E.5 Fusion and stars: IB Physics SL exam-style questions
Stars are stable when the outward radiation pressure balances the inward gravitational force, and fusion in the core needs high temperature and density. You need energy-release calculations for fusion and the effect of a star's mass on how it evolves.
The Hertzsprung–Russell diagram, with luminosity against surface temperature, shows main-sequence stars, red giants, supergiants and white dwarfs. You also need stellar parallax d (pc) = 1/p (arcsec), conversions between AU, light years and parsecs, and how to find a star's radius from its luminosity and surface temperature.
26 questions
201 marks
Paper 1A: 9
Paper 1B: 6
Paper 2: 11
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26 practice questions on E.5 Fusion and stars
1E-1A-21
Stellar equilibrium·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksState
A main-sequence star keeps a steady size for billions of years.
Which row identifies the outward and the inward effects that are in equilibrium?
OutwardInward
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Mark
Notes
Step 1Energy released by fusion in the core maintains an outward radiation (and gas) pressure that balances the inward gravitational pull of the star's mass.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThe two effects have been swapped: gravity pulls the star's material inwards, and the pressure from fusion pushes outwards.
BRotation slightly changes the shape of a star, but it is far too small an effect to support the star against gravity.
CNuclei are all positively charged and repel one another; the inward force on the star's material is gravitational.
DCorrect: outward radiation pressure from core fusion balances the inward gravitational force.
Syllabus understandingE.5 — that the stability of stars relies on an equilibrium between outward radiation pressure and inward gravitational forces Command term: State
2E-1A-22
Conditions for fusion·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksExplain
Fusion of hydrogen requires two nuclei to come within about 10−15 m of one another.
Why does this need both a very high temperature and a very high density?
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Mark
Notes
Step 1Nuclei are positively charged and repel, so they need very large kinetic energies — a very high temperature — to get close enough for the strong nuclear force to act.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Even then only a small fraction of collisions succeed, so a very high density is needed to make collisions frequent enough for a significant fusion rate.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the temperature sets how many nuclei have enough energy to get close, and the density sets how often they meet at all.
BNeither the size nor the mass of a nucleus depends on the conditions around it.
CThe nuclei keep their charge; they must arrive with enough kinetic energy to overcome the repulsion.
DThe kinetic energy of the nuclei is set by the temperature, not the density.
Syllabus understandingE.5 — the conditions leading to fusion in stars in terms of density and temperature Command term: Explain
3E-1A-23
Fusion in the Sun·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksDeduce
The Sun radiates energy at a rate L☉. This energy is released by nuclear fusion in its core.
At what rate does the mass of the Sun decrease because of the energy it radiates? (c is the speed of light.)
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Mark
Notes
Step 1Each second the Sun radiates energy L☉; by E = mc2 this corresponds to a mass L☉/c2 converted each second.
✓ 1
Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AThis multiplies by c2 instead of dividing: m = E/c2.
BThe speed of light has not been squared.
CCorrect: mass lost per second = energy radiated per second ÷ c2 = L☉/c2 (about 4 × 109 kg s−1).
DThe expression has been inverted; it has the unit s kg−1, not kg s−1.
Syllabus understandingE.5 — that fusion is a source of energy in stars; E.3 — the mass–energy equivalence as given by E = mc2 in nuclear reactions Command term: Deduce
4E-1A-24
Stellar parallax·E.5 Fusion and stars
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A star has a parallax angle of 0.25 arc-second. (1 pc = 3.09 × 1016 m, 1 ly = 9.46 × 1015 m)
All 2 steps must be completed — there is no mark for a part-answer.
Step 21 pc = 3.09 × 1016/9.46 × 1015 = 3.27 ly, so d = 4.0 × 3.27 = 13 ly.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis quotes the parallax angle as though it were a distance.
BThis divides by 3.27 instead of multiplying: one parsec is longer than one light year.
CThis is the distance in parsecs, not in light years.
DCorrect: d = 1/0.25 = 4.0 pc, and 4.0 × 3.27 = 13 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) Command term: Determine
5E-1A-25
The HR diagram·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
A star lies at the bottom left of a Hertzsprung–Russell diagram: it has a high surface temperature but a very low luminosity.
What type of star is it, and what does its position imply about its radius?
Diagram NOT accurately drawn
Type of starRadius
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Notes
Step 1Bottom left means a high surface temperature with a low luminosity: the white-dwarf region.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2L = 4πR2σT4, so with T large and L small the radius must be very small.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
ARed giants lie at the top right — cool but very luminous, the opposite corner.
BMain-sequence stars run diagonally across the diagram; a hot main-sequence star is very luminous, not faint.
CThe type is right, but a hot star with a very large radius would be extremely luminous, not faint.
DCorrect: L = 4πR2σT4, so a hot star with a small luminosity must have a very small radius — a white dwarf.
Syllabus understandingE.5 — the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions Command term: Deduce
6E-1B-06
Stellar parallax·E.5 Fusion and stars
Paper 1BEasy8 marks
Data-based question8 steps to full marksDetermine
A student uses a star catalogue produced by a space-based astrometry mission to test whether five stars that appear close together in the sky belong to the same star cluster. The catalogue gives the parallax angle p of each star in milliarcseconds (1 mas = 1 × 10−3 arc-second).
(1 pc = 3.26 ly)
Star
p / mas
A
8.62 ± 0.05
B
8.55 ± 0.04
C
8.71 ± 0.06
D
5.08 ± 0.05
E
8.49 ± 0.07
(a)
Show that the distance to star A is about 116 pc.
(1)
(b)
Determine the distance to star E and its absolute uncertainty.
(3)
(c)
Identify the star that is not a member of the cluster, giving a reason.
(1)
(d)
The cluster is thought to be about 5 pc across. Comment on whether the distances of the other four stars are consistent with this.
(1)
(e)
Calculate the distance to star A in light years.
(1)
(f)
Outline why parallax angles measured from a telescope on the Earth's surface are less precise than those measured from space.
(1)
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Notes
Part (a)
d = 1/(8.62 × 10−3) = 116.0 pc
✓ 1
Must see the conversion of mas to arc-seconds and the substitution OR an answer to 4 s.f.
Part (b)
d = 1/(8.49 × 10−3) = 117.8 pc
✓ 1
Percentage uncertainty in d = percentage uncertainty in p = 0.07/8.49 × 100 = 0.82 %
✓ 1
Accept fractional uncertainty 0.0082.
Δd = 1.0 pc, so d = (117.8 ± 1.0) pc
✓ 1
Accept (118 ± 1) pc. MP3 is for matching the precision of value and uncertainty.
Part (c)
D: its distance is 197 pc, about 81 pc further away than the others
✓ 1
Accept: its parallax is much smaller than the others'.
Part (d)
The four distances lie between 114.8 pc and 117.8 pc, a spread of about 3 pc, which is less than 5 pc (and comparable with the uncertainties), so they are consistent with one cluster
✓ 1
Allow ECF from (b).
Part (e)
116.0 × 3.26 = 378 ly
✓ 1
Accept 378–379 ly.
Part (f)
Turbulence in the atmosphere blurs and moves the star's image, so the smallest angle that can be measured is larger
✓ 1
OWTTE
Answers: (a) 116.0 pc · (b) (117.8 ± 1.0) pc · (c) star D · (e) 378 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: the conversion between AU, ly and pc is required. Command term: Determine
7E-1B-07
Stellar radius from L and T·E.5 Fusion and stars
Paper 1BHard12 marks
Data-based question12 steps to full marksDetermine
Two methods of finding the radius of a star are compared. For six stars, an optical interferometer measures the angular diameter, from which the radius Rint is found with an uncertainty of ±3 %. A stellar database also gives the luminosity L of each star (uncertainty ±4 %) and its surface temperature T, found from its spectrum (uncertainty ±2 %). A student uses L and T to calculate a second value of the radius, RLT, and tests the hypothesis that the two methods agree, so that RLT = Rint.
The graph shows RLT against Rint, with error bars, for five of the stars. (σ = 5.67 × 10−8 W m−2 K−4, L☉ = 3.83 × 1026 W, R☉ = 6.96 × 108 m)
Star
T / K
L / L☉
Rint / R☉
RLT / R☉
1
6240
2.89
1.42
1.45
2
9480
44.5
2.55
2.47
3
5060
24.2
6.30
4
4680
46.6
10.6
10.4
5
4380
88.9
15.9
16.4
6
4040
120
22.7
22.4
Graph drawn to scale
(a)
Show that RLT for star 3 is about 6.4 R☉.
(1)
(b)
Determine the percentage uncertainty in RLT.
(2)
(c)
Plot the point for star 3 on the graph, with its error bars.
(1)
(d)
Draw the line of best fit and determine its gradient.
(3)
(e)
Draw lines of maximum and minimum gradient and hence determine the absolute uncertainty in the gradient.
(2)
(f)
Discuss whether the data support the hypothesis.
(1)
(g)
Identify the region of the Hertzsprung–Russell diagram in which star 6 lies. Give reasons for your answer.
The hypothesis predicts a gradient of 1 (and a line through the origin); 1 lies within gradient ± uncertainty, so the data support it
✓ 1
Allow ECF from (d) and (e); a conclusion with no reference to the uncertainty scores 0.
Part (g)
Red giant region (upper right of the diagram)
✓ 1
Accept: giant branch.
T ≈ 4040 K is lower than the Sun's surface temperature, but L ≈ 120 L☉ is much larger than L☉ because the radius (≈ 22 R☉) is very large
✓ 1
Both a temperature and a luminosity/radius comparison needed. OWTTE
Answers: (a) 6.40 R☉ · (b) 6 % · (d) ≈ 1.00 · (e) ≈ ±0.08 (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); the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions; B.1 — the Stefan–Boltzmann law as given by L = σAT4. Command term: Determine
8E-2-12
Fusion in the Sun·E.5 Fusion and stars
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksExplain
In the core of the Sun, deuterium nuclei fuse with protons: 21H + 11H → 32He + γ
Nuclear masses: proton 1.007276 u, deuterium nucleus 2.013553 u, helium-3 nucleus 3.014932 u.
(1 u = 931.5 MeV c−2, 1 MeV = 1.60 × 10−13 J, h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, Wien's displacement constant = 2.9 × 10−3 m K)
(a)
The reaction.
(i)
State why this reaction is an example of nuclear fusion.
(1)
(ii)
Show that the energy released in the reaction is about 5.5 MeV.
(1)
(iii)
Assume that all of this energy is carried away by the gamma-ray photon. Calculate the wavelength of the photon.
(2)
(b)
The complete sequence of reactions in the Sun converts four protons into one helium-4 nucleus and releases 26.7 MeV. The reaction above occurs twice in each sequence. The luminosity of the Sun is 3.8 × 1026 W. Calculate the number of times per second that the reaction above occurs in the Sun.
(3)
(c)
Explain how the energy released by fusion keeps the Sun stable.
(2)
(d)
The surface of the Sun has a temperature of 5800 K.
(i)
Calculate the wavelength at which the radiation from the Sun's surface has its maximum intensity.
(1)
(ii)
Suggest why the Sun emits mostly visible and infrared radiation although the energy released in its core is carried by gamma-ray photons.
(2)
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Notes
Part (a)(i)
Two light nuclei join to form a heavier nucleus «releasing energy»
Energy per sequence = 26.7 × 1.60 × 10−13 = 4.27 × 10−12 J
✓ 1
Sequences per second = 3.8 × 1026/4.27 × 10−12 = 8.9 × 1037 s−1
✓ 1
Reactions per second = 2 × 8.9 × 1037 = 1.8 × 1038 s−1
✓ 1
Award [3] for CNA. Award [2 max] for 8.9 × 1037 s−1 «factor 2 omitted». Award [1 max] for 4.3 × 1038 s−1 «5.49 MeV used in place of 26.7 MeV».
Part (c)
The energy released keeps the core very hot, which produces an outward «radiation and gas» pressure
✓ 1
This outward pressure balances the inward gravitational force on every layer, so the Sun neither expands nor contracts
✓ 1
OWTTE
Part (d)(i)
λmax = 2.9 × 10−3/5800 = 5.0 × 10−7 m
✓ 1
Part (d)(ii)
The gamma photons are absorbed and re-emitted very many times by the matter of the Sun as the energy is transported outwards
✓ 1
The radiation that finally leaves is emitted by the surface as black-body radiation characteristic of 5800 K, which peaks in the visible «at about 500 nm»
✓ 1
OWTTE
Answers: (a)(ii) 5.49 MeV · (a)(iii) 2.3 × 10−13 m · (b) 1.8 × 1038 s−1 · (d)(i) 5.0 × 10−7 m (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; E.3 — the mass–energy equivalence as given by E = mc2 in nuclear reactions; B.1 — the emission spectrum of a black body and Wien's displacement law Command term: Explain
9E-2-13
Conditions for fusion·E.5 Fusion and stars
Paper 2Hard11 marks
Short answer & extended response11 steps to full marksDetermine
The core of a main-sequence star can be modelled as an ideal gas of fully ionized particles (nuclei and electrons). For the core of the Sun: temperature 1.5 × 107 K, density 1.5 × 105 kg m−3, mean mass of a particle 1.4 × 10−27 kg.
(kB = 1.38 × 10−23 J K−1)
(a)
The core as an ideal gas.
(i)
Show that the number of particles per unit volume in the core is about 1 × 1032 m−3.
(1)
(ii)
Calculate the pressure in the core.
(2)
(iii)
Calculate the mean kinetic energy of a particle in the core.
(1)
(b)
Conditions for fusion.
(i)
A protostar contracts under gravity. Explain why its core temperature rises as it contracts, and why the contraction stops once hydrogen fusion begins at a significant rate.
(4)
(ii)
Assume that hydrogen fusion at a significant rate requires a core temperature of at least 1.0 × 107 K. A contracting cloud of gas (a protostar) has a core density of 8.0 × 104 kg m−3 and a core pressure of 6.0 × 1015 Pa, with the same mean particle mass. Determine whether hydrogen fusion can have started in its core.
(2)
(c)
Suggest one limitation of modelling the core of a star as an ideal gas.
(1)
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Notes
Part (a)(i)
n = ρ/m = 1.5 × 105/1.4 × 10−27 = 1.07 × 1032 m−3
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (a)(ii)
PV = NkBT, so P = (N/V)kBT = 1.07 × 1032 × 1.38 × 10−23 × 1.5 × 107
✓ 1
P = 2.2 × 1016 Pa
✓ 1
Award [2] for CNA. Allow ECF from (a)(i): accept 2.1 × 1016 Pa from 1 × 1032 m−3.
As the gas falls inwards, the gravitational potential energy of the protostar decreases «gravity does work on the gas»
✓ 1
This energy becomes random kinetic energy of the particles, so the core temperature rises «Ek = ³⁄₂kBT»
✓ 1
When the core is hot and dense enough, fusion releases energy that keeps the core hot and its pressure high
✓ 1
The outward force due to pressure «of gas and radiation» then balances the inward gravitational force, so the contraction stops «the star joins the main sequence»
7.6 × 106 K < 1.0 × 107 K, so hydrogen fusion has not yet started «at a significant rate»
✓ 1
MP2 needs a comparison with 1.0 × 107 K. Allow a consistent conclusion from an incorrect temperature.
Part (c)
Any one of: the particles are charged and exert electric forces on one another / the pressure due to radiation is ignored / the temperature and density are not uniform throughout the core
✓ 1
OWTTE
Answers: (a)(i) 1.07 × 1032 m−3 · (a)(ii) 2.2 × 1016 Pa · (a)(iii) 3.1 × 10−16 J · (b)(ii) 7.6 × 106 K — fusion has not started (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 the stability of stars relies on an equilibrium between outward radiation pressure and inward gravitational forces; B.3 — the equations governing the behaviour of ideal gases as given by PV = NkBT; B.1 — that Kelvin temperature is a measure of the average kinetic energy of particles as given by Ek = ³⁄₂kBTCommand term: Determine
10E-2-14
Stellar equilibrium·E.5 Fusion and stars
Paper 2Hard12 marks
Short answer & extended response12 steps to full marksExplain
Star W is a main-sequence star of mass 3.0 × 1030 kg, radius 1.2 × 109 m and luminosity 1.9 × 1027 W. W neither expands nor contracts. Figure 1 shows a cross-section of W, its centre C and a small element X of gas inside W.
Draw, on Figure 1, labelled arrows to show the two forces that act on element X.
(2)
(b)
The surface of W.
(i)
Show that the gravitational field strength at the surface of W is about 140 N kg−1.
(1)
(ii)
Hence state the magnitude of the resultant force due to pressure that acts on 1.0 kg of gas at the surface of W.
(1)
(c)
The rate of fusion in the core of a star increases very steeply with the core temperature. Suppose that the rate of fusion in the core of W increases slightly for a short time. Explain how W returns to equilibrium.
(3)
(d)
Suppose instead that fusion in W stopped. Model W as an ideal monatomic gas of 3.0 × 1057 particles at a mean temperature of 6.0 × 106 K.
(i)
Show that the internal energy of W is about 4 × 1041 J.
(1)
(ii)
Estimate the time, in years, for which W could radiate at its present luminosity using only this internal energy.
(2)
(iii)
W will stay on the main sequence for about 2.5 × 109 years. Discuss what your answer to (d)(ii) shows about the source of the energy radiated by W.
(2)
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Notes
Part (a)
Arrow from X towards C labelled gravitational force «weight / gravity»
✓ 1
Arrow from X directly away from C, labelled force due to «gas and radiation» pressure, of the same length as the gravitational arrow
✓ 1
Award [1 max] for two correctly directed and labelled arrows of clearly different lengths.
Part (b)(i)
g = GM/R2 = 6.67 × 10−11 × 3.0 × 1030/(1.2 × 109)2 = 139 N kg−1
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
139 N «≈ 140 N», outwards: the gas is in equilibrium, so the force due to pressure balances its weight
✓ 1
Allow ECF from (b)(i).
Part (c)
More energy is released, so the temperature and the pressure in the core increase
✓ 1
The outward force due to pressure is now greater than the inward gravitational force, so the core expands
✓ 1
As it expands the core cools «and its density falls», so the rate of fusion falls back until the forces balance again
Must see full substitution OR answer to 3 s.f. «3.73 × 1041 J».
Part (d)(ii)
t = 3.73 × 1041/1.9 × 1027 = 1.96 × 1014 s
✓ 1
t = 1.96 × 1014/3.16 × 107 = 6.2 × 106 years
✓ 1
Accept 6.0–6.7 × 106 years «6.7 × 106 years from 4 × 1041 J». Award [2] for CNA.
Part (d)(iii)
About 6 × 106 years is several hundred times shorter than the main-sequence lifetime of 2.5 × 109 years
✓ 1
Allow ECF from (d)(ii).
So the stored internal energy cannot supply the luminosity for the main-sequence lifetime: the energy must be released continuously by nuclear fusion «of hydrogen into helium» in the core
✓ 1
OWTTE
Answers: (b)(i) 139 N kg−1 · (b)(ii) 139 N · (d)(i) 3.7 × 1041 J · (d)(ii) 6.2 × 106 years (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; that fusion is a source of energy in stars; the conditions leading to fusion in stars in terms of density and temperature; D.1 — gravitational field strength g = GM/r2; B.3 — the internal energy U of an ideal monatomic gas as given by U = ³⁄₂NkBTCommand term: Explain
11E-2-15
Stellar parallax·E.5 Fusion and stars
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksCalculate
The diagram shows the Earth's orbit around the Sun and a nearby star V, viewed from above the plane of the orbit. The position of V against the very distant background stars is measured when the Earth is at E1 and again six months later when it is at E2.
The parallax angle of V is 0.0420 arc-second. (1 pc = 3.26 ly)
Diagram not to scale
(a)
Annotate the diagram to show the parallax angle p of star V.
(1)
(b)
The distance to V.
(i)
Calculate the distance to V in parsecs.
(1)
(ii)
Calculate the distance to V in light years, and hence state the time taken for light from V to reach the Earth.
(2)
(c)
Show that one parsec is about 2.06 × 105 AU.
(1)
(d)
Outline why the two measurements are made six months apart.
(1)
(e)
A telescope on the Earth's surface cannot measure angles smaller than 0.01 arc-second. Determine whether it can be used to measure the distance to a star 150 pc away.
(2)
(f)
The absolute uncertainty in a measured parallax angle is about the same for all stars. Explain why the percentage uncertainty in the distance is greater for more distant stars.
(2)
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Notes
Part (a)
Angle p marked at V, between the line from V to the Sun and the line from V to E1 «or E2»
✓ 1
Accept half of the angle between the lines from V to E1 and from V to E2. Do not accept an angle marked at the Earth or at the Sun.
Part (b)(i)
d = 1/0.0420 = 23.8 pc
✓ 1
Part (b)(ii)
23.8 × 3.26 = 77.6 ly
✓ 1
Allow ECF from (b)(i). Accept 77–78 ly.
77.6 years
✓ 1
Allow ECF from the candidate's distance in ly.
Part (c)
1 arc-second = 1/3600° = 4.85 × 10−6 rad, so 1 pc = 1 AU/4.85 × 10−6 = 2.06 × 105 AU
✓ 1
Must see the conversion of 1 arc-second to radians OR answer to 4 s.f. «2.063 × 105 AU».
Part (d)
The Earth is then on opposite sides of its orbit, giving the largest baseline «2 AU» and so the largest shift in the apparent position of V
✓ 1
OWTTE
Part (e)
p = 1/150 = 0.0067 arc-second
✓ 1
This is smaller than 0.01 arc-second, so the telescope cannot measure it
✓ 1
MP2 only scores with a correct comparison.
Part (f)
The parallax angle is smaller for a more distant star «p = 1/d»
✓ 1
So a fixed absolute uncertainty is a larger fraction of p, and the percentage uncertainty in d equals that in p
✓ 1
OWTTE
Answers: (b)(i) 23.8 pc · (b)(ii) 77.6 ly; 77.6 years · (c) 2.06 × 105 AU · (e) p = 0.0067 arc-second — cannot be measured (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, light years and parsecs is required) Command term: Calculate
12E-1A-41
Stellar radius from L and T·E.5 Fusion and stars
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
Two stars have the same luminosity, but star X has twice the surface temperature of star Y.
What is (radius of Y)/(radius of X)?
Show mark scheme
Marking point
Mark
Notes
Step 1L = 4πR2σT4 with L the same for both, so R2T4 is constant and R ∝ 1/T2.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2TY = TX/2, so RY/RX = 22 = 4.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
A16 = 24 is the ratio of the powers radiated per unit area, not of the radii.
BCorrect: R ∝ 1/T2 at fixed luminosity, so halving the temperature makes the radius 4 times larger.
CThis uses R ∝ 1/T instead of 1/T2.
DThis is RX/RY — the ratio has been inverted.
Syllabus understandingE.5 — how to determine stellar radii (using luminosity and surface temperature) Command term: Determine
13E-1A-42
Apparent brightness·E.5 Fusion and stars
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
The luminosity of star X is four times that of star Y. The parallax angle of X is half that of Y.
What is (apparent brightness of X)/(apparent brightness of Y)?
Show mark scheme
Marking point
Mark
Notes
Step 1d = 1/p, so X, with half the parallax angle, is twice as far away as Y.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2b = L/4πd2, so bX/bY = 4 × (1/2)2 = 1.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: X is 4 times more luminous but twice as far away, and 4/22 = 1.
BThis divides the luminosity ratio by the distance ratio, 4/2, instead of by its square.
CThis is the luminosity ratio alone; the greater distance of X has been ignored.
DA smaller parallax angle has been taken to mean a smaller distance: 4 × 22 = 16.
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 concept of apparent brightness b = L/4πd2Command term: Determine
14E-1A-43
Stellar evolution by mass·E.5 Fusion and stars
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDescribe
Two main-sequence stars form at the same time: one of about one solar mass and one of about twenty solar masses.
Which row states which star leaves the main sequence first and the final state of the 20-solar-mass star?
Diagram NOT accurately drawn
Leaves the main sequence firstFinal state of the 20-solar-mass star
Show mark scheme
Marking point
Mark
Notes
Step 1A more massive star has a far greater luminosity and uses its core hydrogen much faster, so it leaves the main sequence first.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2A 20-solar-mass star becomes a red supergiant and ends in a supernova, leaving a neutron star or a black hole.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AA more massive star has more hydrogen, but it is so much more luminous that it uses it up far sooner.
BThe main-sequence lifetime depends strongly on mass, so the two stars do not leave together.
CA white dwarf is the end point of low-mass stars such as the Sun; a 20-solar-mass star ends in a supernova.
DCorrect: the massive star exhausts its core hydrogen first, and after a supernova it leaves a neutron star or a black hole.
Syllabus understandingE.5 — the effect of stellar mass on the evolution of a star Command term: Describe
15E-1A-44
Wien's law·E.5 Fusion and stars
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksIdentify
Two stars X and Y, which behave as black bodies, have the same radius. The surface temperature of X is 4500 K and that of Y is 6000 K.
Each graph shows the variation with wavelength λ of the power P emitted per unit wavelength by X (solid line) and by Y (dashed line). Which graph is correct?
Sketch graphs, NOT to scaleShow mark scheme
Marking point
Mark
Notes
Step 1Wien's law, λmaxT = 2.9 × 10−3 m K: the hotter star Y has its peak at the shorter wavelength.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2At equal radius the hotter star emits more at every wavelength (its total power is (6000/4500)4 ≈ 3.2 times greater), so Y's curve lies above X's everywhere.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: Y peaks at the shorter wavelength and its curve lies above X's at every wavelength.
BY's peak has been drawn at the longer wavelength; by Wien's law the hotter star peaks at the shorter wavelength.
CY's curve has been drawn below X's; at the same radius the hotter star emits more power per unit wavelength, not less.
DThe curves cross, so X emits more than Y at long wavelengths; a hotter black body of the same size emits more at every wavelength.
Syllabus understandingE.5 — (guidance) the surface temperature and composition of a star can be determined from the stellar spectrum; B.1 — the emission spectrum of a black body and the determination of the temperature of the body using Wien's displacement law Command term: Identify
16E-1B-14
Apparent brightness·E.5 Fusion and stars
Paper 1BMedium12 marks
Data-based question12 steps to full marksDetermine
To model how the apparent brightness of a star depends on its distance, a student measures the intensity I of the light from a small filament lamp with a light sensor at different distances d from the filament, in a darkened room. Each distance is measured with a metre rule (uncertainty ±0.5 cm).
The student suggests that I = P/(4πd2), where P is the power emitted by the lamp as light. The graph shows log(I / W m−2) against log(d / m) for six of the readings.
d / cm
I / W m−2
log(d / m)
log(I / W m−2)
20.0
3.65
−0.699
0.562
30.0
1.56
−0.523
0.193
40.0
0.904
−0.398
−0.044
50.0
0.567
−0.301
−0.246
60.0
0.406
−0.222
−0.391
80.0
0.219
100.0
0.145
0.000
−0.839
Graph drawn to scale
(a)
Calculate the percentage uncertainty in d = 20.0 cm.
(1)
(b)
Complete the table for d = 80.0 cm and plot the point on the graph.
(2)
(c)
Draw the line of best fit and determine its gradient.
(3)
(d)
State what the gradient shows about the relationship between I and d.
(1)
(e)
Use the graph to determine P.
(2)
(f)
Suggest one systematic error in this experiment and its effect on the graph.
(1)
(g)
A star has a parallax angle of 0.0200 arc-second and an apparent brightness of 2.4 × 10−10 W m−2. Determine its luminosity in terms of L☉. (1 pc = 3.09 × 1016 m, L☉ = 3.83 × 1026 W)
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.5/20.0 × 100 = 2.5 %
✓ 1
Accept 2.5 % or 3 %.
Part (b)
log(d / m) = −0.097 and log(I / W m−2) = −0.660
✓ 1
Accept 2 or more d.p.
Point plotted correctly to within half a small square
✓ 1
Allow ECF.
Part (c)
Single straight line with points evenly on both sides
✓ 1
Gradient from a large triangle (at least half the line)
✓ 1
Gradient = −2.00
✓ 1
Accept −1.90 to −2.10.
Part (d)
Gradient ≈ −2, so I ∝ d−2: an inverse-square law
✓ 1
Allow ECF from (c).
Part (e)
At log(d / m) = 0 the line gives log(I / W m−2) ≈ −0.845, so I = 0.143 W m−2 at d = 1.00 m, and this equals P/(4π × 1.002)
✓ 1
Accept use of any point on the line with I = P/4πd2.
P = 4π × 0.143 = 1.80 W
✓ 1
Accept 1.7–1.9 W. Allow ECF.
Part (f)
Light reflected from walls/stray light adds to I, most noticeably at large d, so the points at large d lie above the line and the gradient is less steep
✓ 1
Accept: the filament is not a point source, so d to the emitting surface is uncertain at small d; zero error of the sensor. Error and effect both needed.
Part (g)
d = 1/0.0200 = 50 pc = 50 × 3.09 × 1016 = 1.545 × 1018 m
Answers: (a) 2.5 % · (b) −0.097, −0.660 · (c) ≈ −2.00 · (e) ≈ 1.80 W · (g) 19 L☉(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; B.1 — the concept of apparent brightness b and luminosity L of a body as given by b = L/4πd2. Command term: Determine
17E-1B-15
Stellar radius from L and T·E.5 Fusion and stars
Paper 1BHard12 marks
Data-based question12 steps to full marksDetermine
A student uses a database of nearby white dwarfs to test whether white dwarfs of similar mass have the same radius. For six white dwarfs with masses close to 0.6 solar masses, the database gives the luminosity L (uncertainty ±5 %) and the surface temperature T (uncertainty ±2 %).
(σ = 5.67 × 10−8 W m−2 K−4, L☉ = 3.83 × 1026 W, radius of the Earth = 6.37 × 106 m)
T / K
L / 10−3L☉
T4 / 1016 K4
13200
4.40
3.04
15600
8.09
17300
12.9
8.96
18600
16.2
12.0
19700
22.1
20600
24.9
18.0
Graph drawn to scale
(a)
State the percentage uncertainty in T4.
(1)
(b)
Complete the table.
(1)
(c)
Explain why, if the white dwarfs all have the same radius, the graph is a straight line through the origin.
(1)
(d)
Draw the line of best fit and determine its gradient.
(3)
(e)
Determine the radius of the white dwarfs.
(2)
(f)
By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in the radius.
(2)
(g)
Compare the radius with that of the Earth, and suggest why the points do not lie exactly on one line.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
4 × 2 % = 8 %
✓ 1
Part (b)
5.92 and 15.1 (× 1016 K4)
✓ 1
Both needed. Accept more s.f.
Part (c)
L = 4πR2σT4; with R the same for all, 4πR2σ is constant, so L ∝ T4
✓ 1
OWTTE
Part (d)
Straight line through the origin passing through all the error bars
✓ 1
Gradient from a large triangle
✓ 1
Gradient = 1.41 × 10−19L☉ K−4
✓ 1
Accept 1.34–1.48 with correct power of ten and unit.
R = √(5.39 × 107/(4π × 5.67 × 10−8)) = 8.7 × 106 m
✓ 1
Allow ECF from (d). Award [2] for CNA.
Part (f)
Steepest and shallowest lines through all the error bars give gradients of about 1.65 and 1.22 (× 10−19), i.e. ±15 %
✓ 1
R ∝ √gradient, so the percentage uncertainty in R is half: ΔR ≈ ±7 × 105 m
✓ 1
MP2 is for halving the percentage uncertainty. Accept ±3 × 105 m to ±8 × 105 m.
Part (g)
R ≈ 1.4 × the Earth's radius: white dwarfs are about the size of the Earth
✓ 1
The white dwarfs do not have exactly the same mass, so their radii differ slightly
✓ 1
Accept random uncertainties in L and T. OWTTE
Answers: (b) 5.92 and 15.1 · (d) 1.41 × 10−19L☉ K−4 · (e) 8.7 × 106 m · (f) ±7 × 105 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 is required); the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions. Command term: Determine
18E-1B-16
Fusion in the Sun·E.5 Fusion and stars
Paper 1BMedium9 marks
Data-based question9 steps to full marksDetermine
Radiometers on satellites measure the solar constant S, the power per unit area received from the Sun above the atmosphere at the mean Earth–Sun distance of 1.50 × 1011 m. The table gives the annual mean values of S from an eight-year satellite record.
In the core of the Sun, the overall effect of fusion is to convert four hydrogen-1 atoms (mass 1.007825 u each) into one helium-4 atom (mass 4.002602 u). (1 u = 931.5 MeV c−2, 1 MeV = 1.60 × 10−13 J)
Year
S / W m−2
Year 1
1360.9
Year 2
1361.4
Year 3
1360.6
Year 4
1361.2
Year 5
1361.8
Year 6
1361.0
Year 7
1360.7
Year 8
1361.3
(a)
Calculate the mean value of S and its absolute uncertainty.
(2)
(b)
Show that the luminosity of the Sun is about 3.8 × 1026 W.
(1)
(c)
Determine the energy released, in MeV, when one helium-4 atom is formed.
(2)
(d)
Determine the number of helium-4 nuclei formed in the Sun each second.
(2)
(e)
About 2 % of the energy released by fusion is carried out of the Sun by neutrinos, which are not detected by the radiometers. Explain the effect of this on your answer to (d).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Mean = 10888.9/8 = 1361.1 W m−2
✓ 1
Uncertainty = (1361.8 − 1360.6)/2 = ±0.6 W m−2
✓ 1
Accept (1361.1 ± 0.6) W m−2; value and uncertainty to the same precision.
Part (b)
L = 4πd2S = 4π × (1.50 × 1011)2 × 1361.1 = 3.85 × 1026 W
✓ 1
Must see the substitution OR an answer to at least 3 s.f.
Part (c)
Δm = 4 × 1.007825 − 4.002602 = 0.028698 u
✓ 1
E = 0.028698 × 931.5 = 26.7 MeV
✓ 1
Award [2] for CNA.
Part (d)
Energy per helium nucleus = 26.7 × 1.60 × 10−13 = 4.28 × 10−12 J
✓ 1
Number per second = 3.85 × 1026/4.28 × 10−12 = 9.0 × 1037 s−1
✓ 1
Allow ECF from (b) and (c). Award [2] for CNA.
Part (e)
The luminosity measured by the radiometers does not include the neutrino energy, so it is only about 98 % of the power released by fusion
✓ 1
So the true rate is about 2 % larger than in (d), ≈ 9.2 × 1037 s−1
✓ 1
Accept "(d) is an underestimate" with the reason.
Answers: (a) (1361.1 ± 0.6) W m−2 · (b) 3.85 × 1026 W · (c) 26.7 MeV · (d) 9.0 × 1037 s−1(the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — that fusion is a source of energy in stars (guidance: energy release calculations are required); B.2 — the solar constant S. Command term: Determine
19E-1B-19
Wien's law·E.5 Fusion and stars
Paper 1BMedium12 marks
Data-based question12 steps to full marksDetermine
Before analysing the spectra of stars, a student uses a computer simulation of black-body radiation. For each temperature T the simulation draws the spectrum, and the student places a cursor on the peak to read the peak wavelength λmax. Each reading has an uncertainty of ±10 nm.
The student tests the hypothesis that λmax ∝ Tn with n = −1. The graph shows log(λmax / nm) against log(T / K), with error bars, for six of the temperatures.
T / K
λmax / nm
log(T / K)
log(λmax / nm)
3000
970
3.477
2.987
4000
720
3.602
2.857
5000
580
3.699
2.763
6000
480
3.778
2.681
8000
360
10000
290
4.000
2.462
12000
240
4.079
2.380
Graph drawn to scale
(a)
Calculate the percentage uncertainty in λmax at T = 12 000 K.
(1)
(b)
Complete the table for T = 8000 K and plot the point on the graph.
(2)
(c)
Draw the line of best fit and determine its gradient.
(3)
(d)
State what your answer to (c) shows about the hypothesis.
(1)
(e)
Using a point on your line, determine the value of λmaxT and compare it with the data-booklet value of 2.9 × 10−3 m K.
(2)
(f)
Suggest how the student could reduce the uncertainty in the readings of λmax at high temperatures.
(1)
(g)
The spectrum of a star peaks at (414 ± 10) nm. Determine the surface temperature of the star and its absolute uncertainty.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
10/240 × 100 = 4.2 %
✓ 1
Accept 4 %.
Part (b)
log(T / K) = 3.903 and log(λmax / nm) = 2.556
✓ 1
Accept 2 or more d.p.
Point plotted correctly to within half a small square
✓ 1
Allow ECF.
Part (c)
Single straight line through all the error bars
✓ 1
Gradient from a large triangle
✓ 1
Gradient = −1.01
✓ 1
Accept −0.95 to −1.05.
Part (d)
The gradient is −1 (within the precision of the data), so λmax ∝ 1/T: the hypothesis is supported
✓ 1
Allow ECF from (c).
Part (e)
e.g. at log(T / K) = 3.70, log(λmax / nm) = 2.761, so λmax = 577 nm and T = 5012 K
✓ 1
λmaxT = 2.89 × 10−3 m K, which agrees with 2.9 × 10−3 m K
✓ 1
Accept 2.8–3.0 × 10−3 m K. Allow ECF.
Part (f)
Zoom in on the wavelength axis near the peak (use a finer wavelength scale) so that the peak can be located more precisely
✓ 1
Accept: read the two wavelengths at which the curve has equal heights either side of the peak and take their mean. OWTTE
Part (g)
T = 2.9 × 10−3/414 × 10−9 = 7.0 × 103 K
✓ 1
Percentage uncertainty = 10/414 × 100 = 2.4 %, so T = (7.0 ± 0.2) × 103 K
✓ 1
MP2 is for matching the precision of value and uncertainty. Accept (7000 ± 170) K.
Answers: (a) 4.2 % · (b) 3.903, 2.556 · (c) ≈ −1.01 · (e) ≈ 2.89 × 10−3 m K · (g) (7.0 ± 0.2) × 103 K (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — guidance: the surface temperature and composition of a star can be determined from the stellar spectrum; B.1 — the emission spectrum of a black body and the determination of the temperature of the body using Wien's displacement law as given by λmaxT = 2.9 × 10−3 m K. Command term: Determine
20E-2-28
Conditions for fusion·E.5 Fusion and stars
Paper 2Hard14 marks
Short answer & extended response14 steps to full marksExplain
Hydrogen fuses in the core of a main-sequence star at a temperature of about 1.5 × 107 K. When the hydrogen in the core has been used up, the core contracts until its temperature reaches about 1 × 108 K. Helium nuclei can then fuse in the triple-alpha process: 3 42He → 126C.
The table gives some atomic masses. (1 u = 931.5 MeV c−2, k = 8.99 × 109 N m2 C−2, e = 1.60 × 10−19 C)
Atom
Mass / u
hydrogen-1
1.007825
helium-4
4.002603
carbon-12
12.000000
(a)
Outline why the core of the star contracts when the hydrogen in the core has been used up.
(1)
(b)
(i)
Calculate the electric force between two helium nuclei whose centres are 3.0 × 10−15 m apart.
(1)
(ii)
Explain why helium fusion needs a much higher temperature than hydrogen fusion.
(2)
(c)
Show that the energy released in one triple-alpha process is about 7.3 MeV.
(1)
(d)
On the main sequence, four hydrogen-1 nuclei are converted into one helium-4 nucleus and 26.7 MeV is released. Calculate the ratio energy released per kilogram of hydrogen fused/energy released per kilogram of helium fused in the triple-alpha process.
(3)
(e)
Suggest two reasons why the helium-fusion stage in the life of a star is much shorter than its main-sequence stage.
(2)
(f)
A student suggests that the temperature needed for two nuclei to fuse is proportional to the product of their charges.
(i)
Using the student's model and the temperature for hydrogen fusion, predict the temperature needed for two carbon-12 nuclei to fuse, and comment on your answer given that carbon fusion is observed to need about 6 × 108 K.
(2)
(ii)
Carbon fusion takes place only in stars with a mass greater than about eight times the mass of the Sun. Explain why carbon fusion will never take place in the core of the Sun.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
fusion stops, so the outward «radiation and gas» pressure falls and no longer balances the inward gravitational force
✓ 1
OWTTE
Part (b)(i)
F = 8.99 × 109 × (2 × 1.60 × 10−19)2/(3.0 × 10−15)2 = 1.0 × 102 N
✓ 1
Award [0] if the charge of a helium nucleus is taken as e.
Part (b)(ii)
a helium nucleus has twice the charge of a proton, so at the same separation the repulsive force is four times as large
✓ 1
Allow ECF from (b)(i).
so the nuclei need much more kinetic energy to come close enough for the strong nuclear force to act, and the mean kinetic energy of the nuclei is proportional to the temperature
✓ 1
Part (c)
Δm = 3 × 4.002603 − 12.000000 = 0.007809 u AND E = 0.007809 × 931.5 «= 7.274 MeV»
✓ 1
Must see full substitution OR answer to 3 s.f. (7.27 MeV). The electrons balance «6 on each side».
Part (d)
hydrogen: 26.7/(4 × 1.007825) = 6.62 MeV u−1
✓ 1
Accept masses taken as nucleon numbers throughout «6.68 and 0.61 MeV u−1».
helium: 7.27/(3 × 4.002603) = 0.606 MeV u−1
✓ 1
Allow ECF from (c).
ratio = 11
✓ 1
Accept 10.5–11. Award [3] for CNA. Award [2 max] for the inverse ratio «0.09».
Part (e)
helium fusion releases far less energy per kilogram of fuel «about 1/11», so the fuel is used up sooner for the same power
✓ 1
Allow ECF from (d).
the star is much more luminous in this stage «a red giant», so it uses its fuel faster / only the helium in the core is available as fuel
✓ 1
Accept any one.
Part (f)(i)
carbon nuclei have charge 6e: 6 × 6 = 36 times the product for two protons, so T ≈ 36 × 1.5 × 107 = 5.4 × 108 K
✓ 1
this is close to 6 × 108 K «within about 10 %», so the model gives a reasonable estimate
✓ 1
Accept a valid comment that the model predicts the correct order of magnitude. MP2 only scores if a value is calculated.
Part (f)(ii)
carbon fusion needs a much higher core temperature «6 × 108 K» than helium fusion
✓ 1
the Sun's mass is too small for gravity to compress its core enough to reach this temperature; the core becomes a white dwarf instead
✓ 1
OWTTE
Answers: (b)(i) 1.0 × 102 N · (c) 7.274 MeV · (d) 11 · (f)(i) 5.4 × 108 K (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; the effect of stellar mass on the evolution of a star; E.3 — nuclear binding energy and mass defect (guidance: masses in u); D.2 — Coulomb's law Command term: Explain
21E-2-29
Stellar radius from L and T·E.5 Fusion and stars
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine whether
Star Z is a pulsating star. Its luminosity and its surface temperature rise and fall regularly with a period of 0.57 days. The table gives the luminosity L and the surface temperature T of Z at maximum and at minimum luminosity. The HR diagram shows the main sequence and the Sun, S.
(Surface temperature of the Sun = 5800 K, L☉ = 3.83 × 1026 W, R☉ = 6.96 × 108 m, σ = 5.67 × 10−8 W m−2 K−4, Wien constant = 2.90 × 10−3 m K)
L / L☉
T / K
maximum luminosity
62
7300
minimum luminosity
38
6200
Diagram drawn to scale (logarithmic axes)
(a)
State what is meant by the luminosity of a star.
(1)
(b)
On the HR diagram, draw crosses to show the positions of Z at maximum and at minimum luminosity.
(2)
(c)
State the name of the region of the HR diagram in which pulsating stars such as Z are found.
(1)
(d)
Deduce, using the HR diagram, whether Z is a main-sequence star.
(2)
(e)
(i)
Determine the radius of Z at maximum luminosity, in terms of R☉.
(2)
(ii)
Determine whether Z is larger at maximum luminosity than at minimum luminosity.
(2)
(f)
Calculate the wavelength at which the spectrum of Z is most intense at maximum luminosity, and state how the colour of Z changes as its luminosity falls.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the total power radiated by the star «energy emitted per unit time»
✓ 1
Do not accept 'brightness'.
Part (b)
cross at 7300 K and between 10 and 100 L☉ «about 60 L☉»
✓ 1
Accept within 0.2 of a decade of luminosity and within 300 K.
cross at 6200 K and about 40 L☉, to the right of and below the first cross
✓ 1
Part (c)
the instability strip
✓ 1
Part (d)
main-sequence stars at these temperatures have luminosities of only about 2–6 L☉ / both crosses lie well above the main-sequence band
✓ 1
Allow ECF from the candidate's crosses.
so Z is not a main-sequence star «it is larger, an evolved star»
✓ 1
MP2 only scores if MP1 scores.
Part (e)(i)
R/R☉ = √(L/L☉) × (T☉/T)2 = √62 × (5800/7300)2
✓ 1
Or a full calculation with σ and L☉.
= 5.0 R☉
✓ 1
Award [2] for CNA. Accept 4.9–5.0 R☉.
Part (e)(ii)
at minimum luminosity R = √38 × (5800/6200)2 = 5.4 R☉
✓ 1
Accept 5.3–5.4 R☉.
no: Z is smaller at maximum luminosity; its higher temperature «T4 rises by a factor of 1.9» more than makes up for its smaller surface area
✓ 1
Allow ECF from (e)(i). The conclusion must be consistent with the two radii.
Part (f)
λ = 2.90 × 10−3/7300 = 4.0 × 10−7 m «397 nm»
✓ 1
the peak moves to longer wavelengths «468 nm at minimum», so Z becomes redder «less blue-white»
✓ 1
OWTTE
Answers: (e)(i) 5.0 R☉ · (e)(ii) 5.4 R☉ at minimum — Z is smaller at maximum luminosity · (f) 4.0 × 10−7 m (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — how to determine stellar radii; the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions (guidance: the instability strip; the surface temperature of a star can be determined from the stellar spectrum); B.1 — the Stefan–Boltzmann law and Wien's displacement law Command term: Determine whether
22E-2-30
Apparent brightness·E.5 Fusion and stars
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine
Star K has a luminosity of 0.45 L☉ and a stellar parallax angle of 0.0625 arc-second. A planet orbits star K at a distance of 0.60 AU.
(L☉ = 3.83 × 1026 W, 1 AU = 1.50 × 1011 m, 1 pc = 3.09 × 1016 m, 1 ly = 9.46 × 1015 m, σ = 5.67 × 10−8 W m−2 K−4)
(a)
(i)
Determine the distance from the Earth to star K, in light years.
(2)
(ii)
Outline why the stellar parallax method can only be used for relatively nearby stars.
(1)
(b)
Calculate the apparent brightness of star K at the Earth.
(2)
(c)
The planet has an albedo of 0.30 and no atmosphere. Its surface may be treated as a perfect emitter of infrared radiation (emissivity 1).
(i)
Show that the intensity of the radiation from star K at the distance of the planet is about 1700 W m−2.
(1)
(ii)
Determine the mean equilibrium temperature of the surface of the planet.
(3)
(iii)
Discuss whether your answer to (c)(ii) shows that liquid water cannot exist on the surface of any planet at this distance from star K.
(2)
(iv)
Suggest how the albedo of the planet would change if its surface became covered in ice, and the effect of this on its temperature.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
d = 1/0.0625 = 16 pc
✓ 1
16 × 3.09 × 1016/9.46 × 1015 = 52 ly
✓ 1
Award [2] for CNA. Accept 52–53 ly.
Part (a)(ii)
for distant stars the parallax angle is too small to be measured accurately «the baseline, the Earth's orbit, is fixed»
✓ 1
OWTTE
Part (b)
b = 0.45 × 3.83 × 1026/(4π × (4.94 × 1017)2)
✓ 1
Allow ECF from (a)(i).
= 5.6 × 10−11 W m−2
✓ 1
Part (c)(i)
I = 1.72 × 1026/(4π × (0.60 × 1.50 × 1011)2) «= 1690 W m−2»
✓ 1
Must see full substitution OR answer to 3 s.f. (1690 W m−2).
Part (c)(ii)
mean intensity over the whole surface = 1690/4 «= 423 W m−2»
✓ 1
The factor 4 is the ratio of the surface area to the cross-sectional area.
0.70 × 423 = 296 W m−2 = σT4
✓ 1
T = 269 K
✓ 1
Award [3] for CNA. Accept 268–270 K. Award [2] max for 380 K «factor of 4 omitted».
Part (c)(iii)
269 K is below 273 K, so on this model any water on the surface would be frozen
✓ 1
Allow ECF from (c)(ii).
but a planet with an atmosphere containing greenhouse gases would have a higher surface temperature «as the Earth does, by about 30 K», so liquid water could exist; the model does not rule it out
✓ 1
OWTTE
Part (c)(iv)
ice reflects more of the incident radiation, so the albedo would increase and the temperature would fall further
✓ 1
OWTTE
Answers: (a)(i) 52 ly · (b) 5.6 × 10−11 W m−2 · (c)(i) 1690 W m−2 · (c)(ii) 269 K (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; B.2 — albedo; the mean incoming intensity S/4 Command term: Determine
23E-2-31
Stellar evolution by mass·E.5 Fusion and stars
Paper 2Easy11 marks
Short answer & extended response11 steps to full marksSketch
Stars X and Y are both on the main sequence. Star X has a mass of 1 M☉ and a luminosity of 1 L☉. Star Y has a mass of 15 M☉ and a luminosity of 2 × 104L☉. Their positions are shown on the HR diagram.
(M☉ is the mass of the Sun and L☉ is the luminosity of the Sun.)
Diagram drawn to scale (logarithmic axes)
(a)
State the source of the energy radiated by both stars while they are on the main sequence.
(1)
(b)
Explain why star Y stays on the main sequence for a much shorter time than star X.
(2)
(c)
The main-sequence lifetime of X is about 1 × 1010 years. Estimate the main-sequence lifetime of Y.
(2)
(d)
Sketch, on the HR diagram, the evolutionary path of star X after it leaves the main sequence. Label the final state of X.
(3)
(e)
Outline how the evolution of Y after it leaves the main sequence differs from that of X.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
nuclear fusion of hydrogen into helium in the core
✓ 1
Do not accept 'burning'.
Part (b)
Y has about 15 times as much hydrogen «fuel» as X
✓ 1
but radiates energy 2 × 104 times faster, so it uses up its fuel much sooner «lifetime ∝ M/L»
✓ 1
Part (c)
tY = 1 × 1010 × 15/(2 × 104)
✓ 1
Assumes the lifetime is proportional to M/L «the same fraction of the mass is fused».
≈ 8 × 106 years
✓ 1
Award [2] for CNA. Accept 7–8 × 106 years.
Part (d)
path from X moving up and to the right «cooler and more luminous» into the red-giant region
✓ 1
then moving to the lower left, ending below the main sequence
✓ 1
final state labelled white dwarf «hot, faint, lower left»
✓ 1
Part (e)
Y becomes a red supergiant «moving to the right across the top of the diagram»
✓ 1
it ends in a supernova explosion
✓ 1
leaving a neutron star or a black hole, not a white dwarf
✓ 1
Accept either final state.
Answers: (c) ≈ 8 × 106 years (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — the effect of stellar mass on the evolution of a star; that fusion is a source of energy in stars; the main regions of the Hertzsprung–Russell (HR) diagram and how to describe the main properties of stars in these regions Command term: Sketch
24E-2-32
Wien's law·E.5 Fusion and stars
Paper 2Hard12 marks
Short answer & extended response12 steps to full marksDraw
Stars P and Q are supergiants at the same distance from the Earth. They have the same luminosity, 1.2 × 105L☉. The surface temperature of P is 3500 K and that of Q is 11 000 K.
The graph shows the variation with wavelength λ of the intensity of the radiation received from Q per unit wavelength, in arbitrary units.
(L☉ = 3.83 × 1026 W, σ = 5.67 × 10−8 W m−2 K−4, Wien constant = 2.90 × 10−3 m K, 1 AU = 1.50 × 1011 m)
Graph drawn to scale
(a)
(i)
Calculate the peak wavelength in the spectrum of P.
(1)
(ii)
Explain why P appears red even though the peak of its spectrum lies in the infrared.
(2)
(b)
Draw, on the graph, a curve to show the variation with wavelength of the intensity of the radiation received from P.
(3)
(c)
Determine the ratio radius of P/radius of Q.
(2)
(d)
Determine the radius of P and compare it with the radius of the Earth's orbit, 1 AU.
(3)
(e)
State the region of the HR diagram in which each star lies.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
λmax = 2.90 × 10−3/3500 = 8.3 × 10−7 m «829 nm»
✓ 1
Part (a)(ii)
across the visible range «400–700 nm» the intensity of the radiation from P increases with wavelength
✓ 1
so much more red light than blue or violet light is emitted
✓ 1
Part (b)
peak at about 830 nm
✓ 1
Allow ECF from (a)(i). Accept 780–880 nm.
peak much lower than the peak for Q «about a third of its height»
✓ 1
Accept a peak between one fifth and one half of Q's peak.
curve below Q's curve at short wavelengths and above it at long wavelengths «crossing near 630 nm», because the areas under the two curves are equal «same luminosity, same distance»
✓ 1
Award MP3 for a curve that crosses Q's curve and is above it at long wavelengths.
Part (c)
L = 4πR2σT4 with L the same, so RP/RQ = (TQ/TP)2
✓ 1
= (11 000/3500)2 = 9.9
✓ 1
Award [2] for CNA.
Part (d)
L = 1.2 × 105 × 3.83 × 1026 = 4.6 × 1031 W
✓ 1
R = √(4.6 × 1031/(4π × 5.67 × 10−8 × 35004)) = 6.6 × 1011 m
✓ 1
≈ 4.4 AU: P is larger than the Earth's orbit «it would engulf the inner planets»
✓ 1
Allow ECF from MP2.
Part (e)
both at the top of the diagram «supergiants»: P at the top right and Q at the top left
✓ 1
Answers: (a)(i) 8.3 × 10−7 m · (c) 9.9 · (d) 6.6 × 1011 m ≈ 4.4 AU (the remaining parts are explanations — see the table above)
Syllabus understandingE.5 — how to determine stellar radii; the main regions of the Hertzsprung–Russell (HR) diagram (guidance: the surface temperature of a star can be determined from the stellar spectrum); B.1 — the black-body spectrum and Wien's displacement law Command term: Draw
25E-2-36
The HR diagram·E.5 Fusion and stars
Paper 2Easy8 marks
Short answer & extended response8 steps to full marksAnnotate
The diagram shows the axes of a Hertzsprung–Russell (HR) diagram. The position of the Sun is marked S.
Diagram drawn to scale (logarithmic axes)
(a)
Annotate the diagram to show the regions occupied by main-sequence stars, red giants, red supergiants and white dwarfs.
(3)
(b)
Annotate the diagram to show the position of the instability strip.
(1)
(c)
Star D and star E have the same surface temperature. The luminosity of E is 104 times the luminosity of D.
(i)
State and explain which star has the larger radius.
(2)
(ii)
Calculate the ratio radius of E/radius of D.
(1)
(d)
Describe how the radius and the luminosity of a white dwarf compare with those of the Sun.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
main sequence: a band from the top left to the bottom right, passing through S
✓ 1
red giants above the main sequence on the right «about 10–103 L☉, 3000–5000 K» AND red supergiants at the top right «above about 104 L☉»
✓ 1
white dwarfs at the lower left, below the main sequence «hotter than about 6000 K, about 10−2–10−4 L☉»
✓ 1
Part (b)
a narrow, almost vertical band between the main sequence and the supergiants, at about 6000–7500 K
✓ 1
Accept a band a little hotter than the Sun, crossing from the main sequence towards the top of the diagram.
Part (c)(i)
E
✓ 1
MP1 only scores if a reason is given.
at the same temperature each square metre emits the same power «σT4», so the more luminous star must have the larger surface area
✓ 1
Part (c)(ii)
√(104) = 100
✓ 1
Part (d)
much smaller radius «similar to the Earth's» AND much lower luminosity
✓ 1
Both needed.
Answers: (c)(ii) 100 (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 (guidance: main sequence stars, red giants, super giants, white dwarfs, the instability strip); how to determine stellar radii Command term: Annotate
26E-2-37
Stellar radius from L and T·E.5 Fusion and stars
Paper 2Hard13 marks
Short answer & extended response13 steps to full marksDetermine
Star T is a main-sequence star. The table gives some observations of T made from the Earth.
(L☉ = 3.83 × 1026 W, R☉ = 6.96 × 108 m, 1 pc = 3.09 × 1016 m, σ = 5.67 × 10−8 W m−2 K−4, Wien constant = 2.90 × 10−3 m K)
Observation
Value
wavelength at which the continuous spectrum of T is most intense
414 nm
stellar parallax angle of T
0.0125 arc-second
apparent brightness of T
3.0 × 10−11 W m−2
(a)
State two assumptions about star T that are made when its radius is determined from its luminosity and its surface temperature.
(2)
(b)
(i)
Show that the surface temperature of T is about 7000 K.
(1)
(ii)
Determine the luminosity of T in terms of L☉.
(3)
(iii)
Determine the radius of T in terms of R☉.
(2)
(c)
The parallax angle of T is measured with an absolute uncertainty of ±0.0005 arc-second. Ignore the uncertainties in the other observations.
(i)
Calculate the percentage uncertainty in the distance to T.
(1)
(ii)
Deduce the percentage uncertainty in the radius of T.
(2)
(d)
Suggest two reasons why the luminosity found in (b)(ii) may be smaller than the true luminosity of T.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
T emits as a black body «so that Wien’s displacement law and the Stefan–Boltzmann law apply»
✓ 1
T is a sphere whose whole surface is at the same temperature OR T radiates equally in all directions «so that L = 4πd2b»
✓ 1
Accept any two distinct assumptions about the star, one per mark. Do not accept assumptions about the measurements «e.g. no absorption by dust»: these are credited in (d).
Part (b)(i)
T = 2.90 × 10−3/414 × 10−9 «= 7005 K»
✓ 1
Must see full substitution OR answer to 4 s.f. (7005 K).
Part (b)(ii)
d = 1/0.0125 = 80 pc = 80 × 3.09 × 1016 = 2.47 × 1018 m
✓ 1
L = 4πd2b = 4π × (2.47 × 1018)2 × 3.0 × 10−11 = 2.3 × 1027 W
✓ 1
Allow ECF from MP1.
= 6.0 L☉
✓ 1
Award [3] for CNA.
Part (b)(iii)
R = √(2.30 × 1027/(4π × 5.67 × 10−8 × 70004)) = 1.16 × 109 m
✓ 1
Allow ECF from (b)(i) and (b)(ii).
= 1.7 R☉
✓ 1
Accept 1.6–1.7 R☉.
Part (c)(i)
0.0005/0.0125 × 100 = 4 %, and d = 1/p, so the distance is uncertain by 4 %
✓ 1
Accept 4.0 %.
Part (c)(ii)
L = 4πd2b, so the percentage uncertainty in L is 2 × 4 = 8 %
✓ 1
Allow ECF from (c)(i).
R = √(L/4πσT4), so the percentage uncertainty in R is ½ × 8 = 4 % «R ∝ d when b and T are fixed»
✓ 1
Award [1 max] for 8 % or 16 %. Allow ECF from (c)(i).
Part (d)
Some of the radiation from T is absorbed or scattered before it reaches the detector «by the Earth's atmosphere or by interstellar dust and gas», so the measured apparent brightness is too small
✓ 1
The detector may not respond to all wavelengths «e.g. not to the ultraviolet or infrared», so not all of the power received is measured
✓ 1
Accept any two distinct reasons, one per mark. Do not accept random uncertainty in the parallax angle.
Answers: (b)(i) 7.0 × 103 K · (b)(ii) d = 2.47 × 1018 m; L = 6.0 L☉ · (b)(iii) 1.7 R☉ · (c)(i) 4 % · (c)(ii) 4 % (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; how to determine stellar radii; B.1 — the concept of apparent brightness b; the Stefan–Boltzmann law and Wien’s displacement law; Tools 3 — propagate uncertainties through calculations involving addition, subtraction, multiplication, division and raising to a power Command term: Determine
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