distance d to celestial bodies as given by d(parsec) = 1/p(arc-second); B.1 — the apparent brightness b = L/4πd²; conversion between metres and parsecs Command term: Deduce
8E-1A-52
Nuclear fusion·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Three changes to the core of a star are proposed. Each change is made on its own, with everything else kept the same:
I. increasing the temperature of the core II. increasing the density of the core III. replacing the hydrogen nuclei by nuclei of larger proton number
Which changes would increase the rate of fusion in the core?
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Step 1I: at a higher temperature the nuclei have greater kinetic energies, so more of them overcome the electric repulsion and get close enough to fuse.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: at a higher density the nuclei collide more often, so more fusion reactions occur per second.
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Step 3III: nuclei with more protons repel each other more strongly, so at the same temperature fewer get close enough to fuse — the rate decreases.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
ADensity also matters: more nuclei per unit volume means more frequent collisions.
BTemperature also matters: it decides what fraction of collisions overcome the electric repulsion.
CCorrect: higher temperature and higher density both raise the rate; a larger nuclear charge lowers it.
DIII would reduce the rate, because the electric repulsion between the nuclei is larger.
Syllabus understandingE.5 — the conditions leading to fusion in stars in terms of density and temperature; D.2 — the electric repulsion between like charges (Coulomb's law) Command term: Deduce
9E-1A-53
Binding energy and fusion·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
In a young star, deuterium fuses with protons: 21H + 11H → 32He + γ.
The binding energy per nucleon is 1.11 MeV for 21H and 2.57 MeV for 32He.
What is the energy released in one reaction?
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Step 1Total binding energy = (binding energy per nucleon) × (number of nucleons). A single proton has zero binding energy.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 3Energy released = increase in total binding energy = 7.71 − 2.22 = 5.49 MeV.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis is 2.57 − 1.11, the difference of the values per nucleon; the totals depend on the nucleon numbers (3 and 2).
BCorrect: 3 × 2.57 − 2 × 1.11 = 5.49 MeV.
CThis is 7.71 − 1.11: the deuteron's binding energy per nucleon has not been multiplied by its 2 nucleons.
DThis is 7.71 + 2.22: the binding energies have been added instead of subtracted.
Syllabus understandingE.5 — that fusion is a source of energy in stars; guidance: energy release calculations are required (E.3 — binding energy and binding energy per nucleon) Command term: Determine
10E-1A-54
Stellar parallax·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
Observed from the Earth six months apart, a nearby star appears to move by a total of 0.032 arc-second against the very distant background stars.
Which row gives the distance to the star?
Distance / pcDistance / ly
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Step 1The parallax angle p is half the total angular shift, because the baseline is 1 AU (the radius, not the diameter, of the Earth's orbit): p = 0.016 arc-second.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 31 pc = 3.26 ly, so d = 62.5 × 3.26 = 204 ly.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AUses the full shift 0.032 as p (giving 31 pc) and then divides by 3.26 instead of multiplying.
BUses the full shift 0.032 arc-second as the parallax angle: 1/0.032 = 31 pc = 102 ly.
CThe distance in parsecs is right but it has been divided by 3.26: 62.5/3.26 = 19.
DCorrect: p = 0.016 arc-second, 62.5 pc = 204 ly.
Syllabus understandingE.5 — the use of stellar parallax as a method to determine the distance d to celestial bodies as given by d(parsec) = 1/p(arc-second); guidance: conversion between AU, ly and pc Command term: Determine
11E-1A-55
Stellar parallax·E.5 Fusion and stars
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A proposed space telescope would orbit the Sun at a radius of 5.2 AU. It measures the parallax angle of a star, using the radius of its own orbit as the baseline, as 0.13 arc-second.
What is the distance to the star?
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Notes
Step 1By definition, a star at 1 pc shows a parallax of 1 arc-second for a baseline of 1 AU. For small angles p = (baseline)/d.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2With a baseline of 5.2 AU, the parallax of a star at distance d pc is 5.2/d arc-second.
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Step 3d = 5.2/0.13 = 40 pc.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis is 1/(0.13 × 5.2): the baseline has been used to divide rather than multiply — a longer baseline gives a larger angle for the same star.
BThis is 1/0.13: the Earth-based relation assumes a 1 AU baseline, so it underestimates the distance by the factor 5.2.
CThis is 5.2/(2 × 0.13): the parallax angle has been treated as a full shift and halved again.
Syllabus understandingE.5 — the use of stellar parallax as a method to determine the distance d to celestial bodies as given by d(parsec) = 1/p(arc-second); guidance: the conversion between astronomical units (AU), light years (ly) and parsecs (pc) is required Command term: Deduce
12E-1A-56
Stellar evolution·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
A main-sequence star has a mass of 15 solar masses.
Which sequence describes the evolution of the star after its core hydrogen is used up?
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Step 1A star of large mass becomes a red supergiant after the main sequence, fusing progressively heavier elements in its core.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2When the core can no longer release energy by fusion, it collapses and the star explodes as a supernova.
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Step 3The remnant core is a neutron star (or, for the most massive cores, a black hole).
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis is the path of a star of about one solar mass, such as the Sun.
BA supernova of a single massive star leaves a neutron star or black hole, not a white dwarf; low-mass stars do not become supernovae.
CCorrect: the path of a high-mass star.
DA planetary nebula is the gentle ejection of the envelope of a low-mass star; a neutron star forms only in the core collapse of a supernova.
Syllabus understandingE.5 — the effect of stellar mass on the evolution of a star Command term: Identify
13E-1A-57
The Hertzsprung–Russell diagram·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The Hertzsprung–Russell diagram shows the Sun and four stars W, X, Y and Z.
Which star has the largest radius?
Hertzsprung–Russell diagram (logarithmic scales on both axes) showing the Sun and stars W, X, Y and Z.Show mark scheme
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Step 1From L = 4πR²σT⁴, R ∝ √L/T².
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Reading the diagram: W ≈ 104L☉ at 25 000 K gives R ≈ √104 × (5800/25 000)² ≈ 5 R☉; X ≈ 103L☉ at 3500 K gives R ≈ √103 × (5800/3500)² ≈ 90 R☉.
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Step 3Y (≈ 0.01 R☉) and Z (≈ 0.3 R☉) are much smaller, so X — a red giant at the upper right — is the largest.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AW is the most luminous star, but it is very hot, so its high luminosity comes mainly from the T⁴ factor; its radius is only about 5 R☉.
BCorrect: cool yet luminous, so its surface area must be very large (about 90 R☉).
CY is the hottest star, but it is a white dwarf (about 0.01 R☉): high temperature with very low luminosity means a tiny surface.
DZ is the coolest star, but it is a faint main-sequence star (about 0.3 R☉); low temperature alone does not mean a large radius.
Syllabus understandingE.5 — the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions; how to determine stellar radii (B.1 — L = σAT⁴) Command term: Deduce
14E-1A-58
The Hertzsprung–Russell diagram·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
The Hertzsprung–Russell diagram shows four regions, P, Q, R and S.
In which region are stars whose outer layers expand and contract periodically, so that their luminosity varies?
Hertzsprung–Russell diagram (logarithmic scales; temperature decreases to the right) with four regions P, Q, R and S.Show mark scheme
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Step 1Stars whose outer layers pulsate lie in the instability strip.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The instability strip is a narrow, nearly vertical band that crosses the main sequence and extends up through the giants and supergiants.
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Step 3This is region S.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AP is the main sequence, where stars fuse hydrogen in a stable equilibrium and their luminosity is constant.
BQ contains red giants and supergiants: large and cool, but not, in general, pulsating.
CR contains white dwarfs: small, hot remnants in which fusion has stopped; they cool steadily.
DCorrect: S is the instability strip.
Syllabus understandingE.5 — the main regions of the Hertzsprung–Russell diagram and how to describe the main properties of stars in these regions (including the instability strip) Command term: Identify
15E-1A-59
The Hertzsprung–Russell diagram·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
On a Hertzsprung–Russell diagram, luminosity is plotted on a logarithmic scale against surface temperature on a logarithmic scale that decreases to the right. Lines of constant radius are straight lines on this diagram.
Which row describes the lines of constant radius?
Change in luminosity when the temperature halves along one lineLines for larger radii lie towards the
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Step 1At constant R, L = 4πR²σT⁴ gives L ∝ T⁴: halving T divides L by 2⁴ = 16.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2At a fixed temperature, a larger radius gives a larger luminosity (L ∝ R²), so the line lies higher.
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Step 3Higher on the diagram and, equivalently, towards the cool (right-hand) side: larger radii lie towards the top right, where the giants and supergiants are.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AUses L ∝ T² (confusing the dependence on radius with that on temperature) and places large stars with the white dwarfs.
BThe direction is right but the factor uses L ∝ T² instead of T⁴.
CThe factor is right, but the bottom left is where the smallest stars (white dwarfs) lie.
DCorrect: L ∝ T⁴ along a line; larger radii towards the top right.
Syllabus understandingE.5 — how to determine stellar radii; guidance: the sketching and interpretation of HR diagrams, including lines of constant radius Command term: Deduce
16E-1A-70
Stellar spectra and luminosity·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
Two stars X and Y have the same radius. The wavelength at which the spectrum of Y peaks is half the wavelength at which the spectrum of X peaks.
What is (luminosity of Y)/(luminosity of X)?
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Step 1Wien's law: λmaxT = constant, so halving λmax doubles the surface temperature.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2L = σ4πR²T⁴ with equal radii, so LY/LX = 2⁴ = 16.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis takes temperature to be proportional to the peak wavelength, inverting Wien's law.
BThis takes luminosity to be proportional to temperature instead of to T⁴.
CThis takes luminosity to be proportional to T², confusing the R² and T⁴ factors.
DCorrect: TY = 2TX and L ∝ T⁴ at fixed radius.
Syllabus understandingE.5 — guidance: the surface temperature of a star can be determined from the stellar spectrum; B.1 — Wien's displacement law λmaxT = 2.9 × 10−3 m K and the Stefan–Boltzmann law L = σAT⁴ Command term: Determine
17E-1A-71
The Hertzsprung–Russell diagram·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Stars P and Q are both on the main sequence. The mass of Q is greater than the mass of P.
Which row compares the surface temperature and the radius of Q with those of P?
Surface temperature of QRadius of Q
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Step 1On the main sequence the luminosity rises steeply with mass, so Q lies higher up the main sequence, towards the upper left of the HR diagram.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Further to the left means a higher surface temperature: Q is hotter than P.
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Step 3The main sequence crosses the lines of constant radius, from about 0.1 R☉ at its faint, cool end to about 10 R☉ at its luminous, hot end (L = σ4πR²T⁴ with L rising faster than T⁴), so Q also has the larger radius.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: a more massive main-sequence star is hotter, larger and more luminous.
BThis links a high surface temperature with a small radius, as for white dwarfs at the lower left; along the main sequence both increase with mass.
CThis places the more massive star in the cool, luminous red-giant region at the upper right, which main-sequence stars do not occupy.
DThis reverses the main sequence, taking the more massive stars to lie at its cool, faint lower end.
Syllabus understandingE.5 — the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions; the effect of stellar mass on the evolution of a star; guidance: lines of constant radius on HR diagrams Command term: Deduce
18E-1A-124
Chemical composition from a stellar spectrum·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The diagram shows the absorption spectrum of a star and the emission spectra of four elements, W, X, Y and Z, measured in a laboratory. All the spectra are drawn to the same wavelength scale.
Which elements are shown by these spectra to be present in the outer layers of the star?
Absorption spectrum of the star (top) and laboratory emission spectra of the elements W, X, Y and Z (same wavelength scale).Show mark scheme
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Step 1A dark line is produced when atoms in the cooler outer layers of the star absorb photons whose energies equal the differences between their energy levels, so an element absorbs at the same wavelengths at which it emits.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Every line of X (434, 486 and 656 nm) and every line of Y (447, 502 and 588 nm) appears as a dark line in the spectrum of the star: X and Y are present.
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Step 3Only one of the three lines of Z (486 nm) matches a dark line, and that line is already accounted for by X; none of the lines of W appears. Neither W nor Z is shown to be present.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis ignores the three fainter dark lines. An element is identified by the wavelengths of its lines, not by how dark they are, and all three lines of Y are present.
BCorrect: the whole line pattern of X and the whole line pattern of Y appear as dark lines; Z and W do not match.
CThis counts Z because one of its lines coincides with a dark line. That line is accounted for by X, and the other two lines of Z are missing, so the pattern of Z is not present.
DThis inverts the meaning of the dark lines, treating them as wavelengths that the elements in the star cannot produce, and so chooses the elements whose line patterns are not matched by the dark lines.
Syllabus understandingE.5 — guidance: the surface temperature and the chemical composition of a star can be determined from the stellar spectrum; E.1 — that the spectra of elements can be used to identify their presence (emission and absorption spectra) Command term: Deduce
19E-1A-125
Pressure inside a star in equilibrium·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A main-sequence star is in equilibrium. A thin spherical shell of gas lies deep inside the star.
Which row compares the pressure on the inner surface of the shell with the pressure on its outer surface, and gives the direction of the resultant force exerted on the shell by these pressures?
Pressure on inner surface compared with outer surfaceDirection of resultant force due to the pressures
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Step 1The gravitational force on the shell, due to the mass inside it, acts inwards towards the centre of the star.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The shell is in equilibrium, so the resultant force due to the gas and radiation pressure must be equal and opposite: it acts outwards.
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Step 3An outward resultant force needs a greater pressure on the inner surface than on the outer surface: the pressure increases towards the centre of the star.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the pressure increases inwards, so the resultant pressure force on the shell acts outwards and balances its weight.
BWith equal pressures there is no resultant pressure force, so nothing balances the inward gravitational force. Equilibrium needs zero resultant force on the shell, not zero pressure difference.
CA smaller pressure on the inner surface gives an inward pressure force that adds to gravity, so the shell would fall inwards.
DThe pressure comparison is right, but the resultant pressure force acts from the region of higher pressure towards the region of lower pressure, which is outwards here.
Syllabus understandingE.5 — that the stability of stars relies on an equilibrium between outward radiation pressure and inward gravitational forces Command term: Deduce
20E-1A-126
Temperature needed for helium fusion·E.5 Fusion and stars
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A simple model of fusion assumes that two nuclei fuse when their surfaces touch, and that the core temperature needed for fusion is proportional to the electric potential energy of two such nuclei. A nucleus of nucleon number A has radius R0A1/3.
What, according to this model, is (temperature needed for two 42He nuclei to fuse)/(temperature needed for two 11H nuclei to fuse)?
Show mark scheme
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Notes
Step 1Ep = kq1q2/r, where r is the sum of the two radii. Two protons: q1q2 = e², r = 2R0.
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All 3 steps must be completed — there is no mark for a part-answer.
AThis gives a charge of 2e to only one of the helium nuclei: 2/41/3 = 1.26.
BCorrect: the product of the charges is 4 times larger and the separation 41/3 times larger, so the ratio is 42/3 = 2.5.
CThis includes the fourfold product of the charges but ignores the greater separation of the larger helium nuclei.
DThis multiplies by the separation ratio 41/3 instead of dividing by it, as though the potential energy increased with separation: 4 × 41/3 = 6.35.
Syllabus understandingE.5 — the conditions leading to fusion in stars in terms of density and temperature; E.1 (HL) — the relationship between the radius and the nucleon number for a nucleus as given by R = R0A1/3; D.2 — electric potential energy Ep = kq1q2/rCommand term: Deduce
21E-1A-127
Neutrinos from fusion in the Sun·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
In the core of the Sun the net result of hydrogen fusion is 411H → 42He + 2e+ + 2ν, and 26.7 MeV of energy is released for each helium nucleus formed. Ignore the energy carried away by the neutrinos.
The luminosity of the Sun is 3.85 × 1026 W and the distance from the Sun to the Earth is 1.50 × 1011 m.
What is the number of neutrinos from the Sun that pass through one square metre at the Earth, facing the Sun, every second?
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Notes
Step 1Number of helium nuclei formed per second = L/E = 3.85 × 1026/(26.7 × 1.60 × 10−13) = 9.0 × 1037 s−1.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Two neutrinos are emitted for each helium nucleus: 1.8 × 1038 neutrinos per second, spreading out uniformly in all directions.
—
Step 3Number per square metre per second = 1.8 × 1038/(4π × (1.50 × 1011)²) = 6.4 × 1014 m−2 s−1.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis counts one neutrino for each helium nucleus formed; the equation shows two.
BCorrect: two neutrinos per reaction, spread over a sphere of radius 1 AU.
CThis counts one neutrino for each of the four protons fused, four per reaction instead of two.
DThis divides by d² instead of by the area 4πd² of the sphere over which the neutrinos spread.
Syllabus understandingE.5 — that fusion is a source of energy in stars; guidance: energy release calculations are required; E.3 — neutrinos; B.1 — apparent brightness b = L/4πd² (linking question: where do inverse-square relationships appear?) Command term: Determine
22E-1A-128
Ages of star clusters·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The Hertzsprung–Russell diagram shows the stars of two clusters, P and Q. All the stars in a cluster formed at about the same time.
Which row identifies the older cluster and gives the reason?
Hertzsprung–Russell diagram of the stars in clusters P and Q (both axes logarithmic; temperature decreases to the right). The shaded band is the main sequence.
Older clusterReason
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Notes
Step 1Main-sequence stars of greater mass are more luminous and use up the hydrogen in their cores much faster, so a cluster loses its main-sequence stars from the top (upper left) downwards as it ages.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2In P the main sequence still extends to hot stars of several thousand L☉, and there are no giants: even its most massive stars are still on the main sequence.
—
Step 3In Q the main sequence ends at about 2 L☉ and the stars above that point have become red giants, so Q is older.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis assumes that stars become more luminous as they age on the main sequence; very luminous main-sequence stars are massive stars, which have short lives, so their presence shows that P is young.
BThis reads the absence of red giants as the end of evolution; the massive stars of P are still on the main sequence, so none has yet become a giant.
CCorrect: Q has lost its massive, luminous main-sequence stars, which have become red giants, so it is older.
DThe older cluster is right, but main-sequence stars do not slide down the main sequence as they age; the upper main sequence of Q is missing because those stars have left it.
Syllabus understandingE.5 — the effect of stellar mass on the evolution of a star; the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions; guidance: the sketching and interpretation of HR diagrams, including the location of main sequence stars, red giants, super giants and white dwarfs, and lines of constant radius Command term: Deduce
23E-1A-129
Mean density of a red giant·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
A red giant has a mass of 2.0 M☉ and a radius of 50 R☉.
What is (mean density of the red giant)/(mean density of the Sun)?
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Notes
Step 1ρ = M/(4/3πR³), so ρ ∝ M/R³.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Ratio = 2.0/50³ = 2.0/(1.25 × 105).
—
Step 3= 1.6 × 10−5: a red giant has a very low mean density.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis inverts the mass ratio: 1/(2.0 × 50³) = 4.0 × 10−6.
BCorrect: ρ ∝ M/R³.
CThis uses R² (as for an area) instead of R³: 2.0/50² = 8.0 × 10−4.
DThis takes the density to be inversely proportional to the radius: 2.0/50 = 4.0 × 10−2.
Syllabus understandingE.5 — the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions (red giants); B.3 — density ρ = m/VCommand term: Determine
24E-1A-130
Properties of white dwarfs·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksIdentify
A white dwarf has a mass similar to that of the Sun.
Which row describes the source of the energy that the white dwarf radiates, and its mean density compared with that of the Sun?
Source of the energy radiatedMean density compared with the Sun
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Notes
Step 1A white dwarf is the exposed core left when a star of low mass has lost its outer layers; fusion has stopped in it.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2It radiates the internal energy stored in it, so it slowly cools.
—
Step 3A mass similar to the Sun's within a radius about 100 times smaller (similar to the Earth's) means a mean density about 106 times greater than the Sun's.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis describes a main-sequence star, and a white dwarf is far denser than the Sun, not less dense.
BThe density is right, but helium fusion takes place in the cores of red giants; a white dwarf no longer fuses any element.
CThe energy source is right, but a hot star of very low luminosity must be very small, and a solar mass in an Earth-sized volume is extremely dense. Low mean density is a property of red giants.
DCorrect: no fusion, so it radiates stored internal energy, and it is extremely dense.
Syllabus understandingE.5 — the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions (white dwarfs); the effect of stellar mass on the evolution of a star Command term: Identify
25E-1A-131
Stellar radius from parallax, brightness and spectrum·E.5 Fusion and stars
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The parallax angle of a star is p arc-second, its apparent brightness is b and its spectrum peaks at the wavelength λmax. One parsec is P metres, the constant in Wien's displacement law is W (λmaxT = W) and σ is the Stefan–Boltzmann constant.
Which expression gives the radius of the star in metres?
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Notes
Step 1d = 1/p parsec = P/p metres, and T = W/λmax.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2L = 4πd²b and L = 4πR²σT4, so R² = d²b/(σT4): the 4π cancels.
—
Step 3R = (d/T²)√(b/σ) = (Pλmax²/(pW²))√(b/σ).
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis takes the distance as Pp metres, multiplying by the parallax angle instead of dividing by it.
BThis inverts Wien's law, taking T = λmax/W; the hotter the star, the smaller λmax.
CCorrect: d = P/p, T = W/λmax and R = (d/T²)√(b/σ).
DThis omits the 4π from the surface area of the star (L = σR²T4), so the 4π from L = 4πd²b no longer cancels.
Syllabus understandingE.5 — how to determine stellar radii; the use of stellar parallax as a method to determine the distance d to celestial bodies as given by d (parsec) = 1/p (arc-second); B.1 — apparent brightness b = L/4πd²; B.1 — the Stefan–Boltzmann law as given by L = σAT4; B.1 — Wien's displacement law λmaxT = 2.9 × 10−3 m K Command term: Deduce
26E-1A-132
Angular separation of a planet and its star·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
A planet moves around a star in a circular orbit of radius a astronomical units. The orbit is seen face-on from the Earth. The parallax angle of the star is p arc-second.
What is the angle, in arc-second, between the directions of the planet and the star as seen from the Earth?
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Notes
Step 1By the definition of the parsec, a length of 1 AU seen face-on from a distance of 1 pc subtends an angle of 1 arc-second.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2For small angles the angle is proportional to (length)/(distance), so the angle in arc-second = a (AU)/d (pc).
—
Step 3d = 1/p pc, so the angle = ap arc-second.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis uses p as the distance in parsec; the distance is 1/p parsec.
BThis inverts the small-angle relation, taking the angle as (distance)/(length).
CThis uses the diameter of the orbit, 2a; the planet is a distance a from the star.
DCorrect: angle = a/d(pc) = ap, using the definition of the parsec.
Syllabus understandingE.5 — the use of stellar parallax as a method to determine the distance d to celestial bodies as given by d (parsec) = 1/p (arc-second); guidance: the conversion between astronomical units (AU), light years (ly) and parsecs (pc) is required Command term: Deduce
27E-1A-133
Long-term change in a stellar core·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The core of a main-sequence star is a fully ionised gas. In the fusion of hydrogen, four protons and four electrons are replaced by one helium nucleus and two electrons (the two positrons emitted annihilate with two electrons).
Which statement describes the long-term effect of this fusion on the core?
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Notes
Step 1Each conversion replaces 8 particles (4 protons and 4 electrons) by 3 particles (1 helium nucleus and 2 electrons).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the same volume and temperature, PV = NkBT gives a lower pressure, so the pressure no longer balances the inward gravitational force on the core.
—
Step 3The core contracts; gravitational potential energy is transferred to the particles, so the temperature and density of the core rise until the pressure again balances gravity (and the fusion rate rises slowly).
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: fewer particles give less pressure, so gravity compresses the core, which heats up.
BThe fall in the number of particles is right, but a lower pressure lets gravity compress the core: it contracts and heats up rather than expanding.
CThis assumes that the heavier helium nuclei or the energy released add particles; in fact 8 particles become 3, and the core contracts.
DConservation of mass–energy does not mean conservation of the number of particles: 8 particles become 3.
Syllabus understandingE.5 — that the stability of stars relies on an equilibrium between outward radiation pressure and inward gravitational forces; that fusion is a source of energy in stars; B.3 — the equation of state PV = NkBT for an ideal gas Command term: Deduce
28E-1A-134
A protostar approaching the main sequence·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
The Hertzsprung–Russell diagram shows the path of a protostar of one solar mass. It moves from P to Q and then from Q to S, where it joins the main sequence.
Which statement describes the protostar as it moves from P to Q?
Hertzsprung–Russell diagram (both axes logarithmic; temperature decreases to the right) showing the path P → Q → S of a protostar. The shaded band is the main sequence.Show mark scheme
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Notes
Step 1From P to Q the surface temperature stays almost constant (about 4200 K) while the luminosity falls from about 30 L☉ to 1 L☉.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2L = 4πR²σT4 at almost constant T gives R ∝ √L: the radius decreases, by a factor of about √30 ≈ 5.
—
Step 3Hydrogen fusion starts only when the core has become hot and dense enough, when the star reaches the main sequence at S. Before that, the energy radiated comes from the decrease in gravitational potential energy as the protostar contracts.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThe radius does decrease, but hydrogen fusion begins only when the star reaches the main sequence at S.
BThis reads the fall in luminosity at constant temperature the wrong way: L ∝ R², so a lower luminosity means a smaller radius.
CThis treats a vertical line on the diagram as a line of constant radius; lines of constant radius slope down to the right (L ∝ T4), and fusion has not yet started.
DCorrect: a smaller radius at the same temperature, powered by gravitational contraction before fusion begins.
Syllabus understandingE.5 — the conditions leading to fusion in stars in terms of density and temperature; the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions; guidance: the sketching and interpretation of HR diagrams, including the location of main sequence stars, red giants, super giants and white dwarfs, and lines of constant radius Command term: Deduce
29E-1A-135
Hydrogen fusion in the Sun·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark3 steps to full marksDeduce
Hydrogen fusion takes place in the core of the Sun, where the temperature is about 1.5 × 107 K.
Which statements are correct?
I. The total rest mass of the products of the fusion is less than the total rest mass of the reactants.
II. Fusion takes place because the mean kinetic energy of the protons is greater than the electric potential energy of two protons in contact.
III. Part of the energy released leaves the Sun as the energy of neutrinos.
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Notes
Step 1I is correct: the energy released corresponds to a decrease in rest mass, E = Δmc².
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2II is incorrect: the mean kinetic energy, (3/2)kBT ≈ 2 keV, is hundreds of times smaller than the electric potential energy of two touching protons (about 0.6 MeV). Only the very small fraction of protons with kinetic energies far above the mean can fuse.
—
Step 3III is correct: neutrinos produced in the reactions carry away some of the energy and leave the Sun almost without interacting.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis rejects III, but the neutrinos carry kinetic energy out of the Sun (a few per cent of the energy released).
BThis accepts the misconception in II: the mean kinetic energy is far too small, and fusion relies on the few protons with much higher energies. It also omits III.
CCorrect: I and III are true; the mean kinetic energy of the protons is far below the electric potential energy at contact.
DII is incorrect: at about 2 keV the mean kinetic energy is far below the electric potential energy at contact.
Syllabus understandingE.5 — that fusion is a source of energy in stars; the conditions leading to fusion in stars in terms of density and temperature; E.3 — mass–energy equivalence E = mc² and the neutrino; B.1 — Ek = (3/2)kBTCommand term: Deduce
30E-1A-136
An object that never starts fusion·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark3 steps to full marksExplain
A cloud of gas of mass 0.04 M☉ contracts under gravity. Its core heats up, but hydrogen fusion never starts, and the object slowly cools.
What is the best explanation?
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Notes
Step 1Fusion needs a high temperature, so that protons have enough kinetic energy to come close together against their electric repulsion, and a high density, so that such collisions are frequent.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Both come from gravitational contraction: the smaller the mass, the smaller the gravitational compression and the lower the core temperature that is reached.
—
Step 3For 0.04 M☉ the core never becomes hot enough, so hydrogen fusion does not start.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: a small mass cannot compress and heat its core enough for fusion.
BRadiation pressure is significant only where a large power is being released inside the object; it is not what stops fusion in an object of small mass.
CEven this object contains vastly more hydrogen than fusion would need; whether fusion starts depends on the temperature and density of the core, not on the amount of fuel.
DThis inverts the role of density: a greater density makes collisions more frequent. The speeds of the protons depend on the temperature.
Syllabus understandingE.5 — the conditions leading to fusion in stars in terms of density and temperature; the effect of stellar mass on the evolution of a star Command term: Explain
31E-1B-05
Stellar parallax·E.5 Fusion and stars
Paper 1BEasy6 marks
Data-based question4 steps to full marksDetermine
An astronomer studies six stars that move together through space as a group. All six have identical spectra, so she proposes that they have the same luminosity L. For each star a space telescope measures the parallax angle p and a calibrated photometer measures the apparent brightness b. If the luminosity is the same for every star, then b = Lp²/(4πK²), where p is in arc-seconds and K = 3.09 × 1016 m is the number of metres in one parsec.
The graph shows b against p² with a line of best fit through the origin. (1 pc = 3.26 ly)
p / arcsec
p² / 10−4 arcsec²
b / 10−10 W m−2
0.045
20.25
1.706
0.036
12.96
1.065
0.030
9.00
0.760
0.024
5.76
0.477
0.018
3.24
0.275
0.012
1.44
0.119
Apparent brightness against p² (graph drawn to scale)
(a)
(i)
Test the astronomer's proposal using the data for three of the stars.
(2)
(b)
(i)
Determine L using the gradient of the graph.
(2)
(c)
(i)
A seventh star with the same spectrum has b = 0.85 × 10−10 W m−2. Assuming that it belongs to the group, predict its distance in light-years.
(2)
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Notes
Part (a)(i)
b/p² for three stars, e.g. 1.706/20.25 = 0.0842; 0.760/9.00 = 0.0844; 0.119/1.44 = 0.0826 (× 10−6 W m−2 arcsec−2)
✓ 1
Any three stars, preferably spread across the range.
The values agree to within about 2 %, so the data support equal luminosity
✓ 1
The conclusion must be consistent with the candidate's ratios.
Part (b)(i)
Gradient = 8.4 × 10−8 W m−2 arcsec−2
✓ 1
Accept 8.2–8.6 × 10−8. Allow ECF from (a) if the mean ratio is used.
L = 4πK² × gradient = 4π × (3.09 × 1016)² × 8.4 × 10−8 = 1.0 × 1027 W
✓ 1
Allow ECF from the gradient. Accept 0.98–1.04 × 1027 W.
Allow ECF from (b). ALT: d = √(L/4πb) = 9.7 × 1017 m.
d = 1/p = 31 pc = 102 ly
✓ 1
Accept 98–105 ly.
Answers: (b)(i) 1.0 × 1027 W · (c)(i) 102 ly (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the use of stellar parallax as a method to determine the distance d to celestial bodies as given by d(parsec) = 1/p(arc-second); guidance: conversion between AU, ly and pc; B.1 — apparent brightness b = L/4πd²; Tools 3 — testing a proportional relationship Command term: Determine
32E-1B-07
Stefan–Boltzmann law for stars·E.5 Fusion and stars
Paper 1BHard7 marks
Data-based question6 steps to full marksDetermine
At noon on a clear day a student points a thin copper disc, blackened on its front face, directly at the Sun. The disc has a mass of 25.0 g and a diameter of 4.00 cm, and its back face is insulated. A thermocouple fixed to the disc records its temperature θ every minute, starting when the disc is at the temperature of the surrounding air. The graph shows θ against time t with a curve of best fit.
(Specific heat capacity of copper = 385 J kg−1 K−1; Earth–Sun distance = 1.50 × 1011 m; R☉ = 6.96 × 108 m; σ = 5.67 × 10−8 W m−2 K−4)
Temperature of the disc against time (graph drawn to scale)
(a)
(i)
Explain why the initial rate of rise of temperature should be used, and determine it.
(2)
(b)
(i)
Determine the intensity of the sunlight absorbed by the disc.
(2)
(c)
(i)
Determine the surface temperature of the Sun implied by your answer to (b).
(2)
(d)
(i)
The accepted surface temperature of the Sun is 5.8 × 103 K. Suggest one reason for the difference between this value and your answer to (c).
(1)
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Part (a)(i)
At t = 0 the disc is at air temperature, so it loses no energy to its surroundings: all the absorbed power raises its temperature
✓ 1
Later the gradient falls as the losses grow.
Tangent to the curve at t = 0: gradient = 7.5 K min−1
✓ 1
Accept 7.0–8.0 K min−1. A chord from 0 to 2 min (6.6 K min−1) is not a tangent: 0 for this mark.
Part (b)(i)
Power absorbed = mc Δθ/Δt = 0.0250 × 385 × 7.5/60 = 1.20 W
✓ 1
Allow ECF from (a).
I = 1.20/(π × 0.0200²) = 9.6 × 102 W m−2
✓ 1
Accept 9.0–10.2 × 102 W m−2.
Part (c)(i)
L = 4πd²I = 4π × (1.50 × 1011)² × 957 = 2.7 × 1026 W
✓ 1
Allow ECF from (b).
T = (L/4πR☉²σ)1/4 = 5.3 × 103 K
✓ 1
Accept 5.2–5.4 × 103 K. ALT: T = (Id²/σR☉²)1/4 in one step.
Part (d)(i)
The atmosphere absorbs and scatters part of the sunlight (or the disc is not a perfect absorber), so the intensity reaching the disc is less than the intensity at this distance from the Sun, giving a lower T
✓ 1
Allow ECF from (c). The reason must reduce the measured intensity, consistent with the candidate's comparison. Do not accept "heat losses" unless linked to underestimating the initial gradient.
Answers: (a)(i) 7.5 K min−1 · (b)(i) 9.6 × 102 W m−2 · (c)(i) 5.3 × 103 K (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — how to determine stellar radii; guidance: the determination of stellar radii using luminosity and surface temperature is required; B.1 — quantitative analysis of thermal energy transfers Q with Q = mcΔT; the Stefan–Boltzmann law as given by L = σAT⁴; luminosity L of a body as given by b = L/4πd²; Tools 3 — interpret features of graphs including gradient; Inquiry 3 — compare the outcomes of an investigation to the accepted scientific context Command term: Determine
33E-1B-09
Wien's law from black-body spectra·E.5 Fusion and stars
Paper 1BMedium6 marks
Data-based question4 steps to full marksDetermine
A student tests Wien's displacement law using a 12 V tungsten filament lamp. The potential difference V across the lamp and the current I in it are measured with digital meters, and an infrared spectrometer records the wavelength λmax at which the emitted intensity is greatest. The temperature T of the filament is found from its resistance R using T = T0(R/R0)0.83, where R0 = 0.500 Ω is the resistance at T0 = 293 K. Each value of λmax is uncertain by ±0.02 μm.
V / V
I / A
λmax / μm
2.0
0.794
2.46
4.0
1.111
1.81
6.0
1.373
1.55
8.0
1.600
1.40
10.0
1.805
1.26
12.0
1.993
1.19
(a)
(i)
Show that the temperature of the filament when V = 8.0 V is about 2000 K.
(2)
(b)
(i)
Wien's law predicts that λmaxT is constant. Test this prediction using the data for 4.0 V, 8.0 V and 12.0 V.
(2)
(c)
(i)
Compare your results with the value of the constant in the data booklet, and suggest a reason for any difference.
(2)
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Part (a)(i)
R = V/I = 8.0/1.600 = 5.00 Ω
✓ 1
T = 293 × (5.00/0.500)0.83 = 1981 K
✓ 1
At least 3 s.f. required.
Part (b)(i)
T = 1508 K at 4.0 V and 2311 K at 12.0 V
✓ 1
Same method as (a); Allow ECF from (a).
λmaxT = 2.73, 2.77, 2.75 (× 10−3 m K): constant to within about 1 %, which is less than the ≈ 1–2 % uncertainty in λmax, so the prediction is supported
✓ 1
Three products and a conclusion consistent with them are needed.
Part (c)(i)
Mean ≈ 2.75 × 10−3 m K, about 5 % below 2.9 × 10−3 m K — more than the experimental uncertainty
✓ 1
Allow ECF from (b).
The glass bulb absorbs the longer infrared wavelengths, or tungsten is not a perfect black body (it emits relatively more strongly at short wavelengths), so the observed peak is shifted to a shorter wavelength
✓ 1
Also accept: the given T–R relation overestimates T. Do not accept "random error" or "human error".
Answers: (a)(i) 1981 K (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — guidance: the surface temperature of a star can be determined from the stellar spectrum; B.1 — the emission spectrum of a black body and Wien's displacement law λmaxT = 2.9 × 10−3 m K; B.5 — resistance R = V/I; Tools 3 — testing a relationship with three data points; comparing with an accepted value Command term: Determine
34E-1B-18
Stellar properties·E.5 Fusion and stars
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
An astronomer determines the luminosity L and the surface temperature T of six white dwarfs of similar mass. The luminosities come from measurements of apparent brightness and parallax, and the temperatures from the stellar spectra. She suspects that the six stars have the same radius, and plots lg(L/L☉) against lg(T/K). The graph shows the data with a line of best fit.
(L☉ = 3.85 × 1026 W, R☉ = 6.96 × 108 m, σ = 5.67 × 10−8 W m−2 K−4)
T / K
L / 10−3L☉
lg(T/K)
lg(L/L☉)
6 000
0.20
3.778
−3.70
8 000
0.56
3.903
−3.25
11 000
2.3
4.041
−2.64
15 000
7.1
4.176
−2.15
20 000
25
4.301
−1.60
27 000
74
4.431
−1.13
lg(L/L☉) against lg(T/K) (graph drawn to scale)
(a)
(i)
Determine the gradient of the line of best fit.
(2)
(b)
(i)
Explain what your answer to (a) suggests about the radii of the white dwarfs.
(1)
(c)
(i)
Determine, using the line, the common radius of the white dwarfs.
(2)
(d)
(i)
Each luminosity is uncertain by ±10 % and each temperature by ±3 %. Determine the absolute uncertainty in a radius calculated from the data for a single star.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Two well-separated points on the line, e.g. (3.75, −3.83) and (4.45, −1.04)
✓ 1
Points must be read from the line, not the table.
Gradient = 4.0
✓ 1
Accept 3.9–4.1; no unit.
Part (b)(i)
L = 4πR²σT⁴ gives lg L = 4 lg T + lg(4πR²σ): a gradient of 4 (with all points on one line) means lg(4πR²σ) is the same for every star, i.e. the radii are equal
✓ 1
Allow ECF from (a). Conclusion must match the candidate's gradient.
Part (c)(i)
At lg(T/K) = 4.00 the line gives lg(L/L☉) = −2.83, so L = 1.48 × 10−3 × 3.85 × 1026 = 5.7 × 1023 W
✓ 1
Any point on the line. Using a single table point instead of the line: [1 max]. Allow ECF from (a) if the line's equation is used.
R = √(L/4πσT⁴) = 8.9 × 106 m (0.013 R☉)
✓ 1
Accept 8.4–9.4 × 106 m.
Part (d)(i)
R ∝ L½T−2, so %R = ½ × 10 + 2 × 3 = 11 %
✓ 1
ΔR = 0.11 × 8.9 × 106 = ±1 × 106 m
✓ 1
Allow ECF from (c). 1 s.f. required, i.e. (9 ± 1) × 106 m.
Answers: (a)(i) 4.0 · (c)(i) 8.9 × 106 m · (d)(i) ±1 × 106 m (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — how to determine stellar radii; guidance: the determination of stellar radii using luminosity and surface temperature; lines of constant radius on the HR diagram; B.1 — the Stefan–Boltzmann law L = σAT⁴; Tools 3 — linearising a power law with logarithms, propagation of uncertainties Command term: Determine
35E-1B-19
Stellar parallax·E.5 Fusion and stars
Paper 1BMedium6 marks
Data-based question4 steps to full marksDetermine
A nearby star lies in the plane of the Earth's orbit. Over six months an astronomer measures its angular position x relative to a reference point fixed by very distant background stars; the time t is measured in months from the moment when x is greatest. The motion is modelled by x = x0 + p cos(2πt/12), where x0 is a constant and p is the parallax angle. Each value of x is uncertain by ±0.8 × 10−3 arc-second.
The graph shows x against cos(2πt/12) with the line of best fit (solid) and the lines of maximum and minimum gradient (dashed).
t / month
0
1
2
3
4
5
6
cos(2πt/12)
1.000
0.866
0.500
0.000
−0.500
−0.866
−1.000
x / 10−3 arcsec
14.8
12.4
8.4
1.4
−4.3
−8.8
−11.1
Angular position against cos(2πt/12) (graph drawn to scale)
(a)
(i)
Determine p from the line of best fit.
(2)
(b)
(i)
Use the dashed lines to determine the absolute uncertainty in p.
(2)
(c)
(i)
Determine the distance to the star in parsec, with its absolute uncertainty.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient of x against cos(2πt/12) is p: e.g. (14.5 − (−10.8))/2.00
✓ 1
The intercept is x0.
p = 12.7 × 10−3 arcsec
✓ 1
Accept 12.4–12.9 × 10−3 arcsec.
Part (b)(i)
Gradients of the dashed lines: 13.2 and 12.1 (× 10−3 arcsec)
✓ 1
Accept ±0.2 on each.
Δp = (13.2 − 12.1)/2 = ±0.5 × 10−3 arcsec
✓ 1
Allow ECF from (a) if the larger difference from the best-fit value is used instead. Accept ±0.4 to ±0.6 × 10−3.
Part (c)(i)
d = 1/p = 1/(12.65 × 10−3) = 79 pc
✓ 1
Allow ECF from (a).
Δd/d = Δp/p = 4 %, so d = (79 ± 3) pc
✓ 1
Allow ECF from (b). Uncertainty to 1 s.f.; accept ±2 to ±4 pc.
Answers: (a)(i) 12.7 × 10−3 arcsec · (b)(i) ±0.5 × 10−3 arcsec · (c)(i) (79 ± 3) pc (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the use of stellar parallax as a method to determine the distance d to celestial bodies as given by d(parsec) = 1/p(arc-second); Tools 3 — linearising data, lines of maximum and minimum gradient, uncertainty in a gradient, propagation of uncertainties Command term: Determine
36E-1B-20
Stellar spectra·E.5 Fusion and stars
Paper 1BMedium6 marks
Data-based question4 steps to full marksDetermine
A spectrometer on a telescope records the continuous spectrum of a nearby star. The relative intensity is measured at 13 wavelengths λ between 350 nm and 950 nm, and a smooth best-fit curve is drawn through the points. The broad, flat top of the curve makes the peak hard to locate precisely. Interferometer measurements give the radius of the star as (0.84 ± 0.03) R☉.
The graph shows the data and the curve. (R☉ = 6.96 × 108 m; σ = 5.67 × 10−8 W m−2 K−4)
λ / nm
350
400
450
500
550
600
650
700
750
800
850
900
950
Relative intensity
53
72
88
96
100
100
95
91
83
77
70
64
57
Continuous spectrum of the star with a best-fit curve (graph drawn to scale)
(a)
(i)
Estimate, using the graph, the wavelength at which the intensity is greatest, with its uncertainty.
(2)
(b)
(i)
Determine the surface temperature of the star, with its absolute uncertainty.
(2)
(c)
(i)
Determine the luminosity of the star, with its absolute uncertainty.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
λmax ≈ 560 nm (centre of the flat top of the curve)
✓ 1
Accept 550–575 nm.
Uncertainty ±20 nm (the range over which the curve is level within the scatter)
✓ 1
Accept ±10 to ±30 nm.
Part (b)(i)
T = 2.9 × 10−3/(560 × 10−9) = 5.2 × 103 K
✓ 1
Allow ECF from (a).
ΔT/T = Δλ/λ = 3.6 %, so ΔT = ±2 × 102 K, i.e. (5.2 ± 0.2) × 103 K
✓ 1
Allow ECF from (a). Uncertainty to 1 s.f.
Part (c)(i)
L = 4π(0.84 × 6.96 × 108)² × 5.67 × 10−8 × 5179⁴ = 1.8 × 1026 W
✓ 1
Allow ECF from (b).
%L = 2 × 3.6 + 4 × 3.6 = 21 %, so ΔL = ±4 × 1025 W
✓ 1
Allow ECF from (b). 1 s.f. required: (1.8 ± 0.4) × 1026 W. Adding the percentages without the powers: [1 max].
Answers: (a)(i) 560 ± 20 nm · (b)(i) (5.2 ± 0.2) × 103 K · (c)(i) (1.8 ± 0.4) × 1026 W (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — guidance: the surface temperature of a star can be determined from the stellar spectrum; B.1 — Wien's displacement law λmaxT = 2.9 × 10−3 m K and the Stefan–Boltzmann law L = σAT⁴; Tools 3 — uncertainty in a value read from a graph, propagation of uncertainties through a power law Command term: Determine
37E-1B-36
Calibrating a stellar spectrograph·E.5 Fusion and stars
Paper 1BEasy6 marks
Data-based question5 steps to full marksDeduce
A spectrograph attached to a telescope spreads the light from a star along a row of pixels of a detector. Light of wavelength λ falls on the pixel whose number x is counted from one end of the row. To calibrate the spectrograph, an astronomer records the emission spectrum of a helium lamp, whose wavelengths are known accurately, and notes the pixel number of each line. The graph shows λ against x with the line of best fit.
The spectrum of the star shows dark absorption lines at x = 360, 923 and 1842. Each pixel number is uncertain by ±2 pixels.
λ / nm
447.1
471.3
492.2
501.6
587.6
667.8
x / pixel
447
610
748
810
1384
1920
Wavelength λ of the helium lines against pixel number x, with the line of best fit (drawn to scale).
(a)
(i)
Determine the gradient of the line of best fit. State its unit.
(2)
(b)
(i)
The line of best fit passes through the point x = 748, λ = 492.2 nm. Calculate the wavelength of the absorption line at x = 923.
(1)
(c)
(i)
Determine the absolute uncertainty in your answer to (b).
(1)
(d)
(i)
In this part of the spectrum, iron absorbs at 516.9 nm and magnesium absorbs at 518.4 nm. Deduce which of these elements produces the absorption line at x = 923.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (679.9 − 380.1)/(2000 − 0) = 0.150
✓ 1
Accept 0.148–0.152. Adjacent data points only: [0] for this mark.
Unit: nm pixel−1 (nm per pixel)
✓ 1
Accept m pixel−1 with the value converted (1.50 × 10−10).
Part (b)(i)
λ = 492.2 + 0.150 × (923 − 748) = 518.4 nm
✓ 1
Allow ECF from (a). Accept 518.1–518.8 nm. Reading the graph directly cannot give this precision but accept an answer in range.
Part (c)(i)
Δλ = 2 pixels × 0.150 nm pixel−1 = ±0.3 nm
✓ 1
Allow ECF from (a). Accept ±0.3 nm; ±0.6 nm (adding an uncertainty for the calibration point) also accepted.
Part (d)(i)
Range of the measured wavelength: 518.1 to 518.7 nm
✓ 1
Allow ECF from (b) and (c).
518.4 nm lies inside the range and 516.9 nm lies well outside it, so the line is produced by magnesium (iron is ruled out)
✓ 1
The conclusion must be consistent with the candidate's range. "Closer to magnesium" without reference to the uncertainty: [1 max].
Answers: (a)(i) 0.150 nm pixel−1 · (b)(i) 518.4 nm · (c)(i) ±0.3 nm · (d)(i) magnesium (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — guidance: the surface temperature and the chemical composition of a star can be determined from the stellar spectrum; E.1 — emission and absorption spectra; that the spectra of elements can be used to identify their presence; Tools — gradient of a graph with its unit, uncertainty in a calculated value, comparing a result with accepted values Command term: Deduce
38E-1B-37
Distance to a star cluster·E.5 Fusion and stars
Paper 1BMedium7 marks
Data-based question6 steps to full marksDetermine
An astronomer studies six main-sequence stars in a star cluster. All the stars in the cluster may be taken to be at the same distance d from the Earth. For each star she finds the surface temperature T from its spectrum and measures its apparent brightness b.
For main-sequence stars close to the Sun, whose distances are known from parallax, the luminosity L is related to T by lg(L/W) = 6.4 lg(T/K) + 2.50.
The graph shows lg(b/W m−2) against lg(T/K) for the six cluster stars, with the line of best fit. 1 pc = 3.08 × 1016 m.
T / K
lg(T/K)
lg(b/W m−2)
4 800
3.681
−12.33
5 400
3.732
−12.10
6 100
3.785
−11.68
6 900
3.839
−11.31
7 800
3.892
−11.06
8 800
3.944
−10.68
lg(b/W m−2) against lg(T/K) for the cluster stars, with the line of best fit (drawn to scale).
(a)
(i)
Show that, if the cluster stars obey the same relation as the stars close to the Sun, the graph is a straight line of gradient 6.4.
(1)
(ii)
Determine the gradient of the line of best fit and comment on whether the cluster stars obey the relation.
(2)
(b)
(i)
Use the line of best fit at lg(T/K) = 3.80 to determine d.
(2)
(c)
(i)
Calculate the parallax angle of the cluster.
(1)
(d)
(i)
Dust between the Earth and the cluster absorbs some of the light from the stars. State and explain the effect of this on the value of d found in (b).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
b = L/4πd² gives lg b = lg L − lg(4πd²) = 6.4 lg T + 2.50 − lg(4πd²); d is the same for every star, so lg(4πd²) is a constant and the gradient is 6.4
✓ 1
The constancy of d must be stated or implied.
Part (a)(ii)
Gradient from two well-separated points on the line = 6.35
✓ 1
Accept 6.1–6.6.
Within about 1 % of 6.4 (and the points lie close to a straight line), so the cluster stars do obey the relation
✓ 1
Allow ECF from (a)(i). The comment must be consistent with the candidate's gradient.
Accept −11.65 to −11.55 for the read value. Allow ECF from (a)(i).
d = √(1038.42/4π) = 4.6 × 1018 m
✓ 1
Accept 4.3 × 1018–4.8 × 1018 m. Allow ECF.
Part (c)(i)
d = 4.6 × 1018/3.08 × 1016 = 149 pc, so p = 1/149 = 6.7 × 10−3 arc-second
✓ 1
Allow ECF from (b). Accept 6.3–7.2 × 10−3 arc-second.
Part (d)(i)
Every measured b is too small, so the line is too low and lg(4πd²) = lg L − lg b is too large: d is overestimated
✓ 1
Both the direction and the reason are needed. "Makes it inaccurate" scores [0].
Answers: (a)(ii) 6.35 · (b)(i) 4.6 × 1018 m · (c)(i) 6.7 × 10−3 arc-second (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions; the use of stellar parallax as a method to determine the distance d to celestial bodies as given by d (parsec) = 1/p (arc-second); B.1 — apparent brightness b = L/4πd²; Tools — linearising a relationship with logarithms, interpreting gradient and intercept, systematic error Command term: Determine
39E-1B-38
Fusion power and plasma density·E.5 Fusion and stars
Paper 1BHard7 marks
Data-based question5 steps to full marksDetermine
In an experimental fusion reactor, a plasma of equal numbers of deuterium and tritium nuclei is held at a constant temperature. The number density n of the nuclei is varied, and the fusion power P is found from the rate at which neutrons are detected. Each value of P is uncertain by ±10 %; the uncertainty in n is negligible.
It is suggested that P = kn², where k is a constant. The graph shows lg(P/MW) against lg(n/1019 m−3) with error bars and the line of best fit.
n / 1019 m−3
P / MW
lg(n/1019 m−3)
lg(P/MW)
1.5
0.215
0.176
−0.668
2.0
0.342
0.301
−0.466
2.5
0.585
0.398
−0.233
3.2
0.894
0.505
−0.049
4.0
1.48
0.602
0.170
5.0
2.16
0.699
0.334
6.0
3.30
0.778
0.519
lg(P/MW) against lg(n/1019 m−3) with error bars and the line of best fit (drawn to scale).
(a)
(i)
Outline why the rate of fusion reactions per unit volume is expected to be proportional to n², and why a graph of lg P against lg n tests this.
(2)
(b)
(i)
Determine the gradient of the line of best fit.
(1)
(ii)
Draw the lines of steepest and shallowest gradient that pass through all the error bars. Use them to deduce whether the data support the suggestion.
(2)
(c)
(i)
Predict, using the line of best fit, the fusion power when n = 1.0 × 1020 m−3.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Each reaction needs two nuclei to collide: the collision rate of one nucleus is proportional to n, and the number of nuclei per unit volume is proportional to n, so the rate per unit volume ∝ n²
✓ 1
Accept: rate ∝ (number density of D) × (number density of T), each ∝ n.
lg P = lg k + 2 lg n: if the suggestion is correct the graph is a straight line of gradient 2
✓ 1
Part (b)(i)
Gradient = 1.98
✓ 1
Accept 1.90–2.06. Points must be read from the line.
Part (b)(ii)
Steepest ≈ 2.1 and shallowest ≈ 1.86, so gradient = 1.98 ± 0.1
✓ 1
Accept 2.05–2.12 and 1.83–1.90; uncertainty ±0.08 to ±0.15. Allow ECF from (b)(i).
The value 2 lies within this range, so the data support P ∝ n²
✓ 1
The conclusion must be consistent with the candidate's range.
Part (c)(i)
lg(n/1019 m−3) = 1.00, so lg(P/MW) = 0.95 (extending the line)
✓ 1
Allow ECF from (b)(i). Using the hypothesis with one data point: e.g. 3.30 × (10/6.0)² = 9.2 MW is accepted.
P = 100.95 ≈ 8.8 MW
✓ 1
Accept 8.0–10 MW.
Answers: (b)(i) 1.98 · (b)(ii) 1.98 ± 0.1; supported · (c)(i) 8.8 MW (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the conditions leading to fusion in stars in terms of density and temperature; that fusion is a source of energy in stars; Tools — linearising a power law with logarithms, lines of maximum and minimum gradient, uncertainty in a gradient, testing a hypothesis, extrapolation Command term: Determine
40E-2-04
Stellar parallax·E.5 Fusion and stars
Paper 2Medium10 marks
Short answer & extended response7 steps to full marksDeduce
The diagram shows the Earth at two positions, E₁ and E₂, six months apart in its orbit of radius 1 AU about the Sun S. The parallax angle p of the nearby star Q, marked on the diagram, is 0.0400 arc-second.
(1 AU = 1.50 × 1011 m, 1 pc = 3.26 ly, L☉ = 3.85 × 1026 W)
Parallax of star Q (not to scale)
(a)
(i)
One parsec is the distance of a star whose parallax angle is 1 arc-second. Show, using the diagram, that 1 pc is about 3.1 × 1016 m.
(2)
(ii)
Determine the distance to Q in light-years.
(1)
(b)
(i)
The apparent brightness of Q is 2.9 × 10−10 W m−2. Determine the luminosity of Q in terms of L☉.
(2)
(ii)
The parallax angle of Q is measured to ±0.0010 arc-second. Determine the percentage uncertainty in the luminosity of Q.
(2)
(c)
(i)
The smallest parallax angle that can be measured with this telescope is 0.0050 arc-second. Star R has the same luminosity as Q and an apparent brightness of 1.0 × 10−12 W m−2. Deduce whether the distance to R can be measured by parallax with this telescope.
(2)
(ii)
Suggest why parallax angles measured with a telescope in space are more precise than those measured from the Earth's surface.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
1 arc-second = (1/3600)° = 4.85 × 10−6 rad; from the diagram tan p = (1 AU)/d
✓ 1
The conversion to radians (or use of tan with the angle in degrees) must be seen.
d = 1.50 × 1011/4.848 × 10−6 = 3.09 × 1016 m
✓ 1
Must see the full substitution or an answer to at least 3 s.f. (3.09 × 1016 m).
Part (a)(ii)
d = 1/p = 1/0.0400 = 25.0 pc = 25.0 × 3.26 = 81.5 ly
✓ 1
Unit required.
Part (b)(i)
d = 25.0 × 3.09 × 1016 = 7.7 × 1017 m; L = 4πd²b = 2.2 × 1027 W
✓ 1
Allow ECF from (a)(i) and (a)(ii).
L = 5.7 L☉
✓ 1
Accept 5.5–5.7 L☉.
Part (b)(ii)
Percentage uncertainty in p, and so in d = 0.0010/0.0400 = 2.5 %
✓ 1
L ∝ d², so the percentage uncertainty in L = 5 %
✓ 1
Allow ECF from (b)(i). The uncertainty in b is ignored.
Part (c)(i)
Same L, so d ∝ 1/√b: dR = 25.0 × √(2.9 × 10−10/1.0 × 10−12) = 4.3 × 102 pc
✓ 1
ALT: d = √(L/4πb) = 1.3 × 1019 m = 4.3 × 102 pc. Allow ECF from (b)(i).
pR = 1/426 = 0.0023 arc-second, smaller than 0.0050 arc-second, so the distance cannot be measured by parallax
✓ 1
ALT: the greatest measurable distance is 1/0.0050 = 200 pc, less than 4.3 × 102 pc. The conclusion must agree with the candidate's own value.
Part (c)(ii)
There is no atmospheric turbulence (refraction) to blur and shift the star images
✓ 1
Answers: (a)(i) 3.09 × 1016 m · (a)(ii) 81.5 ly · (b)(i) 5.7 L☉ · (b)(ii) 5 % · (c)(i) No: p ≈ 0.0023″ (d ≈ 4.3 × 102 pc) (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the use of stellar parallax as a method to determine the distance d to celestial bodies as given by d(parsec) = 1/p(arc-second); guidance: the conversion between astronomical units (AU), light years (ly) and parsecs (pc) is required; B.1 — the concept of apparent brightness b; luminosity L of a body as given by b = L/4πd²; Tools 3 — propagate uncertainties in processed data in calculations involving raising to a power Command term: Deduce
41E-2-08
Conditions for fusion·E.5 Fusion and stars
Paper 2Hard9 marks
Short answer & extended response6 steps to full marksEstimate
In a proposed fusion power station, deuterium and tritium nuclei fuse: ²₁H + ³₁H → ⁴₂He + ¹₀n + 17.6 MeV.
Model each nucleus as a sphere of radius R = R0A1/3, with R0 = 1.20 × 10−15 m, and assume that fusion requires the two nuclei to touch. (k = 8.99 × 109 N m² C−2, e = 1.60 × 10−19 C, kB = 1.38 × 10−23 J K−1, 1 MeV = 1.60 × 10−13 J, 1 u = 1.661 × 10−27 kg)
(a)
(i)
Show that the electric potential energy of a deuterium nucleus and a tritium nucleus that are just touching is about 0.44 MeV.
(2)
(ii)
Estimate the temperature at which the average kinetic energy of the nuclei would equal this potential energy.
(2)
(iii)
The plasma in the power station is heated to about 1.5 × 108 K. Explain why a useful rate of fusion is nevertheless obtained, and why the plasma must also have a high density.
(2)
(b)
(i)
Tritium has a half-life of 12.3 years. Suggest why the tritium fuel must be manufactured.
(1)
(ii)
Determine the mass of tritium used per day by a power station in which the fusion power is 500 MW.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Separation of centres = sum of radii = 1.20 × 10−15 × (21/3 + 31/3) = 3.24 × 10−15 m
✓ 1
Using one radius only scores 0 for MP1.
Ep = ke²/r = 7.10 × 10−14 J = 0.444 MeV
✓ 1
Must see 0.444 MeV or more figures.
Part (a)(ii)
(3/2)kBT = 7.10 × 10−14 J
✓ 1
Allow ECF from (a)(i).
T = 3.4 × 109 K
✓ 1
Accept 3.3–3.5 × 109 K.
Part (a)(iii)
The nuclei have a spread of kinetic energies: a small fraction have energies many times the average, enough to overcome the repulsion and reach the separation at which the strong force acts
✓ 1
Allow ECF from (a)(ii). Do not accept answers that ignore the spread of kinetic energies.
A high density gives a high collision rate, so these rare high-energy collisions happen often enough to release a useful power
✓ 1
Part (b)(i)
Its half-life is very short compared with the age of the Earth, so any tritium formed long ago has decayed: almost none exists naturally
✓ 1
Part (b)(ii)
Reactions per second = 500 × 106/(17.6 × 1.60 × 10−13) = 1.78 × 1020 s−1
✓ 1
Mass per day = 1.78 × 1020 × 86 400 × 3 × 1.661 × 10−27 = 0.076 kg
✓ 1
Accept 0.076–0.077 kg (using 3.016 u gives 0.077 kg). Allow ECF from MP1.
Answers: (a)(ii) 3.4 × 109 K · (b)(ii) 0.076 kg (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the conditions leading to fusion in stars in terms of density and temperature; that fusion is a source of energy; E.1 (HL) — nuclear radius R = R0A1/3; D.2 — electric potential energy Ep = kq1q2/r; B.1 — that Kelvin temperature is a measure of the average kinetic energy of particles as given by Ek = (3/2)kBTCommand term: Estimate
42E-2-11
Neutron stars and nuclear density·E.5 Fusion and stars
Paper 2Hard10 marks
Short answer & extended response6 steps to full marksEstimate
After a supernova, the collapsed core of a massive star forms a neutron star of mass 1.4 M☉. Model the neutron star as a uniform sphere of neutrons packed together at the density of nuclear matter.
Nuclear radii are given by R = R0A1/3 with R0 = 1.20 × 10−15 m. (Mass of a nucleon mn = 1.675 × 10−27 kg, M☉ = 1.99 × 1030 kg, G = 6.67 × 10−11 N m² kg−2, c = 3.00 × 108 m s−1)
(a)
(i)
Show that the density of nuclear matter is ρ = 3mn/(4πR0³), independent of A.
(2)
(ii)
Calculate the density of nuclear matter.
(1)
(b)
(i)
Estimate the radius of the neutron star.
(2)
(ii)
Suggest why the actual radius of a neutron star of this mass is likely to be smaller than your estimate.
(1)
(c)
(i)
Calculate the escape speed from the surface of the neutron star as a fraction of the speed of light.
(2)
(ii)
A collapsing core of greater mass forms a black hole. Deduce the radius to which this neutron star would have to be compressed for the escape speed from its surface to equal c, and comment on your answer.
The neutron star (radius ≈ 14 km) is about 3 times larger than this, so it is not a black hole
✓ 1
Allow ECF from (b)(i).
Answers: (a)(ii) 2.3 × 1017 kg m−3 · (b)(i) 1.4 × 104 m · (c)(i) 0.54c · (c)(ii) 4.1 × 103 m (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the effect of stellar mass on the evolution of a star (neutron stars and black holes as end points); E.1 (HL) — R = R0A1/3 and its implications for nuclear densities; D.1 (HL) — escape speed vesc = √(2GM/r) Command term: Estimate
43E-2-14
The cooling of a white dwarf·E.5 Fusion and stars
Paper 2Hard11 marks
Short answer & extended response7 steps to full marksEstimate
A white dwarf has mass 0.60 M☉, radius 8.7 × 106 m and surface temperature 1.2 × 104 K. It consists mainly of carbon-12 and no fusion takes place in it. Its interior temperature is about 1.0 × 107 K.
Model the internal energy of the white dwarf as the random kinetic energy of its carbon nuclei, treated as the particles of an ideal gas. (M☉ = 1.99 × 1030 kg, L☉ = 3.85 × 1026 W, 1 u = 1.661 × 10−27 kg, kB = 1.38 × 10−23 J K−1, σ = 5.67 × 10−8 W m−2 K−4, 1 year = 3.16 × 107 s)
(a)
(i)
Show that the white dwarf contains about 6 × 1055 carbon nuclei.
(1)
(ii)
Determine the internal energy of the white dwarf.
(2)
(b)
(i)
Show that the luminosity of the white dwarf is about 3 × 10−3L☉.
(2)
(ii)
Estimate the time, in years, for the white dwarf to radiate its internal energy.
(2)
(iii)
State one assumption made in your estimate in (b)(ii) and explain whether the real cooling time is longer or shorter than your estimate.
(2)
(c)
(i)
Describe and explain the path of the white dwarf on a Hertzsprung–Russell diagram as it cools.
Average kinetic energy per nucleus = (3/2)kBT = 1.5 × 1.38 × 10−23 × 1.0 × 107 = 2.07 × 10−16 J
✓ 1
U = 5.99 × 1055 × 2.07 × 10−16 = 1.2 × 1040 J
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
L = 4π(8.7 × 106)² × 5.67 × 10−8 × (1.2 × 104)⁴ = 1.12 × 1024 W
✓ 1
= 2.9 × 10−3L☉
✓ 1
Must see 2.9 × 10−3 or more figures.
Part (b)(ii)
t = U/L = 1.24 × 1040/1.12 × 1024 = 1.1 × 1016 s
✓ 1
Allow ECF from (a)(ii) and (b)(i).
= 3.5 × 108 years
✓ 1
Accept 3–4 × 108 years.
Part (b)(iii)
Assumption: the luminosity (surface temperature) stays constant while the energy is radiated
✓ 1
Accept also: no energy is gained, e.g. from contraction.
In reality the surface temperature, and so L ∝ T⁴, falls as the star cools, so energy is radiated ever more slowly: the real time is longer
✓ 1
MP2 depends on a relevant assumption.
Part (c)(i)
No fusion and no contraction: the radius stays (almost) constant while the temperature falls, so L falls as T⁴
✓ 1
It moves down and to the right, along a line of constant radius, in the region below the main sequence
✓ 1
Answers: (a)(ii) 1.2 × 1040 J · (b)(ii) 3.5 × 108 years (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the effect of stellar mass on the evolution of a star; the main regions of the HR diagram (white dwarfs, lines of constant radius); B.1 — that Kelvin temperature is a measure of the average kinetic energy of particles as given by Ek = (3/2)kBT; B.3 — the internal energy U = (3/2)NkBT of an ideal monatomic gas; B.1 — the Stefan–Boltzmann law L = σAT⁴ Command term: Estimate
44E-2-16
An eclipsing binary star·E.5 Fusion and stars
Paper 2Hard12 marks
Short answer & extended response7 steps to full marksDeduce
Stars A and B, both on the main sequence, orbit their common centre of mass in circular orbits. The Earth lies in the plane of the orbits, so each star passes in turn in front of the other. B has a smaller radius than A and is completely hidden when it passes behind A.
The graph shows the total apparent brightness b of the pair, as a fraction of its normal value, over 6 days. Treat the disc of each star as uniformly bright. (Wien constant = 2.9 × 10−3 m K, c = 3.00 × 108 m s−1)
Brightness of the binary star against time (graph drawn to scale)
(a)
(i)
Determine the orbital period of the pair.
(1)
(ii)
Explain why the two eclipses produce different decreases in brightness.
(2)
(iii)
Show that (decrease in brightness in the deep dip)/(decrease in brightness in the shallow dip) = (TA/TB)⁴.
(2)
(iv)
The spectrum of A peaks at a wavelength of 387 nm. Determine the surface temperature of B.
(2)
(b)
(i)
The hydrogen line of laboratory wavelength 656.280 nm in the spectrum of A varies periodically between 656.103 nm and 656.457 nm. Determine the orbital speed of A.
(2)
(ii)
The same line in the spectrum of B is shifted by up to ±0.301 nm. Deduce the ratio of the masses MA/MB.
(2)
(iii)
Explain why your answer to (b)(ii) is consistent with your answer to (a)(iv).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Time between successive deep (or shallow) dips = 2.5 days
✓ 1
Accept 2.4–2.6 days. Half this value scores 0.
Part (a)(ii)
Deep dip: the smaller star B passes in front of A and hides part of A's disc; shallow dip: B is completely hidden behind A
✓ 1
In both cases the hidden area is the same (πRB²), so the different losses mean that the two stars emit different powers per unit area (different surface temperatures)
✓ 1
Part (a)(iii)
In each eclipse an area πRB² of a stellar surface is hidden; the power per unit area emitted is σT⁴
✓ 1
Allow ECF from (a)(ii).
Deep: Δb ∝ πRB²σTA⁴; shallow: Δb ∝ πRB²σTB⁴; the ratio is (TA/TB)⁴
✓ 1
Part (a)(iv)
TA = 2.9 × 10−3/387 × 10−9 = 7.5 × 103 K
✓ 1
From the graph the decreases are 0.28 and 0.08: TB = 7.5 × 103 × (0.08/0.28)1/4 = 5.5 × 103 K
✓ 1
Allow ECF from (a)(iii). Accept 5.3–5.7 × 103 K. Using the minimum brightness values (0.72, 0.92) instead of the decreases scores [1].
Part (b)(i)
Maximum shift Δλ = 0.177 nm (half the range)
✓ 1
Using the whole range (0.354 nm) scores [1] max.
vA = cΔλ/λ = 3.00 × 108 × 0.177/656.280 = 8.1 × 104 m s−1
✓ 1
Part (b)(ii)
The total momentum about the centre of mass is zero (the stars are always on opposite sides with the same period), so MAvA = MBvB
✓ 1
MA/MB = vB/vA = 0.301/0.177 = 1.70
✓ 1
Allow ECF from (b)(i). Accept 1.7. An inverted ratio (0.59) scores [1].
Part (b)(iii)
On the main sequence a more massive star is hotter (and more luminous); A is both more massive and hotter than B
✓ 1
Allow ECF from (a)(iv) and (b)(ii).
Answers: (a)(i) 2.5 days · (a)(iv) 5.5 × 103 K · (b)(i) 8.1 × 104 m s−1 · (b)(ii) 1.70 (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the main properties of stars in the regions of the HR diagram (main sequence: mass, temperature and luminosity); guidance: the surface temperature of a star can be determined from the stellar spectrum; B.1 — the Stefan–Boltzmann law and Wien's displacement law; C.5 — the Doppler effect for light, Δλ/λ ≈ v/c; A.2 — conservation of momentum Command term: Deduce
45E-2-27
The Sun as a red giant·E.5 Fusion and stars
Paper 2Medium10 marks
Short answer & extended response6 steps to full marksDetermine
The Hertzsprung–Russell diagram shows the predicted evolutionary track of the Sun. S marks the Sun today, P the top of its red-giant stage and W the white dwarf that it eventually becomes.
(L☉ = 3.85 × 1026 W, σ = 5.67 × 10−8 W m−2 K−4, 1 AU = 1.50 × 1011 m)
Predicted evolutionary track of the Sun (both axes logarithmic)
(a)
(i)
Outline why the Sun will leave the main sequence and move towards P.
(2)
(ii)
State the luminosity of the Sun at P.
(1)
(b)
(i)
Determine the intensity of the Sun's radiation at the Earth's orbit when the Sun is at P.
(2)
(ii)
A planet of radius r orbits at distance d from a star of luminosity L. It absorbs a fraction (1 − α) of the radiation incident on it, where α is its albedo, and radiates as a black body at a uniform temperature T from its whole surface. Show that T⁴ = (1 − α)L/(16πσd²).
(2)
(iii)
Determine the equilibrium temperature of the Earth when the Sun is at P. Take α = 0.30.
(2)
(iv)
Deduce, without substituting values, the ratio (equilibrium temperature of Mars)/(equilibrium temperature of the Earth). Mars orbits at 1.52 AU and has the same albedo as the Earth.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
When the hydrogen in the core is used up, fusion in the core stops and the core contracts and heats up
✓ 1
Hydrogen fusion continues in a shell around the core; the outer layers expand greatly and cool, so the luminosity rises while the surface temperature falls
✓ 1
Part (a)(ii)
≈ 2.5 × 103L☉
✓ 1
Accept 2.0–3.0 × 103L☉.
Part (b)(i)
L = 2.5 × 103 × 3.85 × 1026 = 9.6 × 1029 W
✓ 1
Allow ECF from (a)(ii).
b = L/4π(1.50 × 1011)² = 3.4 × 106 W m−2
✓ 1
Accept 2.7–4.1 × 106 W m−2.
Part (b)(ii)
Power absorbed = (1 − α) × L/(4πd²) × πr² (the planet intercepts radiation on its cross-section)
✓ 1
Power emitted = 4πr²σT⁴; equating, r cancels and T⁴ = (1 − α)L/(16πσd²)
Answers: (a)(ii) 2.5 × 103L☉ · (b)(i) 3.4 × 106 W m−2 · (b)(iii) 1.8 × 103 K · (b)(iv) 0.81 (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the main regions of the HR diagram; the effect of stellar mass on the evolution of a star; B.1 — apparent brightness b = L/4πd²; B.2 — albedo; guidance: the estimation of equilibrium temperature of a body using energy balance between incoming and outgoing radiation intensity, including albedo Command term: Determine
46E-2-28
Nuclear fusion·E.5 Fusion and stars
Paper 2Medium13 marks
Short answer & extended response6 steps to full marksExplain
When the hydrogen in the core of a star of two solar masses has been used up, the core contracts until helium fusion begins at a temperature of about 108 K. Hydrogen fusion needs only about 1.5 × 107 K. Helium fusion occurs in two steps:
3 42He → 126C and then 126C + 42He → 168O
Atomic masses: 1H 1.007 825 u, 4He 4.002 602 u, 12C 12.000 000 u, 16O 15.994 915 u. 1 u = 931.5 MeV c−2 = 1.66 × 10−27 kg, 1 MeV = 1.60 × 10−13 J, L☉ = 3.85 × 1026 W, solar mass = 1.99 × 1030 kg, 1 year = 3.16 × 107 s.
(a)
Starting helium fusion.
(i)
When hydrogen fusion in the core stops, the core contracts. Explain, in terms of energy, why the temperature of the core rises as it contracts.
(2)
(b)
Energy released.
(i)
Show that the energy released when three helium nuclei form one carbon nucleus is about 7.3 MeV.
(2)
(ii)
Calculate the energy released when a carbon nucleus captures a helium nucleus to form oxygen.
(1)
(iii)
Hydrogen fusion releases 26.7 MeV for every four hydrogen atoms converted into one helium atom. Determine the ratio (energy released per kilogram of hydrogen fused to helium) / (energy released per kilogram of helium fused to carbon).
(2)
(c)
The helium-burning stage.
(i)
During this stage helium fusion to carbon supplies 30 L☉. Calculate the mass of helium fused per second.
(2)
(ii)
The core contains 0.30 solar masses of helium. Estimate the duration of the helium-burning stage, in years.
(2)
(iii)
Suggest two reasons why the helium-burning stage is much shorter than the star's main-sequence lifetime.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
As the core contracts, the gravitational potential energy of its matter decreases (the gravitational forces do work on the core material)
✓ 1
Do not accept an answer that only states that the pressure no longer balances gravity: the question asks about energy.
This energy is transferred to the random kinetic energy of the nuclei and electrons; since their average kinetic energy is proportional to the absolute temperature (Ek = (3/2)kBT), the temperature rises
✓ 1
The link between mean kinetic energy and temperature is needed for MP2. Accept "some of the energy is radiated away, the rest heats the core".
Part (b)(i)
Δm = 3 × 4.002 602 − 12.000 000 = 0.007 806 u
✓ 1
Atomic masses can be used: the electron masses cancel.
Mass of helium = 0.30 × 1.99 × 1030 = 6.0 × 1029 kg; t = 6.0 × 1029/1.98 × 1014 = 3.0 × 1015 s
✓ 1
≈ 1 × 108 years (9.5 × 107 years)
✓ 1
Accept 9–10 × 107 years. Allow ECF from (c)(i).
Part (c)(iii)
Helium fusion releases about 11 times less energy per kilogram of fuel than hydrogen fusion (from (b)(iii))
✓ 1
Any two of: (1) about 11 times less energy per kg (from (b)(iii)); (2) only the core helium, a small fraction of the star's mass, is available; (3) the star is more luminous than on the main sequence, so it uses fuel faster. [1] each, max 2. Allow ECF from (b)(iii).
Only the core helium (a small fraction of the star's mass) is available; the star is more luminous than on the main sequence, so it uses its fuel faster
✓ 1
Answers: (b)(i) 7.27 MeV · (b)(ii) 7.16 MeV · (b)(iii) ≈ 11 · (c)(i) 2.0 × 1014 kg s−1 · (c)(ii) ≈ 1 × 108 years (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — that fusion is a source of energy in stars; that the stability of stars relies on an equilibrium between outward radiation pressure and inward gravitational forces; the effect of stellar mass on the evolution of a star; guidance: energy release calculations are required; D.1 — gravitational potential energy; B.1 — that Kelvin temperature is a measure of the average kinetic energy of particles as given by Ek = (3/2)kBTCommand term: Explain
47E-2-29
The Hertzsprung–Russell diagram·E.5 Fusion and stars
Paper 2Medium11 marks
Short answer & extended response8 steps to full marksDetermine
The Hertzsprung–Russell diagram shows the Sun and three other stars S, V and W. The dashed lines are lines of constant radius. Stars in the shaded band have a luminosity that varies periodically.
(R☉ = 6.96 × 108 m, L☉ = 3.85 × 1026 W, σ = 5.67 × 10−8 W m−2 K−4, 1 AU = 1.50 × 1011 m)
HR diagram showing the Sun and stars S, V and W. Dashed lines: constant stellar radius. Shaded: a narrow band that crosses the main sequence. Both axes logarithmic.
(a)
Reading the diagram.
(i)
State the name of the shaded band.
(1)
(ii)
Identify the type of star represented by S and by W.
(2)
(iii)
Use the diagram to estimate the radius of W.
(1)
(b)
Star S has a surface temperature of 3600 K and a luminosity of 1.0 × 105L☉.
(i)
Show that the radius of S is about 800 R☉.
(2)
(ii)
Express this radius in AU and comment on your answer.
(1)
(c)
Star V lies in the shaded band. Its luminosity varies between 60 L☉, when its surface temperature is 7200 K, and 40 L☉, when its surface temperature is 6200 K.
(i)
Determine the radius of V at maximum and at minimum luminosity, in terms of R☉.
(2)
(ii)
Explain, using your answers, why V is most luminous when it is smallest.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Instability strip
✓ 1
Part (a)(ii)
S: (red) supergiant — very luminous and cool, at the top right
✓ 1
Accept red supergiant; do not accept red giant alone.
W: white dwarf — hot but very faint, at the lower left below the main sequence
✓ 1
Part (a)(iii)
W lies just above the 0.01 R☉ line, so its radius is about 0.01 R☉
R = 5.7 × 1011/1.50 × 1011 = 3.8 AU; S would extend well beyond the orbit of Mars (about 1.5 AU) — larger than the orbits of all the inner planets
✓ 1
Accept 3.7–3.9 AU with a sensible comment.
Part (c)(i)
Maximum: R = √60 × (5800/7200)² = 5.0 R☉
✓ 1
Accept 4.9–5.1 R☉; SI route acceptable.
Minimum: R = √40 × (5800/6200)² = 5.5 R☉
✓ 1
Accept 5.4–5.6 R☉.
Part (c)(ii)
L ∝ R²T⁴: at maximum light R² is smaller by a factor (5.0/5.5)² ≈ 0.83 but T⁴ is larger by (7200/6200)⁴ ≈ 1.8
✓ 1
The temperature effect dominates, so the net luminosity is 0.83 × 1.8 ≈ 1.5 times larger (60/40) even though the star is smallest
✓ 1
Answers: (a)(iii) ≈ 0.01 R☉ · (b)(i) ≈ 820 R☉ · (b)(ii) 3.8 AU · (c)(i) 5.0 R☉ and 5.5 R☉(the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the main regions of the HR diagram and how to describe the main properties of stars in these regions (supergiants, white dwarfs, the instability strip, lines of constant radius); how to determine stellar radii; guidance: the determination of stellar radii using luminosity and surface temperature; conversion between AU and m; B.1 — the Stefan–Boltzmann law L = σAT⁴ Command term: Determine
48E-2-30
Stellar evolution·E.5 Fusion and stars
Paper 2Medium10 marks
Short answer & extended response5 steps to full marksExplain
Two stars A and B formed at the same time in the same cluster. Star A has a mass of 20 solar masses and a main-sequence luminosity of 4.0 × 104L☉. Star B has a mass of 1.0 solar mass and a luminosity of 1.0 L☉; its main-sequence lifetime is about 1.0 × 1010 years.
(Binding energy per nucleon: ²⁸Si 8.448 MeV, ⁵⁶Ni 8.643 MeV, ⁵⁶Fe 8.790 MeV — close to the maximum of the binding-energy curve.)
(a)
The main sequence.
(i)
Explain, in terms of the equilibrium of the star, why the core of A must be hotter than the core of B.
(2)
(ii)
Assume that each star fuses the same fraction of its mass while on the main sequence. Estimate the main-sequence lifetime of A.
(2)
(iii)
Explain why A has the shorter lifetime even though it has 20 times more fuel.
(1)
(b)
The end of fusion in A. In the final stage of core fusion in A, silicon fuses: 2 2814Si → 5628Ni.
(i)
Calculate the energy released in this reaction.
(2)
(ii)
The core of A eventually becomes iron. Explain why the core then collapses.
(2)
(c)
After the collapse, a supernova occurs and the remnant of A is a neutron star.
(i)
State the remnant that would be formed if the collapsing core had a much greater mass.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
A has much more mass, so the inward gravitational force on its layers (the weight of the overlying layers) is much greater
✓ 1
To balance it a much greater outward pressure is needed in the core, which requires a higher core temperature (and fusion rate)
Accept 4–6 × 106 years. A ratio inverted (M/L the wrong way up) scores [1].
Part (a)(iii)
The fusion rate rises very steeply with core temperature, so A's luminosity (rate of fuel use) is 4 × 104 times B's while its fuel is only 20 times greater
✓ 1
Allow ECF from (a)(ii).
Part (b)(i)
Increase in total binding energy = 56 × 8.643 − 2 × 28 × 8.448 = 56 × 0.195
✓ 1
= 10.9 MeV
✓ 1
Accept 10.9–11 MeV.
Part (b)(ii)
Iron is near the peak of the binding-energy curve, so fusing it would absorb rather than release energy: fusion in the core stops
✓ 1
No energy is released to maintain the outward (radiation and gas) pressure, so gravity is no longer balanced and the core collapses
✓ 1
Part (c)(i)
A black hole
✓ 1
Answers: (a)(ii) ≈ 5 × 106 years · (b)(i) 10.9 MeV (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the effect of stellar mass on the evolution of a star; that the stability of stars relies on an equilibrium between outward radiation pressure and inward gravitational forces; that fusion is a source of energy in stars (energy release calculations); E.3 — binding energy per nucleon Command term: Explain
49E-2-31
Fusion in the Sun·E.5 Fusion and stars
Paper 2Medium13 marks
Short answer & extended response10 steps to full marksDeduce
In the nineteenth century physicists debated what could power the Sun. The Sun has mass 1.99 × 1030 kg, radius 6.96 × 108 m and luminosity 3.85 × 1026 W. Geological and radioactive dating show that it has been shining for about 4.6 × 109 years at roughly its present luminosity.
(G = 6.67 × 10−11 N m2 kg−2, c = 3.00 × 108 m s−1, 1 year = 3.16 × 107 s; atomic masses: 1H 1.007 825 u, 4He 4.002 602 u)
(a)
Energy needed.
(i)
Calculate the total energy radiated by the Sun so far.
(1)
(b)
Two rejected hypotheses.
(i)
The best chemical fuels release about 5 × 107 J per kilogram. Determine how long the Sun could shine at its present luminosity if it were made entirely of such fuel.
(2)
(ii)
The energy released by the gravitational contraction of the Sun to its present size is of order GM²/R. Determine how long this energy could supply the present luminosity, and comment on your answers to (b)(i) and (b)(ii).
(2)
(c)
Fusion.
(i)
The net result of hydrogen fusion in the Sun is that four hydrogen atoms become one helium atom. Show that about 0.7 % of the mass of the hydrogen is converted into energy.
(2)
(ii)
Determine the total mass of hydrogen that has been fused in the Sun so far.
(2)
(iii)
Only the hydrogen in the core, about 10 % of the Sun's mass, is ever hot and dense enough to fuse while the Sun is on the main sequence. Deduce how far the Sun is through its main-sequence life.
(2)
(iv)
Suggest why the Sun's luminosity has changed little over 4.6 × 109 years even though 0.44 of its core hydrogen has been used.
Both are far shorter than 4.6 × 109 years (by factors of about 6 × 105 and 150), so neither chemical reactions nor contraction can be the Sun's main energy source
Mass converted to radiated energy = E/c² = 5.6 × 1043/(3.00 × 108)² = 6.2 × 1026 kg
✓ 1
Mass of hydrogen fused = 6.2 × 1026/7.1 × 10−3 = 8.7 × 1028 kg
✓ 1
Accept 8.5–8.9 × 1028 kg; ecf from (a).
Part (c)(iii)
Hydrogen available in the core = 0.10 × 1.99 × 1030 = 2.0 × 1029 kg; fraction used = 8.7 × 1028/2.0 × 1029 = 0.44
✓ 1
The Sun is a little less than halfway through its main-sequence life (total ≈ 4.6 × 109/0.44 ≈ 1 × 1010 years)
✓ 1
Accept "about halfway".
Part (c)(iv)
The fusion rate is self-regulated by the equilibrium between the inward gravitational force and the outward (radiation and gas) pressure
✓ 1
Any small increase (or decrease) in core temperature or density makes the core expand (or contract), which returns the fusion rate to its equilibrium value — negative feedback, so the luminosity stays nearly constant
✓ 1
Answers: (a)(i) 5.6 × 1043 J · (b)(i) ≈ 8 × 103 years · (b)(ii) ≈ 3 × 107 years · (c)(i) 0.71 % · (c)(ii) 8.7 × 1028 kg · (c)(iii) ≈ 0.44 of the way (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — that fusion is a source of energy in stars; that the stability of stars relies on an equilibrium between outward radiation pressure and inward gravitational forces; guidance: energy release calculations are required; E.3 — mass defect and mass–energy equivalence E = mc²; D.1 — gravitational potential energy Command term: Deduce
50E-2-57
Power output of the solar core·E.5 Fusion and stars
Paper 2Easy13 marks
Short answer & extended response8 steps to full marksDetermine
Nearly all of the Sun's energy is released by fusion in its core, a sphere of radius 0.25 R☉. The net result of the fusion is that four protons become one helium nucleus, releasing 26.7 MeV.
The intensity of the Sun's radiation at the Earth (above the atmosphere) is 1.36 × 103 W m−2. (Earth–Sun distance = 1.50 × 1011 m; R☉ = 6.96 × 108 m; mass of a proton = 1.673 × 10−27 kg; 1 MeV = 1.60 × 10−13 J; 1 year = 3.16 × 107 s)
(a)
(i)
Show that the luminosity of the Sun is about 3.8 × 1026 W.
(2)
(ii)
Determine the average power released per unit volume in the core.
(2)
(iii)
A resting human body releases about 100 W and has a volume of 0.070 m³. Compare the power per unit volume of the body with your answer to (a)(ii).
(1)
(b)
(i)
Calculate the number of helium nuclei formed in the Sun each second.
(2)
(ii)
The core contains about 1.5 × 1029 kg of hydrogen. Estimate the fraction of this hydrogen that is fused each year.
(3)
(c)
(i)
Explain why the Sun is so luminous even though its core releases so little power per unit volume.
(1)
(ii)
A fusion reactor on the Earth contains a plasma whose density is about 10−12 of the density in the Sun's core, and it must release a power per unit volume far greater than the Sun's core. Explain why the plasma must be heated to a temperature about ten times greater than that of the Sun's core.
Allow ECF from (a)(i). Accept 8.9–9.0 × 1037 s−1 (8.9 × 1037 from the show-that value 3.8 × 1026 W).
Part (b)(ii)
Mass of hydrogen fused per second = 4 × 1.673 × 10−27 × 9.0 × 1037 = 6.0 × 1011 kg s−1
✓ 1
Allow ECF from (b)(i). Using one proton per reaction: [0] for this mark.
Per year: 6.0 × 1011 × 3.16 × 107 = 1.9 × 1019 kg
✓ 1
Fraction = 1.9 × 1019/1.5 × 1029 = 1.3 × 10−10
✓ 1
Accept 1.2–1.4 × 10−10. Allow ECF.
Part (c)(i)
The core has an enormous volume (about 1025 m³), so the small power per unit volume adds up to a very large total power
✓ 1
Allow ECF from (a)(ii).
Part (c)(ii)
The collision rate per unit volume is proportional to the square of the density, so at the low density of the reactor the collisions are far less frequent
✓ 1
Do not accept "to make fusion happen" without reference to density or rate.
A higher temperature means a much larger fraction of the nuclei have enough kinetic energy to overcome the electric repulsion, so a much larger fraction of collisions lead to fusion, compensating for the low density
✓ 1
Accept: a greater proportion of high-energy nuclei. Allow ECF from (a)(ii).
Answers: (a)(ii) 17 W m−3 · (a)(iii) body ≈ 82 times greater · (b)(i) 9.0 × 1037 s−1 · (b)(ii) 1.3 × 10−10(the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — that fusion is a source of energy in stars; the conditions leading to fusion in stars in terms of density and temperature; guidance: energy release calculations are required; B.1 — apparent brightness b = L/4πd² Command term: Determine
51E-2-58
Radius of a red giant from its angular diameter·E.5 Fusion and stars
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine
An interferometer, which combines the light collected by several telescopes, measures the angular diameter of a nearby red giant as (0.0190 ± 0.0005) arc-second. The parallax angle of the star is (0.0500 ± 0.0010) arc-second, and its spectrum peaks at a wavelength of (690 ± 10) nm.
(1 pc = 3.08 × 1016 m; 1 arc-second = 4.85 × 10−6 rad; R☉ = 6.96 × 108 m; L☉ = 3.85 × 1026 W; σ = 5.67 × 10−8 W m−2 K−4; Wien constant = 2.9 × 10−3 m K)
(a)
(i)
Show that the distance to the star is about 6.2 × 1017 m.
(1)
(ii)
Determine the radius of the star in terms of R☉.
(2)
(b)
(i)
Calculate the surface temperature of the star.
(1)
(ii)
Determine the luminosity of the star in terms of L☉.
(3)
(iii)
Explain, using your answers, where the star lies on a Hertzsprung–Russell diagram.
(2)
(c)
(i)
Determine the absolute uncertainty in the luminosity.
(3)
(d)
(i)
Suggest why the radii of very few stars can be found by this method.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
d = 1/0.0500 = 20.0 pc = 20.0 × 3.08 × 1016 = 6.16 × 1017 m
✓ 1
Must see 20 pc and the conversion, or 6.16 × 1017 m.
Part (a)(ii)
Angle in radians = 0.0190 × 4.85 × 10−6 = 9.21 × 10−8 rad; R = d × (angle)/2
✓ 1
Using the whole angle (diameter) as the radius: [1 max] (82 R☉).
R = 6.16 × 1017 × 9.21 × 10−8/2 = 2.8 × 1010 m = 41 R☉
Accept 440–490 L☉. ALT: L/L☉ = (R/R☉)²(T/T☉)4 with T☉ = 5800 K.
Part (b)(iii)
It is cool (about 4200 K, cooler than the Sun) but several hundred times more luminous than the Sun
✓ 1
Allow ECF from (b)(i) and (b)(ii).
So it lies above the main sequence, in the upper right region of red giants, because its radius (about 40 R☉) is very large
✓ 1
Do not accept "top right" alone without a reason linked to L or R.
Part (c)(i)
% uncertainty in R = % in p + % in angle = 2.0 % + 2.6 % = 4.6 %
✓ 1
% uncertainty in T = % in λ = 1.4 %; % in L = 2 × 4.6 + 4 × 1.4 = 15 %
✓ 1
Adding without the powers 2 and 4: [0] for this mark.
ΔL = 0.15 × 470 ≈ ±70 L☉
✓ 1
Allow ECF from (b)(ii). Accept ±60 to ±80 L☉; uncertainty to 1 s.f.
Part (d)(i)
The angular diameter is 2R/d, which is extremely small for all but large, nearby stars (the Sun at 10 pc would subtend only about 0.001 arc-second), too small to measure
✓ 1
Accept any answer linking a tiny angle to small radius or large distance.
Answers: (a)(ii) 41 R☉ · (b)(i) 4.2 × 103 K · (b)(ii) 470 L☉ · (c)(i) ±70 L☉(the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — how to determine stellar radii; the use of stellar parallax as a method to determine the distance d to celestial bodies as given by d (parsec) = 1/p (arc-second); the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions; guidance: the determination of stellar radii using luminosity and surface temperature is required; B.1 — Wien's displacement law λmaxT = 2.9 × 10−3 m K; B.1 — the Stefan–Boltzmann law as given by L = σAT4; Tools — propagating uncertainties through a power law Command term: Determine
52E-2-59
Estimating the temperature of a stellar core·E.5 Fusion and stars
Paper 2Hard13 marks
Short answer & extended response8 steps to full marksShow that
A simple model of a star of mass M and radius R assumes that the star has a uniform density and is in equilibrium. Inside a uniform sphere, the gravitational field strength at a distance r from the centre is g = GMr/R³.
The pressure at the centre of the star is taken to be equal to the weight of a column of gas of unit cross-sectional area that reaches from the centre to the surface.
(G = 6.67 × 10−11 N m² kg−2; mass of the Sun = 1.99 × 1030 kg; radius of the Sun = 6.96 × 108 m; mass of a proton = 1.673 × 10−27 kg; kB = 1.38 × 10−23 J K−1)
(a)
(i)
Show that the pressure at the centre is Pc = 3GM²/(8πR4).
(2)
(ii)
The star is a fully ionised gas of hydrogen, so its particles (protons and electrons) have a mean mass m̄ = mp/2, where mp is the mass of a proton. Show that the model gives a central temperature Tc = GMm̄/(2kBR).
(2)
(b)
(i)
Calculate Tc for the Sun.
(2)
(ii)
The accepted value of the central temperature of the Sun is 1.5 × 107 K. Suggest why the model gives a smaller value.
(1)
(c)
(i)
A main-sequence star has a mass of 10 M☉ and a radius of 5.0 R☉. Assuming that the ratio (actual central temperature)/(model central temperature) is the same as for the Sun, estimate its central temperature.
(2)
(ii)
In stars of large mass, radiation pressure provides a large fraction of the outward pressure. Explain the effect of this on the estimate in (c)(i).
(2)
(d)
(i)
Use the model to explain why a contracting protostar heats up, and why its contraction stops when fusion begins.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
g rises linearly from 0 at the centre to GM/R² at the surface, so its mean value along the column is GM/(2R²)
✓ 1
The mean field (or an integral of ρg over the column) must be seen.
Pc = ρ × (mean g) × R with ρ = 3M/(4πR³): Pc = (3M/4πR³)(GM/2R²)R = 3GM²/(8πR4)
✓ 1
Part (a)(ii)
PV = NkBT gives P = (N/V)kBT = ρkBT/m̄, since the number of particles per unit volume is ρ/m̄
The real density is much greater near the centre than the mean density (mass is concentrated towards the centre), so the gravitational field and the weight of the overlying gas, and hence the central pressure and temperature, are greater than in the uniform model
✓ 1
Accept: the model underestimates the central pressure because density is not uniform. Allow ECF from (b)(i).
Part (c)(i)
Tc ∝ M/R, so the ratio to the Sun is 10/5.0 = 2.0
✓ 1
Allow ECF from (a)(ii).
Tc ≈ 2.0 × 1.5 × 107 = 3 × 107 K
✓ 1
Using the model value instead (1.2 × 107 K): [1 max].
Part (c)(ii)
The model assumes all of the central pressure is gas pressure, ρkBT/m̄; if part of the pressure needed to balance gravity is radiation pressure, the gas pressure needed is smaller
✓ 1
So a smaller temperature is needed: the actual central temperature is less than the estimate (the estimate is too high)
✓ 1
MP2 depends on MP1. Allow ECF from (c)(i).
Part (d)(i)
Tc ∝ M/R: at constant mass, as R decreases the central temperature increases (gravitational potential energy is transferred to the particles); the density increases too
✓ 1
Allow ECF from (a)(ii).
When the core is hot and dense enough, fusion releases energy that maintains the pressure (gas and radiation) needed to balance gravity, so the star stops contracting and is in equilibrium
✓ 1
Do not accept "fusion pushes outwards" without reference to pressure balancing gravity.
Answers: (b)(i) 5.8 × 106 K · (c)(i) 3 × 107 K (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — that the stability of stars relies on an equilibrium between outward radiation pressure and inward gravitational forces; the conditions leading to fusion in stars in terms of density and temperature; the effect of stellar mass on the evolution of a star; D.1 — gravitational field strength g = GM/r² (inside a uniform sphere given); B.3 — the equation of state PV = NkBT for an ideal gas; pressure P = F/ACommand term: Show that
53E-2-60
The life and death of a massive star·E.5 Fusion and stars
Paper 2Medium13 marks
Short answer & extended response8 steps to full marksDetermine
The Hertzsprung–Russell diagram shows the track of a star of 15 solar masses from the end of its main-sequence stage at A to a later stage at B. The surface temperature is 30 000 K at A and 3500 K at B.
(G = 6.67 × 10−11 N m² kg−2; M☉ = 1.99 × 1030 kg; L☉ = 3.85 × 1026 W; 1 year = 3.16 × 107 s; c = 3.00 × 108 m s−1)
Hertzsprung–Russell diagram (both axes logarithmic) showing the track A → B of a star of 15 solar masses after it leaves the main sequence. Grid lines are drawn at each power of ten of luminosity.
(a)
(i)
State the type of star at B.
(1)
(ii)
Determine, using the diagram, the ratio (radius of the star at B)/(radius of the star at A).
(2)
(b)
Eventually the iron core of the star, of mass 1.4 M☉ and radius 3.0 × 106 m, collapses to form a neutron star of radius 1.2 × 104 m. The gravitational potential energy of a uniform sphere of mass M and radius R is −3GM²/(5R).
(i)
Determine the energy released by the collapse.
(3)
(ii)
Compare your answer to (b)(i) with the energy radiated by the Sun during its main-sequence lifetime of 1.0 × 1010 years.
(1)
(c)
Before the collapse the iron core rotates once every 1500 s. Treat the core and the neutron star as uniform spheres, each with moment of inertia (2/5)MR², and assume that no mass is lost from the core.
(i)
Determine the period of rotation of the neutron star.
(3)
(ii)
Calculate the speed of a point on the equator of the neutron star, as a fraction of c.
(1)
(d)
(i)
The core of a star of one solar mass becomes a white dwarf made mainly of carbon and oxygen. Explain, in terms of the conditions for fusion, why carbon does not fuse in this core.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Red supergiant
✓ 1
Do not accept red giant alone.
Part (a)(ii)
Luminosities read from the graph: A ≈ 2 × 104L☉ and B ≈ 5 × 104L☉
✓ 1
Accept 1.5–2.5 × 104 and 4–6 × 104L☉.
R ∝ √L/T², so ratio = √(5/2) × (30 000/3500)² = 116
✓ 1
Accept 90–150, consistent with the candidate's readings. Inverted temperature ratio: [1 max].
Part (b)(i)
ΔE = (3GM²/5)(1/R2 − 1/R1)
✓ 1
Sign errors are not penalised if the answer is given as a positive energy released.
Accept 2.5–2.6 × 1046 J (ignoring the 1/R1 term gives 2.6 × 1046 J, also accepted).
Part (b)(ii)
Sun: 3.85 × 1026 × 1.0 × 1010 × 3.16 × 107 = 1.2 × 1044 J, so the collapse releases about 210 times more
✓ 1
Allow ECF from (b)(i). Accept 200–220 times.
Part (c)(i)
Angular momentum is conserved because there is no external torque: I1ω1 = I2ω2
✓ 1
I ∝ R², and ω = 2π/T, so T ∝ R²: T2 = T1(R2/R1)²
✓ 1
T2 = 1500 × (1.2 × 104/3.0 × 106)² = 0.024 s
✓ 1
Using T ∝ R instead of R²: [2 max] (6.0 s).
Part (c)(ii)
v = 2πR/T = 2π × 1.2 × 104/0.024 = 3.1 × 106 m s−1 = 0.010c
✓ 1
Allow ECF from (c)(i).
Part (d)(i)
Carbon nuclei have a much larger charge (6e), so their electric repulsion is much greater and a much higher temperature (greater kinetic energy) is needed for them to get close enough to fuse
✓ 1
Accept a reference to a higher density as well as temperature.
The gravitational force of a core of small mass cannot compress it enough to reach this temperature, whereas the massive core of the 15 M☉ star can
✓ 1
Do not accept "it is too small" without reference to compression or temperature.
Answers: (a)(ii) ≈ 120 · (b)(i) 2.6 × 1046 J · (b)(ii) about 210 times greater · (c)(i) 0.024 s · (c)(ii) 0.010c(the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the effect of stellar mass on the evolution of a star; the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions; how to determine stellar radii; the conditions leading to fusion in stars in terms of density and temperature; guidance: the sketching and interpretation of HR diagrams, including the location of main sequence stars, red giants, super giants and white dwarfs, and lines of constant radius; D.1 — gravitational potential energy; A.4 (HL) — conservation of angular momentum L = Iω; A.2 — speed in circular motion Command term: Determine
54E-2-61
A white dwarf in a binary system·E.5 Fusion and stars
Paper 2Easy10 marks
Short answer & extended response7 steps to full marksDetermine
A main-sequence star of 1.5 solar masses has a companion that is a white dwarf. The two stars formed at the same time. The white dwarf has a luminosity of 0.020 L☉, a surface temperature of 2.0 × 104 K and a mass of 0.90 M☉.
(L☉ = 3.85 × 1026 W; M☉ = 1.99 × 1030 kg; σ = 5.67 × 10−8 W m−2 K−4; G = 6.67 × 10−11 N m² kg−2; radius of the Earth = 6.37 × 106 m)
(a)
(i)
State the region of the Hertzsprung–Russell diagram in which the white dwarf lies, and why it lies there.
(1)
(ii)
Show that the radius of the white dwarf is about 8 × 106 m.
(2)
(iii)
Compare the radius of the white dwarf with the radius of the Earth.
(1)
(b)
(i)
Calculate the mean density of the white dwarf.
(2)
(ii)
Determine the gravitational field strength at the surface of the white dwarf, as a multiple of the field strength at the surface of the Earth (9.81 N kg−1).
(2)
(c)
(i)
The white dwarf is the remnant of a star that was more massive than its companion. Explain why the more massive star of the pair is the one that has already left the main sequence.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Lower left, below the main sequence: it is hot (high surface temperature) but has a very low luminosity
✓ 1
Both the region and the reason are needed.
Part (a)(ii)
L = 0.020 × 3.85 × 1026 = 7.7 × 1024 W and R = √(L/4πσT4)
✓ 1
R = √(7.7 × 1024/(4π × 5.67 × 10−8 × (2.0 × 104)4)) = 8.22 × 106 m
✓ 1
Must see the full substitution or 8.22 × 106 m.
Part (a)(iii)
8.2 × 106/6.37 × 106 = 1.3: the white dwarf is only a little larger than the Earth
Accept 7.5–8.0 × 108 kg m−3; using the show-that value 8 × 106 m gives 8.4 × 108, also accepted.
Part (b)(ii)
g = GM/R² = 6.67 × 10−11 × 1.79 × 1030/(8.2 × 106)² = 1.8 × 106 N kg−1
✓ 1
Allow ECF from (a)(ii).
1.8 × 106/9.81 ≈ 1.8 × 105 times the Earth's value
✓ 1
Accept 1.7–1.9 × 105.
Part (c)(i)
A main-sequence star of greater mass has a much greater luminosity, so it uses the hydrogen in its core at a much greater rate
✓ 1
Accept: a hotter, denser core with a much greater fusion rate.
Its fuel increases only in proportion to its mass, so its main-sequence lifetime is shorter: it has finished its core hydrogen, become a red giant and lost its outer layers, while the 1.5 M☉ star is still on the main sequence
✓ 1
Both stars formed at the same time must be used in the argument.
Answers: (a)(iii) about 1.3 × the Earth's radius · (b)(i) 7.7 × 108 kg m−3 · (b)(ii) 1.8 × 105 × gEarth(the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions; the effect of stellar mass on the evolution of a star; guidance: the determination of stellar radii using luminosity and surface temperature is required; B.1 — the Stefan–Boltzmann law as given by L = σAT4; D.1 — gravitational field strength g = GM/r² Command term: Determine
55E-2-62
A transiting exoplanet·E.5 Fusion and stars
Paper 2Hard19 marks
Short answer & extended response12 steps to full marksDetermine
The main-sequence star K has a mass of 1.00 M☉. A planet, of much smaller mass, moves around K in a circular orbit of period 3.52 days. The orbit is seen edge-on, so once in every orbit the planet passes in front of K (a transit).
The graph shows the apparent brightness b of K, as a fraction of its normal value, during one transit. Take the duration of the transit to be the time between the two points at which the brightness is half-way down the dip. Treat the disc of K as uniformly bright.
(G = 6.67 × 10−11 N m² kg−2, M☉ = 1.99 × 1030 kg, R☉ = 6.96 × 108 m, radius of the Earth = 6.37 × 106 m)
Brightness of star K during a transit (graph drawn to scale)
(a)
(i)
Show that the radius of the planet's orbit is about 7 × 109 m.
(2)
(ii)
Calculate the orbital speed of the planet.
(1)
(b)
(i)
Determine, using the graph, the radius of K in terms of R☉.
(2)
(ii)
State one assumption made in your answer to (b)(i).
(1)
(iii)
Show that the radius of the planet is r = R√(Δb/b), where Δb/b is the fractional decrease in brightness during the transit and R is the radius of K.
(2)
(iv)
Determine the radius of the planet, giving your answer in units of the Earth's radius.
(2)
(v)
The brightness takes 19 minutes to fall from its normal value to its minimum value. Show that this is consistent with your answer to (b)(iv).
(2)
(c)
(i)
Measurements of the motion of K show that the mass of the planet is 1.3 × 1027 kg. Determine the mean density of the planet.
(2)
(ii)
The mean densities of rocky planets lie between about 3000 kg m−3 and 5500 kg m−3. Suggest what your answer to (c)(i) indicates about the planet.
(1)
(d)
(i)
The fractional decrease in brightness is (0.0165 ± 0.0005) and the duration of the transit is (2.50 ± 0.05) h. The mass of K and the period are known precisely. Determine the percentage uncertainty in the radius of the planet.
(2)
(ii)
Suggest two reasons why most of the planets discovered by the transit method are large planets in small orbits.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The gravitational force provides the centripetal force: GMm/a² = m4π²a/T², so a³ = GMT²/4π²
✓ 1
a = (6.67 × 10−11 × 1.99 × 1030 × (3.52 × 86 400)²/4π²)1/3 = 6.78 × 109 m
✓ 1
Must see the full substitution or 6.78 × 109 m. The period must be in seconds.
Part (a)(ii)
v = 2πa/T = 2π × 6.78 × 109/(3.52 × 86 400) = 1.40 × 105 m s−1
✓ 1
Allow ECF from (a)(i).
Part (b)(i)
Duration of the transit from the graph = 2.5 h = 9.0 × 103 s
✓ 1
Accept 2.4–2.6 h.
The planet moves a distance 2R in this time: R = vt/2 = 1.40 × 105 × 9.0 × 103/2 = 6.3 × 108 m = 0.91 R☉
✓ 1
Allow ECF from (a)(ii). Accept 0.87–0.94 R☉. The answer must be expressed in R☉ for MP2.
Part (b)(ii)
The planet crosses along a diameter of the disc of K (or: the planet moves in a straight line at constant speed across the disc / the motion of K itself is negligible)
✓ 1
Any one.
Part (b)(iii)
The planet hides an area πr² of the disc of K, of area πR², which is uniformly bright
✓ 1
So the fraction of the light blocked is Δb/b = πr²/πR², giving r = R√(Δb/b)
✓ 1
Part (b)(iv)
From the graph Δb/b = 0.0165
✓ 1
Accept 0.016–0.017.
r = 6.3 × 108 × √0.0165 = 8.1 × 107 m = 13 Earth radii
✓ 1
Allow ECF from (b)(i) and (b)(iii). Accept 12–13.5 Earth radii; the answer must be expressed in Earth radii for MP2.
Part (b)(v)
While the brightness falls, the planet moves from first touching the edge of the disc to lying wholly inside it: a distance equal to its own diameter 2r
✓ 1
2r = 1.40 × 105 × 19 × 60 = 1.6 × 108 m, so r = 8.0 × 107 m, in agreement with (b)(iv)
✓ 1
Allow ECF from (a)(ii) and (b)(iv). A comparison with the answer to (b)(iv) is needed.
Its density is far below that of rock (and even below that of water), so it is a gas giant made mainly of hydrogen and helium, not a rocky planet
✓ 1
Allow ECF from (c)(i).
Part (d)(i)
With M and T exact, R ∝ t: 0.05/2.50 = 2.0 %; the fractional dip contributes ½ × (0.0005/0.0165) = ½ × 3.0 % = 1.5 %
✓ 1
Percentage uncertainty in r = 2.0 % + 1.5 % = 3.5 %
✓ 1
Allow ECF from (b). Adding 3.0 % without halving (5.0 %) scores [1].
Part (d)(ii)
The dip Δb/b ∝ r², so only a large planet blocks enough light to be detected above the scatter of the measurements (an Earth-sized planet would give a dip of only about 1 × 10−4 here)
✓ 1
Any two of: (1) the r² dependence of the dip; (2) a short period means many transits can be observed and confirmed in a given observing time; (3) a transit is seen only if the orbit is close to edge-on, and the range of orientations for which this happens is larger when the orbit is small compared with the star. [1] each, [2 max].
A planet in a small orbit has a short period, so its transits repeat often and can be confirmed quickly; also its transits are seen over a wider range of orbit orientations
✓ 1
Answers: (a)(ii) 1.40 × 105 m s−1 · (b)(i) 0.91 R☉ · (b)(iv) 13 Earth radii · (c)(i) 5.8 × 102 kg m−3 · (d)(i) 3.5 % (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — how to determine stellar radii (here from the transit of a planet across a main-sequence star); D.1 — orbital motion: the gravitational force provides the centripetal force, T² ∝ r³; A.2 — speed in uniform circular motion; B.1 — density ρ = m/V; Tools 3 — propagation of uncertainties through a power law Command term: Determine
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