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IB Physics Standard Level · Topic E.5

E.5 Fusion and stars

135 free, high-quality past-paper-style questions on Fusion and stars for IB Physics SL, each with a full mark scheme. Covers: Stellar equilibrium; Fusion in the Sun; Stellar parallax; and more. For when you have finished the past papers, or want to practise one topic at a time.E.5 at Higher Level →Read the E.5 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.685 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1E-1A-21Stellar equilibrium Paper 1AEasy1 mark

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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Marking pointMarkNotes
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.✓ 1Answer 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.
2E-1A-23Fusion in the Sun Paper 1AEasy1 mark

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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Step 1Each second the Sun radiates energy L☉; by E = mc2 this corresponds to a mass L☉/c2 converted each second.✓ 1Answer 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.
3E-1A-24Stellar parallax Paper 1AEasy1 mark

A star has a parallax angle of 0.25 arc-second. (1 pc = 3.09 × 1016 m, 1 ly = 9.46 × 1015 m)

What is the distance to the star in light years?

The Earth at two points six months apart in its orbit, with the parallax angle p subtended at a nearby star.SunJanuaryJulystarp1 AUd (parsec) = 1 / p (arc-second)
Diagram NOT accurately drawn
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Step 1d (parsec) = 1/p (arc-second) = 1/0.25 = 4.0 pc.—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.✓ 1Answer 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.
4E-1A-60Stellar spectra Paper 1AEasy1 mark

The spectrum of a star consists of a continuous spectrum crossed by dark absorption lines.

Which row gives the property of the star that can be determined from each feature of the spectrum?

Wavelength at which the continuous spectrum is most intenseWavelengths of the dark absorption lines
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Step 1By Wien's displacement law, λmaxT = 2.9 × 10−3 m K, so the peak wavelength gives the surface temperature.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Each element absorbs at its own set of wavelengths, so the dark lines identify the elements in the star's outer layers.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe two features have been swapped: the peak gives the temperature and the line pattern gives the composition.
  • BThe composition is right, but the position of the peak depends only on the temperature, not on the size of the star.
  • CCorrect: the peak wavelength gives T through Wien's law, and the absorption lines are the fingerprints of the elements present.
  • DThe luminosity depends on the radius as well as the temperature, so it cannot be found from the peak wavelength; the lines give the composition, not the temperature.
5E-1A-99The HR diagram Paper 1AEasy1 mark

Some stars are unstable: their radius, surface temperature and luminosity rise and fall regularly. The Hertzsprung–Russell diagram shows the main sequence and four regions, A, B, C and D.

Which region is the instability strip?

Hertzsprung–Russell diagram with logarithmic axes: luminosity from 10⁻⁴ to 10⁶ solar luminosities, surface temperature from 40 000 K on the left to 2 500 K on the right. The main sequence is a dashed line. Region A: 9 000–25 000 K, 3 × 10⁻⁴ to 10⁻² L☉. Region B: 3 400–4 800 K, 15 to 1 000 L☉. Region C: 9 000–25 000 K, 10⁵ to 6 × 10⁵ L☉. Region D: a narrow, nearly vertical band from about 6 600–7 800 K at 3 L☉ to 5 200–6 200 K at 10⁴ L☉.40 00020 00010 0005 0002 500surface temperature / K10⁻⁴10⁻²110²10⁴10⁶luminosity / L☉ABCDmain sequence
Diagram drawn to scale (logarithmic axes)
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Step 1The instability strip is a narrow, almost vertical band: its stars have a small range of surface temperature (about 5000–8000 K) but a wide range of luminosity, from a few L☉ up to about 10⁴ L☉.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2It starts near the main sequence and runs upwards between the main sequence and the red giants: region D.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis links “unstable” with stars at the end of their lives: region A holds white dwarfs, which are hot but very faint. The instability strip lies above the main sequence.
  • BRegion B is the red-giant region. The instability strip lies at higher surface temperatures, between the main sequence and the red giants.
  • CThis takes “unstable” to mean the hottest, most luminous stars: region C holds hot supergiants. The strip is nearly vertical, at about 5000–8000 K, not a horizontal band.
  • DCorrect: the instability strip is the narrow, nearly vertical band at about 5000–8000 K that rises from the main sequence towards the giants and supergiants.
6E-1A-102Stellar parallax Paper 1AEasy1 mark

A star is 1.24 × 106 AU from the Sun. What is this distance in parsecs? (1 AU = 1.50 × 1011 m, 1 ly = 9.46 × 1015 m, 1 pc = 3.09 × 1016 m)

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Step 1d = 1.24 × 106 × 1.50 × 1011 = 1.86 × 1017 m.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2d = 1.86 × 1017/3.09 × 1016 = 6.0 pc.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • A1 AU has been taken as 1.50 × 108 m, which is its value in kilometres; the answer is 1000 times too small.
  • BThe conversion has been inverted: (metres in 1 pc)/(metres in the distance) is not a distance.
  • CCorrect: 1.24 × 106 × 1.50 × 1011 m ÷ 3.09 × 1016 m pc−1 = 6.0 pc.
  • DThe distance has been converted into light years (20 ly), not parsecs.
7E-1A-103The HR diagram Paper 1AEasy1 mark

Which row describes the axes of a Hertzsprung–Russell (HR) diagram?

Vertical axisHorizontal axis
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Step 1An HR diagram plots luminosity on the vertical axis against surface temperature on the horizontal axis, with the temperature increasing from right to left (hot stars on the left).✓ 1Answer D

Answer: D  ·  1 stage of work, one mark

Every option, and why

  • AThe quantities are right, but on an HR diagram the hottest stars are on the left: the temperature scale increases to the left, the opposite of an ordinary graph.
  • BThe two quantities have been swapped: luminosity is on the vertical axis.
  • CMass is not plotted on an HR diagram; the mass of a star increases along the main sequence, but the vertical axis is luminosity.
  • DCorrect: luminosity on the vertical axis; surface temperature on the horizontal axis, increasing to the left.
8IBP2S-042-02-01Stellar parallax Paper 1AEasy1 mark

An astrometry satellite measures the parallax angle of a red dwarf star as 0.025 arcsec. What is the distance to the star?

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With the parallax angle p in arcseconds, the distance in parsecs is d = 1/p = 1/0.025 = 40 pc. Option A takes d = p. Option B misplaces the decimal point in the reciprocal. Option D treats the angle as if it were in units ten times smaller.✓ 1Answer C

Answer: C

Exam tip: d (pc) = 1/p only works when p is in arcseconds. Convert milliarcseconds first.

9IBP2S-042-02-03Stellar parallax Paper 1AEasy1 mark

Why can the stellar parallax method only be used for relatively nearby stars?

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The parallax angle is inversely proportional to distance. Beyond a certain distance the angle falls below the smallest angle the instrument can resolve. Option C is wrong because there is always some shift, even if it is too small to detect. Options A and D describe effects that have nothing to do with measuring an angular shift.✓ 1Answer B

Answer: B

Exam tip: Link the limit of the method to the smallest measurable angle, not to brightness or spectra.

10IBP2S-042-02-05Stellar spectra Paper 1AEasy1 mark

Dark absorption lines appear in the spectrum of a star. How are they formed?

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The hot photosphere emits a continuous spectrum. Atoms in the cooler gas above it absorb photons whose energies equal the differences between their energy levels. They re-emit in random directions, so fewer photons of those wavelengths reach the observer. Option A describes emission, which produces bright lines. Option C would dim all wavelengths, not produce lines. Option D wrongly puts the cause in the photosphere's own emission.✓ 1Answer B

Answer: B

Exam tip: Absorption lines need two things: a hot continuous source and cooler gas in front of it.

11IBP2S-042-02-10Stellar parallax Paper 1AEasy1 mark

Which of the following defines the parsec?

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One parsec is the distance at which a baseline of 1 AU (the Earth–Sun distance) subtends 1 arcsecond. This gives d (pc) = 1/p (arcsec). Option A defines the light-year. Option C uses the full diameter of the orbit. Option D uses the wrong angular unit.✓ 1Answer B

Answer: B

Exam tip: The parallax angle is half the total angular shift seen over six months, because the baseline is 1 AU, not 2 AU.

12IBP2S-042-04-01Wien's law Paper 1AEasy1 mark

The tungsten filament of an old-style incandescent lamp operates at a temperature of 2900 K. Treating the filament as a black body, at what wavelength is its emitted intensity a maximum?

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Wien's displacement law gives λmax = 2.9×10−3 m K ÷ 2900 K = 1.0×10−6 m, which is in the near infrared. This is why such lamps waste most of their energy as heat. 8.4 m comes from multiplying the constant by T instead of dividing. 1.0×10−3 m comes from treating the constant as 2.9 m K. 1.0×10−9 m is a power-of-ten slip.✓ 1Answer B

Answer: B

Common error: Wien's law is λmaxT = constant, so λmax = b/T. A hotter body has a shorter peak wavelength, and you must divide by T, not multiply.

13IBP2S-042-04-05Stellar evolution by mass Paper 1AEasy1 mark

A star similar to the Sun exhausts the hydrogen in its core and becomes a red giant. Which element is produced by fusion in its core during the red giant stage?

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Once core hydrogen runs out, the core contracts and heats up until helium fusion begins. Helium nuclei fuse to produce carbon (and some oxygen), so carbon is correct. Helium is the product of hydrogen fusion. In a red giant this continues only in a shell around the core, not in the core itself. Iron is produced only in the cores of very massive stars. Uranium and other elements heavier than iron form in supernovae and neutron-star mergers, not by core fusion.✓ 1Answer B

Answer: B

Common error: Do not assume that every star fuses all the way to iron. A low-mass star stops at carbon and oxygen, because its core never becomes hot enough for further fusion.

14IBP2S-042-04-08Stellar evolution by mass Paper 1AEasy1 mark

Why can fusion in a stellar core not continue to release energy once iron has formed?

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The binding energy per nucleon peaks near iron and nickel. Fusion releases energy only if the products have a higher binding energy per nucleon than the reactants. Fusing iron would therefore absorb energy rather than release it. The second option is wrong because higher temperatures always raise nuclear speeds. The third option gets the binding energy curve the wrong way round. The fourth option is irrelevant, because a stellar core is a plasma, not a solid.✓ 1Answer A

Answer: A

Exam tip: Link any 'why does fusion stop at iron' answer to the peak of the binding energy per nucleon curve.

15IBP2S-042-04-09The HR diagram Paper 1AEasy1 mark

On a Hertzsprung–Russell diagram, luminosity increases upwards and surface temperature decreases from left to right. How does the position of a Sun-like star change as it leaves the main sequence to become a red giant?

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A red giant is more luminous than its main-sequence progenitor, so it moves upwards. It is also cooler (redder), and on the HR diagram cooler means further to the right. 'Upwards and to the left' comes from reading the temperature axis as increasing to the right. 'Downwards and to the left' describes the later move to a white dwarf. 'Downwards and to the right' assumes that a cooler star must be less luminous, which ignores the huge increase in radius.✓ 1Answer B

Answer: B

Common error: The temperature axis of an HR diagram runs backwards, with hot stars on the left. Check the axis direction before describing any movement.

16IBP2S-042-03-01The HR diagram Paper 1AEasy1 mark

Two stars X and Y have the same surface temperature. The radius of X is three times the radius of Y. What is \(\frac{L_X}{L_Y}\), the ratio of the luminosity of X to the luminosity of Y?

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L = 4πσR2T4. With T the same for both stars, L ∝ R2, so LX/LY = 32 = 9. Option A treats L as proportional to R. Option C uses R3, as if luminosity depended on volume. Option D applies the fourth power, which belongs to temperature, not radius.✓ 1Answer B

Answer: B

Common error: Luminosity depends on surface area, so radius enters as R². The fourth power belongs only to temperature.

17IBP2S-042-03-02Stellar parallax Paper 1AEasy1 mark

Which of the following is the apparent brightness of a star?

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Apparent brightness b = L/(4πd2) is the power received per unit area, measured perpendicular to the direction of the light, at the observer's location. Its unit is W m−2. Option A defines luminosity. Option B is the surface flux σT4, which also has the unit W m−2 but is measured at the star. Option D depends on the size of the telescope, so it is not a property of the star and its distance.✓ 1Answer C

Answer: C

Exam tip: Both σT⁴ and b have the unit W m⁻². Ask where the area is: at the star's surface, or at the observer.

18IBP2S-042-03-08Stellar radius from L and T Paper 1AEasy1 mark

An astronomer wants to determine the radius of a distant star using the Stefan–Boltzmann law. Which set of measurements is sufficient?

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The peak wavelength gives T through Wien's law. Apparent brightness and distance give L = 4πd2b. Then R = √(L/(4πσT4)). Option A has no way to find L without the distance. Option C gives d and T but not L. Option D gives L but not T.✓ 1Answer B

Answer: B

Exam tip: To find a radius you always need both L and T. Check that each measurement listed leads to one of them.

19IBP2S-042-00-01The HR diagram Paper 1AEasy1 mark

An astronomy app shows a Hertzsprung–Russell (HR) diagram. What is plotted along the horizontal axis, and in which direction does it change?

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The horizontal axis of an HR diagram shows surface temperature, with hot stars on the left and cool stars on the right. The scale is reversed compared with ordinary graphs, so B is correct. Luminosity goes on the vertical axis. Mass is not an axis, although mass does vary along the main sequence.✓ 1Answer B

Answer: B

Common error: The temperature axis runs backwards: hot (blue) stars are on the left. It is also logarithmic, so equal spacings represent equal ratios, not equal differences.

20IBP2S-042-00-02Conditions for fusion Paper 1AEasy1 mark

A solar-farm engineer notes that the energy collected by her panels was originally released in the core of the Sun. Which process releases this energy in the Sun's core today?

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The Sun is a main-sequence star. Its core fuses hydrogen into helium, mainly through the proton–proton chain, so C is correct. Helium fusion starts only after the core hydrogen is used up, in the red giant stage. Fission is not a significant energy source in stars. Gravitational contraction powers protostars, not a stable main-sequence star.✓ 1Answer C

Answer: C

Exam tip: Link the stage of a star's life to its fuel: main sequence means hydrogen → helium in the core, and red giant means helium fusion in the core.

21IBP2S-042-00-09The HR diagram Paper 1AEasy1 mark

A red giant and the Sun are compared. The red giant lies above and to the right of the Sun on an HR diagram. Compared with the Sun, the red giant is:

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A position to the right means a lower surface temperature. A position higher up means a greater luminosity. A cooler star can only be more luminous if it has a much larger surface area, so the red giant is cooler and larger. B is correct.✓ 1Answer B

Answer: B

Exam tip: Read the two axes separately: left/right gives temperature, up/down gives luminosity. Then use L = 4πR²σT⁴ to deduce the size.

22IBP2S-042-01-01Conditions for fusion Paper 1AEasy1 mark

Engineers at a tokamak research facility heat a deuterium–tritium plasma to about 108 K. What is the main reason such a high temperature is needed for fusion to occur?

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Nuclei are positively charged and repel electrostatically. Only nuclei with very high kinetic energy, and so a very high temperature, can get close enough (about 10−15 m) for the attractive strong nuclear force to bind them. The strong force is attractive at these ranges, not repulsive. Temperature does not increase nuclear mass. Fusion does not remove neutrons.✓ 1Answer A

Answer: A

Common error: The barrier to fusion is the electrostatic (Coulomb) repulsion. The strong force is what holds the fused nucleus together once the nuclei are close enough.

23IBP2S-042-01-02Stellar equilibrium Paper 1AEasy1 mark

A main-sequence star keeps a constant radius for billions of years. Which pair of effects is balanced in the star?

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Stellar equilibrium is a balance between gravity, which pulls matter inward, and outward pressure. That pressure comes from the radiation and hot gas produced by fusion in the core. The strong and electrostatic forces act within nuclei, not on the star as a whole. Rotation does not hold a star up. Radiation pressure acts outward, not inward.✓ 1Answer C

Answer: C

Exam tip: Always state the direction of each effect: gravity inward, radiation (and gas) pressure outward.

24IBP2S-042-01-10Conditions for fusion Paper 1AEasy1 mark

A deuterium nucleus and a tritium nucleus fuse to form helium-4 and a neutron, releasing about 17.6 MeV. Where does this energy come from?

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Helium-4 has a much higher binding energy per nucleon than deuterium or tritium. The products therefore have less total rest mass than the reactants, and this mass difference is released as energy (E = Δmc2). If the products had more mass, energy would have to be supplied, not released. The initial kinetic energy only allows the nuclei to get close; the 17.6 MeV comes from the mass difference. This is a nuclear process, not a chemical one.✓ 1Answer B

Answer: B

Common error: In energy-releasing nuclear reactions, the binding energy increases and the total mass decreases.

25E-1A-22Conditions for fusion Paper 1AMedium1 mark

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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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.✓ 1Answer 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.
26E-1A-25The HR diagram Paper 1AMedium1 mark

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?

A Hertzsprung–Russell diagram with logarithmic scales: luminosity from 10^−4 to 10^6 L☉ on the vertical axis and surface temperature from about 40 000 K to 2500 K, decreasing to the right, on the horizontal axis. The main sequence runs from hot luminous stars at the top left to cool faint stars at the bottom right, with red giants above it at the right and white dwarfs below it at the left.40 00020 00010 0005000250010⁻⁴10⁻²10⁰10²10⁴10⁶Surface temperature (K) — increasing to the leftLuminosity (L / L☉)main sequencered giantswhite dwarfsSun
Diagram NOT accurately drawn
Type of starRadius
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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.✓ 1Answer 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.
27E-1A-43Stellar evolution by mass Paper 1AMedium1 mark

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?

A Hertzsprung–Russell diagram with logarithmic scales: luminosity from 10^−4 to 10^6 L☉ on the vertical axis and surface temperature from about 40 000 K to 2500 K, decreasing to the right, on the horizontal axis. The main sequence runs from hot luminous stars at the top left to cool faint stars at the bottom right, with red giants above it at the right and white dwarfs below it at the left.40 00020 00010 0005000250010⁻⁴10⁻²10⁰10²10⁴10⁶Surface temperature (K) — increasing to the leftLuminosity (L / L☉)main sequencered giantswhite dwarfsSun
Diagram NOT accurately drawn
Leaves the main sequence firstFinal state of the 20-solar-mass star
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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.✓ 1Answer 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.
28E-1A-61Fusion in the Sun Paper 1AMedium1 mark

In the core of a main-sequence star, hydrogen is fused into helium. A fraction f of the mass of the hydrogen that is fused is converted into energy, and all of this energy eventually leaves the star as radiation. The luminosity of the star is L.

What mass of hydrogen is fused each second? (c is the speed of light.)

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Step 1Mass converted into energy each second = L/c2.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2This is only a fraction f of the mass fused, so the mass fused each second = L/(fc2).✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe fraction has been applied the wrong way: fL/c2 is smaller than the mass converted, but the mass fused must be larger.
  • BL/c2 is the mass converted into energy each second, not the mass of hydrogen that takes part in fusion.
  • CThe speed of light has not been squared: E = mc2.
  • DCorrect: (L/c2)/f; for hydrogen fusing into helium, f is only about 0.007, so the mass fused is about 140 times the mass converted.
29E-1A-62Conditions for fusion Paper 1AMedium1 mark

A protostar is a cloud of gas that is contracting under its own gravity. Nuclear fusion has not yet started in its core.

Which row describes the protostar as it contracts?

Temperature of the coreSource of the energy that the protostar radiates
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Step 1As the gas falls inwards its gravitational potential energy decreases; part of this energy becomes random kinetic energy of the particles, so the core temperature rises.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The rest is radiated away, so before fusion starts the radiated energy comes from the decrease in gravitational potential energy.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe temperature does rise, but fusion has not yet started; the energy comes from the gravitational contraction.
  • BThe energy source is right, but only part of the energy released by contraction is radiated; the rest heats the gas, so the core gets hotter.
  • CThis assumes that radiating energy cools the core and that fusion has already begun; contraction heats the core, and fusion starts only once the core is hot and dense enough.
  • DCorrect: contraction converts gravitational potential energy into thermal energy, heating the core towards the temperature and density needed for fusion, and the protostar radiates part of this energy.
30E-1A-97Fusion in the Sun Paper 1AMedium1 mark

The overall result of hydrogen fusion in the core of the Sun is

4 ¹₁p → ⁴₂He + 2e+ + 2ν

The mass of a proton is mp, the mass of a helium-4 nucleus is mα and the mass of a positron is me. The neutrinos have negligible mass. What is the energy released in this reaction? Ignore any later annihilation of the positrons. (c is the speed of light.)

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Step 1Energy released = (total mass before − total mass after) × c²; the products are one helium-4 nucleus, two positrons and two neutrinos of negligible mass.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2So E = [4mp − (mα + 2me)]c² = (4mp − mα − 2me)c² (about 24.7 MeV with real masses).✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe two positrons are products of the reaction and have mass; their mass must be subtracted from the mass before the reaction.
  • BOnly one positron has been included: the equation produces two, so 2me must be subtracted.
  • CCorrect: mass before − mass after = 4mp − mα − 2me, and this mass difference × c² is the energy released.
  • DThis is mass after − mass before, which is negative: the expression gives the energy that would have to be supplied, with the wrong sign.
31E-1A-98Conditions for fusion Paper 1AMedium1 mark

In the Sun, hydrogen nuclei fuse only in the core, the region within about a quarter of the Sun’s radius from its centre. About 70 % of the mass of the Sun’s outer layers is also hydrogen.

Which statement best explains why no fusion takes place in the outer layers?

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Marking pointMarkNotes
Step 1Fusion needs nuclei to approach closely against their electrostatic repulsion (a very high temperature) and to collide very often (a very high density).—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Both the temperature and the density fall steeply with distance from the centre, so outside the core the fusion rate is negligible even though hydrogen is present.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • ARadiation pressure acts on the gas as a whole and balances gravity; it does not stop individual nuclei from colliding. Nuclei fail to fuse because they move too slowly, not because they are pushed apart.
  • BCorrect: fusion depends on the temperature (kinetic energy of the nuclei) and on the density (collision rate), and both are too low outside the core.
  • CNuclei that fuse are held together by the strong nuclear force; gravity between two nuclei is negligible everywhere in the Sun.
  • DIt is the core, not the outer layers, where hydrogen has been converted into helium; the outer layers keep almost their original composition.
32E-1A-100Stellar evolution by mass Paper 1AMedium1 mark

The Hertzsprung–Russell diagram shows four main-sequence stars P, Q, R and S.

Which row identifies the star with the greatest mass and the star that will stay on the main sequence for the longest time?

Hertzsprung–Russell diagram with logarithmic axes. The main sequence is drawn as a dashed band. Four main-sequence stars are marked: P at 15 000 K and 400 solar luminosities, Q at 9 500 K and 40, R at 5 200 K and 0.4, S at 3 800 K and 0.06.40 00020 00010 0005 0002 500surface temperature / K10⁻⁴10⁻²110²10⁴10⁶luminosity / L☉PQRSmain sequence
Diagram drawn to scale (logarithmic axes)
Greatest massLongest time on the main sequence
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Marking pointMarkNotes
Step 1Along the main sequence, the more luminous (and hotter) a star is, the greater its mass: P, at the upper left, has the greatest mass.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The main-sequence lifetime depends on the fuel available (∝ mass) divided by the rate at which it is used (∝ luminosity); the luminosity rises much faster than the mass, so S, the faintest, lasts longest.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: P is the most massive; S is the least massive and least luminous, so it uses its hydrogen most slowly and stays longest on the main sequence.
  • BP has the most hydrogen, but its luminosity is about 7000 times that of S, so it uses its fuel far sooner: more fuel does not mean a longer life.
  • CThis is the reverse of the correct answer: the faint, cool star at the lower right has the smallest mass.
  • DS does live longest, but the faint, cool end of the main sequence holds the least massive stars, not the most massive.
33E-1A-105The HR diagram Paper 1AMedium1 mark

A red giant and a red supergiant have the same surface temperature.

Which row compares the red supergiant with the red giant?

Radius of the supergiantMass of the supergiant when it was on the main sequence
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Marking pointMarkNotes
Step 1Supergiants lie above the giants on the HR diagram: at the same surface temperature a supergiant is far more luminous.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2L = 4πR²σT⁴ with T the same gives R ∝ √L, so the supergiant has the larger radius.—
Step 3Only stars of much greater mass (more than about eight solar masses) become supergiants; stars of low mass, such as the Sun, become red giants.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThe radius is right, but a supergiant is not the late stage of a less massive star: only stars that were much more massive on the main sequence become supergiants.
  • BCorrect: at the same temperature the more luminous supergiant must have a larger surface area (R ∝ √L), and it formed from a star of much greater mass.
  • CThe mass is right, but equal temperatures do not mean equal radii: the supergiant is far more luminous, and at the same T its radius is proportional to √L.
  • DThis treats stars of equal temperature as equal in size and the supergiant as the late stage of a smaller star: both comparisons are wrong.
34E-1A-106Stellar radius from L and T Paper 1AMedium1 mark

An astronomer determines the radius R of a star by using L = 4πR²σT⁴.

Which row identifies the observations of the star from which T and L are determined?

T is determined fromL is determined from
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Marking pointMarkNotes
Step 1Wien’s displacement law, λmaxT = 2.9 × 10−3 m K, gives T from the peak wavelength of the continuous spectrum.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2L = 4πd²b, so the apparent brightness b and the distance d = 1/p (from the parallax angle) are both needed.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: the peak wavelength gives T; the apparent brightness and the distance from the parallax angle give L = 4πd²b.
  • BThe dark absorption lines identify the elements in the outer layers of the star (its composition); the temperature comes from the peak of the continuous spectrum.
  • CA faint star may be nearby and dim or far away and very luminous: without the distance, the apparent brightness does not give L.
  • DThe parallax angle gives the distance, not the temperature, and the peak wavelength cannot give L without the radius, which is the unknown.
35E-1A-107Stellar equilibrium Paper 1AMedium1 mark

The hydrogen in the core of a main-sequence star is used up and fusion in the core stops. Hydrogen continues to fuse in a shell around the core.

Which row describes what happens next to the core and to the outer layers of the star?

CoreOuter layers
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Marking pointMarkNotes
Step 1With no fusion in the core, the outward pressure there no longer balances the inward gravitational force, so the core contracts; the gravitational compression heats it.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The hotter core and the fusion in the shell around it release energy faster, so the pressure on the outer layers rises: they expand and their surface cools — the star becomes a red giant.✓ 1Answer C

Answer: C  ·  2 stages of work, one mark

Every option, and why

  • AThe outer layers do expand and cool, but the core cannot expand: without fusion its pressure falls, so gravity compresses it.
  • BBoth responses are the wrong way round: the core contracts and the outer layers expand.
  • CCorrect: the core contracts and heats up because gravity is no longer balanced; the outer layers expand and cool, and the star becomes a red giant.
  • DThe core does contract, but the whole star does not collapse: the extra energy released pushes the outer layers outwards, so they expand and cool.
36E-1A-108The HR diagram Paper 1AMedium1 mark

Stars X and Y are both main-sequence stars. Their continuous spectra are most intense at the same wavelength. The apparent brightness of X is 1/400 of the apparent brightness of Y.

Which row gives (surface temperature of X)/(surface temperature of Y) and (distance to X)/(distance to Y)?

TX/TYdX/dY
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Marking pointMarkNotes
Step 1The spectra peak at the same wavelength, so by Wien’s displacement law (λmaxT = constant) the surface temperatures are equal.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Main-sequence stars with the same surface temperature occupy the same point of the HR diagram, so they have the same luminosity.—
Step 3b = L/4πd² with L the same, so dX/dY = √(bY/bX) = √400 = 20.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: equal peak wavelengths give equal temperatures, equal temperatures on the main sequence give equal luminosities, and a star 400 times fainter is √400 = 20 times further away.
  • BThe square root has not been taken: apparent brightness is inversely proportional to the square of the distance.
  • CThe difference in brightness has been put down to a difference in temperature, as if the stars were at the same distance; the equal peak wavelengths show that the temperatures are equal.
  • DThe ratio has been inverted: the fainter star X is the more distant one.
37IBP2S-042-02-02Stellar parallax Paper 1AMedium1 mark

Seen from Earth, a star has a parallax angle of 0.010 arcsec. A proposed spacecraft would orbit the Sun in a circular orbit of radius 1.5 AU. It would observe the same star from two diametrically opposite points of its orbit. The star lies perpendicular to the line joining these points. What parallax angle would the spacecraft measure?

Sunposition 1position 2star1.5 AU?
Diagram not to scale
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Marking pointMarkNotes
The parallax angle is half the total angular shift. For small angles it is p ≈ (orbital radius)/d. The star's distance d is unchanged and the orbital radius is 1.5 times larger, so p = 1.5 × 0.010 = 0.015 arcsec (C). Option A divides by 1.5 instead of multiplying. Option B assumes parallax depends only on distance. Option D squares the factor of 1.5.✓ 1Answer C

Answer: C

Common error: The relation d (pc) = 1/p (arcsec) assumes a baseline radius of 1 AU. A longer baseline gives a proportionally larger parallax angle for the same star.

38IBP2S-042-02-04Wien's law Paper 1AMedium1 mark

An ultraviolet space telescope finds that the spectrum of a star peaks at a wavelength of 290 nm. Wien's displacement constant is 2.9×10−3 m K. What is the surface temperature of the star?

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Marking pointMarkNotes
From Wien's law, T = b/λmax = 2.9×10−3/(290×10−9) = 1.0×104 K. Option A forgets to convert nm to m. Option B inverts the relation. Option D uses a wavelength ten times too small.✓ 1Answer C

Answer: C

Common error: Always convert nanometres to metres (×10⁻⁹) before using Wien's law.

39IBP2S-042-02-06Wien's law Paper 1AMedium1 mark

Stars X and Y have the same luminosity. The peak wavelength in the spectrum of X is 400 nm and the peak wavelength in the spectrum of Y is 800 nm. What is the ratio \(\frac{\text{radius of Y}}{\text{radius of X}}\)?

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Marking pointMarkNotes
By Wien's law T ∝ 1/λmax, so TX = 2TY. Since L = 4πσR2T4 is the same for both stars, R2 ∝ 1/T4, which gives R ∝ 1/T2. Therefore RY/RX = (TX/TY)2 = 4. Option D forgets the square root. Option B uses R ∝ 1/T. Option A inverts the ratio.✓ 1Answer C

Answer: C

Common error: After using L ∝ R²T⁴, take the square root to get R, not R².

40IBP2S-042-02-07Apparent brightness Paper 1AMedium1 mark

Two stars have the same luminosity. Star X has a parallax angle of 0.04 arcsec and star Y has a parallax angle of 0.01 arcsec. What is the ratio \(\frac{\text{apparent brightness of X}}{\text{apparent brightness of Y}}\)?

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Marking pointMarkNotes
The distance is d ∝ 1/p, so dY = 4dX. Brightness is b = L/(4πd2), so bX/bY = (dY/dX)2 = 16. Option C forgets the inverse-square law. Options A and B assume a larger parallax means a greater distance.✓ 1Answer D

Answer: D

Exam tip: Larger parallax means closer star, so higher brightness for the same luminosity.

41IBP2S-042-02-08Stellar radius from L and T Paper 1AMedium1 mark

The parallax angle of a star is measured as (0.020 ± 0.002) arcsec. The star's apparent brightness has negligible uncertainty. What is the percentage uncertainty in the luminosity calculated for the star?

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Marking pointMarkNotes
The percentage uncertainty in p is 0.002/0.020 = 10 %. Since d = 1/p, the percentage uncertainty in d is also 10 %. Luminosity is L = 4πbd2, so its percentage uncertainty is 2 × 10 % = 20 %. Option B ignores the square. Option A halves instead of doubling. Option D doubles twice.✓ 1Answer C

Answer: C

Exam tip: Taking a reciprocal leaves the percentage uncertainty unchanged. Raising to a power n multiplies it by n.

42IBP2S-042-04-04Wien's law Paper 1AMedium1 mark

Two stars A and B have the same radius. The peak wavelength in the spectrum of A is three times the peak wavelength in the spectrum of B. What is the ratio \(\frac{L_A}{L_B}\) of their luminosities?

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Marking pointMarkNotes
By Wien's law T ∝ 1/λmax, so TA = TB/3. With equal radii, L = 4πR2σT4 gives L ∝ T4. So LA/LB = (1/3)4 = 1/81. The option 81 treats a longer wavelength as meaning a hotter star. The options 1/3 and 1/9 use the wrong power of T.✓ 1Answer D

Answer: D

Exam tip: Chain the two laws. Wien's law gives the temperature ratio, then the Stefan–Boltzmann law raises that ratio to the fourth power when the radii are equal.

43IBP2S-042-04-06Wien's law Paper 1AMedium1 mark

The surface temperature of a black body is doubled. What happens to the wavelength at which its emission peaks, and to the power it emits per unit area?

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Marking pointMarkNotes
Wien's law gives λmax ∝ 1/T, so the peak wavelength halves. The power per unit area is σT4, so it increases by 24 = 16. The option ×4 confuses T4 with the R2 dependence of luminosity. The option in which the peak wavelength doubles treats λmax as proportional to T. The option ×2 assumes the power is proportional to T.✓ 1Answer C

Answer: C

Exam tip: Learn the two temperature dependences together. The peak wavelength varies as T⁻¹, and the power per unit area varies as T⁴.

44IBP2S-042-04-07Stellar evolution by mass Paper 1AMedium1 mark

A main-sequence star has a mass of 20 solar masses. Consider these statements.
I. Fusion in its core eventually produces iron.
II. It will end its life in a supernova.
III. It will spend longer on the main sequence than the Sun.
Which statements are correct?

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Marking pointMarkNotes
A star this massive becomes a red supergiant. Its core fuses successively heavier elements up to iron. Fusing iron releases no energy, so the core collapses and the star explodes as a supernova, so I and II are correct. III is wrong: lifetime ∝ M−2.5, so a 20 solar mass star lasts only a few million years, compared with about 1010 years for the Sun.✓ 1Answer B

Answer: B

Common error: A larger fuel supply does not mean a longer life. High-mass stars have the shortest main-sequence lifetimes.

45IBP2S-042-03-03Apparent brightness Paper 1AMedium1 mark

Star A has twice the surface temperature of star B and half the radius of star B. What is \(\frac{L_A}{L_B}\)?

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Marking pointMarkNotes
L ∝ R2T4, so LA/LB = (1/2)2 × 24 = (1/4) × 16 = 4. Option A assumes the two factors of 2 cancel. Option C uses R instead of R2. Option D ignores the change in radius.✓ 1Answer B

Answer: B

Exam tip: For ratio questions, write L ∝ R²T⁴ and substitute each factor separately. The constants 4π and σ cancel.

46IBP2S-042-03-04Apparent brightness Paper 1AMedium1 mark

Stars P and Q have the same luminosity. P is four times as far from Earth as Q. What is \(\frac{b_P}{b_Q}\), the ratio of their apparent brightnesses?

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Marking pointMarkNotes
b = L/(4πd2). With L the same for both stars, b ∝ 1/d2, so bP/bQ = 1/42 = 1/16. Option C uses an inverse proportion instead of an inverse square. Option B uses 1/√d. Option A inverts the relationship.✓ 1Answer D

Answer: D

Common error: Apparent brightness follows an inverse-square law. Doubling the distance quarters the brightness; it does not halve it.

47IBP2S-042-03-05Apparent brightness Paper 1AMedium1 mark

An engineer designs solar panels for a rover on Mars. Earth orbits the Sun at 1.0 AU and Mars at 1.5 AU. What is the ratio \(\frac{\text{intensity of sunlight at Mars}}{\text{intensity of sunlight at Earth}}\)?

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Marking pointMarkNotes
The intensity of sunlight is the Sun's apparent brightness, b = L/(4πd2). The ratio is (1.0/1.5)2 = 1/2.25 ≈ 0.44. Option B uses 1/d instead of 1/d2. Option C assumes the intensity depends only on the source, not on the distance from it. Option D inverts the ratio.✓ 1Answer A

Answer: A

Exam tip: The ratio of the distances in AU can be used directly, because the AU cancels in a ratio.

48IBP2S-042-03-06The HR diagram Paper 1AMedium1 mark

A red giant and a blue main-sequence star have the same luminosity. The surface temperature of the blue star is four times the surface temperature of the red giant. What is \(\frac{R_{\text{giant}}}{R_{\text{blue}}}\)?

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Marking pointMarkNotes
Equal luminosities give Rg2Tg4 = Rb2Tb4. So Rg/Rb = (Tb/Tg)2 = 42 = 16. Option D is R2, not R. Option B treats R as proportional to 1/T. Option A uses a square root in the wrong place.✓ 1Answer C

Answer: C

Common error: After setting the luminosities equal, take the square root of both sides: R ∝ 1/T² when L is constant.

49IBP2S-042-03-07Apparent brightness Paper 1AMedium1 mark

Two stars A and B have the same surface temperature and the same apparent brightness when viewed from Earth. Star A is twice as far from Earth as star B. What is \(\frac{R_A}{R_B}\)?

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Marking pointMarkNotes
L = 4πd2b. With b the same for both stars, L ∝ d2, so LA = 4LB. With T the same, L ∝ R2, so RA/RB = √4 = 2. Option C stops at the luminosity ratio. Option A takes the square root of the distance ratio. Option D squares the luminosity ratio instead of taking its square root.✓ 1Answer B

Answer: B

Exam tip: With equal temperature and equal brightness, R ∝ d. Linking the two inverse-square relationships gives this result quickly.

50IBP2S-042-03-09Stellar radius from L and T Paper 1AMedium1 mark

A space telescope is upgraded so that the diameter of its mirror is doubled. It then observes the same star as before. Which row correctly describes the apparent brightness of the star and the power collected by the telescope?

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Marking pointMarkNotes
Apparent brightness depends only on L and d, so it is unchanged. The power collected is b × A. Doubling the diameter multiplies the area πr2 by 4, so the power collected increases by a factor of 4. Option A assumes the area is proportional to the diameter. Options B and D wrongly treat brightness as a property of the detector.✓ 1Answer C

Answer: C

Common error: A bigger telescope collects more power, but the star's apparent brightness at that location is unchanged.

51IBP2S-042-00-03The HR diagram Paper 1AMedium1 mark

A space telescope finds that two stars X and Y have the same surface temperature. The luminosity of X is 100 times the luminosity of Y. What is \(\frac{\text{radius of X}}{\text{radius of Y}}\)?

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Marking pointMarkNotes
\(L = 4\pi R^2\sigma T^4\). With T the same for both stars, \(L \propto R^2\), so \(R_X/R_Y = \sqrt{100} = 10\). A is correct. 100 assumes L ∝ R. 10 000 squares the ratio when it should be square-rooted. 0.1 inverts the ratio.✓ 1Answer A

Answer: A

Common error: Luminosity depends on surface area, so it varies as R², not R. Take a square root to go from a luminosity ratio to a radius ratio.

52IBP2S-042-00-04Fusion in the Sun Paper 1AMedium1 mark

Which equation correctly summarises the overall result of the proton–proton chain in the Sun?

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Marking pointMarkNotes
Four protons (total charge +4e) become one helium-4 nucleus (charge +2e). Charge conservation therefore requires two positrons, and lepton number conservation requires two electron neutrinos. D is correct.
A does not conserve charge. B gives the wrong sign of charge and uses antineutrinos. C has only 2 nucleons on the left but 4 on the right.
✓ 1Answer D

Answer: D

Exam tip: Check any nuclear equation for conservation of nucleon number, charge and lepton number, in that order. It quickly rules out wrong options.

53IBP2S-042-00-05Conditions for fusion Paper 1AMedium1 mark

Engineers building fusion reactors on Earth need plasma temperatures of order 108 K. Why is a very high temperature needed for nuclear fusion, in the Sun's core or in a reactor?

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Marking pointMarkNotes
Nuclei are positively charged and repel each other. They must approach to within about 10−15 m for the short-range strong force to bind them. Only very high temperatures give nuclei kinetic energies large enough to overcome the Coulomb repulsion (helped by quantum tunnelling in the Sun). B is correct.
Fusion does not break the strong force; it relies on it. Bare nuclei still repel each other. The strong force is attractive at these ranges.
✓ 1Answer B

Answer: B

Exam tip: In a fusion explanation, name both forces: electrostatic repulsion must be overcome so that the attractive strong nuclear force can act.

54IBP2S-042-00-06The HR diagram Paper 1AMedium1 mark

A planetarium display places stars on an HR diagram. A white dwarf has a surface temperature of about 20 000 K and a luminosity of about 10−3 times that of the Sun. Where is it plotted?

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Marking pointMarkNotes
A high temperature places the star on the left of the diagram. A very low luminosity places it near the bottom. White dwarfs therefore lie in the lower-left region, so C is correct. The upper right holds red giants and supergiants. The lower right holds cool red dwarfs on the main sequence. The upper left holds hot, luminous main-sequence stars.✓ 1Answer C

Answer: C

Common error: Hot does not always mean luminous. A white dwarf is hot but tiny, so its small surface area gives it a low luminosity.

55IBP2S-042-00-07Stellar equilibrium Paper 1AMedium1 mark

The Sun has kept an almost constant radius for billions of years. Which statement best explains this stability?

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Marking pointMarkNotes
The Sun is in hydrostatic equilibrium. Energy released by fusion keeps the interior hot, so it produces outward gas and radiation pressure that balances gravitational compression. A is correct.
The strong force acts only over nuclear distances. Solar rotation is far too slow to matter. D wrongly compares a reaction rate with a mass flow; in equilibrium there is no net inward flow of mass.
✓ 1Answer A

Answer: A

Exam tip: Describe stellar equilibrium as a balance of pressure (outward) against gravity (inward), and state that fusion is what maintains the high temperature and pressure.

56IBP2S-042-00-10Fusion in the Sun Paper 1AMedium1 mark

Fusion in the Sun converts about 4.3×109 kg of mass into energy every second. Which value is closest to the Sun's luminosity? (c = 3.0×108 m s−1)

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Marking pointMarkNotes
Power equals the rate of mass conversion multiplied by c2: \(L = \frac{\Delta m}{\Delta t}c^2 = 4.3\times10^{9} \times 9.0\times10^{16} \approx 3.9\times10^{26}\) W. A is correct.
B multiplies by c instead of c2. C wrongly includes a factor of ½, as in kinetic energy. D treats the mass rate itself as the power.
✓ 1Answer A

Answer: A

Common error: E = mc² has no factor of ½. It is not a kinetic energy formula.

57IBP2S-042-01-03Conditions for fusion Paper 1AMedium1 mark

In a fusion experiment, the temperature of a deuterium plasma is raised from 50 MK to 200 MK. By what factor does the root-mean-square speed of the deuterons increase?

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Marking pointMarkNotes
Mean kinetic energy = (3/2)kT, so ½mv2 ∝ T and v ∝ √T. The temperature rises by a factor of 4, so the speed rises by a factor of √4 = 2. A factor of 4 is the change in kinetic energy, not speed. A factor of 16 squares the temperature ratio instead of taking its root.✓ 1Answer B

Answer: B

Common error: Kinetic energy is proportional to T, but speed is proportional to √T. Read carefully which quantity is asked for.

58IBP2S-042-01-04Stellar equilibrium Paper 1AMedium1 mark

A space telescope observes a star whose core has just run out of hydrogen. What happens to the core immediately afterwards?

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Marking pointMarkNotes
When hydrogen fusion in the core stops, less energy is released there, so the outward thermal and radiation pressure falls. Gravity is no longer balanced and the core contracts. As it contracts, gravitational potential energy is transferred to thermal energy, so the core heats up. It can eventually become hot enough for helium fusion.
Expansion and cooling describes the outer layers of a red giant, not the core. The equilibrium has been broken, so the core cannot stay the same. Only some very massive stars end as black holes, and only much later.
✓ 1Answer B

Answer: B

Exam tip: Link each step: less fusion → lower outward pressure → contraction → loss of gravitational potential energy → higher temperature.

59IBP2S-042-01-05Conditions for fusion Paper 1AMedium1 mark

In laser inertial confinement fusion, powerful lasers squeeze a tiny fuel pellet to a very small volume. At a given temperature, why does this compression increase the rate of fusion reactions?

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Marking pointMarkNotes
Fusion needs both a high temperature and a high density. Compression raises the number density, so nuclei collide more often and more fusion reactions happen per second. Compression does not change the charge of the nuclei, so the repulsion between any pair is unchanged. It does not change the rest mass of the nuclei or convert neutrons into protons.✓ 1Answer A

Answer: A

Exam tip: Temperature sets whether a collision can lead to fusion. Density sets how often collisions happen. Both conditions are needed.

60IBP2S-042-01-07Stellar equilibrium Paper 1AMedium1 mark

In a main-sequence star, the rate of fusion in the core increases slightly. Which sequence describes how the star returns to equilibrium?

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Marking pointMarkNotes
More fusion releases more energy, which raises the outward pressure above what is needed to balance gravity. The core expands, and expansion cools it. The lower temperature reduces the fusion rate, so the original balance is restored. This self-regulation keeps main-sequence stars stable. A rise in pressure causes expansion, not contraction.✓ 1Answer B

Answer: B

Exam tip: Describe stellar stability as a feedback loop. Any change sets off a response that cancels it.

61IBP2S-042-01-08Conditions for fusion Paper 1AMedium1 mark

Main-sequence stars of greater mass have higher core temperatures. Which statement best explains this?

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In a more massive star, gravity pulls inward more strongly. Equilibrium then requires a larger outward pressure, which needs a hotter and denser core with a higher fusion rate. The energy per reaction does not depend on how much fuel there is. Stars emit energy rather than absorb it from space. All main-sequence stars fuse hydrogen into helium.✓ 1Answer C

Answer: C

Common error: Having more fuel does not make a star hotter. The core temperature is set by how much pressure is needed to balance gravity.

62E-1A-41Stellar radius from L and T Paper 1AHard1 mark

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)?

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Marking pointMarkNotes
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.✓ 1Answer 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.
63E-1A-42Apparent brightness Paper 1AHard1 mark

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)?

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Marking pointMarkNotes
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.✓ 1Answer 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.
64E-1A-44Wien's law Paper 1AHard1 mark

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?

Four sketch graphs A to D of power emitted per unit wavelength against wavelength for star X (4500 K, solid) and star Y (6000 K, dashed) of the same radius.A.Pλ0B.Pλ0C.Pλ0D.Pλ0
Sketch graphs, NOT to scale
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Marking pointMarkNotes
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.✓ 1Answer 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.
65E-1A-58The HR diagram Paper 1AHard1 mark

The Hertzsprung–Russell diagram shows four stars P, Q, R and S.

Which row identifies the star with the largest radius and the star with the smallest radius?

Hertzsprung–Russell diagram: luminosity from 10^−4 to 10^6 solar luminosities on a logarithmic vertical axis; surface temperature from 40 000 K on the left to 2500 K on the right on a logarithmic horizontal axis. P: 20 000 K, 10^4; Q: 3500 K, 10^3; R: 10 000 K, 10^−2; S: 3200 K, 10^−3.40 00020 00010 0005 0002 500surface temperature / K10⁻⁴10⁻²110²10⁴10⁶luminosity / L☉PQRS
Diagram drawn to scale (logarithmic axes)
Largest radiusSmallest radius
Show mark scheme
Marking pointMarkNotes
Step 1L = 4πR2σT4, so R ∝ √L/T2: a large radius needs a high luminosity for its temperature.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Relative to the Sun: P ≈ 8 R☉, Q ≈ 87 R☉, R ≈ 0.03 R☉, S ≈ 0.1 R☉; Q (a cool, luminous giant) is largest and R (a hot, faint white dwarf) is smallest.✓ 1Answer A

Answer: A  ·  2 stages of work, one mark

Every option, and why

  • ACorrect: Q is very luminous although it is cool, so it must be very large; R is faint although it is hot, so it must be very small.
  • BThis ranks the stars by luminosity alone. P is the most luminous because it is very hot; for its temperature it is much smaller than Q.
  • CQ is right, but S has been chosen as smallest because it is the least luminous; S is also much cooler than R, so it is about three times larger than R.
  • DThis ranks the stars by temperature alone, as if cooler stars were always larger; luminosity must be taken into account as well.
66E-1A-59Stellar parallax Paper 1AHard1 mark

A proposed space telescope would orbit the Sun in a circle of radius 5.0 AU. It would measure the parallax angle of a star from two positions on opposite sides of its orbit, in the same way as parallax angles are measured from the Earth.

What is (parallax angle measured from the telescope)/(parallax angle measured from the Earth) for the same star?

Sunposition 1position 2star5.0 AU?
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Step 1The parallax angle is the angle subtended at the star by the radius of the observer's orbit: for small angles p ≈ (orbit radius)/d.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2The radius is 5.0 times larger and d is the same, so the angle is 5.0 times larger.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AA longer baseline gives a larger shift in the star's apparent position, not a smaller one: the ratio has been inverted.
  • BCorrect: the parallax angle is proportional to the radius of the orbit used as the baseline, so it is 5.0 times larger.
  • CThis compares the diameter of the telescope's orbit (10 AU) with the radius of the Earth's orbit (1 AU); the same quantity must be used for both.
  • DThe parallax angle is proportional to the baseline, not to its square.
67E-1A-101The HR diagram Paper 1AHard1 mark

The Hertzsprung–Russell diagram has logarithmic scales on both axes. Four straight lines, A, B, C and D, are drawn through the position of the Sun.

Which line passes through the positions of all stars that have the same radius as the Sun?

Hertzsprung–Russell diagram with logarithmic axes: luminosity from 10^−4 to 10^6 solar luminosities, surface temperature from 40 000 K on the left to 2 500 K on the right. Four straight lines A, B, C and D pass through the Sun at 5 800 K and 1 solar luminosity. A falls to the right by two decades of luminosity for each decade of temperature; B falls to the right by four decades per decade; C rises to the right by four decades per decade; D is horizontal.40 00020 00010 0005 0002 500surface temperature / K10⁻⁴10⁻²110²10⁴10⁶luminosity / L☉ABCDSun
Diagram drawn to scale (logarithmic axes)
Show mark scheme
Marking pointMarkNotes
Step 1With R fixed, L = 4πR²σT⁴ gives L ∝ T⁴: each factor of 10 in temperature changes the luminosity by a factor of 10⁴, so on log–log axes the line has a gradient of magnitude 4.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Temperature increases to the left, so cooler stars of the same radius (to the right) are less luminous: the line falls to the right by four decades for each decade of temperature — line B (for example 16 L☉ at 11 600 K).✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • ALine A falls by only two decades per decade of temperature, as if L ∝ T²; the luminosity depends on the fourth power of the temperature.
  • BCorrect: at constant radius L ∝ T⁴, a straight line on log–log axes that falls four decades of luminosity for each decade of temperature towards the cooler, right-hand side.
  • CLine C has the right steepness but slopes the wrong way: it treats the temperature as increasing to the right. On an HR diagram the temperature increases to the left.
  • DLine D joins stars of equal luminosity; a cooler star with the same luminosity as the Sun must be larger than the Sun.
68E-1A-104Stellar parallax Paper 1AHard1 mark

The apparent position of a nearby star is measured against very distant background stars on two dates six months apart, when the Earth is on opposite sides of its orbit. Between the two dates the star appears to move through a total angle of 0.0500 arc-second.

What is the distance to the star? (1 pc = 3.26 ly)

Sunposition 1position 2star1 AU0.0500″
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Step 1The parallax angle is the angle subtended at the star by the radius of the Earth’s orbit (1 AU); the shift between opposite sides of the orbit is subtended by 2 AU, so p = 0.0500/2 = 0.0250 arc-second.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2d = 1/0.0250 = 40.0 pc = 40.0 × 3.26 = 130 ly.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThe distance in parsecs, 40.0 pc, has been divided by 3.26 instead of multiplied: a parsec is longer than a light year.
  • BThe total shift has been doubled instead of halved; the parallax angle is half of the total shift.
  • CThe total shift has been used as the parallax angle. It is subtended by the diameter of the Earth’s orbit, so it is twice the parallax angle.
  • DCorrect: p = half the total shift = 0.0250 arc-second, d = 40.0 pc = 130 ly.
69IBP2S-042-02-09Stellar spectra Paper 1AHard1 mark

The hydrogen absorption lines in the visible spectrum (Balmer lines) are weak in the hottest stars (surface temperature above about 30 000 K), even though these stars are mostly hydrogen. What is the best explanation?

Show mark scheme
Marking pointMarkNotes
Balmer absorption needs hydrogen atoms with electrons in the n = 2 level. At very high temperatures most hydrogen is ionized, so few atoms are able to absorb these wavelengths and the lines are weak. Option A is wrong because the atmosphere is still mostly hydrogen. Option C is wrong because hot stars still emit plenty of visible light, even though the peak is in the ultraviolet. Option D is wrong because thermal motion broadens the lines slightly but does not shift them out of the visible region.✓ 1Answer B

Answer: B

Common error: Weak lines of an element do not mean the element is absent. Line strength depends strongly on temperature.

70IBP2S-042-04-10The HR diagram Paper 1AHard1 mark

In the future, the Sun's luminosity will increase by a factor of 1000 while its surface temperature halves. By approximately what factor will its radius increase?

Show mark scheme
Marking pointMarkNotes
From L = 4πR2σT4, R ∝ √(L/T4). This gives Rnew/Rold = √(1000 × 24) = √16 000 ≈ 126. The option 63 uses T2 instead of T4. The option 2000 treats R as proportional to L/T. The option 16 000 forgets to take the square root.✓ 1Answer B

Answer: B

Exam tip: Rearrange the Stefan–Boltzmann law for R before substituting any ratios. Halving T contributes a factor of 16 inside the square root.

71IBP2S-042-03-10Stellar radius from L and T Paper 1AHard1 mark

As a star evolves off the main sequence, its luminosity increases by a factor of 100 and its surface temperature halves. By what factor does its radius increase?

Show mark scheme
Marking pointMarkNotes
R ∝ √L / T2, so R2/R1 = √100 / (1/2)2 = 10 × 4 = 40. Option D is the ratio of R2. Option A ignores the temperature change. Option B uses T instead of T2 and divides rather than multiplies (10 × 2 = 20 would be the result of using T alone; 25 comes from treating the factors as 100/4).✓ 1Answer C

Answer: C

Exam tip: Rearrange to R ∝ L¹/²T⁻² before substituting the ratios. A fall in temperature increases the radius needed for a given luminosity.

72IBP2S-042-01-06Conditions for fusion Paper 1AHard1 mark

At the temperature of the Sun's core, the mean kinetic energy of a proton is a few hundred times smaller than the energy needed to overcome the electrostatic repulsion of another proton. Why does fusion still happen in the Sun?

Show mark scheme
Marking pointMarkNotes
The proton speeds follow a distribution, so a small fraction have energies well above the mean. Quantum tunnelling also lets protons fuse with less energy than the full barrier height. The huge number of protons in the core makes the overall reaction rate large. Gravity confines the core as a whole but is negligible between two protons. The strong force acts only over about 10−15 m. The core is a fully ionised plasma, not neutral atoms.✓ 1Answer C

Answer: C

Common error: Gravity provides the high density and pressure in a stellar core. It does not provide the force that joins two individual nuclei.

73IBP2S-042-01-09Stellar equilibrium Paper 1AHard1 mark

A simple model gives the pressure needed at the centre of a star in equilibrium as proportional to M2/R4, where M is the mass and R the radius of the star. Star Y has twice the mass and twice the radius of star X. What is the ratio (central pressure of Y)/(central pressure of X)?

Show mark scheme
Marking pointMarkNotes
Ratio = 22/24 = 4/16 = 1/4. The answer 1 comes from assuming the doublings cancel. The answer 4 ignores the change in radius. The answer 1/2 uses M/R2 instead of the given relationship.✓ 1Answer A

Answer: A

Exam tip: For proportionality questions, raise each factor to its own power before combining them.

74E-1B-06Stellar parallax Paper 1BEasy8 marks

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)

Starp / mas
A8.62 ± 0.05
B8.55 ± 0.04
C8.71 ± 0.06
D5.08 ± 0.05
E8.49 ± 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)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = 1/(8.62 × 10−3) = 116.0 pc✓ 1Must 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 %✓ 1Accept fractional uncertainty 0.0082.
Δd = 1.0 pc, so d = (117.8 ± 1.0) pc✓ 1Accept (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✓ 1Accept: 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✓ 1Allow ECF from (b).
Part (e)
116.0 × 3.26 = 378 ly✓ 1Accept 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✓ 1OWTTE

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)

75E-1B-44Stellar parallax Paper 1BEasy10 marks

A space telescope in orbit around the Sun with the Earth measures the position of a nearby star against very distant galaxies at regular intervals for 1.5 years. The steady drift of the star caused by its own motion through space has been subtracted. The star lies in the plane of the Earth’s orbit, so the remaining offset Δx of its apparent position is along a straight line.

The graph shows Δx, in milliarcseconds (1 mas = 10−3 arc-second), against time t. Each measurement has an uncertainty of ±1.5 mas. (1 pc = 3.26 ly)

Offset of the star's apparent position against time over 1.5 years: thirteen points with error bars of ±1.5 mas, oscillating between about +21 mas and −21 mas with a period of one year00.250.50.7511.251.5time t / year-30-20-100102030offset Δx / mas
Graph drawn to scale
(a)

Outline why the offset varies with a period of one year.

(1)
(b)

Draw the curve of best fit for the data.

(1)
(c)

Use your curve to determine the parallax angle p of the star, in mas.

(2)
(d)

The uncertainty in p is ±0.8 mas. Determine the distance to the star, in parsecs, and its absolute uncertainty.

(3)
(e)

Calculate the distance to the star in light years.

(1)
(f)

A star catalogue gives the distance to this star as (45.9 ± 0.2) pc. Comment on your value.

(1)
(g)

Suggest one way in which the uncertainty in p could be reduced.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
the telescope «with the Earth» goes once round the Sun each year, so the direction from which the star is viewed changes and returns to its starting value every year✓ 1OWTTE
Part (b)
smooth sinusoidal curve of period 1 year passing through «almost» all the error bars✓ 1Do not accept straight lines joining the points.
Part (c)
p = half of the peak-to-peak offset OR the maximum offset from Δx = 0 «the star moves ±p about its mean position»✓ 1
p = 21.4 mas✓ 1Accept 20–23 mas. Award [1 max] for the peak-to-peak value «≈ 43 mas» given as p.
Part (d)
d = 1/(21.4 × 10−3) = 46.7 pc✓ 1Allow ECF from (c).
fractional uncertainty = 0.8/21.4 = 0.037 «3.7 %»✓ 1The fractional uncertainty in d equals that in p.
d = (46.7 ± 1.7) pc✓ 1Accept (47 ± 2) pc. MP3 is for matching the precision of value and uncertainty.
Part (e)
46.7 × 3.26 = 152 ly✓ 1Allow ECF from (d). Accept 150–155 ly.
Part (f)
the catalogue value lies inside the range 45.0–48.4 pc, so the values agree «within the uncertainty»✓ 1Allow ECF from (d). A reference to the uncertainty range is needed.
Part (g)
measure over several more years «or more often» and fit one curve to all the data, so that random errors in individual positions average out✓ 1Accept: use a telescope with a larger aperture/better resolution to reduce the uncertainty of each position. Do not accept 'repeat the readings' alone.

Answers: (c) p ≈ 21.4 mas  ·  (d) (46.7 ± 1.7) pc  ·  (e) 152 ly (the remaining parts are explanations — see the table above)

76E-1B-14Apparent brightness Paper 1BMedium12 marks

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 / cmI / W m−2log(d / m)log(I / W m−2)
20.03.65−0.6990.562
30.01.56−0.5230.193
40.00.904−0.398−0.044
50.00.567−0.301−0.246
60.00.406−0.222−0.391
80.00.219
100.00.1450.000−0.839
log I against log d for the lamp−0.8−0.7−0.6−0.5−0.4−0.3−0.2−0.10.00.1log(d / m)−1.0−0.8−0.6−0.4−0.20.00.20.40.60.8log(I / W m⁻²)
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 pointMarkNotes
Part (a)
0.5/20.0 × 100 = 2.5 %✓ 1Accept 2.5 % or 3 %.
Part (b)
log(d / m) = −0.097 and log(I / W m−2) = −0.660✓ 1Accept 2 or more d.p.
Point plotted correctly to within half a small square✓ 1Allow 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✓ 1Accept −1.90 to −2.10.
Part (d)
Gradient ≈ −2, so I ∝ d−2: an inverse-square law✓ 1Allow 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)✓ 1Accept use of any point on the line with I = P/4πd2.
P = 4π × 0.143 = 1.80 W✓ 1Accept 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✓ 1Accept: 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✓ 1
L = 4πd2b = 4π × (1.545 × 1018)2 × 2.4 × 10−10 = 7.2 × 1027 W = 19 L☉✓ 1Award [2] for CNA.

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)

77E-1B-16Fusion in the Sun Paper 1BMedium9 marks

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)

YearS / W m−2
Year 11360.9
Year 21361.4
Year 31360.6
Year 41361.2
Year 51361.8
Year 61361.0
Year 71360.7
Year 81361.3
(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 pointMarkNotes
Part (a)
Mean = 10888.9/8 = 1361.1 W m−2✓ 1
Uncertainty = (1361.8 − 1360.6)/2 = ±0.6 W m−2✓ 1Accept (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✓ 1Must 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✓ 1Award [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✓ 1Allow 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✓ 1Accept "(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)

78E-1B-19Wien's law Paper 1BMedium12 marks

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 / Kλmax / nmlog(T / K)log(λmax / nm)
30009703.4772.987
40007203.6022.857
50005803.6992.763
60004803.7782.681
8000360
100002904.0002.462
120002404.0792.380
log of peak wavelength against log T from a black-body simulation3.43.53.63.73.83.94.04.1log(T / K)2.32.42.52.62.72.82.93.0log(λmax / nm)
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 pointMarkNotes
Part (a)
10/240 × 100 = 4.2 %✓ 1Accept 4 %.
Part (b)
log(T / K) = 3.903 and log(λmax / nm) = 2.556✓ 1Accept 2 or more d.p.
Point plotted correctly to within half a small square✓ 1Allow ECF.
Part (c)
Single straight line through all the error bars✓ 1
Gradient from a large triangle✓ 1
Gradient = −1.01✓ 1Accept −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✓ 1Allow 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✓ 1Accept 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✓ 1Accept: 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✓ 1MP2 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)

79E-1B-27Conditions for fusion Paper 1BMedium11 marks

A student uses a published computer model of the core of a Sun-like star. The model gives the power ε released by hydrogen fusion per kilogram of core material. In Table 1 the density is fixed at 1.5 × 105 kg m−3 and the temperature T is varied. In Table 2 the temperature is fixed at 1.5 × 107 K and the density ρ is varied. The student tests the hypothesis that ε = kρTn, where k and n are constants.

The graph shows log(ε/W kg−1) against log(T/K) for five of the temperatures in Table 1. The values calculated by the model have no random uncertainty.

Table 1: T / 107 Kε / 10−3 W kg−1log(T / K)log(ε / W kg−1)
1.00.3167.000−3.500
1.20.6557.079−3.184
1.41.217.146−2.917
1.62.07
1.83.327.255−2.479
2.05.067.301−2.296
Table 2: ρ / 105 kg m−30.501.001.502.002.50
ε / 10−3 W kg−10.5331.071.602.132.67
Graph of log(ε / W kg^−1) against log(T / K) for five of the six temperatures; points only. Horizontal axis 6.95 to 7.35, vertical axis −3.6 to −2.0.6.957.007.057.107.157.207.257.307.35log(T / K)−3.6−3.2−2.8−2.4−2.0log(ε / W kg⁻¹)
Graph drawn to scale
(a)

Outline why ε increases as the temperature of the core increases.

(2)
(b)

Show that the data in Table 2 are consistent with ε being proportional to ρ.

(1)
(c)

Complete Table 1 for T = 1.6 × 107 K and plot the point on the graph.

(2)
(d)

Draw the line of best fit and determine n.

(3)
(e)

Predict the percentage increase in ε when the core temperature rises by 10 %, at constant density.

(1)
(f)

Suppose the core of the star is compressed slightly, so that its temperature and density increase. Explain, using your results, how the core returns to equilibrium.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
At a higher temperature the nuclei have a greater mean kinetic energy «Ek ∝ T»✓ 1
so a larger fraction of them have enough kinetic energy to overcome the electrostatic repulsion and come close enough to fuse✓ 1OWTTE
Part (b)
ε/ρ = 1.07 × 10−8 W m3 kg−2 for every pair «e.g. doubling ρ from 1.00 to 2.00 doubles ε from 1.07 to 2.13»✓ 1Must see at least two ratios OR a doubling compared. Units not required.
Part (c)
log(1.6 × 107) = 7.204 and log(2.07 × 10−3) = −2.684✓ 1Both needed. Accept 2 or more decimal places.
Point plotted to within half a small square✓ 1Allow ECF.
Part (d)
Single straight line passing through (or very close to) all the points✓ 1
Gradient calculated from a large triangle✓ 1
n = gradient = 4.0✓ 1Accept 3.8–4.2.
Part (e)
1.104 = 1.46, so ε increases by about 46 %✓ 1Allow ECF from (d).
Part (f)
ε rises steeply «as T4 and as ρ», so the rate of energy release in the core increases, raising the pressure in the core✓ 1
The outward force due to pressure then exceeds the inward gravitational force, so the core expands and cools until ε falls back and the forces balance again✓ 1OWTTE

Answers: (c) 7.204, −2.684  ·  (d) n ≈ 4.0  ·  (e) ≈ 46 % (the remaining parts are explanations — see the table above)

80E-1B-46Stellar spectra Paper 1BMedium12 marks

A student analyses the continuous spectrum of a star recorded by a spectrometer on a telescope. The spectrometer gives the relative intensity of the light at fourteen wavelengths, each with an uncertainty of ±0.03. The graph shows the results.

The luminosity of the star is known to be 45 L☉ with an uncertainty of ±6 %. (Wien’s displacement constant = 2.9 × 10−3 m K, σ = 5.67 × 10−8 W m−2 K−4, L☉ = 3.83 × 1026 W, R☉ = 6.96 × 108 m)

Relative intensity of the star's continuous spectrum against wavelength from 350 nm to 1000 nm: fourteen points with error bars of ±0.03, rising to a broad maximum of about 1.0 near 600 nm and falling slowly at longer wavelengths3004005006007008009001000wavelength λ / nm00.20.40.60.811.2relative intensity
Graph drawn to scale
(a)

Draw the curve of best fit for the data.

(1)
(b)

Determine, using your curve, the wavelength λmax at which the intensity is greatest, and estimate its absolute uncertainty.

(2)
(c)

Show that the surface temperature of the star is about 4800 K.

(1)
(d)

Determine the radius of the star in terms of R☉.

(2)
(e)

Determine the absolute uncertainty in the radius.

(3)
(f)

A star catalogue gives the surface temperature of the star as (4950 ± 100) K. Comment on whether your value of the surface temperature agrees with the catalogue value.

(1)
(g)

The Earth’s atmosphere absorbs short wavelengths more strongly than long wavelengths. The student did not correct the readings for this. Explain the effect on the value of the radius found in (d).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
single smooth curve through «almost» all the error bars, with one broad maximum✓ 1Do not accept dot-to-dot lines.
Part (b)
λmax = 600 nm✓ 1Accept 580–620 nm.
uncertainty ±20 nm, because the curve is very flat near its maximum✓ 1Accept ±10 nm to ±40 nm with a reason.
Part (c)
T = 2.9 × 10−3/600 × 10−9 = 4833 K✓ 1Must see full substitution OR answer to 3 s.f. Allow ECF from (b).
Part (d)
R = √(45 × 3.83 × 1026/(4π × 5.67 × 10−8 × 4833⁴)) = 6.66 × 109 m✓ 1
= 9.6 R☉✓ 1Accept 9.7 R☉ from √45 × (5800/T)². Award [2] for CNA.
Part (e)
percentage uncertainty in T = 20/600 × 100 = 3.3 %✓ 1Allow ECF from (b).
R ∝ L1/2T−2, so percentage uncertainty in R = ½ × 6 + 2 × 3.3 = 9.7 %✓ 1
R = (9.6 ± 0.9) R☉✓ 1Accept (10 ± 1) R☉. MP3 is for matching the precision of value and uncertainty.
Part (f)
T = 4830 K ± 3.3 % = (4830 ± 160) K; the range 4670–4990 K overlaps the catalogue range 4850–5050 K, so the values agree «within their uncertainties»✓ 1Allow ECF from (b), (c) and (e). A comparison of the ranges (or of the difference with the combined uncertainty) is needed.
Part (g)
the measured intensities at short wavelengths are too small, so the peak appears at a longer wavelength and the temperature found is too low✓ 1
R ∝ 1/T², so the radius found is too large✓ 1MP2 only scores if MP1 scores. OWTTE.

Answers: (b) (600 ± 20) nm  ·  (c) 4830 K  ·  (d) 9.6 R☉  ·  (e) ± 0.9 R☉ (the remaining parts are explanations — see the table above)

81E-1B-47Conditions for fusion Paper 1BMedium11 marks

A student uses a published computer model of a protostar of mass 1 M☉ that is contracting under its own gravity. The table gives the core temperature T of the protostar at seven values of its radius R. Each value of T has an uncertainty of ±5 %.

The student tests the hypothesis that T = kRn, where k and n are constants. The graph shows log(T/K) against log(R/R☉) for six of the stages, with error bars.

R/R☉T/Klog(R/R☉)log(T/K)
403.06 × 1051.6025.486
254.70 × 1051.3985.672
167.72 × 1051.2045.888
101.19 × 106
6.31.85 × 1060.7996.267
4.03.06 × 1060.6026.486
2.54.80 × 1060.3986.681
log of core temperature against log of radius for six stages of the model protostar, with error bars of ±0.02; the points fall close to a straight line sloping downwards from about (0.40, 6.68) to (1.60, 5.48)00.20.40.60.811.21.41.61.8log(R / R☉)5.45.65.866.26.46.66.877.2log(T / K)
Graph drawn to scale
(a)

State why the graph of log(T/K) against log(R/R☉) is a straight line if the hypothesis is correct.

(1)
(b)

Show that the absolute uncertainty in each value of log(T/K) is about 0.02.

(1)
(c)

Complete the table for R = 10 R☉ and plot the point on the graph.

(2)
(d)

Draw the line of best fit and determine n.

(3)
(e)

Hydrogen fusion starts at a significant rate when the core temperature reaches about 1.0 × 107 K. Predict, using your graph, the radius of the protostar at this stage.

(2)
(f)

In the model the core density is proportional to R−3. When R = 10 R☉ the core density is 140 kg m−3. Determine the core density at the stage in (e) and compare it with the core density of the Sun today, 1.5 × 105 kg m−3.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
log T = n log R + log k, which has the form y = mx + c «gradient n, intercept log k»✓ 1OWTTE
Part (b)
Δlog T = log(1.05T) − log T = log 1.05 = 0.021✓ 1Accept 0.434 × 0.05 = 0.022 or half of log(1.05/0.95). Must see working.
Part (c)
log(R/R☉) = 1.000 AND log(T/K) = 6.076✓ 1Both needed. Accept 2 or more d.p.
point plotted to within half a small square✓ 1Allow ECF.
Part (d)
single straight line passing through all the error bars✓ 1
gradient calculated from a large triangle✓ 1
n = −1.00✓ 1Accept −0.95 to −1.05. Do not accept a positive value.
Part (e)
line extended to log(T/K) = 7.00✓ 1Read-off gives log(R/R☉) ≈ 0.08.
R ≈ 1.2 R☉✓ 1Accept 1.1–1.4 R☉. Allow ECF from the candidate’s line.
Part (f)
ρ = 140 × (10/1.2)³ = 8.1 × 104 kg m−3✓ 1Allow ECF from (e). Accept 5.1 × 104 to 1.1 × 105 kg m−3.
similar to «about half of / the same order of magnitude as» the Sun’s core density, so the core then has a temperature and a density close to those of the Sun’s core «fusion conditions»✓ 1MP2 needs a comparison of the values. OWTTE

Answers: (c) 1.000, 6.076  ·  (d) n = −1.00  ·  (e) ≈ 1.2 R☉  ·  (f) ≈ 8.1 × 104 kg m−3 (the remaining parts are explanations — see the table above)

82E-1B-07Stellar radius from L and T Paper 1BHard12 marks

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)

StarT / KL / L☉Rint / R☉RLT / R☉
162402.891.421.45
2948044.52.552.47
3506024.26.30
4468046.610.610.4
5438088.915.916.4
6404012022.722.4
Radius from luminosity and temperature against radius from interferometry for five stars, with error bars0510152025Rint / R☉0510152025RLT / R☉
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.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R = √(L/4πσT4) = √(24.2 × 3.83 × 1026/(4π × 5.67 × 10−8 × 50604)) = 4.45 × 109 m = 6.40 R☉✓ 1Must see the substitution with L converted to W OR an answer to at least 3 s.f.
Part (b)
R ∝ L1/2T−2, so % uncertainty in R = ½ × (% in L) + 2 × (% in T)✓ 1
= ½ × 4 % + 2 × 2 % = 6 %✓ 1Award [1] max for 4 % + 2 % or 4 % + 2 × 2 %.
Part (c)
Point at (6.30, 6.40) within half a small square, with error bars ±0.2 horizontally and ±0.4 vertically✓ 1Allow ECF from (a) and (b).
Part (d)
Single straight line passing through all the error bars✓ 1
Gradient from a large triangle (at least half the line)✓ 1
Gradient = 1.00✓ 1Accept 0.95–1.05. No unit.
Part (e)
Steepest and shallowest lines drawn through all the error bars✓ 1
Uncertainty = (max − min)/2, e.g. (1.09 − 0.93)/2 = ±0.08✓ 1Accept ±0.05 to ±0.15 from the candidate's lines.
Part (f)
The hypothesis predicts a gradient of 1 (and a line through the origin); 1 lies within gradient ± uncertainty, so the data support it✓ 1Allow 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)✓ 1Accept: 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✓ 1Both 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)

83E-1B-15Stellar radius from L and T Paper 1BHard12 marks

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 / KL / 10−3 L☉T4 / 1016 K4
132004.403.04
156008.09
1730012.98.96
1860016.212.0
1970022.1
2060024.918.0
Luminosity against T to the fourth for six white dwarfs, with error bars02468101214161820T⁴ / 10¹⁶ K⁴051015202530L / 10⁻³ L☉
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 pointMarkNotes
Part (a)
4 × 2 % = 8 %✓ 1
Part (b)
5.92 and 15.1 (× 1016 K4)✓ 1Both needed. Accept more s.f.
Part (c)
L = 4πR2σT4; with R the same for all, 4πR2σ is constant, so L ∝ T4✓ 1OWTTE
Part (d)
Straight line through the origin passing through all the error bars✓ 1
Gradient from a large triangle✓ 1
Gradient = 1.41 × 10−19 L☉ K−4✓ 1Accept 1.34–1.48 with correct power of ten and unit.
Part (e)
4πR2σ = gradient × L☉ = 1.41 × 10−19 × 3.83 × 1026 = 5.39 × 107 W K−4✓ 1
R = √(5.39 × 107/(4π × 5.67 × 10−8)) = 8.7 × 106 m✓ 1Allow 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✓ 1MP2 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✓ 1Accept random uncertainties in L and T. OWTTE

Answers: (b) 5.92 and 15.1  ·  (d) 1.41 × 10−19 L☉ K−4  ·  (e) 8.7 × 106 m  ·  (f) ±7 × 105 m (the remaining parts are explanations — see the table above)

84E-1B-26The HR diagram Paper 1BHard12 marks

A student uses a published star catalogue to study eight main-sequence stars in one open cluster. The catalogue gives the surface temperature T of each star (uncertainty ±2 %) and its luminosity L (uncertainty ±6 %) in terms of the luminosity of the Sun, L☉. The student tests the hypothesis that, for main-sequence stars, L = kTn, where k and n are constants.

The graph shows log(L/L☉) against log(T/K), with error bars, for seven of the stars.

T / KL / L☉log(T / K)log(L / L☉)
42000.1153.623−0.939
51000.4023.708−0.396
59001.153.7710.061
68002.863.8330.456
79008.46
930024.33.9681.386
1120090.44.0491.956
138003544.1402.549
Graph of log(L / L_sun) against log(T / K) for seven of the eight main-sequence stars, with error bars; points only. Horizontal axis 3.6 to 4.2, vertical axis −1.6 to 2.8.3.63.73.83.944.14.2log(T / K)−1.6−0.80.00.81.62.4log(L / L☉)
Graph drawn to scale
(a)

Show that the absolute uncertainty in each value of log(L/L☉) is about 0.03.

(1)
(b)

Complete the table for the star with T = 7900 K.

(1)
(c)

Plot the point for this star and draw the line of best fit.

(2)
(d)

Determine n.

(2)
(e)

Draw lines of maximum and minimum gradient and hence determine the absolute uncertainty in n.

(2)
(f)

A second student suggests that all main-sequence stars have the same radius.

(i)

Show that this suggestion predicts n = 4.

(1)
(ii)

Discuss whether the data support this suggestion.

(1)
(g)

Determine, using the table, (radius of the star with T = 13 800 K)/(radius of the star with T = 4200 K).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Δlog L = log(1.06L) − log L = log 1.06 = 0.025 ≈ 0.03✓ 1Accept 0.434 × 0.06 = 0.026, or half of log(1.06/0.94). Must see the working.
Part (b)
log 7900 = 3.898 and log 8.46 = 0.927✓ 1Both needed. Accept 2 or more decimal places.
Part (c)
Point plotted to within half a small square✓ 1Allow ECF from (b).
Single straight line passing through all the error bars✓ 1
Part (d)
Gradient calculated from a large triangle; n = gradient✓ 1
n = 6.8✓ 1Accept 6.5–7.1.
Part (e)
Steepest and shallowest straight lines drawn through all the error bars✓ 1
Uncertainty = (maximum − minimum gradient)/2, e.g. (7.1 − 6.4)/2 = ±0.3✓ 1Accept ±0.2 to ±0.6 from the candidate's lines.
Part (f)(i)
L = 4πR2σT4; with R the same for every star, 4πR2σ is constant, so L ∝ T4 and n = 4✓ 1OWTTE
Part (f)(ii)
n = 6.8 ± 0.3 does not include 4, so the suggestion is not supported: luminosity rises faster than T4, so hotter main-sequence stars are larger✓ 1Allow ECF from (d) and (e). Must refer to the uncertainty.
Part (g)
R ∝ √L/T2, so ratio = √(354/0.115) × (4200/13 800)2✓ 1
Ratio = 5.1✓ 1Accept 4.9–5.3. Award [2] for CNA.

Answers: (a) 0.025  ·  (b) 3.898, 0.927  ·  (d) n ≈ 6.8  ·  (e) ≈ ±0.3  ·  (g) 5.1 (the remaining parts are explanations — see the table above)

85E-1B-45Stellar evolution by mass Paper 1BHard12 marks

A student uses a published table of computer models of main-sequence stars. For each model the table gives the mass M of the star, its mean luminosity L while on the main sequence and its main-sequence lifetime t. The model lifetimes have an uncertainty of ±10 %.

The student tests the hypothesis that t = k(M/L)n, where k is a constant and n = 1. The graph shows log(t/109 years) against log(M/L) for six of the seven models, with error bars. (M☉ = 1.99 × 1030 kg, L☉ = 3.83 × 1026 W, c = 3.00 × 108 m s−1, 1 year = 3.16 × 107 s)

M/M☉L/L☉t/109 yearslog(M/L)log(t/109 years)
0.80.35240.3591.380
1.01.0100.0001.000
1.43.93.8−0.4450.580
2.0141.4−0.8450.146
3.0600.52
5.04000.12−1.903−0.921
8.024000.035−2.477−1.456
log of main-sequence lifetime against log of M/L for six model stars, with error bars of ±0.04 in log t; the points lie close to a straight line rising from about (−2.5, −1.5) to (0.36, 1.37)-3-2.5-2-1.5-1-0.500.5log(M/L) with M in M☉ and L in L☉-2-1.5-1-0.500.511.5log(t / 10⁹ years)
Graph drawn to scale
(a)

Explain why t would be proportional to M/L if every star fused the same fraction of its mass before leaving the main sequence.

(2)
(b)

Complete the table for the star of mass 3.0 M☉.

(1)
(c)

Plot the point for this star on the graph and draw the line of best fit.

(2)
(d)

Determine the gradient of your line.

(2)
(e)
(i)

Draw lines of maximum and minimum gradient and hence determine the absolute uncertainty in the gradient.

(2)
(ii)

State whether the data support the hypothesis.

(1)
(f)

On the main sequence, 0.70 % of the mass of the hydrogen that is fused is converted into energy. Use your line to determine the fraction of its mass that a star fuses while on the main sequence.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the energy available «from fusion» is proportional to the mass fused, so if the same fraction is fused it is proportional to M✓ 1
the rate at which this energy is used «radiated» is L, so t = energy/L ∝ M/L✓ 1OWTTE
Part (b)
log(M/L) = −1.301 AND log(t) = −0.284✓ 1Both needed. Accept 2 or more decimal places.
Part (c)
point plotted to within half a small square, with its error bar✓ 1Allow ECF from (b).
single straight line passing through all the error bars✓ 1
Part (d)
gradient calculated from a large triangle «at least half the line»✓ 1
gradient = 1.00✓ 1Accept 0.96–1.04. No unit.
Part (e)(i)
steepest and shallowest straight lines that pass through all the error bars✓ 1
uncertainty = (max − min)/2, e.g. (1.02 − 0.98)/2 = ±0.02✓ 1Accept ±0.02 to ±0.06 from the candidate’s lines.
Part (e)(ii)
1 lies within the range gradient ± uncertainty, so n = 1 «and t ∝ M/L» is supported✓ 1Allow ECF from (d) and (e)(i). A conclusion with no reference to the uncertainty scores 0.
Part (f)
at log(M/L) = 0 the line gives log(t) ≈ 1.01, so t ≈ 10 × 109 years for M = M☉ and L = L☉ «≈ 3.2 × 1017 s»✓ 1Accept use of any point on the line, or of the Sun’s row in the table.
fraction = tL/(0.0070 Mc²) = 3.2 × 1017 × 3.83 × 1026/(0.0070 × 1.99 × 1030 × (3.00 × 108)²) = 0.10 «about 10 %»✓ 1Accept 0.09–0.11. Allow ECF. Award [2] for CNA.

Answers: (b) −1.301, −0.284  ·  (d) 1.00  ·  (e)(i) ±0.02  ·  (f) ≈ 0.10 (the remaining parts are explanations — see the table above)

86E-1B-48Stellar equilibrium Paper 1BHard10 marks

A student tests whether a simple published computer model of the Sun is in equilibrium. Table 1 gives the pressure P, the density ρ and the gravitational field strength g at seven distances r from the centre, where R☉ is the radius of the Sun. In the model the gravitational force on each thin spherical layer of gas is balanced by the difference between the pressures on its inner and outer surfaces.

For each layer between two neighbouring radii, Table 2 gives −ΔP/Δr and the mean of the two values of ρg. The graph shows −ΔP/Δr against mean ρg for five of the six layers. (R☉ = 6.96 × 108 m)

r/R☉P/1013 Paρ/kg m−3g/N kg−1
0.10915606001290
0.15638463001650
0.20400326001800
0.25232217001780
0.30127138001660
0.3566.484801490
0.4033.851101310
layer (r/R☉)−ΔP/Δr / 107 Pa m−1mean ρg / 107 N m−3
0.10–0.157.967.73
0.15–0.206.846.75
0.20–0.254.834.87
0.25–0.30
0.30–0.351.741.78
0.35–0.400.9370.966
Graph of −ΔP/Δr against the mean value of ρg for five layers of the model Sun; the points lie close to a straight line through the origin with a gradient of about 10246810mean ρg / 10⁷ N m⁻³0246810−ΔP/Δr / 10⁷ Pa m⁻¹
Graph drawn to scale
(a)

A thin layer of gas has area A, thickness Δr and density ρ. Show that, if the layer is in equilibrium, the pressure difference across it is ΔP = −ρgΔr.

(2)
(b)

Complete Table 2 for the layer between 0.25 R☉ and 0.30 R☉.

(2)
(c)

Plot the point for this layer on the graph and draw the line of best fit.

(2)
(d)

Determine the gradient of your line.

(2)
(e)

State what your answer to (d) shows about the model.

(1)
(f)

Suggest why the values of −ΔP/Δr and mean ρg for each layer are not exactly equal, even though the model is in equilibrium.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
weight of the layer = mass × g = ρAΔr × g✓ 1
in equilibrium the net force due to pressure balances it: (P − (P + ΔP))A = ρAΔrg, so ΔP = −ρgΔr «the pressure falls outwards»✓ 1Must see the force balance with A cancelling.
Part (b)
−ΔP/Δr = (232 − 127) × 1013/(0.05 × 6.96 × 108) = 3.02 × 107 Pa m−1✓ 1Accept 2 or more s.f.
mean ρg = (21700 × 1780 + 13800 × 1660)/2 = 3.08 × 107 N m−3✓ 1Accept 2 or more s.f.
Part (c)
point plotted to within half a small square✓ 1Allow ECF from (b).
single straight line through «or very close to» the origin with points evenly on both sides✓ 1
Part (d)
gradient calculated from a large triangle✓ 1
gradient = 1.01✓ 1Accept 0.95–1.10. No unit «Pa m−1 ÷ N m−3».
Part (e)
gradient ≈ 1 «and the line passes through the origin», so −ΔP/Δr = ρg: the pressure difference across each layer balances its weight, so the model Sun is in equilibrium✓ 1Allow ECF from (d).
Part (f)
ρ and g change across each layer «the layers are thick», so the mean of the values at the two boundaries is not exactly the value that applies to the layer; thinner layers would agree better✓ 1Accept: rounding of the tabulated values. OWTTE

Answers: (b) 3.02 × 107 Pa m−1; 3.08 × 107 N m−3  ·  (d) 1.01 (the remaining parts are explanations — see the table above)

87E-2-15Stellar parallax Paper 2Easy10 marks

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)

Plan view: the Earth's circular orbit around the Sun with the Earth at E1 and, six months later, at E2 on the opposite side; dashed sight lines run from E1 and E2 through the nearby star V to the very distant background stars; a grey line joins the Sun to V; the Sun–Earth distance is marked 1 AU.SunE₁E₂star V1 AUvery distant background starsEarth's orbit
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)
Show mark scheme
Marking pointMarkNotes
Part (a)
Angle p marked at V, between the line from V to the Sun and the line from V to E1 «or E2»✓ 1Accept 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✓ 1Allow ECF from (b)(i). Accept 77–78 ly.
77.6 years✓ 1Allow 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✓ 1Must 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✓ 1OWTTE
Part (e)
p = 1/150 = 0.0067 arc-second✓ 1
This is smaller than 0.01 arc-second, so the telescope cannot measure it✓ 1MP2 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✓ 1OWTTE

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)

88E-2-31Stellar evolution by mass Paper 2Easy11 marks

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 × 104 L☉. Their positions are shown on the HR diagram.

(M☉ is the mass of the Sun and L☉ is the luminosity of the Sun.)

HR diagram, luminosity in solar units from 10^-4 to 10^6 on a log scale against surface temperature from 40 000 K to 2500 K on a reversed log scale, showing the main-sequence band, star X at 5800 K and 1 solar luminosity and star Y at 30 000 K and 2 × 10^4 solar luminosities.40 00020 00010 0005 0002 500surface temperature / K10⁻⁴10⁻³10⁻²10⁻¹11010²10³10⁴10⁵10⁶luminosity / L☉main sequenceXY
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 pointMarkNotes
Part (a)
nuclear fusion of hydrogen into helium in the core✓ 1Do 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)✓ 1Assumes the lifetime is proportional to M/L «the same fraction of the mass is fused».
≈ 8 × 106 years✓ 1Award [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✓ 1Accept either final state.

Answers: (c) ≈ 8 × 106 years (the remaining parts are explanations — see the table above)

89E-2-36The HR diagram Paper 2Easy8 marks

The diagram shows the axes of a Hertzsprung–Russell (HR) diagram. The position of the Sun is marked S.

Blank HR diagram: luminosity in solar units from 10^-4 to 10^6 on a log scale against surface temperature from 40 000 K to 2500 K on a reversed log scale, with only the Sun S marked at 5800 K and 1 solar luminosity.40 00020 00010 0005 0002 500surface temperature / K10⁻⁴10⁻³10⁻²10⁻¹11010²10³10⁴10⁵10⁶luminosity / L☉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 pointMarkNotes
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✓ 1Accept a band a little hotter than the Sun, crossing from the main sequence towards the top of the diagram.
Part (c)(i)
E✓ 1MP1 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✓ 1Both needed.

Answers: (c)(ii) 100 (the remaining parts are explanations — see the table above)

90E-2-49The HR diagram Paper 2Easy12 marks

The stars of an open star cluster formed at about the same time from the same cloud of gas. The table gives data for five stars of the cluster. Stars P, Q and T have been plotted on the HR diagram. (L☉ = 3.83 × 1026 W, surface temperature of the Sun = 5800 K, 1 pc = 3.09 × 1016 m)

StarSurface temperature / KLuminosity / L☉
P950030
Q60001.3
T45000.2
R4200100
W12 0003.0 × 10−3
HR diagram, luminosity from 10^-4 to 10^6 solar luminosities on a log scale against surface temperature from 40 000 K to 2500 K on a reversed log scale, showing the main-sequence band and three cluster stars: P at 9500 K and 30 solar luminosities, Q at 6000 K and 1.3 solar luminosities, and T at 4500 K and 0.2 solar luminosities.40 00020 00010 0005 0002 500surface temperature / K10⁻⁴10⁻³10⁻²10⁻¹11010²10³10⁴10⁵10⁶luminosity / L☉main sequencePQT
Diagram drawn to scale (logarithmic axes)
(a)

Draw crosses on the HR diagram to show the positions of stars R and W. Label each cross.

(2)
(b)

Identify the type of star that R is and the type of star that W is. Give a reason for each answer.

(2)
(c)

Determine the radius of R in terms of the radius of the Sun, R☉.

(2)
(d)

There are no main-sequence stars in the cluster with surface temperatures above about 10 000 K. Explain this observation.

(3)
(e)

A main-sequence star with a surface temperature of about 10 000 K stays on the main sequence for about 1 × 109 years. Estimate the age of the cluster.

(1)
(f)

The parallax angle of the cluster is 2.5 × 10−3 arc-second. Calculate the apparent brightness of star R at the Earth.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R plotted at 4200 K and 100 L☉, above the main sequence on the right✓ 1Accept ±500 K in temperature and ±0.3 of a decade in luminosity.
W plotted at 12 000 K and 3 × 10−3 L☉, well below the main sequence on the left✓ 1Accept ±1500 K in temperature and ±0.3 of a decade in luminosity.
Part (b)
R: red giant — cooler than the Sun but 100 times more luminous, so it must have a very large radius «it lies above the main sequence»✓ 1Accept 'giant'.
W: white dwarf — hotter than the Sun but far less luminous, so it must be very small «it lies below the main sequence»✓ 1Award [1 max] for both types with no reasons.
Part (c)
L = 4πR2σT4, so R/R☉ = √(L/L☉) × (5800/T)2 = √100 × (5800/4200)2✓ 1Accept a full calculation using σ and L☉.
= 19 «R☉»✓ 1Award [2] for CNA. Accept 18–20.
Part (d)
main-sequence stars that are hotter are more massive and much more luminous✓ 1
they use up the hydrogen in their cores much faster, so they have much shorter main-sequence lifetimes✓ 1
the cluster is older than the main-sequence lifetime of these stars, so they have already left the main sequence «become giants/supergiants and then white dwarfs, neutron stars or black holes»✓ 1OWTTE
Part (e)
about 1 × 109 years «the age equals the lifetime of the hottest stars still on the main sequence»✓ 1Accept 'at least 1 × 109 years'.
Part (f)
d = 1/2.5 × 10−3 = 400 pc = 1.24 × 1019 m✓ 1
b = L/4πd2 = 100 × 3.83 × 1026/(4π × (1.24 × 1019)2) = 2.0 × 10−11 W m−2✓ 1Award [2] for CNA.

Answers: (c) 19 R☉  ·  (e) ≈ 1 × 109 years  ·  (f) 2.0 × 10−11 W m−2 (the remaining parts are explanations — see the table above)

91E-2-87Stellar parallax Paper 2Easy11 marks

An interstellar probe is 160 AU from the Sun and is moving away from it at a constant speed of 17 km s−1. It is heading towards star Q, which has a stellar parallax angle of 0.400 arc-second and a luminosity of 0.60 L☉.

(1 AU = 1.50 × 1011 m, 1 pc = 3.09 × 1016 m, 1 ly = 9.46 × 1015 m, c = 3.00 × 108 m s−1, 1 year = 3.16 × 107 s, L☉ = 3.83 × 1026 W)

(a)

Calculate the time, in hours, taken for a radio signal from the probe to reach the Sun.

(2)
(b)
(i)

Calculate the distance from the Sun to Q in parsecs and in astronomical units.

(2)
(ii)

Calculate the distance from the Sun to Q in light years.

(1)
(c)

Determine whether the probe could reach Q within 100 000 years.

(3)
(d)

Show that the apparent brightness of the Sun at the probe is about 0.05 W m−2.

(1)
(e)

Q and the Sun appear equally bright at one point on the straight line between them. Determine the distance of this point from the Sun, as a fraction of the Sun–Q distance.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
160 × 1.50 × 1011 = 2.4 × 1013 m AND t = 2.4 × 1013/3.00 × 108 = 8.0 × 104 s✓ 1
= 22 hours✓ 1Award [2] for CNA.
Part (b)(i)
d = 1/0.400 = 2.50 pc✓ 1
1 pc = 3.09 × 1016/1.50 × 1011 = 2.06 × 105 AU, so d = 5.15 × 105 AU✓ 1Accept 5.2 × 105 AU from 1 pc = 2.06 × 105 AU. Allow ECF.
Part (b)(ii)
2.50 × 3.09 × 1016/9.46 × 1015 = 8.17 ly✓ 1Allow ECF from (b)(i).
Part (c)
distance = 7.73 × 1016 m «− 2.4 × 1013 m, negligible»✓ 1Allow ECF from (b)(i).
t = 7.73 × 1016/1.7 × 104 = 4.5 × 1012 s = 1.4 × 105 years✓ 1
this is longer than 100 000 years, so the probe could not reach Q in that time✓ 1MP3 needs a comparison consistent with the candidate’s time.
Part (d)
b = 3.83 × 1026/(4π × (2.4 × 1013)²) = 0.053 W m−2✓ 1Must see full substitution OR answer to 2 s.f. or more.
Part (e)
L☉/x² = 0.60L☉/(D − x)², so (D − x)/x = √0.60✓ 1
x = D/(1 + √0.60) = 0.56D✓ 1Award [2] for CNA. Accept 0.56.

Answers: (a) 22 hours  ·  (b)(i) 2.50 pc, 5.15 × 105 AU  ·  (b)(ii) 8.17 ly  ·  (c) 1.4 × 105 years — no  ·  (e) 0.56 (the remaining parts are explanations — see the table above)

92E-2-90Stellar evolution by mass Paper 2Easy10 marks

A computer model follows the evolution of a star of mass 5 M☉ from the time it joins the main sequence. The table gives the age of the star, its surface temperature T and its luminosity L at four stages A, B, C and D. Stage A is plotted on the HR diagram. (Surface temperature of the Sun = 5800 K)

StageAge / 106 yearsT / KL / L☉
A017 000600
B9413 5001 100
C956 000900
D964 3001 600
Hertzsprung–Russell diagram with logarithmic axes: surface temperature from 40 000 K on the left to 2 500 K on the right, luminosity from 0.1 to 100 000 solar luminosities. The main sequence is a dashed band. Stage A is plotted at 17 000 K and 600 solar luminosities.40 00020 00010 0005 0002 500surface temperature / K10⁻¹11010²10³10⁴10⁵luminosity / L☉Amain sequence
Diagram drawn to scale (logarithmic axes)
(a)

Plot stages B, C and D on the HR diagram, and join A, B, C and D in order.

(2)
(b)

State the source of the energy radiated by the star between stages A and B.

(1)
(c)
(i)

Show that the radius of the star at stage A is about 2.9 times the radius of the Sun.

(1)
(ii)

Determine (radius of the star at stage D)/(radius of the star at stage A).

(2)
(d)

Suggest why very few stars are observed in the region of the HR diagram between stages B and C.

(2)
(e)

State and explain the final stage in the evolution of this star.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B, C and D plotted within half a square «±0.1 of a decade in L» of the correct positions✓ 1Award [1] for two points correct.
A → B → C → D joined in order: B up and to the right of A, C far to the right, D up and to the right of C✓ 1
Part (b)
fusion of hydrogen into helium in the core✓ 1Do not accept 'burning'.
Part (c)(i)
R/R☉ = √600 × (5800/17 000)² = 2.85✓ 1Must see full substitution OR answer to 3 s.f.
Part (c)(ii)
RD/R☉ = √1600 × (5800/4300)² = 72.8✓ 1
ratio = 72.8/2.85 = 26✓ 1Accept 25–26. Award [2] for CNA. Accept a direct calculation √(1600/600) × (17 000/4300)².
Part (d)
the star crosses this region in only about 1 × 106 years, about 1 % of its main-sequence time✓ 1
so at any time only a very small fraction of stars are in this stage «and are seen there»✓ 1OWTTE
Part (e)
white dwarf✓ 1MP1 only scores if a reason is given.
its mass is too small for a supernova «less than about 8 M☉»: its outer layers are ejected «as a planetary nebula» and the core remains as a white dwarf✓ 1OWTTE

Answers: (c)(i) 2.85 R☉  ·  (c)(ii) 26 (the remaining parts are explanations — see the table above)

93IBP2S-042-04-11Wien's law Paper 2Easy6 marks

A medical thermal-imaging camera is used to look for areas of inflammation on a patient's skin. Healthy skin has a surface temperature of 34 °C. Treat the skin as a black body.

(a)

State Wien's displacement law.

(1)
(b)

Calculate the wavelength at which the emission from healthy skin peaks.

(2)
(c)

State the region of the electromagnetic spectrum in which the camera must be sensitive.

(1)
(d)

An inflamed region is a few kelvin warmer than the surrounding skin. Explain why it shows up on the camera image.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The wavelength at which black-body emission peaks is inversely proportional to the absolute temperature: λmaxT = 2.9×10−3 m K✓ 1
Part (b)
T = 34 + 273 = 307 K✓ 1
λmax = 2.9×10−3/307 = 9.4×10−6 m✓ 1
Part (c)
Infrared . Allow error carried forward from (b).✓ 1
Part (d)
A higher temperature means a greater total power emitted per unit area, since this is proportional to T4✓ 1
The black-body curve is higher at every wavelength, including the wavelengths the camera detects, so the region appears brighter than the surrounding skin. Also accept: the peak shifts slightly to a shorter wavelength and the intensity there is greater.✓ 1

Common error: Always convert to kelvin before using Wien's law. Using 34 instead of 307 gives an answer about nine times too large.

94IBP2S-042-03-11Stellar parallax Paper 2Easy5 marks

A solar-powered space probe is designed to fly close to the Sun.
Luminosity of the Sun = 3.83×1026 W
1 AU = 1.50×1011 m

(a)

State what is meant by apparent brightness.

(1)
(b)

Calculate the apparent brightness of the Sun at a distance of 1.0 AU.

(2)
(c)

The probe's closest approach to the Sun is 0.30 AU. Calculate the apparent brightness of the Sun at this point.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The power received per unit area, measured perpendicular to the direction of the light, at the observer.✓ 1
Part (b)
b = L/(4πd2) = 3.83×1026/(4π × (1.50×1011)2)✓ 1
b = 1.35×103 W m−2✓ 1
Part (c)
Use of the inverse-square law: b = 1.35×103/0.302, or a full recalculation with d = 4.5×1010 m✓ 1
b = 1.5×104 W m−2✓ 1
Allow error carried forward from the previous part.—

Exam tip: Once you know b at one distance, scaling by (d₁/d₂)² is faster and less error-prone than recalculating from L.

95IBP2S-042-03-12Wien's law Paper 2Easy5 marks

A hot star has a luminosity of 4.0×1028 W. Its spectrum peaks at a wavelength of 290 nm.
Wien's constant = 2.9×10−3 m K
σ = 5.67×10−8 W m−2 K−4
Solar radius R☉ = 6.96×108 m

(a)

Show that the surface temperature of the star is about 1.0×104 K.

(2)
(b)

Calculate the radius of the star.

(2)
(c)

Calculate the radius of the star in units of R☉.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
T = 2.9×10−3/λmax✓ 1
T = 2.9×10−3/290×10−9 = 1.0×104 K✓ 1
Part (b)
R = √(L/(4πσT4)) = √(4.0×1028/(4π × 5.67×10−8 × (1.0×104)4))✓ 1
R = 2.4×109 m✓ 1
Part (c)
R = 2.37×109/6.96×108 = 3.4 R☉✓ 1
ECF from (b).—

Common error: Convert nm to m (×10⁻⁹) before using Wien's law. Forgetting this gives a temperature that is wrong by a factor of 10⁹.

96IBP2S-042-00-11Fusion in the Sun Paper 2Easy5 marks

A satellite in orbit near Earth measures the intensity of sunlight above the atmosphere as 1361 W m−2. The mean Earth–Sun distance is 1.50×1011 m.

(a)

State what is meant by the apparent brightness of a star.

(1)
(b)

Calculate the luminosity of the Sun.

(2)
(c)

A probe in orbit around Mars, 2.28×1011 m from the Sun, carries solar panels. Calculate the intensity of sunlight at the probe.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The power received per unit area perpendicular to the direction of the light (at the observer)✓ 1
Part (b)
\(L = 4\pi d^2 b = 4\pi(1.50\times10^{11})^2 \times 1361\)✓ 1
L = 3.85×1026 W✓ 1
Part (c)
\(b = \frac{L}{4\pi d^2}\), or use of the inverse-square ratio \(1361\times(1.50/2.28)^2\)✓ 1
b ≈ 589 W m−2 (accept 590 W m−2)✓ 1
ECF from (b)—

Exam tip: An inverse-square ratio avoids recalculating L and reduces rounding errors. The intensity scales as (d₁/d₂)².

97IBP2S-042-00-12Fusion in the Sun Paper 2Easy5 marks

One step of the proton–proton chain in the Sun is
\(^{2}_{1}\text{H} + \,{}^{1}_{1}\text{H} \rightarrow\, {}^{3}_{2}\text{He} + \gamma\)
The table gives atomic masses.

AtomMass / u
¹H1.007825
²H2.014102
³He3.016029
1 u = 931.5 MeV c−2; 1 MeV = 1.60×10−13 J.

(a)

Outline what is meant by nuclear fusion.

(2)
(b)

Show that the energy released in this reaction is about 5.5 MeV.

(2)
(c)

Calculate this energy in joules.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Two light nuclei join to form a heavier nucleus✓ 1
The product has a greater binding energy per nucleon (or the products have a lower total mass), so energy is released✓ 1
Part (b)
Mass defect = (2.014102 + 1.007825) − 3.016029 = 0.005898 u✓ 1
Energy = 0.005898 × 931.5 = 5.494 MeV ≈ 5.5 MeV✓ 1
Part (c)
5.494 × 1.60×10−13 = 8.79×10−13 J (accept 8.78×10−13 J from using 5.49 MeV, or 8.8×10−13 J from 5.5 MeV)✓ 1
ECF from (b)—

Exam tip: Atomic masses can be used directly here because the number of electrons is the same on both sides (two each). Carry at least six decimal places in the mass defect, and keep the unrounded energy for later steps.

98IBP2S-042-01-11Conditions for fusion Paper 2Easy6 marks

A tokamak fusion reactor confines a deuterium–tritium plasma with strong magnetic fields. In normal operation the plasma temperature is 1.5×108 K.
Boltzmann constant k = 1.38×10−23 J K−1; e = 1.60×10−19 C.

(a)

Explain why the plasma must be at a very high temperature for fusion to occur.

(2)
(b)

Outline why the plasma is held away from the reactor walls by magnetic fields.

(2)
(c)

Calculate, in keV, the mean kinetic energy of a deuteron in the plasma.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Deuterium and tritium nuclei are both positively charged, so they repel each other electrostatically .✓ 1
They need very high kinetic energies to get close enough (about 10−15 m) for the strong nuclear force to act; a very high temperature gives the nuclei these high kinetic energies .✓ 1
Part (b)
Contact with the walls would cool the plasma below the temperature needed for fusion, and could damage the walls .✓ 1
The plasma consists of moving charged particles, so a magnetic field exerts a force on them, making them spiral around the field lines and keeping them away from the walls .✓ 1
Part (c)
E = (3/2)kT = 1.5 × 1.38×10−23 × 1.5×108 = 3.1×10−15 J✓ 1
Dividing by 1.60×10−19 J eV−1 gives 1.94×104 eV = 19.4 keV ≈ 19 keV✓ 1

Exam tip: To convert joules to electronvolts, divide by 1.60×10⁻¹⁹. Then divide by 1000 to get keV.

99E-2-12Fusion in the Sun Paper 2Medium12 marks

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)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Two light nuclei join to form a heavier nucleus «releasing energy»✓ 1
Part (a)(ii)
Δm = 1.007276 + 2.013553 − 3.014932 = 0.005897 u; E = 0.005897 × 931.5 = 5.49 MeV✓ 1Must see full substitution OR answer to 3 s.f.
Part (a)(iii)
E = 5.49 × 1.60 × 10−13 = 8.78 × 10−13 J✓ 1
λ = hc/E = 6.63 × 10−34 × 3.00 × 108/8.78 × 10−13 = 2.3 × 10−13 m✓ 1Award [2] for CNA. Allow ECF from (a)(ii).
Part (b)
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✓ 1Award [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✓ 1OWTTE
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»✓ 1OWTTE

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)

100E-2-29Stellar radius from L and T Paper 2Medium12 marks

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 luminosity627300
minimum luminosity386200
HR diagram, luminosity in solar units from 10^-4 to 10^6 on a log scale against surface temperature from 40 000 K to 2500 K on a reversed log scale, showing the main-sequence band and the Sun S at 5800 K and 1 solar luminosity.40 00020 00010 0005 0002 500surface temperature / K10⁻⁴10⁻³10⁻²10⁻¹11010²10³10⁴10⁵10⁶luminosity / L☉main sequenceS
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 pointMarkNotes
Part (a)
the total power radiated by the star «energy emitted per unit time»✓ 1Do not accept 'brightness'.
Part (b)
cross at 7300 K and between 10 and 100 L☉ «about 60 L☉»✓ 1Accept 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✓ 1Allow ECF from the candidate's crosses.
so Z is not a main-sequence star «it is larger, an evolved star»✓ 1MP2 only scores if MP1 scores.
Part (e)(i)
R/R☉ = √(L/L☉) × (T☉/T)2 = √62 × (5800/7300)2✓ 1Or a full calculation with σ and L☉.
= 5.0 R☉✓ 1Award [2] for CNA. Accept 4.9–5.0 R☉.
Part (e)(ii)
at minimum luminosity R = √38 × (5800/6200)2 = 5.4 R☉✓ 1Accept 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✓ 1Allow 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»✓ 1OWTTE

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)

101E-2-30Apparent brightness Paper 2Medium12 marks

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 pointMarkNotes
Part (a)(i)
d = 1/0.0625 = 16 pc✓ 1
16 × 3.09 × 1016/9.46 × 1015 = 52 ly✓ 1Award [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»✓ 1OWTTE
Part (b)
b = 0.45 × 3.83 × 1026/(4π × (4.94 × 1017)2)✓ 1Allow 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»✓ 1Must 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»✓ 1The 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✓ 1Award [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✓ 1Allow 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✓ 1OWTTE
Part (c)(iv)
ice reflects more of the incident radiation, so the albedo would increase and the temperature would fall further✓ 1OWTTE

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)

102E-2-50Stellar parallax Paper 2Medium12 marks

Star P is one of the stars nearest to the Sun. Its stellar parallax angle, measured from the Earth, is 0.768 arc-second. A spacecraft in the outer Solar System and a telescope on the Earth photograph P against the background of very distant galaxies at the same time. The spacecraft is 47 AU from the Earth, and the line joining them is perpendicular to the direction of P (Figure 1).

(1 pc = 3.26 ly = 3.09 × 1016 m, L☉ = 3.83 × 1026 W, Wien's displacement constant = 2.9 × 10−3 m K)

Figure 1: the Earth and a spacecraft 47 AU apart both observe star P; dashed lines show the two lines of sight to P. Very distant galaxies lie far beyond P.SunEarthspacecraftstar P47 AUvery distant galaxies· · ·· · ·
Figure 1 — diagram not to scale
(a)

Calculate the distance to P in parsecs and in light years.

(2)
(b)

Deduce, in arc-seconds, the angle between the direction of P seen from the Earth and the direction of P seen from the spacecraft.

(2)
(c)

Explain why the distant galaxies appear in the same positions in both photographs.

(1)
(d)

Suggest why the two photographs must be taken at almost the same time.

(1)
(e)

The apparent brightness of P is 3.0 × 10−11 W m−2. Calculate the luminosity of P in terms of L☉.

(2)
(f)

The continuous spectrum of P is most intense at a wavelength of 1.0 μm. Determine the surface temperature of P, and state and explain where P lies on the HR diagram.

(3)
(g)

Suggest why a spacecraft far from the Sun could measure the distances of stars that are too far away for the parallax method to be used from the Earth.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = 1/0.768 = 1.30 pc✓ 1
1.30 × 3.26 = 4.24 ly✓ 1Award [2] for CNA. Allow ECF.
Part (b)
by definition of the parallax angle, a baseline of 1 AU perpendicular to the line of sight subtends 0.768 arc-second at P✓ 1Award [1 max] for 18″ «2 AU taken as the parallax baseline» or 72″.
so 47 AU subtends 47 × 0.768 = 36.1 arc-second✓ 1Accept 36″. Award [2] for CNA.
Part (c)
the galaxies are so far away that the angle subtended at them by a 47 AU baseline is far too small to measure✓ 1OWTTE
Part (d)
the Earth and the spacecraft keep moving «the Earth round its orbit», so the baseline and its direction would change between the photographs✓ 1Accept: P itself moves across the sky.
Part (e)
L = 4πd2b = 4π × (1.30 × 3.09 × 1016)2 × 3.0 × 10−11 = 6.1 × 1023 W✓ 1Allow ECF from (a).
= 1.6 × 10−3 L☉✓ 1Award [2] for CNA. Accept 1.5–1.7 × 10−3 L☉.
Part (f)
T = 2.9 × 10−3/1.0 × 10−6 = 2900 K✓ 1
P is on the main sequence, at its lower right end✓ 1Do not accept 'white dwarf'.
it is cool «redder than the Sun» and has a very low luminosity, and main-sequence stars at 2900 K have luminosities of about 10−3 L☉✓ 1MP3 needs a reference to both temperature and luminosity. Allow ECF from (e).
Part (g)
the baseline is much longer than 2 AU, so the shift in direction of a distant star is much larger and can be measured for stars much further away✓ 1OWTTE

Answers: (a) 1.30 pc, 4.24 ly  ·  (b) 36.1 arc-second  ·  (e) 1.6 × 10−3 L☉  ·  (f) 2900 K (the remaining parts are explanations — see the table above)

103E-2-82Fusion in the Sun Paper 2Medium12 marks

The overall result of the fusion reactions in the core of the Sun is

4 ¹₁H → ⁴₂He + 2e+ + 2ν

Each time this happens 26.7 MeV is released and two neutrinos are produced. Neutrinos interact so weakly with matter that almost all of them pass straight out of the Sun. A neutrino observatory deep underground counts solar neutrinos to measure the number arriving per unit area per second, the neutrino flux.

(L☉ = 3.83 × 1026 W, 1 AU = 1.50 × 1011 m, 1 MeV = 1.60 × 10−13 J, c = 3.00 × 108 m s−1)

Blank axes for a sketch: neutrino flux at the Earth, from 5.8 to 6.8 × 10^14 per square metre per second, against time from 1 January (0 months) to the next 1 January (12 months)024681012time after 1 January / months5.866.26.46.66.8flux / 10¹⁴ m⁻² s⁻¹
Axes for (c)(ii)
(a)
(i)

Show that about 1.8 × 1038 neutrinos are produced in the Sun each second. Ignore the energy carried away by the neutrinos.

(2)
(ii)

Calculate the neutrino flux at the Earth, at a distance of 1.00 AU from the Sun.

(2)
(b)

The energy released by fusion in the core takes about 105 years to reach the surface of the Sun, but the neutrinos leave the Sun almost immediately.

(i)

Calculate the time taken for a neutrino to travel from the Sun to the Earth. Assume that neutrinos travel at the speed of light.

(1)
(ii)

Explain why measurements of the neutrino flux, but not of the Sun’s luminosity, show the rate of fusion in the core of the Sun at present.

(2)
(c)

The distance between the Earth and the Sun varies from 0.983 AU in early January to 1.017 AU in early July.

(i)

Determine (greatest neutrino flux at the Earth)/(smallest neutrino flux at the Earth).

(2)
(ii)

Sketch, on the axes, a graph to show the variation of the neutrino flux at the Earth during one year.

(2)
(iii)

Suggest why the observed variation is evidence that the detected neutrinos come from the Sun.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
energy per reaction = 26.7 × 1.60 × 10−13 = 4.27 × 10−12 J AND reactions per second = 3.83 × 1026/4.27 × 10−12 = 8.97 × 1037 s−1✓ 1
neutrinos per second = 2 × 8.97 × 1037 = 1.79 × 1038 s−1✓ 1Must see the factor 2 OR answer to 3 s.f.
Part (a)(ii)
flux = 1.79 × 1038/(4π × (1.50 × 1011)²)✓ 1Allow ECF from (a)(i).
= 6.3 × 1014 m−2 s−1✓ 1Award [2] for CNA. Accept 6.4 × 1014 m−2 s−1 from 1.8 × 1038.
Part (b)(i)
t = 1.50 × 1011/3.00 × 108 = 500 s «≈ 8.3 minutes»✓ 1
Part (b)(ii)
neutrinos reach the Earth about 8 minutes after they are produced, so the flux measured now depends on the rate of fusion now✓ 1
the energy that leaves the surface today «as light» was released by fusion about 105 years ago «the radiation is absorbed and re-emitted many times on its way out», so the luminosity shows the fusion rate of long ago✓ 1OWTTE
Part (c)(i)
flux ∝ 1/d², so ratio = (1.017/0.983)²✓ 1
= 1.070✓ 1Award [2] for CNA. Accept 1.07. Award [1 max] for 1.035 «distance ratio not squared» or 0.934 «inverted».
Part (c)(ii)
smooth curve with a maximum at 0 and 12 months «January» and a minimum at about 6 months «July»✓ 1Do not accept a curve that is a maximum in July.
maximum about 7 % greater than the minimum, e.g. from about 6.56 to 6.13 × 1014 m−2 s−1, and the curve never falls to zero✓ 1Allow ECF from (a)(ii) and (c)(i).
Part (c)(iii)
the flux follows the inverse-square law for the Earth–Sun distance «greatest when the Earth is closest to the Sun», which would not happen for neutrinos from another source✓ 1OWTTE

Answers: (a)(i) 1.8 × 1038 s−1  ·  (a)(ii) 6.3 × 1014 m−2 s−1  ·  (b)(i) 500 s  ·  (c)(i) 1.070 (the remaining parts are explanations — see the table above)

104E-2-85Stellar evolution by mass Paper 2Medium12 marks

Planetary nebula N is a shell of glowing gas surrounding a very hot central star C. Seen from the Earth, N is a ring of angular radius 30 arc-second. The stellar parallax angle of C is 1.40 × 10−3 arc-second. Measurements of the spectrum of N show that the shell is expanding at a speed of 22 km s−1.

C has a surface temperature of 1.1 × 105 K and a luminosity of 2500 L☉. No fusion takes place in C. Before N formed, C was the core of red giant G, shown on the HR diagram.

(1 pc = 3.09 × 1016 m, 1 year = 3.16 × 107 s, L☉ = 3.83 × 1026 W, R☉ = 6.96 × 108 m, σ = 5.67 × 10−8 W m−2 K−4)

Hertzsprung–Russell diagram with logarithmic axes: surface temperature from 200 000 K on the left to 2 000 K on the right, luminosity from 10^−4 to 10^6 solar luminosities. The main sequence is a dashed band. The Sun S is at 5 800 K and 1 solar luminosity. The red giant G, the earlier stage of the central star, is at 3 200 K and 4 000 solar luminosities.200 000100 00050 00020 00010 0005 0002 000surface temperature / K10⁻⁴10⁻²110²10⁴10⁶luminosity / L☉GSmain sequence
Diagram drawn to scale (logarithmic axes)
(a)

Outline how N was formed.

(2)
(b)
(i)

Calculate the distance from the Earth to N, in metres.

(1)
(ii)

Show that the radius of the shell is about 3 × 1015 m.

(1)
(iii)

Estimate the time, in years, since the shell was ejected.

(2)
(iv)

State one assumption that you made in (b)(iii).

(1)
(c)
(i)

Determine the radius of C in terms of R☉.

(2)
(ii)

On the HR diagram, plot the position of C. Draw the path of the star from G through C to its final stage, and label the final stage.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
when its core fusion ended, the red giant «a star of low or intermediate mass» became unstable✓ 1
and its outer layers were ejected into space, leaving the hot core exposed✓ 1OWTTE. Do not accept 'supernova'.
Part (b)(i)
d = 1/1.40 × 10−3 = 714 pc = 2.21 × 1019 m✓ 1
Part (b)(ii)
r = dθ = 2.21 × 1019 × 30 × 4.85 × 10−6 = 3.21 × 1015 m✓ 1Must see the conversion of 30 arc-second to radians OR answer to 3 s.f.
Part (b)(iii)
t = 3.21 × 1015/2.2 × 104 = 1.5 × 1011 s✓ 1Allow ECF from (b)(ii).
= 4.6 × 103 years✓ 1Accept 4.3–4.7 × 103 years «4.3 × 103 years from 3 × 1015 m». Award [2] for CNA.
Part (b)(iv)
the shell has expanded at a constant speed «since it was ejected»✓ 1Accept: the shell was ejected from a point/negligible initial radius; the expansion is the same in all directions.
Part (c)(i)
L = 2500 × 3.83 × 1026 = 9.58 × 1029 W AND R = √(L/4πσT⁴) = 9.6 × 107 m✓ 1Accept R/R☉ = √2500 × (5800/1.1 × 105)².
= 0.14 R☉✓ 1Accept 0.13–0.14 R☉. Award [2] for CNA.
Part (c)(ii)
C plotted at 1.1 × 105 K and 2500 L☉ «far to the left, about the same height as G»✓ 1Accept ±0.2 of a decade in luminosity and ±10 000 K in temperature.
path from G to C almost horizontally to the left «temperature rises at roughly constant luminosity»✓ 1
path then from C downwards «and to the right» to below the main sequence, ending in a region labelled white dwarf✓ 1Do not accept 'neutron star' or 'black hole'.

Answers: (b)(i) 2.21 × 1019 m  ·  (b)(ii) 3.2 × 1015 m  ·  (b)(iii) 4.6 × 103 years  ·  (c)(i) 0.14 R☉ (the remaining parts are explanations — see the table above)

105E-2-88The HR diagram Paper 2Medium11 marks

A star can be seen with the unaided eye on a dark night if its apparent brightness is greater than about 1.0 × 10−10 W m−2. About 40 % of the stars that can be seen in this way are giants or supergiants. However, there are no giants or supergiants among the stars within 10 pc of the Sun; most of these stars are faint, cool main-sequence stars.

(L☉ = 3.83 × 1026 W, 1 pc = 3.09 × 1016 m, surface temperature of the Sun = 5800 K)

Hertzsprung–Russell diagram with logarithmic axes: surface temperature from 40 000 K on the left to 2 500 K on the right, luminosity from 10^−4 to 10^6 solar luminosities. The main sequence is a dashed band and the Sun S is marked at 5 800 K and 1 solar luminosity.40 00020 00010 0005 0002 500surface temperature / K10⁻⁴10⁻²110²10⁴10⁶luminosity / L☉Smain sequence
Diagram drawn to scale (logarithmic axes)
(a)

Show that a star identical to the Sun could be seen with the unaided eye up to a distance of about 18 pc.

(2)
(b)

Red giant K has a luminosity of 300 L☉ and a surface temperature of 4000 K.

(i)

Calculate the greatest distance at which K could be seen with the unaided eye.

(1)
(ii)

Deduce (volume of space within which K could be seen)/(volume of space within which the Sun could be seen).

(1)
(iii)

Determine the radius of K in terms of the radius of the Sun.

(2)
(c)

Explain why giants make up a large fraction of the stars seen with the unaided eye although they are rare.

(2)
(d)

A main-sequence star has a luminosity of 0.010 L☉ and is 3.0 pc from the Sun. Determine whether it can be seen with the unaided eye.

(2)
(e)

On the HR diagram, draw a ring around the region in which most of the stars within 10 pc of the Sun lie.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = √(L/4πb) = √(3.83 × 1026/(4π × 1.0 × 10−10)) = 5.5 × 1017 m✓ 1
= 5.5 × 1017/3.09 × 1016 = 17.9 pc✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)(i)
d ∝ √L, so d = √300 × 17.9 = 310 pc✓ 1Allow 18 × √300 = 310 pc. Allow ECF from (a).
Part (b)(ii)
(√300)³ = 3001.5 = 5.2 × 103✓ 1Accept 5000 or 5200.
Part (b)(iii)
R/R☉ = √(L/L☉) × (5800/T)² = √300 × (5800/4000)²✓ 1Or a full calculation with σ, after finding R☉ from L☉.
= 36✓ 1Accept 35–37. Award [2] for CNA.
Part (c)
giants are so luminous that they can be seen from much greater distances, so they are seen within a volume of space thousands of times larger «(b)(ii)»✓ 1Allow ECF from (b)(ii).
faint main-sequence stars are much more common but can only be seen if they are very close, so few of them are visible✓ 1OWTTE
Part (d)
b = 0.010 × 3.83 × 1026/(4π × (3.0 × 3.09 × 1016)²) = 3.5 × 10−11 W m−2✓ 1
this is less than 1.0 × 10−10 W m−2, so it cannot be seen with the unaided eye✓ 1MP2 needs a comparison consistent with the candidate’s value. Accept a comparison of distances: the greatest distance is 1.8 pc < 3.0 pc.
Part (e)
ring around the lower right part of the main sequence «cooler than about 5000 K and less luminous than about 0.1 L☉»✓ 1Do not accept a ring that includes the Sun’s position or the giant region.

Answers: (a) 17.9 pc  ·  (b)(i) 310 pc  ·  (b)(ii) 5.2 × 103  ·  (b)(iii) 36 R☉  ·  (d) 3.5 × 10−11 W m−2 — no (the remaining parts are explanations — see the table above)

106E-2-89Stellar spectra Paper 2Medium13 marks

Technetium (Tc) has no stable isotopes. The technetium isotope produced in stars, ⁹⁹₄₃Tc, has a half-life of 2.1 × 105 years and decays by β− emission to ruthenium (Ru).

The spectrum of red giant G contains dark absorption lines of technetium, one of which has a wavelength of 403.2 nm. G is about 1 × 109 years old, has a surface temperature of 3300 K and a luminosity of 4000 L☉.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, L☉ = 3.83 × 1026 W, σ = 5.67 × 10−8 W m−2 K−4, 1 AU = 1.50 × 1011 m, Wien’s displacement constant = 2.9 × 10−3 m K)

(a)

Outline how the absorption lines in the spectrum of G are formed.

(2)
(b)

Calculate, in eV, the energy of a photon of wavelength 403.2 nm.

(1)
(c)

Write down the nuclear equation for the decay of technetium-99.

(2)
(d)

Show that less than one millionth of the technetium-99 nuclei present at a given time remain after 20 half-lives, and state this time in years.

(2)
(e)

Deduce what the technetium lines show about the nuclear processes in G.

(2)
(f)

Determine the radius of G, in AU.

(3)
(g)

Calculate the wavelength at which the continuous spectrum of G is most intense.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
radiation with a continuous spectrum from deeper, hotter layers passes through the cooler outer layers of G✓ 1
atoms there absorb photons whose energies equal the differences between their energy levels, so these wavelengths are missing «the re-emitted photons go in all directions»✓ 1OWTTE
Part (b)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/(403.2 × 10−9 × 1.60 × 10−19) = 3.08 eV✓ 1Accept 3.1 eV.
Part (c)
⁹⁹₄₃Tc → ⁹⁹₄₄Ru + …✓ 1Ru with nucleon number 99 and proton number 44.
… + ⁰₋₁e + ν̄✓ 1Accept β− for the electron. The antineutrino must be shown.
Part (d)
(1/2)20 = 9.5 × 10−7 < 1 × 10−6✓ 1Must see the calculation.
20 × 2.1 × 105 = 4.2 × 106 years✓ 1
Part (e)
G is about 1 × 109 years old, very much longer than 4.2 × 106 years, so any technetium in the gas from which G formed has decayed✓ 1Allow ECF from (d).
so technetium must have been produced recently by nuclear reactions inside G and carried to its surface: elements are being made in the star✓ 1OWTTE
Part (f)
L = 4000 × 3.83 × 1026 = 1.53 × 1030 W✓ 1
R = √(L/4πσT⁴) = √(1.53 × 1030/(4π × 5.67 × 10−8 × 3300⁴)) = 1.35 × 1011 m✓ 1
= 1.35 × 1011/1.50 × 1011 = 0.90 AU✓ 1Accept 0.90–0.91 AU. Award [3] for CNA.
Part (g)
λ = 2.9 × 10−3/3300 = 8.8 × 10−7 m✓ 1Accept 880 nm.

Answers: (b) 3.08 eV  ·  (d) 9.5 × 10−7; 4.2 × 106 years  ·  (f) 0.90 AU  ·  (g) 8.8 × 10−7 m (the remaining parts are explanations — see the table above)

107E-2-91Conditions for fusion Paper 2Medium13 marks

Jupiter is a ball of mainly hydrogen and helium of mass 1.90 × 1027 kg and radius 7.0 × 107 m. It orbits the Sun at a distance of 5.2 AU and is sometimes called a ‘failed star’.

In a simple model of a ball of gas held together by its own gravity, the core temperature is proportional to M/R, where M is the mass and R is the radius. For the Sun, M = 1.99 × 1030 kg, R = 6.96 × 108 m and the core temperature is 1.5 × 107 K. Hydrogen fusion needs a core temperature of at least about 1.0 × 107 K.

(Solar constant at the Earth = 1360 W m−2, σ = 5.67 × 10−8 W m−2 K−4, Wien’s displacement constant = 2.9 × 10−3 m K)

(a)

State why a very high temperature is needed for hydrogen nuclei to fuse.

(1)
(b)

Determine whether hydrogen fusion could take place in the core of Jupiter, according to the model.

(3)
(c)

Estimate, in terms of the mass of Jupiter, the smallest mass that a body of the same radius as Jupiter would need for hydrogen fusion to start in its core.

(2)
(d)

Jupiter has an albedo of 0.34. It emits infrared radiation like a black body at a temperature of 125 K.

(i)

Show that the intensity of the Sun’s radiation at Jupiter is about 50 W m−2.

(1)
(ii)

Calculate the power absorbed by Jupiter from the Sun.

(2)
(iii)

Determine (power emitted by Jupiter)/(power absorbed by Jupiter).

(2)
(iv)

Suggest the source of the extra energy that Jupiter radiates.

(1)
(e)

Calculate the wavelength at which the radiation emitted by Jupiter is most intense.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
nuclei are positively charged and repel, so they need very large kinetic energies «high T» to get close enough for the strong nuclear force to act✓ 1OWTTE
Part (b)
TJ = 1.5 × 107 × (1.90 × 1027/1.99 × 1030) × (6.96 × 108/7.0 × 107)✓ 1
= 1.4 × 105 K✓ 1
this is far below 1.0 × 107 K, so hydrogen fusion cannot take place✓ 1MP3 needs a comparison consistent with the candidate’s value.
Part (c)
M ∝ T at constant R: M = 1.0 × 107/1.4 × 105 × MJ✓ 1Allow ECF from (b).
≈ 70 MJ «1.3 × 1029 kg»✓ 1Accept 65–75 MJ.
Part (d)(i)
1360/5.2² = 50.3 W m−2✓ 1Must see the inverse-square ratio OR answer to 3 s.f.
Part (d)(ii)
power absorbed = (1 − 0.34) × 50.3 × π(7.0 × 107)²✓ 1Do not award MP1 if 4πR² is used.
= 5.1 × 1017 W✓ 1Award [2] for CNA.
Part (d)(iii)
power emitted = 5.67 × 10−8 × 125⁴ × 4π(7.0 × 107)² = 8.5 × 1017 W✓ 1
ratio = 1.7✓ 1Allow ECF from (d)(ii). Award [2] for CNA.
Part (d)(iv)
Jupiter is slowly contracting, so gravitational potential energy is transferred to internal energy and radiated «as in a protostar»✓ 1Accept: internal energy left over from its formation. Do not accept 'fusion'.
Part (e)
λ = 2.9 × 10−3/125 = 2.3 × 10−5 m✓ 1

Answers: (b) 1.4 × 105 K — no  ·  (c) ≈ 70 MJ  ·  (d)(i) 50.3 W m−2  ·  (d)(ii) 5.1 × 1017 W  ·  (d)(iii) 1.7  ·  (e) 2.3 × 10−5 m (the remaining parts are explanations — see the table above)

108IBP2S-042-02-12Stellar radius from L and T Paper 2Medium8 marks

A student is given the following data for three stars.

Star | Peak wavelength / nm | Parallax / arcsec | Apparent brightness / W m−2
P | 480 | 0.12 | 3.1×10−9
Q | 725 | 0.050 | 2.0×10−10
R | 310 | 0.0081 | 6.5×10−11

Wien's constant = 2.9×10−3 m K; σ = 5.67×10−8 W m−2 K−4; 1 pc = 3.09×1016 m.

(a)

State which star has the highest surface temperature. Explain your answer.

(2)
(b)

Show that the surface temperature of star Q is about 4000 K.

(1)
(c)

Calculate the luminosity of star Q.

(3)
(d)

Determine the radius of star Q.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Star R✓ 1
It has the shortest peak wavelength, and by Wien's law T is inversely proportional to λmax✓ 1
Part (b)
T = 2.9×10−3/(725×10−9) = 4.0×103 K✓ 1
Part (c)
d = 1/0.050 = 20 pc✓ 1
d = 20 × 3.09×1016 = 6.18×1017 m✓ 1
L = 4πd2b = 4π × (6.18×1017)2 × 2.0×10−10 = 9.6×1026 W✓ 1
Part (d)
R = √(L/(4πσT4)) = √(9.6×1026/(4π × 5.67×10−8 × (4.0×103)4))✓ 1
R = 2.3×109 m✓ 1
Allow error carried forward from the temperature and luminosity of star Q found in the earlier parts.—

Exam tip: Convert the parallax distance from parsecs to metres before using L = 4πd²b. Keep extra significant figures until the final step, because T is raised to the fourth power and d is squared.

109IBP2S-042-02-13Stellar spectra Paper 2Medium7 marks

A spectrometer attached to a small observatory telescope records the spectrum of a nearby star. The spectrum shows a continuous background crossed by dark lines. One of the dark lines is at 656 nm, the same wavelength as a bright line seen when hydrogen gas in a laboratory discharge tube is excited.
h = 6.63×10−34 J s; c = 3.00×108 m s−1; e = 1.60×10−19 C.

(a)

Outline how the dark lines in the star's spectrum are produced.

(3)
(b)

Explain why the dark line at 656 nm allows astronomers to conclude that hydrogen is present in the star's atmosphere.

(2)
(c)

Calculate, in eV, the energy difference between the two hydrogen energy levels involved in the 656 nm line.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The hot, dense surface (photosphere) of the star emits a continuous spectrum✓ 1
Atoms in the cooler outer atmosphere absorb photons whose energy equals the difference between two of their energy levels✓ 1
The absorbed energy is re-emitted in all directions, so fewer photons of those wavelengths travel towards the observer, giving dark lines✓ 1
Part (b)
Each element has its own unique set of energy levels, so its line wavelengths are characteristic of that element✓ 1
Absorption and emission involve the same energy difference ΔE, so a dark line at the same wavelength as the hydrogen emission line (656 nm) shows that hydrogen is absorbing✓ 1
Part (c)
E = hc/λ = (6.63×10−34 × 3.00×108)/(656×10−9) = 3.03×10−19 J✓ 1
E = 3.03×10−19/1.60×10−19 = 1.90 eV✓ 1
(accept 1.89 eV, which results from rounding the intermediate value)—

Exam tip: In an absorption explanation, say that the photon energy must exactly match an energy-level difference. Saying that atoms 'absorb light' is not enough.

110IBP2S-042-02-14Stellar parallax Paper 2Medium6 marks

A telescope observing from Earth can reliably measure parallax angles down to 0.010 arcsec. Astronomers want the distance to an open star cluster that is roughly 1500 light-years away.
1 pc = 3.26 light-years.

(a)

Calculate the maximum distance, in parsecs, that this telescope can determine by parallax.

(1)
(b)

Estimate the parallax angle of the cluster, as measured from Earth.

(2)
(c)

State whether the telescope can measure the parallax of the cluster.

(1)
(d)

A future probe in a circular orbit of radius 5.2 AU around the Sun observes the cluster from two opposite points of its orbit, half an orbital period apart. Determine the parallax angle it would measure.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = 1/0.010 = 100 pc✓ 1
Part (b)
d = 1500/3.26 = 460 pc✓ 1
p = 1/460 = 2.2×10−3 arcsec✓ 1
Part (c)
No: 2.2×10−3 arcsec is below the 0.010 arcsec limit (equivalently, 460 pc > 100 pc)✓ 1
ECF from (a) and (b)—
Part (d)
For small angles the parallax angle is proportional to the orbit radius (baseline), so the angle is 5.2 times the Earth-based value✓ 1
p = 5.2 × 2.2×10−3 ≈ 1.1×10−2 arcsec✓ 1
ECF from (b)—

Common error: The relation d (pc) = 1/p (arcsec) only holds for a 1 AU orbit radius. For a larger orbit, the parallax angle scales with the orbit radius in AU. Do not apply d = 1/p directly.

111IBP2S-042-04-12Stellar evolution by mass Paper 2Medium9 marks

A space-based survey records the following data for three stars.

Star | Peak wavelength / μm | Luminosity / W
A | 1.16 | 1.5×1031
B | 0.29 | 1.0×1024
C | 0.50 | 3.8×1026

Treat each star as a black body.

(a)

Calculate the surface temperature of star B.

(2)
(b)

Determine the radius of star B.

(3)
(c)

Identify, with a reason, which star is a red supergiant.

(2)
(d)

Outline the likely final stages in the evolution of the star identified in (c).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
T = 2.9×10−3/λmax✓ 1
T = 2.9×10−3/0.29×10−6 = 1.0×104 K✓ 1
Part (b)
L = 4πR2σT4 rearranged to R = √(L/(4πσT4))✓ 1
R = √(1.0×1024/(4π × 5.67×10−8 × (1.0×104)4))✓ 1
R = 1.2×107 m. ECF from (a).✓ 1
Part (c)
Star A✓ 1
Its low surface temperature (about 2500 K) combined with an enormous luminosity means it must have a very large radius✓ 1
Part (d)
Its core fuses elements up to iron, then collapses, and the star explodes as a supernova✓ 1
The remnant is a neutron star or a black hole, depending on the mass of the core✓ 1

Exam tip: Compare the radius in (b) with the Earth's radius (6.4×10⁶ m). A hot, Earth-sized, faint star is a white dwarf, and noticing this is a quick check that your answer is reasonable.

112IBP2S-042-03-13Stellar radius from L and T Paper 2Medium7 marks

The data below were collected by an astronomy outreach team for two stars, K and M.

Star K: apparent brightness 1.8×10−10 W m−2; distance 2.0×1017 m; surface temperature 4500 K
Star M: apparent brightness 7.2×10−11 W m−2; distance 6.0×1017 m; surface temperature 9000 K

σ = 5.67×10−8 W m−2 K−4

(a)

Calculate the luminosity of star K.

(2)
(b)

Calculate the radius of star K.

(2)
(c)

A student says that star M is more luminous than star K, so it must have the larger radius. Determine whether the student is correct.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
L = 4πd2b = 4π × (2.0×1017)2 × 1.8×10−10✓ 1
L = 9.05×1025 W ≈ 9.0×1025 W✓ 1
Part (b)
R = √(L/(4πσT4)) = √(9.05×1025/(4π × 5.67×10−8 × 45004))✓ 1
R = 5.6×108 m✓ 1
Allow error carried forward from the luminosity calculated in the previous part.—
Part (c)
LM/LK = (dM2bM)/(dK2bK) = 3.6, so M is more luminous✓ 1
The temperature ratio of 2 gives a factor of 24 = 16 in power per unit area, or R ∝ √L/T2 is used✓ 1
RM/RK = √3.6/4 ≈ 0.47, so star M is smaller and the student is incorrect✓ 1
Allow an equivalent method that calculates RM directly (≈2.6×108 m) and compares it with RK.—

Exam tip: A star can be more luminous yet smaller if it is much hotter. Always account for the T⁴ factor before comparing radii.

113IBP2S-042-03-14Stellar radius from L and T Paper 2Medium7 marks

A space agency is planning a solar-powered lander for a rocky planet orbiting a red dwarf star. The star has a surface temperature of 3200 K and a radius of 0.25 R☉.
R☉ = 6.96×108 m
σ = 5.67×10−8 W m−2 K−4
Wien's constant = 2.9×10−3 m K

(a)

Calculate the luminosity of the red dwarf.

(2)
(b)

The lander's solar panels were originally designed for sunlight, which peaks at about 500 nm. Outline, using a calculation, why the panels may be less effective on this planet.

(2)
(c)

The panels need an incident intensity of 1360 W m−2, equal to that of sunlight at Earth. Determine the orbital radius of a planet that would receive this intensity from the red dwarf.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
R = 0.25 × 6.96×108 = 1.74×108 m, and L = 4πσR2T4 = 4π × 5.67×10−8 × (1.74×108)2 × 32004✓ 1
L = 2.3×1024 W✓ 1
Part (b)
λmax = 2.9×10−3/3200 = 9.1×10−7 m, or about 910 nm✓ 1
Most of the star's radiation is in the infrared, away from the visible wavelengths the panels are optimised for, so a smaller fraction of the incident power is converted✓ 1
Part (c)
b = L/(4πd2), rearranged to d = √(L/(4πb))✓ 1
d = √(2.26×1024/(4π × 1360))✓ 1
d = 1.2×1010 m; accept 1.1×1010 m to 1.2×1010 m, including 1.15×1010 m✓ 1
Allow error carried forward from the luminosity calculated in the first part.—

Common error: When you rearrange b = L/(4πd²) for d, remember to take the square root at the end.

114IBP2S-042-00-13The HR diagram Paper 2Medium8 marks

A software team is adding four stars to a planetarium program. L☉ is the Sun's luminosity. The Sun's surface temperature is 5800 K.
Star P: luminosity 1.0×104 L☉, surface temperature 3500 K
Star Q: luminosity 1.0×10−2 L☉, surface temperature 10 000 K
Star R: luminosity 1.0 L☉, surface temperature 5800 K
Star S: luminosity 1.0×10−3 L☉, surface temperature 3000 K
Wien's constant = 2.9×10−3 m K.

(a)

Identify which star is a white dwarf and which star is a red supergiant.

(2)
(b)

Calculate the wavelength at which the emission from star Q is a maximum.

(1)
(c)

Star Q is hotter than the Sun but has a much lower luminosity. Explain how this is possible.

(2)
(d)

Determine the radius of star P in terms of the Sun's radius R☉.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
White dwarf: Q (hot and low luminosity)✓ 1
Red supergiant: P (cool and very luminous)✓ 1
Part (b)
λmax = 2.9×10−3 / 10 000 = 2.9×10−7 m✓ 1
Part (c)
L = 4πR2σT4, so luminosity depends on surface area as well as temperature✓ 1
Q must have a much smaller radius or surface area than the Sun, which outweighs its higher T4✓ 1
Part (d)
\(\frac{R_P}{R_\odot} = \sqrt{\frac{L_P}{L_\odot}}\left(\frac{T_\odot}{T_P}\right)^2\)✓ 1
\(= 100 \times (5800/3500)^2\)✓ 1
≈ 2.7×102 R☉ (accept 270 to 275)✓ 1

Common error: When rearranging L = 4πR²σT⁴ for R, the temperature ratio must be squared, not raised to the fourth power, because R ∝ √L / T².

115IBP2S-042-00-14Stellar evolution by mass Paper 2Medium6 marks

A science journalist wants to estimate how long the Sun will remain on the main sequence. She assumes that:
• only 10% of the Sun's mass is hydrogen that will ever be fused in the core;
• 0.70% of the mass of hydrogen fused is converted into energy;
• the luminosity stays constant at 3.83×1026 W.
Mass of Sun = 1.99×1030 kg; c = 3.00×108 m s−1; 1 year = 3.16×107 s.

(a)

Outline why only a small fraction of the Sun's hydrogen is available for fusion during the main-sequence stage.

(2)
(b)

Show that the total energy released while the Sun is on the main sequence is about 1.3×1044 J.

(2)
(c)

Estimate the main-sequence lifetime of the Sun in years.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Fusion needs very high temperature and density, and these exist only in the core✓ 1
The outer layers are too cool, and their material does not mix into the core, so their hydrogen is not fused✓ 1
Part (b)
Mass converted = 0.10 × 0.0070 × 1.99×1030 = 1.39×1027 kg✓ 1
E = mc2 = 1.39×1027 × (3.00×108)2 = 1.25×1044 J ≈ 1.3×1044 J✓ 1
Part (c)
t = E/L = 1.25×1044 / 3.83×1026 = 3.27×1017 s✓ 1
t = 3.27×1017 / 3.16×107 = 1.04×1010 years ≈ 1.0×1010 years✓ 1
Accept 1.1×1010 years if 1.3×1044 J is used; allow ECF from (b).—

Exam tip: In a 'show that' part, give your answer to one more significant figure than the value quoted, so it is clear you calculated it rather than copied it.

116IBP2S-042-01-13Conditions for fusion Paper 2Medium7 marks

A student uses a stellar-modelling app to find the approximate core temperatures of main-sequence stars of different masses. M☉ is the mass of the Sun.
Mass / M☉: 0.5 | 1.0 | 2.0 | 5.0
Core temperature / 106 K: 9 | 15 | 20 | 27
k = 1.38×10−23 J K−1; e = 1.60×10−19 C.

0123456051015202530Mass / M☉Core temperature / 10⁶ K
Plotted from the table in the question
(a)

Describe the trend shown by the data.

(1)
(b)

Explain, with reference to stellar equilibrium, why more massive stars need higher core temperatures.

(2)
(c)

Calculate, in eV, the mean kinetic energy of a proton in the core of the 2.0 M☉ star.

(2)
(d)

The energy needed for two protons to overcome their electrostatic repulsion is several hundred keV. Suggest why fusion still occurs in this star.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Core temperature increases as stellar mass increases, but not in proportion: increasing the mass by a factor of 10 increases the temperature by a factor of only about 3 .✓ 1
Part (b)
A larger mass produces a greater inward gravitational compression . A greater outward pressure is needed to balance it, which requires a hotter core and a higher fusion rate .✓ 2
Part (c)
E = (3/2)kT = 1.5 × 1.38×10−23 × 2.0×107 = 4.1×10−16 J✓ 1
4.14×10−16 ÷ 1.60×10−19 = 2.6×103 eV✓ 1
Part (d)
The proton energies form a distribution, so a small fraction of protons have energies far above the mean . Quantum tunnelling allows protons to fuse without having the full barrier energy, and the very large number of protons gives a significant overall reaction rate .✓ 2

Exam tip: When describing a trend from data, quote values to show whether the relationship is proportional or not.

117IBP2S-042-01-15Conditions for fusion Paper 2Medium6 marks

Data from a space observatory follow a star as it leaves the main sequence. Its radius increases from 1.2 times to 60 times the Sun's radius, and its surface temperature falls from 5800 K to 3500 K.

(a)

Explain why the core of the star contracts when its hydrogen is used up.

(2)
(b)

Helium fusion later begins in the core. Explain why helium fusion needs a higher temperature than hydrogen fusion.

(2)
(c)

Calculate the ratio (luminosity after expansion)/(luminosity before expansion).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Hydrogen fusion in the core stops, so less energy is released and the outward radiation and gas pressure falls . Gravity is now greater than the outward pressure, so the core contracts .✓ 2
Part (b)
A helium nucleus has charge +2e, so the electrostatic repulsion between two helium nuclei is greater than between two protons . The nuclei therefore need greater kinetic energy to get close enough to fuse, which requires a higher temperature .✓ 2
Part (c)
L = σ4πR2T4, so L2/L1 = (R2/R1)2(T2/T1)4 = (60/1.2)2 × (3500/5800)4✓ 1
= 2500 × 0.133 = 3.3×102✓ 1

Exam tip: Ratio methods cancel σ and 4π. Remember to square the radius ratio and raise the temperature ratio to the fourth power.

118IBP2S-042-01-16Stellar equilibrium Paper 2Medium6 marks

A solar-sail spacecraft is pushed by the radiation pressure of sunlight. This is the same effect that helps support stars against gravity. The sail is perfectly reflecting, has area 800 m2, faces the Sun directly, and is 1.50×1011 m from the Sun.
Luminosity of the Sun = 3.83×1026 W; c = 3.00×108 m s−1. For a perfectly reflecting surface, radiation pressure = 2I/c, where I is the intensity.

(a)

Outline the role of radiation pressure in keeping a main-sequence star in equilibrium.

(2)
(b)

Calculate the intensity of sunlight at the sail.

(2)
(c)

Determine the force exerted by sunlight on the sail.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Photons released by fusion in the core transfer momentum to the outer layers, producing an outward pressure . Together with gas pressure, this balances the inward gravitational force, so the star neither collapses nor expands .✓ 2
Part (b)
I = L/(4πd2) = 3.83×1026 / (4π × (1.50×1011)2)✓ 1
= 1.35×103 W m−2✓ 1
Part (c)
Pressure = 2I/c = 2 × 1355 / 3.00×108 = 9.0×10−6 Pa, ECF from (b)✓ 1
F = pressure × area = 9.03×10−6 × 800 = 7.2×10−3 N✓ 1

Common error: Brightness uses the area of a sphere, 4πd², not πd². Using πd² makes the intensity four times too large.

119E-2-13Conditions for fusion Paper 2Hard11 marks

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)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
n = ρ/m = 1.5 × 105/1.4 × 10−27 = 1.07 × 1032 m−3✓ 1Must 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✓ 1Award [2] for CNA. Allow ECF from (a)(i): accept 2.1 × 1016 Pa from 1 × 1032 m−3.
Part (a)(iii)
Ek = ³⁄₂kBT = 1.5 × 1.38 × 10−23 × 1.5 × 107 = 3.1 × 10−16 J✓ 1
Part (b)(i)
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»✓ 1OWTTE
Part (b)(ii)
T = Pm/(ρkB) = 6.0 × 1015 × 1.4 × 10−27/(8.0 × 104 × 1.38 × 10−23) = 7.6 × 106 K✓ 1
7.6 × 106 K < 1.0 × 107 K, so hydrogen fusion has not yet started «at a significant rate»✓ 1MP2 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✓ 1OWTTE

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)

120E-2-14Stellar equilibrium Paper 2Hard12 marks

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.

(G = 6.67 × 10−11 N m2 kg−2, kB = 1.38 × 10−23 J K−1, 1 year = 3.16 × 107 s)

Cross-section of star W: a large circle with its centre C marked. A small square element of gas X lies in an outer part of the star, to the right of C at about two thirds of the radius; a dashed line joins X to C.CXstar W (cross-section)surface of W
Figure 1 — diagram not to scale
(a)

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)
Show mark scheme
Marking pointMarkNotes
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✓ 1Award [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✓ 1Must 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✓ 1Allow 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✓ 1OWTTE
Part (d)(i)
U = ³⁄₂NkBT = 1.5 × 3.0 × 1057 × 1.38 × 10−23 × 6.0 × 106 = 3.7 × 1041 J✓ 1Must 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✓ 1Accept 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✓ 1Allow 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✓ 1OWTTE

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)

121E-2-28Conditions for fusion Paper 2Hard14 marks

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)

AtomMass / u
hydrogen-11.007825
helium-44.002603
carbon-1212.000000
(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 pointMarkNotes
Part (a)
fusion stops, so the outward «radiation and gas» pressure falls and no longer balances the inward gravitational force✓ 1OWTTE
Part (b)(i)
F = 8.99 × 109 × (2 × 1.60 × 10−19)2/(3.0 × 10−15)2 = 1.0 × 102 N✓ 1Award [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✓ 1Allow 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»✓ 1Must 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✓ 1Accept 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✓ 1Allow ECF from (c).
ratio = 11✓ 1Accept 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✓ 1Allow 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✓ 1Accept 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✓ 1Accept 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✓ 1OWTTE

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)

122E-2-32Wien's law Paper 2Hard12 marks

Stars P and Q are supergiants at the same distance from the Earth. They have the same luminosity, 1.2 × 105 L☉. 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 of the intensity received from star Q per unit wavelength, in arbitrary units, against wavelength from 0 to 2500 nm; the curve peaks at 9.0 units at about 264 nm.05001000150020002500wavelength λ / nm0246810intensity / arbitrary unitsQ
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 pointMarkNotes
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✓ 1Allow ECF from (a)(i). Accept 780–880 nm.
peak much lower than the peak for Q «about a third of its height»✓ 1Accept 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»✓ 1Award 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✓ 1Award [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»✓ 1Allow 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)

123E-2-37Stellar radius from L and T Paper 2Hard13 marks

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)

ObservationValue
wavelength at which the continuous spectrum of T is most intense414 nm
stellar parallax angle of T0.0125 arc-second
apparent brightness of T3.0 × 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 pointMarkNotes
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»✓ 1Accept 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»✓ 1Must 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✓ 1Allow ECF from MP1.
= 6.0 L☉✓ 1Award [3] for CNA.
Part (b)(iii)
R = √(2.30 × 1027/(4π × 5.67 × 10−8 × 70004)) = 1.16 × 109 m✓ 1Allow ECF from (b)(i) and (b)(ii).
= 1.7 R☉✓ 1Accept 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 %✓ 1Accept 4.0 %.
Part (c)(ii)
L = 4πd2b, so the percentage uncertainty in L is 2 × 4 = 8 %✓ 1Allow 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»✓ 1Award [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✓ 1Accept 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)

124E-2-51Stellar spectra Paper 2Hard13 marks

The spectra of stars X, Y and Z show dark absorption lines on a continuous background. The three stars have almost the same composition: about 70 % of their mass is hydrogen. One hydrogen line, of wavelength 656 nm, is produced only by hydrogen atoms that are already in the −3.40 eV level (Figure 1). The table gives the wavelength at which each continuous spectrum is most intense and the strength of the 656 nm line.

(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, kB = 1.38 × 10−23 J K−1, Wien's displacement constant = 2.9 × 10−3 m K)

StarWavelength of maximum intensity / nm656 nm absorption line
X830very weak
Y290very strong
Z97very weak
Figure 1: energy levels of the hydrogen atom drawn to scale: −13.6 eV (ground state), −3.40 eV, −1.51 eV, −0.85 eV and 0.energy0−0.85 eV−1.51 eV−3.40 eV−13.6 eV (ground state)
Figure 1 — energy levels drawn to scale
(a)

Show that the surface temperature of Y is about 1.0 × 104 K.

(1)
(b)

The 656 nm absorption line.

(i)

Calculate, in eV, the energy of a photon of wavelength 656 nm.

(1)
(ii)

Draw, on Figure 1, an arrow to show the transition that produces the 656 nm absorption line.

(2)
(c)

Star X.

(i)

Calculate, in eV, the mean kinetic energy of a particle in the outer layers of X.

(2)
(ii)

Hence explain why the 656 nm line is very weak in the spectrum of X, although X contains a large amount of hydrogen.

(3)
(d)

Suggest why the 656 nm line is also very weak in the spectrum of Z.

(2)
(e)

The spectrum of X shows strong lines of calcium and iron. A student concludes that X contains more calcium than hydrogen. Evaluate this conclusion.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
T = 2.9 × 10−3/290 × 10−9 «= 10 000 K»✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)(i)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/(656 × 10−9 × 1.60 × 10−19) = 1.90 eV✓ 1Accept 1.89–1.90 eV.
Part (b)(ii)
arrow between the −3.40 eV and −1.51 eV levels «1.89 eV = −1.51 − (−3.40)»✓ 1Allow ECF from (b)(i).
arrow pointing upwards «absorption raises the atom to the higher level»✓ 1MP2 only scores if MP1 scores.
Part (c)(i)
T = 2.9 × 10−3/830 × 10−9 = 3494 K✓ 1
Ek = 3/2 kBT = 1.5 × 1.38 × 10−23 × 3494/1.60 × 10−19 = 0.45 eV✓ 1Award [2] for CNA.
Part (c)(ii)
to reach the −3.40 eV level from the ground state an atom needs 10.2 eV✓ 1
0.45 eV is far less than 10.2 eV, so collisions «and the mostly infrared photons» raise very few atoms out of the ground state: almost all the hydrogen atoms are in the ground state✓ 1
only atoms in the −3.40 eV level can absorb 656 nm photons, so there are very few absorbers and the line is weak✓ 1OWTTE. Allow ECF from (c)(i).
Part (d)
Z is so hot «about 3 × 104 K, mean kinetic energy ≈ 3.9 eV» and emits so many ultraviolet photons that most hydrogen atoms are ionized✓ 1
an ionized hydrogen atom has no bound electron, so it cannot absorb at 656 nm; few atoms are left in the −3.40 eV level✓ 1OWTTE
Part (e)
the strength of a line depends on how many atoms are in the level from which the absorption starts, which depends on the temperature, as well as on how much of the element is present✓ 1
X is cool, so its hydrogen lines are weak even though hydrogen is the most abundant element: the conclusion is not justified without taking the temperature into account✓ 1OWTTE

Answers: (a) 10 000 K  ·  (b)(i) 1.90 eV  ·  (c)(i) 0.45 eV (the remaining parts are explanations — see the table above)

125E-2-52Conditions for fusion Paper 2Hard15 marks

In a tokamak, a plasma of deuterium and tritium nuclei and electrons is held away from the walls of a ring-shaped vessel by a magnetic field of flux density 5.0 T. The vessel has a minor radius of 2.0 m. The reaction is 21H + 31H → 42He + 10n. The plasma has a temperature of 1.5 × 108 K and contains about 1020 nuclei per m3. For comparison, the core of the Sun has a temperature of 1.5 × 107 K and contains about 1032 nuclei per m3.

To come within range of the strong nuclear force, a deuterium nucleus and a tritium nucleus colliding head-on need a total kinetic energy of about 0.4 MeV. (nuclear masses: 2H 2.013553 u, 3H 3.015501 u, 4He 4.001506 u; neutron 1.008665 u; 1 u = 931.5 MeV c−2; mass of a deuterium nucleus = 3.34 × 10−27 kg; mass of a helium-4 nucleus = 6.64 × 10−27 kg; kB = 1.38 × 10−23 J K−1; e = 1.60 × 10−19 C; 1 MeV = 1.60 × 10−13 J)

(a)

Show that the energy released in one reaction is about 17.6 MeV.

(1)
(b)

The plasma.

(i)

Show that the mean kinetic energy of a nucleus in the plasma is about 20 keV.

(1)
(ii)

Explain why fusion occurs at a significant rate in the plasma, although the mean kinetic energy of the nuclei is much less than 0.4 MeV.

(2)
(c)

Magnetic confinement.

(i)

A deuterium nucleus with the mean kinetic energy moves at right angles to the magnetic field. Calculate the radius of its circular path.

(3)
(ii)

Each helium-4 nucleus is produced with a kinetic energy of 3.5 MeV. Determine whether the helium-4 nuclei can be held inside the vessel by the field, so that their energy heats the plasma.

(3)
(iii)

State why the neutrons produced are not held by the magnetic field.

(1)
(d)

Explain why a fusion reactor on the Earth must work at a much higher temperature than the core of the Sun.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
Δm = 2.013553 + 3.015501 − 4.001506 − 1.008665 = 0.018883 u; E = 0.018883 × 931.5 = 17.6 MeV✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)(i)
Ek = 3/2 kBT = 1.5 × 1.38 × 10−23 × 1.5 × 108 = 3.10 × 10−15 J «= 19.4 keV»✓ 1Must see full substitution AND conversion to eV OR answer to 3 s.f.
Part (b)(ii)
the nuclei have a wide range of kinetic energies; a small fraction have energies many times the mean, enough to come within range of the strong force✓ 1
with about 1020 nuclei per m3 there are so many collisions per second that even this small fraction gives a significant rate of fusion✓ 1OWTTE
Part (c)(i)
v = √(2Ek/m) = √(2 × 3.1 × 10−15/3.34 × 10−27) = 1.4 × 106 m s−1✓ 1Allow ECF from (b)(i).
qvB = mv2/r, so r = mv/qB✓ 1
r = 3.34 × 10−27 × 1.36 × 106/(1.60 × 10−19 × 5.0) = 5.7 × 10−3 m «≈ 6 mm»✓ 1Award [3] for CNA.
Part (c)(ii)
v = √(2 × 3.5 × 1.60 × 10−13/6.64 × 10−27) = 1.3 × 107 m s−1✓ 1
r = mv/qB with q = 2e: 6.64 × 10−27 × 1.30 × 107/(3.20 × 10−19 × 5.0) = 0.054 m✓ 1Do not award MP2 for 0.11 m «charge taken as e»; MP1 and MP3 may still be awarded, so [2 max].
0.054 m is much less than 2.0 m, so the helium nuclei spiral round the field lines inside the vessel and can give their energy to the plasma✓ 1The conclusion must be consistent with the radius calculated.
Part (c)(iii)
neutrons are uncharged, so there is no magnetic force on them✓ 1
Part (d)
the rate of fusion per unit volume depends on how often nuclei collide, which increases with the number of nuclei per unit volume, and on the fraction of collisions energetic enough, which increases very steeply with temperature✓ 1
the plasma has about 1012 times fewer nuclei per unit volume than the core of the Sun, so collisions are far less frequent✓ 1
to obtain a useful power from a small volume, a much larger fraction of the collisions must lead to fusion, so a much higher temperature is needed✓ 1
the Sun's core is held together by its own gravity: its very large volume and density give a large total power even though the rate per nucleus is very small «and the Sun lasts for billions of years»✓ 1OWTTE. Award [1 max] for 'the Sun has a higher density' alone.

Answers: (a) 17.6 MeV  ·  (b)(i) 19.4 keV  ·  (c)(i) 5.7 × 10−3 m  ·  (c)(ii) 0.054 m — held inside the vessel (the remaining parts are explanations — see the table above)

126E-2-53Stellar radius from L and T Paper 2Hard13 marks

White dwarf V has a surface temperature of 24 000 K, a luminosity of 0.030 L☉ and a mass of 1.6 × 1030 kg. No fusion takes place in V; as it radiates, it cools while its radius stays constant. V is shown on the HR diagram. (L☉ = 3.83 × 1026 W, σ = 5.67 × 10−8 W m−2 K−4, G = 6.67 × 10−11 N m2 kg−2)

HR diagram, luminosity from 10^-5 to 10^5 solar luminosities on a log scale against surface temperature from 40 000 K to 2500 K on a reversed log scale with marks at 24 000 K, 12 000 K, 6000 K and 3000 K, showing the main-sequence band and white dwarf V at 24 000 K and 0.030 solar luminosities.40 00024 00012 0006 0003 000surface temperature / K10⁻⁵10⁻⁴10⁻³10⁻²10⁻¹11010²10³10⁴10⁵luminosity / L☉main sequenceV
Diagram drawn to scale (logarithmic axes)
(a)

Outline why V continues to radiate although no fusion takes place in it.

(1)
(b)

Determine the radius of V.

(2)
(c)

Calculate the mean density of V.

(2)
(d)

Calculate the gravitational field strength at the surface of V.

(1)
(e)

V cools at constant radius.

(i)

Calculate the luminosity of V when its surface temperature has fallen to 12 000 K.

(2)
(ii)

Draw, on the HR diagram, the path of V as it cools from 24 000 K to 6000 K.

(2)
(iii)

A second white dwarf, U, has a surface temperature of 12 000 K and a luminosity of 7.5 × 10−3 L☉. Deduce, with reference to your line in (e)(ii), how the radius of U compares with the radius of V.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
it is very hot and radiates the internal «thermal» energy it already has, so it gradually cools✓ 1OWTTE. Do not accept 'it is still fusing helium'.
Part (b)
L = 0.030 × 3.83 × 1026 = 1.15 × 1025 W AND L = 4πR2σT4 used✓ 1
R = √(1.15 × 1025/(4π × 5.67 × 10−8 × 24 0004)) = 7.0 × 106 m✓ 1Accept 6.9–7.1 × 106 m «from R/R☉ = √0.030 × (5800/24 000)2».
Part (c)
volume = 4/3 π R3 = 1.4 × 1021 m3✓ 1Allow ECF from (b).
ρ = 1.6 × 1030/1.4 × 1021 = 1.1 × 109 kg m−3✓ 1Award [2] for CNA. Accept 1.1–1.2 × 109 kg m−3.
Part (d)
g = GM/R2 = 6.67 × 10−11 × 1.6 × 1030/(7.0 × 106)2 = 2.2 × 106 N kg−1✓ 1Allow ECF from (b). Accept 2.1–2.3 × 106 N kg−1.
Part (e)(i)
with R constant, L ∝ T4: halving T divides L by 24 = 16✓ 1
L = 0.030/16 = 1.9 × 10−3 L☉✓ 1Award [2] for CNA.
Part (e)(ii)
straight line from V downwards and to the right, through about 1.9 × 10−3 L☉ at 12 000 K✓ 1Allow ECF from (e)(i).
ending at 6000 K at about 1.2 × 10−4 L☉ «L falls by a factor of 16 each time T halves»✓ 1Do not accept a curved line or a path that rises.
Part (e)(iii)
at 12 000 K, U lies above the line of V, so it has a larger radius✓ 1
at the same temperature L ∝ R2; U is 7.5 × 10−3/1.875 × 10−3 = 4.0 times as luminous as V would be at 12 000 K✓ 1Allow ECF from (e)(i).
so the radius of U is √4.0 = 2.0 times the radius of V✓ 1Award [3] for CNA with reasoning.

Answers: (b) 7.0 × 106 m  ·  (c) 1.1 × 109 kg m−3  ·  (d) 2.2 × 106 N kg−1  ·  (e)(i) 1.9 × 10−3 L☉  ·  (e)(iii) RU = 2.0 RV (the remaining parts are explanations — see the table above)

127E-2-83Stellar equilibrium Paper 2Hard12 marks

The Sun has been on the main sequence for 4.6 × 109 years. Computer models show that during this time its luminosity has increased from 0.72 L☉ to 1.00 L☉, and that it will be about 1.30 L☉ when the Sun is 8.0 × 109 years old.

The gas in the core is fully ionized: the hydrogen exists as protons and electrons, and the helium as helium nuclei and electrons. (Mass of a hydrogen atom = 1.0 u, mass of a helium atom = 4.0 u; the mass of an electron is negligible.)

Blank axes for a sketch: luminosity of the Sun in units of the present luminosity, from 0 to 1.6, against time since the Sun joined the main sequence, from 0 to 9 billion years0123456789time / 10⁹ years00.20.40.60.811.21.41.6luminosity / L☉
Axes for (d)
(a)

Show that when a sample of pure hydrogen in the core is converted entirely into helium, the number of particles in the sample falls to about 38 % of its original value.

(2)
(b)

Explain, using PV = NkBT, why the pressure in the core would fall as helium accumulates if the temperature and the density of the core did not change.

(2)
(c)
(i)

Explain why the core slowly contracts and why its temperature rises as it does so.

(3)
(ii)

Hence suggest why the luminosity of the Sun increases during its main-sequence life.

(1)
(d)

Sketch, on the axes, a graph to show the variation of the luminosity of the Sun with time from 0 to 8.0 × 109 years.

(2)
(e)

The models also show that the radius of the Sun 4.6 × 109 years ago was 0.90 of its present value. The present surface temperature of the Sun is 5800 K. Determine the surface temperature of the Sun when it joined the main sequence.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
hydrogen: 2 particles «proton + electron» per 1.0 u; helium: 3 particles «nucleus + 2 electrons» per 4.0 u✓ 1Or any equivalent count, e.g. 4 hydrogen atoms give 8 particles, 1 helium atom gives 3.
ratio = (3/4.0)/(2/1.0) = 0.375 «≈ 38 %»✓ 1Must see the particle counts AND the ratio.
Part (b)
P = (N/V)kBT: at constant density the number of particles per unit volume falls as helium forms✓ 1
so at constant T the pressure is smaller «about 38 % for complete conversion»✓ 1OWTTE
Part (c)(i)
the outward force due to pressure no longer balances the inward gravitational force on the core, so the core contracts✓ 1
gravity does work on the gas «gravitational potential energy decreases», increasing the kinetic energy of the particles, so the temperature rises✓ 1
the higher temperature and density restore the pressure, so the core is in equilibrium again «the contraction is slow and continuous»✓ 1OWTTE
Part (c)(ii)
the rate of fusion increases with temperature and density, so more energy is released in the core each second✓ 1Do not accept 'more hydrogen' or 'more fuel'.
Part (d)
curve rising from 0.72 at t = 0 through 1.00 at 4.6 × 109 years to about 1.3 at 8.0 × 109 years✓ 1Accept ±0.05 L☉.
gradient increasing with time «concave upwards»✓ 1Award this mark only for a curve that rises throughout.
Part (e)
L ∝ R²T⁴, so T0 = 5800 × (0.72)1/4/(0.90)1/2✓ 1
= 5600 K✓ 1Accept 5630 K. Award [2] for CNA. Award [1 max] for 5340 K «change in radius ignored».

Answers: (a) 0.375  ·  (e) 5600 K (the remaining parts are explanations — see the table above)

128E-2-84Stellar evolution by mass Paper 2Hard12 marks

Star S has a mass of 25 M☉. Near the end of its life, nuclei of increasing nucleon number fuse in its core until the core is made of iron. The table gives the binding energy per nucleon of some nuclides. (M☉ = 1.99 × 1030 kg, G = 6.67 × 10−11 N m2 kg−2)

NuclideBinding energy per nucleon / MeV
⁴₂He7.07
¹²₆C7.68
¹⁶₈O7.98
²⁸₁₄Si8.45
⁵⁶₂₆Fe8.79
(a)

State the region of the Hertzsprung–Russell diagram in which S lies at this stage.

(1)
(b)

One of the reactions in the core is ¹⁶₈O + ¹⁶₈O → ²⁸₁₄Si + ⁴₂He.

(i)

Show that the energy released in this reaction is about 9.5 MeV.

(1)
(ii)

Helium-4 has a smaller binding energy per nucleon than oxygen-16. Explain, using the table, why energy is nevertheless released in this reaction.

(2)
(c)

Explain why fusion cannot release energy in the core once it is made of iron.

(2)
(d)

The iron core collapses and the rest of S is blown apart in a supernova. The core becomes a neutron star of mass 1.4 M☉ and radius 12 km.

(i)

Calculate the mean density of the neutron star.

(2)
(ii)

The neutron star rotates 700 times per second. Determine whether the gravitational force on material at its equator is large enough to keep this material in circular motion.

(3)
(e)

State what the core would have become if its mass had been much greater.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
red supergiant region «top right»✓ 1Accept 'supergiants'. Do not accept 'red giant'.
Part (b)(i)
28 × 8.45 + 4 × 7.07 − 32 × 7.98 = 236.6 + 28.28 − 255.36 = 9.52 MeV✓ 1Must see full substitution OR answer to 3 s.f.
Part (b)(ii)
total binding energy of the products = 28 × 8.45 + 4 × 7.07 = 264.9 MeV, greater than that of the reactants, 32 × 7.98 = 255.4 MeV «the gain for the 28 nucleons of silicon outweighs the loss for the 4 nucleons of helium»✓ 1Allow ECF from (b)(i). Accept a comparison of binding energy per nucleon weighted by the nucleon numbers.
a greater total binding energy means that the products have less mass «a greater mass defect» than the reactants; this difference in mass is released as energy «E = mc², kinetic energy of the products»✓ 1OWTTE. Do not accept a statement that binding energy is released without the link to the mass difference.
Part (c)
iron-56 has «about» the greatest binding energy per nucleon✓ 1
fusing iron would form nuclei with a smaller binding energy per nucleon, so energy would have to be supplied «the products would have more mass than the iron»✓ 1OWTTE
Part (d)(i)
V = 4/3 π(1.2 × 104)³ = 7.2 × 1012 m³✓ 1
ρ = 1.4 × 1.99 × 1030/7.2 × 1012 = 3.8 × 1017 kg m−3✓ 1Award [2] for CNA. Accept 3.8–3.9 × 1017 kg m−3.
Part (d)(ii)
g = GM/R² = 6.67 × 10−11 × 2.79 × 1030/(1.2 × 104)² = 1.3 × 1012 N kg−1✓ 1
centripetal acceleration needed = ω²R = (2π × 700)² × 1.2 × 104 = 2.3 × 1011 m s−2✓ 1Accept v = 2πRf = 5.3 × 106 m s−1 and v²/R.
g is greater «about 5.6 times», so gravity can provide the centripetal force: the material stays on the star✓ 1MP3 needs a comparison consistent with the candidate’s values.
Part (e)
a black hole✓ 1

Answers: (b)(i) 9.52 MeV  ·  (d)(i) 3.8 × 1017 kg m−3  ·  (d)(ii) g = 1.3 × 1012 N kg−1 > ω²R = 2.3 × 1011 m s−2 (the remaining parts are explanations — see the table above)

129E-2-86Conditions for fusion Paper 2Hard13 marks

In main-sequence stars more massive than about 1.3 M☉, most of the energy is released by the CNO cycle, a sequence of six reactions shown in the table.

In less massive stars, such as the Sun, most of the energy is released by the proton–proton chain. Near a core temperature of 1.7 × 107 K the two processes release energy at the same rate. Close to this temperature, the rate of energy release is proportional to T4 for the proton–proton chain and to T17 for the CNO cycle, where T is the core temperature.

StepReaction
1¹²₆C + ¹₁H → ¹³₇N + γ
2¹³₇N → ¹³₆C + e+ + ν
3¹³₆C + ¹₁H → ¹⁴₇N + γ
4¹⁴₇N + ¹₁H → ¹⁵₈O + γ
5¹⁵₈O → ¹⁵₇N + e+ + ν
6¹⁵₇N + ¹₁H → ¹²₆C + X
Graph of log of the rate of energy release, in arbitrary units, against log of core temperature from 7.00 to 7.40. A straight line for the proton–proton chain rises with gradient 4 and passes through zero at log T = 7.23 (T = 1.7 × 10^7 K).77.057.17.157.27.257.37.357.4log(T / K)-4-3-2-10123log(rate / arbitrary units)proton–proton chain
Graph drawn to scale (logarithmic axes)
(a)

Identify X.

(1)
(b)

Deduce the overall result of one complete CNO cycle, and state the role of carbon-12.

(2)
(c)

State and explain how the energy released in one CNO cycle compares with the energy released when four protons are converted into helium-4 by the proton–proton chain.

(2)
(d)

Explain why the CNO cycle needs a higher core temperature than the proton–proton chain to proceed at a significant rate.

(2)
(e)

Draw, on the graph, a line to show the variation of log(rate) with log(T/K) for the CNO cycle.

(2)
(f)

The core temperature of a main-sequence star of mass 2 M☉ is 2.1 × 107 K. Calculate (rate of energy release by the CNO cycle)/(rate of energy release by the proton–proton chain) in this core.

(2)
(g)

Explain, with reference to the mass of a star, why the CNO cycle dominates in massive main-sequence stars but not in the Sun.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
⁴₂He «alpha particle»✓ 1Both nucleon number and proton number needed.
Part (b)
4 ¹₁H → ⁴₂He + 2e+ + 2ν «+ energy»✓ 1Accept 4 protons → helium-4 + 2 positrons + 2 neutrinos.
carbon-12 is used in the first reaction and formed again in the last: it is a catalyst «it is not used up»✓ 1OWTTE
Part (c)
the energy released is the same✓ 1
the particles at the start and at the end are the same «the C, N and O nuclei cancel», so the mass difference is the same✓ 1MP1 only scores if a reason is given.
Part (d)
a proton must approach a carbon or nitrogen nucleus, of charge 6e or 7e, so the electric repulsion at a given separation is 6–7 times greater than between two protons✓ 1
so the nuclei need much greater kinetic energies to come close enough to fuse, and the mean kinetic energy of the nuclei is proportional to T✓ 1OWTTE
Part (e)
straight line steeper than the proton–proton line «gradient about 17»✓ 1
crossing the proton–proton line at log(T/K) ≈ 7.23 «1.7 × 107 K»; below the proton–proton line at lower temperatures and above it at higher temperatures✓ 1Accept a crossing between 7.20 and 7.26.
Part (f)
ratio = (2.1/1.7)17 ÷ (2.1/1.7)4 = (2.1/1.7)13✓ 1Equal rates at 1.7 × 107 K.
= 16✓ 1Award [2] for CNA. Accept 15–16. Award [1 max] for 36 «the proton–proton dependence on T ignored».
Part (g)
a more massive star has a stronger gravitational pull, so its core is compressed to a higher temperature «above 1.7 × 107 K», where the CNO rate is far greater✓ 1
the Sun’s core «about 1.5 × 107 K» is below 1.7 × 107 K, so the CNO rate is smaller «about 0.2 of the proton–proton rate»✓ 1Allow a valid argument based on (e).

Answers: (f) 16 (the remaining parts are explanations — see the table above)

130IBP2S-042-02-15Stellar radius from L and T Paper 2Hard8 marks

Stars A and B form a visual binary system with a parallax angle of 0.0080 arcsec. Spectra recorded by a space telescope show that the peak wavelength is 420 nm for star A and 840 nm for star B. The apparent brightness of star A is 3.0 times that of star B. The apparent brightness of star A is 4.0×10−12 W m−2.
1 pc = 3.09×1016 m.

(a)

Explain why the ratio of the luminosities of A and B equals the ratio of their apparent brightnesses.

(2)
(b)

Determine the ratio \(\frac{T_A}{T_B}\) of the surface temperatures.

(1)
(c)

Determine the ratio \(\frac{R_A}{R_B}\) of the radii.

(3)
(d)

Calculate the luminosity of star A.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The two stars are in the same system, so they are at the same distance from Earth✓ 1
b = L/(4πd2), so with d the same, b ∝ L✓ 1
Part (b)
T ∝ 1/λmax, so TA/TB = 840/420 = 2.0✓ 1
Part (c)
L = 4πσR2T4, so R ∝ √L/T2✓ 1
RA/RB = √3.0/(2.0)2✓ 1
= 0.43✓ 1
Allow error carried forward from the temperature ratio found in the previous part.—
Part (d)
d = 1/0.0080 = 125 pc = 3.86×1018 m✓ 1
L = 4πd2b = 4π × (3.86×1018)2 × 4.0×10−12 = 7.5×1026 W✓ 1

Exam tip: Ratio questions are quickest with proportionality. Constants such as σ and 4π cancel and do not need to be substituted.

131IBP2S-042-02-16Stellar parallax Paper 2Hard7 marks

A space-based astrometry instrument tracks a nearby star against very distant background stars. Between two observations made six months apart, the apparent position of the star shifts by a total angle of (0.0624 ± 0.0010) arcsec.

Sunposition 1position 2star1 AU0.0624″
Diagram not to scale
(a)

Determine the distance to the star in parsecs.

(2)
(b)

Determine the absolute uncertainty in the distance.

(2)
(c)

A second star is so distant that its parallax cannot be measured. Outline how its spectrum and apparent brightness could be used to estimate its distance.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Parallax angle = half the total shift = 0.0312 arcsec✓ 1
d = 1/0.0312 = 32 pc (32.05 pc)✓ 1
Part (b)
Fractional uncertainty in d = fractional uncertainty in the angle = 0.0010/0.0624 (equivalently 0.0005/0.0312) = 0.016 (1.6 %)✓ 1
Δd = 0.016 × 32 ≈ 0.5 pc✓ 1
Allow error carried forward from the distance found in the first part.—
Part (c)
Use the spectrum to find the surface temperature, either from the peak wavelength using Wien's law or from the absorption lines, which give the spectral class✓ 1
Assume the star is on the main sequence and read its luminosity L from the HR diagram✓ 1
Measure the apparent brightness b and calculate d = √(L/(4πb))✓ 1

Common error: The parallax angle is half the total six-month shift. Using the full shift halves the distance.

132IBP2S-042-04-14Stellar radius from L and T Paper 2Hard10 marks

An astrometry satellite measures the parallax angle of a main-sequence star as 0.025 arc-seconds. A photometer on the same satellite measures the star's apparent brightness as 5.0×10−10 W m−2. A spectrometer shows that the star's emission peaks at 380 nm.
1 pc = 3.086×1016 m; L☉ = 3.83×1026 W; σ = 5.67×10−8 W m−2 K−4; Wien's displacement constant = 2.9×10−3 m K.

(a)

Calculate the distance to the star in metres.

(2)
(b)

Determine the luminosity of the star.

(2)
(c)

Determine the radius of the star.

(3)
(d)

Estimate the mass of the star in solar masses, using L ∝ M3.5.

(2)
(e)

Predict, with a reason, the final state of this star.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = 1/p = 1/0.025 = 40 pc✓ 1
d = 40 × 3.086×1016 = 1.2×1018 m✓ 1
Part (b)
L = 4πd2b✓ 1
L = 4π × (1.234×1018)2 × 5.0×10−10 = 9.6×1027 W; allow ECF from (a)✓ 1
Part (c)
T = 2.9×10−3 / 380×10−9 = 7.6×103 K✓ 1
R = √(L / (4πσT4))✓ 1
R = 2.0×109 m; allow ECF from (b)✓ 1
Part (d)
L/L☉ = 9.6×1027 / 3.83×1026 ≈ 25✓ 1
M = 251/3.5 ≈ 2.5 M☉; allow ECF from (b)✓ 1
Part (e)
The final state is a white dwarf, because the star's mass (about 2.5 M☉) is below about 8 M☉. It will shed its outer layers as a red giant and planetary nebula, leaving a core below the Chandrasekhar limit of 1.4 M☉. Allow ECF from (d).✓ 1

Common error: The Chandrasekhar limit of 1.4 M☉ applies to the remnant core, not to the star's main-sequence mass. Do not compare 2.5 M☉ directly with 1.4 M☉. Also keep the unrounded distance for (b): because d is squared, rounding it changes L by about 5%.

133IBP2S-042-03-15The HR diagram Paper 2Hard9 marks

A massive main-sequence star has a luminosity of 1.2×1030 W and a surface temperature of 2.0×104 K. It later evolves into a red supergiant. In that phase its surface temperature is 3500 K and its luminosity is 1.5 times its main-sequence value. The star is 6.2×1019 m from Earth.
σ = 5.67×10−8 W m−2 K−4
Wien's constant = 2.9×10−3 m K

(a)

Calculate the radius of the star on the main sequence.

(2)
(b)

Determine the ratio \(\frac{\text{radius as red supergiant}}{\text{radius on main sequence}}\).

(3)
(c)

With reference to Wien's law, explain why the star's power output in visible light can increase by a larger factor than 1.5 when it becomes a red supergiant.

(2)
(d)

Calculate the apparent brightness of the star at Earth during its red supergiant phase.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
R = √(L/(4πσT4)) = √(1.2×1030/(4π × 5.67×10−8 × (2.0×104)4))✓ 1
R = 3.2×109 m✓ 1
Part (b)
R ∝ √L/T2, or a full recalculation of the supergiant radius✓ 1
Ratio = √1.5 × (2.0×104/3500)2✓ 1
Ratio = 40✓ 1
Part (c)
Peak wavelength moves from 2.9×10−3/2.0×104 = 145 nm (ultraviolet) to 2.9×10−3/3500 ≈ 830 nm (red / near infrared)✓ 1
On the main sequence most of the output is ultraviolet, which the eye cannot see. As a supergiant the peak lies close to the visible band, so a larger fraction of the total output is visible light, even though most of it is still infrared. The visible output therefore rises by more than the factor of 1.5 rise in total luminosity✓ 1
Part (d)
L = 1.5 × 1.2×1030 = 1.8×1030 W, and b = 1.8×1030/(4π × (6.2×1019)2)✓ 1
b = 3.7×10−11 W m−2✓ 1

Exam tip: Apparent brightness counts power at all wavelengths. How bright a star looks to the eye depends on how much of that power falls in the visible range.

134IBP2S-042-03-16Stellar parallax Paper 2Hard8 marks

A small robotic telescope has a circular mirror of diameter 0.80 m. When it points at a faint star, the total power of starlight incident on the mirror is 2.0×10−14 W. The parallax angle of the star is 0.025 arcsecond, and its spectrum peaks at 1.0 μm. Assume that no light is absorbed by the atmosphere.
1 pc = 3.09×1016 m
σ = 5.67×10−8 W m−2 K−4
Wien's constant = 2.9×10−3 m K

(a)

Calculate the apparent brightness of the star.

(2)
(b)

Calculate the distance to the star in metres.

(1)
(c)

Determine the luminosity of the star.

(2)
(d)

Determine the radius of the star.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
A = π × 0.402 = 0.503 m2✓ 1
b = 2.0×10−14/0.503 = 4.0×10−14 W m−2✓ 1
Part (b)
d = 1/0.025 = 40 pc = 40 × 3.09×1016 = 1.24×1018 m✓ 1
Part (c)
L = 4πd2b = 4π × (1.24×1018)2 × 3.98×10−14✓ 1
L = 7.6×1023 W✓ 1
ECF from (a) and (b).—
Part (d)
T = 2.9×10−3/1.0×10−6 = 2900 K✓ 1
R = √(L/(4πσT4)) = √(7.64×1023/(4π × 5.67×10−8 × 29004))✓ 1
R = 1.2×108 m✓ 1
ECF from (c).—

Common error: Use the radius of the mirror (0.40 m), not its diameter, in A = πr². Using the diameter makes the brightness and every later answer wrong.

135IBP2S-042-00-16Fusion in the Sun Paper 2Hard7 marks

A neutrino detector deep underground records neutrinos produced by the proton–proton chain in the Sun. Each complete chain releases 26.7 MeV and produces 2 neutrinos. Assume all of this energy contributes to the Sun's luminosity of 3.83×1026 W. Earth–Sun distance = 1.50×1011 m; 1 MeV = 1.60×10−13 J.

(a)

Calculate the number of complete proton–proton chains occurring in the Sun per second.

(2)
(b)

State the number of neutrinos emitted by the Sun per second.

(1)
(c)

Determine the number of solar neutrinos passing through each square metre at Earth per second.

(2)
(d)

Explain why neutrinos give information about the present state of the Sun's core, while sunlight does not.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Energy per chain = 26.7 × 1.60×10−13 = 4.27×10−12 J✓ 1
N = 3.83×1026 / 4.27×10−12 = 8.97×1037 ≈ 9.0×1037 s−1✓ 1
Part (b)
2 × 8.97×1037 = 1.79×1038 ≈ 1.8×1038 s−1✓ 1
Allow error carried forward from (a).—
Part (c)
The neutrinos spread over a sphere of area 4πd2 = 4π(1.50×1011)2 = 2.83×1023 m2✓ 1
Flux = 1.79×1038 / 2.83×1023 = 6.3×1014 m−2 s−1. Accept any value from 6.3×1014 to 6.4×1014; 6.4×1014 follows from the rounded value 1.8×1038.✓ 1
Allow error carried forward from the candidate's value in (b).—
Part (d)
Neutrinos interact very weakly with matter. They therefore escape the core directly and reach Earth about 8 minutes after they are produced, so they show the core as it is now.✓ 1
Photons are repeatedly absorbed and re-emitted in the dense interior. They take a very long time to reach the surface, of order 104–105 years, so sunlight reflects the state of the core long ago.✓ 1

Exam tip: In a chain of linked calculations, keep intermediate values unrounded on your calculator. Errors carried forward are allowed, but early rounding can shift your final answer outside the accepted range.