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IB Physics Higher Level · Topic E.5

E.5 Fusion and stars

99 free, high-quality past-paper-style questions on Fusion and stars for IB Physics HL, each with a full mark scheme. Covers: Stellar evolution; Wien's law from black-body spectra; Apparent brightness; and more. For when you have finished the past papers, or want to practise one topic at a time.E.5 at Standard 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.440 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1E-1A-10Stellar evolution Paper 1AEasy1 mark

The Hertzsprung–Russell diagram shows the Sun on the main sequence and four possible paths, A, B, C and D, that a star might follow after the hydrogen in its core is used up.

Which path represents the evolution of the Sun?

10⁻⁴10⁻²110²10⁴10⁶40 00020 00010 0005 0002 500surface temperature / Kluminosity / L☉ABCDSun
Hertzsprung–Russell diagram (logarithmic scales; temperature decreases to the right) with four possible paths from the Sun.
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Marking pointMarkNotes
Step 1A star of about one solar mass expands and cools at its surface to become a red giant: it moves up and to the right.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2It later loses its outer layers as a planetary nebula; the exposed core is a hot, small, faint white dwarf at the lower left.—
Step 3Only path A goes first to the upper right and then to the lower left.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: main sequence → red giant (upper right) → white dwarf (lower left).
  • BMoving up the main sequence would need a more massive star; when its core hydrogen is used up, a one-solar-mass star swells and its surface cools, so it moves to the right.
  • CThe red-giant stage is right, but the remnant is a hot white dwarf at the lower left, not a cool faint star at the lower right.
  • DThis misses the red-giant stage: the star does not go directly from the main sequence to a white dwarf.
2E-1A-21Wien's law from black-body spectra Paper 1AEasy1 mark

The graph shows the spectra of the light emitted by two stars, X and Y. Both stars may be treated as black bodies.

What is the ratio (surface temperature of X) / (surface temperature of Y)?

05001000150020002500wavelength / nm0.00.20.40.60.81.01.2intensity / arbitrary unitsXY
Intensity against wavelength for the light from stars X and Y (the heights are not to the same scale).
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Marking pointMarkNotes
Step 1Read the peak wavelengths: λmax ≈ 400 nm for X and ≈ 1000 nm for Y.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Wien's law λmaxT = 2.9 × 10−3 m K gives T ∝ 1/λmax.—
Step 3TX/TY = 1000/400 = 2.5 (about 7300 K and 2900 K).✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis takes T ∝ λmax: the ratio is inverted.
  • BThis takes the square root of 2.5, as though T ∝ 1/√λmax.
  • CCorrect: T ∝ 1/λmax.
  • DThis is 2.54, the ratio of the power emitted per unit area (σT⁴), not of the temperatures.
3E-1A-26Apparent brightness Paper 1AEasy1 mark

Star X has luminosity 4L and is at distance 2d from the Earth. Star Y has luminosity L and is at distance d.

What is the ratio (apparent brightness of X)/(apparent brightness of Y)?

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Marking pointMarkNotes
Step 1b = L/4πd².—All 3 steps must be completed — there is no mark for a part-answer.
Step 2For X: 4L/(4π × 4d²) = L/4πd².—
Step 3The same as Y — ratio 1.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AIgnores the factor 4 in luminosity and uses b ∝ 1/d: 1/2.
  • BCorrect: the factor 4 in luminosity is cancelled by (2d)² = 4d².
  • CThis uses 4/2 — inverse-square not applied.
  • DThis ignores the greater distance of X.
4E-1A-56Stellar evolution Paper 1AEasy1 mark

A main-sequence star has a mass of 15 solar masses.

Which sequence describes the evolution of the star after its core hydrogen is used up?

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Marking pointMarkNotes
Step 1A star of large mass becomes a red supergiant after the main sequence, fusing progressively heavier elements in its core.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2When the core can no longer release energy by fusion, it collapses and the star explodes as a supernova.—
Step 3The remnant core is a neutron star (or, for the most massive cores, a black hole).✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis is the path of a star of about one solar mass, such as the Sun.
  • BA supernova of a single massive star leaves a neutron star or black hole, not a white dwarf; low-mass stars do not become supernovae.
  • CCorrect: the path of a high-mass star.
  • DA planetary nebula is the gentle ejection of the envelope of a low-mass star; a neutron star forms only in the core collapse of a supernova.
5E-1A-58The Hertzsprung–Russell diagram Paper 1AEasy1 mark

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

In which region are stars whose outer layers expand and contract periodically, so that their luminosity varies?

10⁻⁴10⁻²110²10⁴10⁶40 00020 00010 0005 0002 500surface temperature / Kluminosity / L☉PQRS
Hertzsprung–Russell diagram (logarithmic scales; temperature decreases to the right) with four regions P, Q, R and S.
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Marking pointMarkNotes
Step 1Stars whose outer layers pulsate lie in the instability strip.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The instability strip is a narrow, nearly vertical band that crosses the main sequence and extends up through the giants and supergiants.—
Step 3This is region S.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AP is the main sequence, where stars fuse hydrogen in a stable equilibrium and their luminosity is constant.
  • BQ contains red giants and supergiants: large and cool, but not, in general, pulsating.
  • CR contains white dwarfs: small, hot remnants in which fusion has stopped; they cool steadily.
  • DCorrect: S is the instability strip.
6E-1A-70Stellar spectra and luminosity Paper 1AEasy1 mark

Two stars X and Y have the same radius. The wavelength at which the spectrum of Y peaks is half the wavelength at which the spectrum of X peaks.

What is (luminosity of Y)/(luminosity of X)?

Show mark scheme
Marking pointMarkNotes
Step 1Wien's law: λmaxT = constant, so halving λmax doubles the surface temperature.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2L = σ4πR²T⁴ with equal radii, so LY/LX = 2⁴ = 16.✓ 1Answer D

Answer: D  ·  2 stages of work, one mark

Every option, and why

  • AThis takes temperature to be proportional to the peak wavelength, inverting Wien's law.
  • BThis takes luminosity to be proportional to temperature instead of to T⁴.
  • CThis takes luminosity to be proportional to T², confusing the R² and T⁴ factors.
  • DCorrect: TY = 2TX and L ∝ T⁴ at fixed radius.
7E-1A-71The Hertzsprung–Russell diagram Paper 1AEasy1 mark

Stars P and Q are both on the main sequence. The mass of Q is greater than the mass of P.

Which row compares the surface temperature and the radius of Q with those of P?

Surface temperature of QRadius of Q
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Marking pointMarkNotes
Step 1On the main sequence the luminosity rises steeply with mass, so Q lies higher up the main sequence, towards the upper left of the HR diagram.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Further to the left means a higher surface temperature: Q is hotter than P.—
Step 3The main sequence crosses the lines of constant radius, from about 0.1 R☉ at its faint, cool end to about 10 R☉ at its luminous, hot end (L = σ4πR²T⁴ with L rising faster than T⁴), so Q also has the larger radius.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: a more massive main-sequence star is hotter, larger and more luminous.
  • BThis links a high surface temperature with a small radius, as for white dwarfs at the lower left; along the main sequence both increase with mass.
  • CThis places the more massive star in the cool, luminous red-giant region at the upper right, which main-sequence stars do not occupy.
  • DThis reverses the main sequence, taking the more massive stars to lie at its cool, faint lower end.
8IBP2H-043-00-01The Hertzsprung–Russell diagram Paper 1AEasy1 mark

Two stars, X and Y, have the same surface temperature. The luminosity of X is 100 times the luminosity of Y. What is \(\frac{R_X}{R_Y}\), the ratio of their radii?

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Marking pointMarkNotes
From L = 4πR2σT4, at equal temperature L ∝ R2, so RX/RY = √100 = 10. Taking the fourth root (3.2) wrongly applies the T4 dependence to radius. 100 assumes L ∝ R. 10 000 squares the ratio instead of square-rooting it.✓ 1Answer B

Answer: B

Exam tip: Hold one quantity in L = 4πR²σT⁴ fixed and write the proportionality before you substitute numbers.

9IBP2H-043-00-02Stellar equilibrium Paper 1AEasy1 mark

What prevents a main-sequence star from collapsing under its own gravity?

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Marking pointMarkNotes
Fusion in the core keeps the gas hot. The resulting thermal gas pressure and radiation pressure push outwards and balance the inward gravitational pressure. The strong force acts only over about 10−15 m, between nucleons, and cannot support a whole star. Electron degeneracy pressure supports white dwarfs, not main-sequence stars. Rotation is far too slow in most stars to matter.✓ 1Answer C

Answer: C

Common error: Do not confuse the support mechanism of a main-sequence star (thermal and radiation pressure) with that of a remnant (degeneracy pressure).

10IBP2H-043-01-03Binding energy and fusion Paper 1AEasy1 mark

Two deuterium nuclei fuse to form a helium-4 nucleus. Which statement explains why energy is released?

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Marking pointMarkNotes
Helium-4 is much more tightly bound than deuterium: its binding energy per nucleon is about 7.1 MeV, compared with about 1.1 MeV for deuterium. The total binding energy therefore increases. The product has less mass than the reactants, and this mass difference is released as energy. Option B reverses the mass change. Option C reverses the change in binding energy per nucleon. Option D is wrong because overcoming the Coulomb barrier costs energy; it is not the source of the energy released.✓ 1Answer A

Answer: A

Common error: Binding energy is not energy stored in the nucleus. A larger binding energy means a lower mass and a more stable nucleus.

11IBP2H-043-01-05Stellar spectra and luminosity Paper 1AEasy1 mark

On an HR diagram, star P lies directly above star Q. Both stars have the same surface temperature, and the luminosity of P is 104 times that of Q. What is \(\frac{R_P}{R_Q}\), the ratio of the radius of P to the radius of Q?

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Marking pointMarkNotes
L = 4πR2σT4. With equal temperatures, L ∝ R2, so RP/RQ = √(104) = 100. Option A takes a fourth root, as if L ∝ R4. Option C assumes L ∝ R. Option D squares the ratio instead of taking its square root.✓ 1Answer B

Answer: B

Exam tip: A vertical line on the HR diagram is a line of constant temperature. Along it, any difference in luminosity is due only to a difference in surface area.

12IBP2H-043-02-01Nuclear fusion Paper 1AEasy1 mark

Hydrogen nuclei in the core of a main-sequence star fuse only because the core temperature is about 107 K. What is the main reason such a high temperature is needed?

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Marking pointMarkNotes
Protons are positively charged and repel each other. Only nuclei with very large kinetic energies, which means a very high temperature, can approach closely enough for the strong nuclear force to act and fusion to occur. Ionisation happens at far lower temperatures. Gravity between nuclei does not depend on temperature. The helium nucleus has less mass than the reactants, so no thermal energy is converted into mass.✓ 1Answer B

Answer: B

Common error: Fusion releases energy because the products have less mass. High temperature is needed to overcome the Coulomb barrier, not to supply mass.

13IBP2H-043-02-02Apparent brightness Paper 1AEasy1 mark

The apparent brightness of the Sun at the Earth's orbit (1.0 AU) is 1.36 kW m−2. A solar probe is designed to operate at 0.50 AU from the Sun. What is the apparent brightness of the Sun at the probe?

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Marking pointMarkNotes
Apparent brightness is b = L/(4πd2). Halving d multiplies b by 22 = 4, giving 4 × 1.36 = 5.44 kW m−2. The value 2.72 assumes b ∝ 1/d. The values 0.34 and 0.68 wrongly assume that brightness falls as the probe moves closer.✓ 1Answer C

Answer: C

Exam tip: Use ratios with the inverse-square law: b₂/b₁ = (d₁/d₂)². The luminosity is not needed.

14IBP3H-E5-20r22-01Stellar parallax Paper 1AEasy1 mark

A robotic survey telescope measures the parallax angle of a nearby star as 0.040 arcseconds. What is the distance to the star?

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Marking pointMarkNotes
Distance in parsecs is 1 divided by the parallax angle in arcseconds, so 1/0.040 = 25 pc. 0.040 pc takes the angle itself as the distance. 2.5 pc and 250 pc come from misplacing the decimal point when dividing.✓ 1Answer B

Answer: B

Common error: The relation d = 1/p works only with d in parsecs and p in arcseconds. Distance is inversely proportional to the angle.

15IBP3H-E5-20r22-02Stellar spectra and luminosity Paper 1AEasy1 mark

Two stars A and B have the same radius. The peak wavelength in the spectrum of A is half the peak wavelength in the spectrum of B. What is the value of \(L_A/L_B\)?

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Marking pointMarkNotes
Wien's law gives T inversely proportional to peak wavelength, so TA = 2TB. With equal radii, L is proportional to T4, so LA/LB = 24 = 16. 1/16 inverts the ratio. 2 treats L as proportional to T. 4 treats L as proportional to T2.✓ 1Answer D

Answer: D

Exam tip: Wien's law links peak wavelength to temperature inversely; the Stefan–Boltzmann law then brings in the fourth power of temperature.

16IBP3H-E5-20r22-03Stellar equilibrium Paper 1AEasy1 mark

Which statement explains the dark absorption lines seen in the spectrum of a star?

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Marking pointMarkNotes
Atoms in the cooler outer atmosphere absorb photons whose energies match transitions between their energy levels, which removes those wavelengths from the continuous spectrum. Fusion in the core does not create the lines. The core emits a continuous spectrum, not selected wavelengths. Rotation can broaden lines but does not create them.✓ 1Answer A

Answer: A

Common error: Absorption lines are produced by electron transitions in atoms, not by nuclear processes.

17IBP3H-E5-20r22-08Stellar equilibrium Paper 1AEasy1 mark

Which statement describes the equilibrium of a main-sequence star?

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Marking pointMarkNotes
Fusion in the core heats the gas and produces photons, giving outward gas and radiation pressure that balances the inward gravitational force. Electrostatic repulsion between individual nuclei is not the bulk balance. The strong force is short range and attractive. Radiation pressure acts outwards, not inwards.✓ 1Answer C

Answer: C

Exam tip: Stellar equilibrium is always inward gravity versus outward pressure; a change in fusion rate shifts the balance.

18E-1A-14The Hertzsprung–Russell diagram Paper 1AMedium1 mark

A red giant and a white dwarf have the same luminosity. The surface temperature of the white dwarf is five times that of the red giant.

What is the ratio (radius of the red giant) / (radius of the white dwarf)?

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Marking pointMarkNotes
Step 1L = σ(4πR²)T⁴; equal luminosities give Rg²Tg⁴ = Rw²Tw⁴.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Rg/Rw = (Tw/Tg)² = 5².—
Step 3= 25: the cool giant must be much larger to radiate the same power.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis is the temperature ratio; radius depends on its square.
  • BCorrect: R ∝ 1/T² at fixed luminosity.
  • CThis is the cube of the temperature ratio.
  • DThis is the fourth power — the ratio of emitted power per unit area, not of radius.
19E-1A-50Stellar equilibrium Paper 1AMedium1 mark

The rate of fusion in the core of a main-sequence star briefly increases.

Which sequence describes how the star returns to equilibrium?

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Marking pointMarkNotes
Step 1More fusion releases energy faster, so the outward radiation (and gas) pressure rises above what is needed to balance gravity.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The core expands; an expanding gas does work and cools, and its density falls.—
Step 3Lower temperature and density reduce the rate of fusion, so the pressure falls back until it balances gravity again (negative feedback).✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AThe extra energy release raises the outward pressure; gravity is unchanged, so it does not "win". This sequence would also run away rather than restore equilibrium.
  • BThe direction of the imbalance is right, but an expanding core cools; a hotter core would make the change grow instead of correcting it.
  • CThe extra energy release raises the outward pressure rather than lowering it, and a contracting core heats rather than cools.
  • DCorrect: expansion lowers the core temperature and density, which slows fusion — a self-regulating balance.
20E-1A-52Nuclear fusion Paper 1AMedium1 mark

Three changes to the core of a star are proposed. Each change is made on its own, with everything else kept the same:

I. increasing the temperature of the core
II. increasing the density of the core
III. replacing the hydrogen nuclei by nuclei of larger proton number

Which changes would increase the rate of fusion in the core?

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Marking pointMarkNotes
Step 1I: at a higher temperature the nuclei have greater kinetic energies, so more of them overcome the electric repulsion and get close enough to fuse.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2II: at a higher density the nuclei collide more often, so more fusion reactions occur per second.—
Step 3III: nuclei with more protons repel each other more strongly, so at the same temperature fewer get close enough to fuse — the rate decreases.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • ADensity also matters: more nuclei per unit volume means more frequent collisions.
  • BTemperature also matters: it decides what fraction of collisions overcome the electric repulsion.
  • CCorrect: higher temperature and higher density both raise the rate; a larger nuclear charge lowers it.
  • DIII would reduce the rate, because the electric repulsion between the nuclei is larger.
21E-1A-53Binding energy and fusion Paper 1AMedium1 mark

In a young star, deuterium fuses with protons: 21H + 11H → 32He + γ.

The binding energy per nucleon is 1.11 MeV for 21H and 2.57 MeV for 32He.

What is the energy released in one reaction?

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Marking pointMarkNotes
Step 1Total binding energy = (binding energy per nucleon) × (number of nucleons). A single proton has zero binding energy.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Reactants: 2 × 1.11 = 2.22 MeV. Product: 3 × 2.57 = 7.71 MeV.—
Step 3Energy released = increase in total binding energy = 7.71 − 2.22 = 5.49 MeV.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis is 2.57 − 1.11, the difference of the values per nucleon; the totals depend on the nucleon numbers (3 and 2).
  • BCorrect: 3 × 2.57 − 2 × 1.11 = 5.49 MeV.
  • CThis is 7.71 − 1.11: the deuteron's binding energy per nucleon has not been multiplied by its 2 nucleons.
  • DThis is 7.71 + 2.22: the binding energies have been added instead of subtracted.
22E-1A-54Stellar parallax Paper 1AMedium1 mark

Observed from the Earth six months apart, a nearby star appears to move by a total of 0.032 arc-second against the very distant background stars.

Which row gives the distance to the star?

Distance / pcDistance / ly
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Marking pointMarkNotes
Step 1The parallax angle p is half the total angular shift, because the baseline is 1 AU (the radius, not the diameter, of the Earth's orbit): p = 0.016 arc-second.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2d(pc) = 1/p(arc-second) = 1/0.016 = 62.5 pc ≈ 63 pc.—
Step 31 pc = 3.26 ly, so d = 62.5 × 3.26 = 204 ly.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AUses the full shift 0.032 as p (giving 31 pc) and then divides by 3.26 instead of multiplying.
  • BUses the full shift 0.032 arc-second as the parallax angle: 1/0.032 = 31 pc = 102 ly.
  • CThe distance in parsecs is right but it has been divided by 3.26: 62.5/3.26 = 19.
  • DCorrect: p = 0.016 arc-second, 62.5 pc = 204 ly.
23E-1A-57The Hertzsprung–Russell diagram Paper 1AMedium1 mark

The Hertzsprung–Russell diagram shows the Sun and four stars W, X, Y and Z.

Which star has the largest radius?

10−410−210010210410640000200001000050002500surface temperature / K (logarithmic scale)luminosity / L☉ (logarithmic scale)main sequenceSunWXYZ
Hertzsprung–Russell diagram (logarithmic scales on both axes) showing the Sun and stars W, X, Y and Z.
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Marking pointMarkNotes
Step 1From L = 4πR²σT⁴, R ∝ √L/T².—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Reading the diagram: W ≈ 104 L☉ at 25 000 K gives R ≈ √104 × (5800/25 000)² ≈ 5 R☉; X ≈ 103 L☉ at 3500 K gives R ≈ √103 × (5800/3500)² ≈ 90 R☉.—
Step 3Y (≈ 0.01 R☉) and Z (≈ 0.3 R☉) are much smaller, so X — a red giant at the upper right — is the largest.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AW is the most luminous star, but it is very hot, so its high luminosity comes mainly from the T⁴ factor; its radius is only about 5 R☉.
  • BCorrect: cool yet luminous, so its surface area must be very large (about 90 R☉).
  • CY is the hottest star, but it is a white dwarf (about 0.01 R☉): high temperature with very low luminosity means a tiny surface.
  • DZ is the coolest star, but it is a faint main-sequence star (about 0.3 R☉); low temperature alone does not mean a large radius.
24E-1A-59The Hertzsprung–Russell diagram Paper 1AMedium1 mark

On a Hertzsprung–Russell diagram, luminosity is plotted on a logarithmic scale against surface temperature on a logarithmic scale that decreases to the right. Lines of constant radius are straight lines on this diagram.

Which row describes the lines of constant radius?

Change in luminosity when the temperature halves along one lineLines for larger radii lie towards the
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Marking pointMarkNotes
Step 1At constant R, L = 4πR²σT⁴ gives L ∝ T⁴: halving T divides L by 2⁴ = 16.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2At a fixed temperature, a larger radius gives a larger luminosity (L ∝ R²), so the line lies higher.—
Step 3Higher on the diagram and, equivalently, towards the cool (right-hand) side: larger radii lie towards the top right, where the giants and supergiants are.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AUses L ∝ T² (confusing the dependence on radius with that on temperature) and places large stars with the white dwarfs.
  • BThe direction is right but the factor uses L ∝ T² instead of T⁴.
  • CThe factor is right, but the bottom left is where the smallest stars (white dwarfs) lie.
  • DCorrect: L ∝ T⁴ along a line; larger radii towards the top right.
25IBP2H-043-00-03Stellar evolution Paper 1AMedium1 mark

For main-sequence stars, luminosity L is proportional to M3.5, where M is the mass. Assume the fraction of a star's mass that undergoes fusion is the same for all main-sequence stars. A star has a mass 4 times that of the Sun. What is the ratio of its main-sequence lifetime to that of the Sun?

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Marking pointMarkNotes
Lifetime ∝ (fuel)/(rate of use) ∝ M/L ∝ M/M3.5 = M−2.5. This gives 4−2.5 = 1/(42 × √4) = 1/32. The option 1/128 = 4−3.5 ignores the larger fuel supply. The option 4 assumes lifetime ∝ fuel only. The option 1/16 uses M−2.✓ 1Answer B

Answer: B

Exam tip: Lifetime ∝ M⁻².⁵ follows from dividing mass by luminosity. Derive it rather than memorising it.

26IBP2H-043-00-04Stellar evolution Paper 1AMedium1 mark

After a supernova, the collapsed core left behind has a mass of 2.5 solar masses. Take the Chandrasekhar limit as 1.4 solar masses and the Oppenheimer–Volkoff limit as about 3 solar masses. What is the remnant?

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Marking pointMarkNotes
The core mass exceeds the Chandrasekhar limit, so electron degeneracy pressure cannot support it. It is below the Oppenheimer–Volkoff limit, so neutron degeneracy pressure can halt the collapse. The remnant is therefore a neutron star. A white dwarf would require a core below 1.4 solar masses. A black hole would require a core above about 3 solar masses. A planetary nebula is ejected gas, not a core remnant.✓ 1Answer B

Answer: B

Common error: Apply the limits to the mass of the remnant core, not to the star's initial mass.

27IBP2H-043-00-05The Hertzsprung–Russell diagram Paper 1AMedium1 mark

A star is plotted in the bottom-left region of a Hertzsprung–Russell diagram. What can be deduced about the star?

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Marking pointMarkNotes
On an HR diagram, temperature increases to the left, so bottom-left means high T and low L. Since L = 4πR2σT4, a low L at high T requires a small R, and so a small surface area. This is a white dwarf. A white dwarf has no fusion and simply radiates stored thermal energy. Low luminosity does not indicate large mass.✓ 1Answer B

Answer: B

Common error: The temperature axis of an HR diagram runs from hot on the left to cool on the right.

28IBP2H-043-00-06Stellar evolution Paper 1AMedium1 mark

Fusion in the core of a massive star produces successively heavier elements, but stops once the core is mainly iron. Why?

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Marking pointMarkNotes
The binding energy per nucleon curve peaks near iron-56. Fusing lighter nuclei increases binding energy per nucleon and releases energy, but fusing iron would decrease it. Iron fusion therefore absorbs energy and cannot support the core. The melting point of iron is irrelevant in a plasma. Hydrogen is often still present in outer shells. Speed alone is not the limiting factor.✓ 1Answer C

Answer: C

Exam tip: Link 'energy released' to an increase in binding energy per nucleon. This explains both why fusion stops at iron and why fission of heavy nuclei releases energy.

29IBP2H-043-01-01Stellar parallax Paper 1AMedium1 mark

A space telescope orbits the Sun in a circular orbit of radius 1.5 AU. Observations made six months apart show that a distant star has a parallax angle of p arcseconds. The parallax angle is half the total angular shift. What is the distance to the star in parsecs?

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Marking pointMarkNotes
The parsec is defined for a baseline of 1 AU, so d(pc) = 1/p only applies to observations from Earth's orbit. For small angles p ≈ (baseline)/d, so with a baseline of 1.5 AU the same star produces a parallax angle 1.5 times larger. The distance is therefore d = 1.5/p parsecs. Option A ignores the larger baseline. Option C reverses the effect of the baseline. Option D wrongly treats distance as proportional to the parallax angle.✓ 1Answer B

Answer: B

Common error: d = 1/p is not a universal law. It is tied to a 1 AU baseline. A larger baseline gives a larger parallax angle for the same star.

30IBP2H-043-01-02Stellar spectra and luminosity Paper 1AMedium1 mark

Star A has a black-body spectrum that peaks at 300 nm. Star B has a black-body spectrum that peaks at 900 nm. The two stars have the same radius. What is \(\frac{L_A}{L_B}\), the ratio of the luminosity of A to the luminosity of B?

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Marking pointMarkNotes
Wien's law gives T ∝ 1/λmax, so TA/TB = 900/300 = 3. For equal radii L = 4πR2σT4 gives L ∝ T4, so LA/LB = 34 = 81. Option A uses L ∝ T. Option B uses L ∝ T2, which confuses the T4 dependence with the R2 dependence. Option D inverts the Wien relation.✓ 1Answer C

Answer: C

Exam tip: A shorter peak wavelength means a hotter star. Find the temperature ratio first, then raise it to the fourth power.

31IBP2H-043-01-04Nuclear fusion Paper 1AMedium1 mark

An interstellar gas cloud of mass M, radius R and temperature T is close to the Jeans criterion for collapse. Which single change makes gravitational collapse more likely?

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Marking pointMarkNotes
The cloud collapses when the magnitude of its gravitational potential energy (∝ GM2/R) exceeds the random kinetic energy of its particles (∝ NkT ∝ MT). Decreasing T reduces the kinetic energy and leaves the gravitational energy unchanged, so the criterion is easier to satisfy. Option A raises the kinetic energy. Option B lowers the gravitational energy (∝ M2) faster than the kinetic energy (∝ M). Option C makes the cloud less dense and lowers the magnitude of the gravitational energy.✓ 1Answer D

Answer: D

Exam tip: Cold, dense, massive clouds collapse. A cloud of fixed mass becomes more likely to collapse if it is cooled or compressed.

32IBP2H-043-01-07Stellar evolution Paper 1AMedium1 mark

After a supernova, the remnant core of a massive star has a mass of 1.8 M☉. No fusion takes place in the remnant. What is the remnant, and what prevents it from collapsing further?

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Marking pointMarkNotes
1.8 M☉ is above the Chandrasekhar limit (about 1.4 M☉), so electron degeneracy pressure cannot support the core. It is below the Oppenheimer–Volkoff limit (commonly quoted as about 3 M☉), so neutron degeneracy pressure can stop further collapse. The remnant is therefore a neutron star, which makes B correct.
Option A ignores the Chandrasekhar limit.
Option C assumes that every supernova core becomes a black hole.
Option D is wrong because the stem states that no fusion occurs in the remnant, so there is no radiation pressure from fusion.
✓ 1Answer B

Answer: B

Exam tip: Classify a stellar remnant by the mass of the remnant core, not the original mass of the star. Compare it with roughly 1.4 M☉ (Chandrasekhar limit) and roughly 3 M☉ (Oppenheimer–Volkoff limit).

33IBP2H-043-01-08Apparent brightness Paper 1AMedium1 mark

Stars X and Y have the same luminosity. The parallax angle of X is four times the parallax angle of Y. What is \(\frac{b_X}{b_Y}\), the ratio of the apparent brightness of X to the apparent brightness of Y?

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Marking pointMarkNotes
Distance is proportional to 1/p, so X is four times closer than Y. Apparent brightness b = L/(4πd2) ∝ 1/d2 ∝ p2, so bX/bY = 42 = 16. Option A treats a larger parallax as a larger distance. Option B makes that error and also uses an inverse rather than an inverse-square relation. Option C forgets to square.✓ 1Answer D

Answer: D

Common error: A larger parallax angle means the star is closer, not further away.

34IBP2H-043-02-03Apparent brightness Paper 1AMedium1 mark

Stars X and Y have the same surface temperature. The radius of X is 3 times the radius of Y. X is 2 times as far from Earth as Y. What is \(\frac{b_X}{b_Y}\), the ratio of the apparent brightness of X to that of Y?

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Marking pointMarkNotes
L = 4πR2σT4. With equal T, LX/LY = 32 = 9. Then b = L/(4πd2), so bX/bY = 9/22 = 9/4. The option 3/4 forgets both squares. The option 9/2 forgets to square the distance. The option 81/4 wrongly uses R4 in place of T4.✓ 1Answer A

Answer: A

Common error: The fourth power in the Stefan–Boltzmann law belongs to temperature. Radius appears squared, because L depends on surface area.

35IBP2H-043-02-04The Hertzsprung–Russell diagram Paper 1AMedium1 mark

The hydrogen Balmer absorption lines are produced when electrons in hydrogen atoms in the n = 2 level absorb photons. These lines are weak in the spectra of the hottest O-type stars (T > 30 000 K). What is the best explanation?

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Marking pointMarkNotes
At such high temperatures collisions and radiation ionise most hydrogen atoms. Very few atoms are left with an electron in n = 2, so Balmer absorption is weak. Hydrogen is still the most abundant element, so the first option is wrong. Electrons sitting in the ground state explains weak Balmer lines in cool stars, not hot ones. A hot star still emits strongly at visible wavelengths, even though its peak is in the ultraviolet.✓ 1Answer D

Answer: D

Exam tip: Balmer lines are strongest at about 10 000 K. Below that, too few electrons are in n = 2. Above that, too many atoms are ionised.

36IBP2H-043-02-05Stellar spectra and luminosity Paper 1AMedium1 mark

Star P has its peak emission at 400 nm. Star Q has its peak emission at 800 nm. The two stars have the same radius. What is \(\frac{L_P}{L_Q}\), the ratio of the luminosity of P to that of Q?

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Marking pointMarkNotes
Wien's law gives λmaxT = constant, so TP = 2TQ. With equal radii, L ∝ T4, so LP/LQ = 24 = 16. The option 2 uses the temperature ratio alone. The option 4 squares the ratio instead of raising it to the fourth power. The option 1/16 assumes temperature is proportional to wavelength.✓ 1Answer B

Answer: B

Common error: A longer peak wavelength means a cooler star. Temperature is inversely proportional to λmax.

37IBP2H-043-02-07Nuclear fusion Paper 1AMedium1 mark

Which statement about the CNO cycle in stars is correct?

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Marking pointMarkNotes
In the CNO cycle, carbon-12 acts as a catalyst. It captures protons, becomes nitrogen and oxygen isotopes, and is returned at the end. The net result is 4 1H → 4He, the same as the proton–proton chain. The heavier nuclei have larger charges, so the Coulomb barrier is higher. This means the cycle needs a higher temperature and dominates in stars more massive than the Sun. The β+ decays of N-13 and O-15 emit positrons and neutrinos.✓ 1Answer A

Answer: A

Common error: The catalyst nuclei in the CNO cycle are not net products. Count what enters and leaves the whole cycle.

38IBP2S-042-04-02Stellar evolution Paper 1AMedium1 mark

After a supernova, the collapsed core left behind by a massive star has a mass of 1.8 solar masses. Take the Chandrasekhar limit as about 1.4 solar masses and the Oppenheimer–Volkoff limit as about 3 solar masses. Which row correctly identifies the remnant and what prevents its further collapse?

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Marking pointMarkNotes
1.8 solar masses is above the Chandrasekhar limit (about 1.4 solar masses), so electron degeneracy pressure cannot support the core. It is below the Oppenheimer–Volkoff limit (about 3 solar masses), so neutron degeneracy pressure can halt the collapse. The remnant is therefore a neutron star.
The other options are wrong:
• A white dwarf would need a core below 1.4 solar masses.
• A black hole forms only when the core is above the Oppenheimer–Volkoff limit.
• No fusion takes place in a neutron star, so radiation pressure from fusion cannot be what supports it.
✓ 1Answer B

Answer: B

Exam tip: Classify a remnant by the mass of the core left behind, compared with the two limits. Do not use the star's original mass.

39IBP2S-042-04-03Stellar evolution Paper 1AMedium1 mark

A main-sequence star has a mass four times that of the Sun. Assume the mass–luminosity relation L ∝ M3.5. What is the ratio of this star's main-sequence lifetime to the Sun's main-sequence lifetime?

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Marking pointMarkNotes
Lifetime is proportional to the fuel available divided by the rate of use, so T ∝ M/L ∝ M/M3.5 = M−2.5. This gives 4−2.5 = 1/(42 × √4) = 1/32. The option 1/4 ignores the steep rise in luminosity. The option 1/128 uses M−3.5 and forgets that the star also has more fuel. The option 4 assumes that more fuel means a longer life.✓ 1Answer B

Answer: B

Common error: More massive stars do have more fuel. However, their luminosity rises as M³.⁵, so they burn through it far faster and live much shorter lives.

40IBP3H-E5-20r22-04Binding energy and fusion Paper 1AMedium1 mark

The binding energy per nucleon is 1.11 MeV for hydrogen-2, 2.83 MeV for hydrogen-3 and 7.07 MeV for helium-4. In the fusion reaction hydrogen-2 + hydrogen-3 → helium-4 + neutron, what is the energy released?

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Marking pointMarkNotes
Total binding energy of helium-4 is 4 × 7.07 = 28.28 MeV; the reactants have 2 × 1.11 + 3 × 2.83 = 10.71 MeV. The release is 28.28 − 10.71 = 17.57 MeV. 3.13 MeV subtracts per-nucleon values without multiplying by nucleon number. 10.71 MeV is the binding energy of the reactants only. 28.28 MeV is the binding energy of helium-4 only.✓ 1Answer C

Answer: C

Common error: Energy released equals the increase in total binding energy, so multiply each binding energy per nucleon by the nucleon number first.

41IBP3H-E5-20r22-05Stellar spectra and luminosity Paper 1AMedium1 mark

A star has the same surface temperature as the Sun and a luminosity of 100 times that of the Sun. What is its radius compared with the Sun's radius?

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Marking pointMarkNotes
At equal temperature L is proportional to R², so R is proportional to the square root of L: √100 = 10. 3.2 comes from using L proportional to R4. 50 and 100 treat L as directly proportional to R.✓ 1Answer B

Answer: B

Exam tip: Use L = 4πσR²T⁴ and cancel whichever quantity is the same for both stars.

42IBP3H-E5-20r22-06Conditions for fusion Paper 1AMedium1 mark

Why must the core of a star reach a very high temperature before hydrogen fusion can occur?

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Marking pointMarkNotes
Nuclei repel each other electrostatically, and only when very close does the short-range strong force dominate; high temperature gives enough kinetic energy to get that close. Nuclei are not split first. The strong force has no temperature threshold. Binding energy per nucleon of the products is a result of fusion, not a requirement.✓ 1Answer D

Answer: D

Common error: Temperature is needed for kinetic energy to beat Coulomb repulsion; it does not change the strong force itself.

43E-1A-09Nuclear fusion Paper 1AHard1 mark

The core of a contracting protostar is modelled as a fixed mass of ideal gas. As the protostar contracts, the radius of the core halves and the temperature of the core becomes three times larger. The number of particles in the core does not change.

By what factor does the pressure in the core change?

Show mark scheme
Marking pointMarkNotes
Step 1PV = NkBT, so P ∝ T/V for a fixed number of particles.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Halving the radius divides the volume by 2³ = 8; tripling T multiplies P by 3.—
Step 3P changes by 3 × 8 = 24: the rise in density and temperature is what makes fusion possible once the core is hot and dense enough.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AThis multiplies by the volume factor (3 × 1/8) instead of dividing by it — pressure taken as proportional to volume.
  • BThis takes the volume as proportional to the radius (3 × 2).
  • CThis takes the volume as proportional to the radius squared, as though it were an area (3 × 4).
  • DCorrect: V ∝ R³, so P rises by 3 × 2³ = 24.
44E-1A-51Stellar parallax Paper 1AHard1 mark

A star has luminosity L and its apparent brightness at the Earth is b. One parsec is equal to P metres.

What parallax angle, in arc-seconds, would be measured for this star from the Earth?

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Marking pointMarkNotes
Step 1b = L/4πd² gives d = √(L/4πb) metres.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2In parsecs: d = √(L/4πb)/P.—
Step 3p (arc-second) = 1/d(pc) = P√(4πb/L).✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: d from the inverse-square law, converted to parsecs, then inverted.
  • BThis is the distance in parsecs; it has not been inverted to give the parallax angle.
  • CThe factor 4π has been omitted from b = L/4πd².
  • DThe square root has been omitted: this is proportional to 1/d², not 1/d.
45E-1A-55Stellar parallax Paper 1AHard1 mark

A proposed space telescope would orbit the Sun at a radius of 5.2 AU. It measures the parallax angle of a star, using the radius of its own orbit as the baseline, as 0.13 arc-second.

What is the distance to the star?

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Marking pointMarkNotes
Step 1By definition, a star at 1 pc shows a parallax of 1 arc-second for a baseline of 1 AU. For small angles p = (baseline)/d.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2With a baseline of 5.2 AU, the parallax of a star at distance d pc is 5.2/d arc-second.—
Step 3d = 5.2/0.13 = 40 pc.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AThis is 1/(0.13 × 5.2): the baseline has been used to divide rather than multiply — a longer baseline gives a larger angle for the same star.
  • BThis is 1/0.13: the Earth-based relation assumes a 1 AU baseline, so it underestimates the distance by the factor 5.2.
  • CThis is 5.2/(2 × 0.13): the parallax angle has been treated as a full shift and halved again.
  • DCorrect: d(pc) = (baseline in AU)/p(arc-second) = 5.2/0.13 = 40 pc.
46IBP2H-043-00-07Apparent brightness Paper 1AHard1 mark

Star A has twice the surface temperature of star B. The luminosity of A is one quarter of the luminosity of B. What is \(\frac{R_A}{R_B}\)?

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Marking pointMarkNotes
R2 ∝ L/T4, so (RA/RB)2 = (1/4)/24 = 1/64. Hence RA/RB = 1/8. The option 1/16 forgets the square root. The option 1/4 uses T2 instead of T4. The option 1/2 ignores the temperature difference.✓ 1Answer C

Answer: C

Exam tip: Rearrange to R ∝ √L / T² first, then substitute the ratios.

47IBP2H-043-00-08Stellar spectra and luminosity Paper 1AHard1 mark

Over a stage of its evolution, a star's luminosity stays constant while the wavelength at which its spectrum peaks increases by 20%. By what factor does its radius change?

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Marking pointMarkNotes
Wien's law gives T ∝ 1/λmax, so the new temperature is T/1.2. At constant L, R ∝ T−2, so R increases by a factor of 1.22 = 1.44. The option 1.2 assumes R ∝ λ. The option 2.07 = 1.24 forgets that R is squared. The option 0.69 inverts the ratio.✓ 1Answer C

Answer: C

Common error: A longer peak wavelength means a cooler star. At fixed luminosity, a cooler star must be larger.

48IBP2H-043-01-06Stellar evolution Paper 1AHard1 mark

For main-sequence stars, luminosity is proportional to M3.5, where M is the mass of the star. The Sun's main-sequence lifetime is about 1×1010 years. Assume the same fraction of each star's mass is fused. What is the approximate main-sequence lifetime of a star of mass 4 M☉?

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Marking pointMarkNotes
Lifetime ∝ (fuel available)/(rate of use) ∝ M/L ∝ M/M3.5 = M−2.5. Since 42.5 = 32, the lifetime is 1×1010/32 ≈ 3.1×108 years. Option A divides by 43.5, which ignores the extra fuel in a heavier star. Option C divides by 41.5, which is an algebra slip in the exponent. Option D assumes the lifetime is inversely proportional to mass.✓ 1Answer B

Answer: B

Common error: A star of greater mass has more fuel, but its luminosity rises far faster than its mass. The lifetime therefore scales as M⁻².⁵, not M⁻³.⁵.

49IBP2H-043-02-06Stellar evolution Paper 1AHard1 mark

For main-sequence stars, luminosity L is related to mass M by L ∝ M3.5. The Sun's main-sequence lifetime is T☉. What is the best estimate of the main-sequence lifetime of a star of mass 2M☉?

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Marking pointMarkNotes
Lifetime is proportional to the fuel available (∝ M) divided by the rate of use (∝ L). So T ∝ M/M3.5 = M−2.5. Then 2−2.5 = 1/(4√2) ≈ 0.18. The option 0.09 uses M−3.5 and ignores the larger fuel supply. The option 0.5 uses T ∝ 1/M. The option 5.7 inverts the relationship.✓ 1Answer C

Answer: C

Exam tip: Write lifetime ∝ (energy available)/(power output) = M/L before substituting the mass–luminosity relation.

50IBP2H-043-02-08Nuclear fusion Paper 1AHard1 mark

An experimental fusion reactor uses the reaction 2H + 3H → 4He + n, which releases 17.6 MeV. The reactants have negligible momentum. Which is the best estimate of the kinetic energy of the neutron?

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Marking pointMarkNotes
The initial momentum is zero, so the helium nucleus and the neutron have equal and opposite momenta p. Since Ek = p2/2m, kinetic energy is inversely proportional to mass. The mass ratio mHe : mn ≈ 4 : 1, so the neutron gets 4/5 of the energy: 0.8 × 17.6 ≈ 14.1 MeV. The value 3.5 MeV is the helium share. The value 8.8 MeV assumes an equal split. The value 17.6 MeV ignores the helium nucleus.✓ 1Answer D

Answer: D

Exam tip: In a two-body break-up from rest, the lighter particle always carries the larger share of the kinetic energy.

51IBP2S-042-00-08Stellar evolution Paper 1AHard1 mark

A main-sequence star has twice the mass of the Sun. For main-sequence stars, \(L \propto M^{3.5}\). Assume the main-sequence lifetime is proportional to the available fuel divided by the luminosity. What is \(\frac{\text{lifetime of the star}}{\text{lifetime of the Sun}}\)?

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Marking pointMarkNotes
Lifetime \(T \propto \frac{M}{L} \propto \frac{M}{M^{3.5}} = M^{-2.5}\).
So \(T/T_\odot = 2^{-2.5} = \frac{1}{4\sqrt{2}} \approx \frac{1}{5.7} \approx 0.18\). D is correct.
5.7 has the sign of the power wrong (more fuel does not mean a longer life). 0.5 assumes L ∝ M. 0.088 uses \(M^{-3.5}\) and ignores the extra fuel.
✓ 1Answer D

Answer: D

Common error: More massive stars have more fuel but burn it far faster. Including the mass as the fuel supply gives T ∝ M⁻².⁵, not M⁻³.⁵.

52IBP3H-E5-20r22-07Apparent brightness Paper 1AHard1 mark

Star X has twice the luminosity of star Y. The parallax angle of X is 0.050 arcseconds and that of Y is 0.020 arcseconds. What is the value of the ratio of the apparent brightness of X to the apparent brightness of Y?

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Marking pointMarkNotes
Brightness b = L/(4πd²) and d is proportional to 1/p, so b is proportional to L p². The ratio is 2 × (0.050/0.020)² = 2 × 6.25 = 12.5. 5.0 forgets to square the distance ratio. 0.32 inverts the parallax ratio. 3.1 divides by the luminosity ratio instead of multiplying.✓ 1Answer A

Answer: A

Common error: A larger parallax angle means a smaller distance, and brightness depends on the inverse square of distance.

53E-1B-05Stellar parallax Paper 1BEasy6 marks

An astronomer studies six stars that move together through space as a group. All six have identical spectra, so she proposes that they have the same luminosity L. For each star a space telescope measures the parallax angle p and a calibrated photometer measures the apparent brightness b. If the luminosity is the same for every star, then b = Lp²/(4πK²), where p is in arc-seconds and K = 3.09 × 1016 m is the number of metres in one parsec.

The graph shows b against p² with a line of best fit through the origin. (1 pc = 3.26 ly)

p / arcsecp² / 10−4 arcsec²b / 10−10 W m−2
0.04520.251.706
0.03612.961.065
0.0309.000.760
0.0245.760.477
0.0183.240.275
0.0121.440.119
0246810121416182022p² / 10⁻⁴ arcsec²0.00.20.40.60.81.01.21.41.61.8b / 10⁻¹⁰ W m⁻²
Apparent brightness against p² (graph drawn to scale)
(a)
(i)

Test the astronomer's proposal using the data for three of the stars.

(2)
(b)
(i)

Determine L using the gradient of the graph.

(2)
(c)
(i)

A seventh star with the same spectrum has b = 0.85 × 10−10 W m−2. Assuming that it belongs to the group, predict its distance in light-years.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
b/p² for three stars, e.g. 1.706/20.25 = 0.0842; 0.760/9.00 = 0.0844; 0.119/1.44 = 0.0826 (× 10−6 W m−2 arcsec−2)✓ 1Any three stars, preferably spread across the range.
The values agree to within about 2 %, so the data support equal luminosity✓ 1The conclusion must be consistent with the candidate's ratios.
Part (b)(i)
Gradient = 8.4 × 10−8 W m−2 arcsec−2✓ 1Accept 8.2–8.6 × 10−8. Allow ECF from (a) if the mean ratio is used.
L = 4πK² × gradient = 4π × (3.09 × 1016)² × 8.4 × 10−8 = 1.0 × 1027 W✓ 1Allow ECF from the gradient. Accept 0.98–1.04 × 1027 W.
Part (c)(i)
p = √(b/gradient) = √(0.85 × 10−10/8.4 × 10−8) = 0.0319 arcsec✓ 1Allow ECF from (b). ALT: d = √(L/4πb) = 9.7 × 1017 m.
d = 1/p = 31 pc = 102 ly✓ 1Accept 98–105 ly.

Answers: (b)(i) 1.0 × 1027 W  ·  (c)(i) 102 ly (the remaining parts are explanations — see the table above)

54E-1B-18Stellar properties Paper 1BMedium7 marks

An astronomer determines the luminosity L and the surface temperature T of six white dwarfs of similar mass. The luminosities come from measurements of apparent brightness and parallax, and the temperatures from the stellar spectra. She suspects that the six stars have the same radius, and plots lg(L/L☉) against lg(T/K). The graph shows the data with a line of best fit.

(L☉ = 3.85 × 1026 W, R☉ = 6.96 × 108 m, σ = 5.67 × 10−8 W m−2 K−4)

T / KL / 10−3 L☉lg(T/K)lg(L/L☉)
6 0000.203.778−3.70
8 0000.563.903−3.25
11 0002.34.041−2.64
15 0007.14.176−2.15
20 000254.301−1.60
27 000744.431−1.13
3.73.83.94.04.14.24.34.44.5lg(T / K)-4.0-3.5-3.0-2.5-2.0-1.5-1.0lg(L / L☉)
lg(L/L☉) against lg(T/K) (graph drawn to scale)
(a)
(i)

Determine the gradient of the line of best fit.

(2)
(b)
(i)

Explain what your answer to (a) suggests about the radii of the white dwarfs.

(1)
(c)
(i)

Determine, using the line, the common radius of the white dwarfs.

(2)
(d)
(i)

Each luminosity is uncertain by ±10 % and each temperature by ±3 %. Determine the absolute uncertainty in a radius calculated from the data for a single star.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Two well-separated points on the line, e.g. (3.75, −3.83) and (4.45, −1.04)✓ 1Points must be read from the line, not the table.
Gradient = 4.0✓ 1Accept 3.9–4.1; no unit.
Part (b)(i)
L = 4πR²σT⁴ gives lg L = 4 lg T + lg(4πR²σ): a gradient of 4 (with all points on one line) means lg(4πR²σ) is the same for every star, i.e. the radii are equal✓ 1Allow ECF from (a). Conclusion must match the candidate's gradient.
Part (c)(i)
At lg(T/K) = 4.00 the line gives lg(L/L☉) = −2.83, so L = 1.48 × 10−3 × 3.85 × 1026 = 5.7 × 1023 W✓ 1Any point on the line. Using a single table point instead of the line: [1 max]. Allow ECF from (a) if the line's equation is used.
R = √(L/4πσT⁴) = 8.9 × 106 m (0.013 R☉)✓ 1Accept 8.4–9.4 × 106 m.
Part (d)(i)
R ∝ L½T−2, so %R = ½ × 10 + 2 × 3 = 11 %✓ 1
ΔR = 0.11 × 8.9 × 106 = ±1 × 106 m✓ 1Allow ECF from (c). 1 s.f. required, i.e. (9 ± 1) × 106 m.

Answers: (a)(i) 4.0  ·  (c)(i) 8.9 × 106 m  ·  (d)(i) ±1 × 106 m (the remaining parts are explanations — see the table above)

55E-1B-19Stellar parallax Paper 1BMedium6 marks

A nearby star lies in the plane of the Earth's orbit. Over six months an astronomer measures its angular position x relative to a reference point fixed by very distant background stars; the time t is measured in months from the moment when x is greatest. The motion is modelled by x = x0 + p cos(2πt/12), where x0 is a constant and p is the parallax angle. Each value of x is uncertain by ±0.8 × 10−3 arc-second.

The graph shows x against cos(2πt/12) with the line of best fit (solid) and the lines of maximum and minimum gradient (dashed).

t / month0123456
cos(2πt/12)1.0000.8660.5000.000−0.500−0.866−1.000
x / 10−3 arcsec14.812.48.41.4−4.3−8.8−11.1
-1.00-0.75-0.50-0.250.000.250.500.751.00cos(2πt/12)-14-10-6-226101418x / 10⁻³ arcsec
Angular position against cos(2πt/12) (graph drawn to scale)
(a)
(i)

Determine p from the line of best fit.

(2)
(b)
(i)

Use the dashed lines to determine the absolute uncertainty in p.

(2)
(c)
(i)

Determine the distance to the star in parsec, with its absolute uncertainty.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Gradient of x against cos(2πt/12) is p: e.g. (14.5 − (−10.8))/2.00✓ 1The intercept is x0.
p = 12.7 × 10−3 arcsec✓ 1Accept 12.4–12.9 × 10−3 arcsec.
Part (b)(i)
Gradients of the dashed lines: 13.2 and 12.1 (× 10−3 arcsec)✓ 1Accept ±0.2 on each.
Δp = (13.2 − 12.1)/2 = ±0.5 × 10−3 arcsec✓ 1Allow ECF from (a) if the larger difference from the best-fit value is used instead. Accept ±0.4 to ±0.6 × 10−3.
Part (c)(i)
d = 1/p = 1/(12.65 × 10−3) = 79 pc✓ 1Allow ECF from (a).
Δd/d = Δp/p = 4 %, so d = (79 ± 3) pc✓ 1Allow ECF from (b). Uncertainty to 1 s.f.; accept ±2 to ±4 pc.

Answers: (a)(i) 12.7 × 10−3 arcsec  ·  (b)(i) ±0.5 × 10−3 arcsec  ·  (c)(i) (79 ± 3) pc (the remaining parts are explanations — see the table above)

56E-1B-20Stellar spectra Paper 1BMedium6 marks

A spectrometer on a telescope records the continuous spectrum of a nearby star. The relative intensity is measured at 13 wavelengths λ between 350 nm and 950 nm, and a smooth best-fit curve is drawn through the points. The broad, flat top of the curve makes the peak hard to locate precisely. Interferometer measurements give the radius of the star as (0.84 ± 0.03) R☉.

The graph shows the data and the curve. (R☉ = 6.96 × 108 m; σ = 5.67 × 10−8 W m−2 K−4)

λ / nm350400450500550600650700750800850900950
Relative intensity5372889610010095918377706457
3004005006007008009001000λ / nm0102030405060708090100110relative intensity
Continuous spectrum of the star with a best-fit curve (graph drawn to scale)
(a)
(i)

Estimate, using the graph, the wavelength at which the intensity is greatest, with its uncertainty.

(2)
(b)
(i)

Determine the surface temperature of the star, with its absolute uncertainty.

(2)
(c)
(i)

Determine the luminosity of the star, with its absolute uncertainty.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
λmax ≈ 560 nm (centre of the flat top of the curve)✓ 1Accept 550–575 nm.
Uncertainty ±20 nm (the range over which the curve is level within the scatter)✓ 1Accept ±10 to ±30 nm.
Part (b)(i)
T = 2.9 × 10−3/(560 × 10−9) = 5.2 × 103 K✓ 1Allow ECF from (a).
ΔT/T = Δλ/λ = 3.6 %, so ΔT = ±2 × 102 K, i.e. (5.2 ± 0.2) × 103 K✓ 1Allow ECF from (a). Uncertainty to 1 s.f.
Part (c)(i)
L = 4π(0.84 × 6.96 × 108)² × 5.67 × 10−8 × 5179⁴ = 1.8 × 1026 W✓ 1Allow ECF from (b).
%L = 2 × 3.6 + 4 × 3.6 = 21 %, so ΔL = ±4 × 1025 W✓ 1Allow ECF from (b). 1 s.f. required: (1.8 ± 0.4) × 1026 W. Adding the percentages without the powers: [1 max].

Answers: (a)(i) 560 ± 20 nm  ·  (b)(i) (5.2 ± 0.2) × 103 K  ·  (c)(i) (1.8 ± 0.4) × 1026 W (the remaining parts are explanations — see the table above)

57E-1B-07Stefan–Boltzmann law for stars Paper 1BHard7 marks

At noon on a clear day a student points a thin copper disc, blackened on its front face, directly at the Sun. The disc has a mass of 25.0 g and a diameter of 4.00 cm, and its back face is insulated. A thermocouple fixed to the disc records its temperature θ every minute, starting when the disc is at the temperature of the surrounding air. The graph shows θ against time t with a curve of best fit.

(Specific heat capacity of copper = 385 J kg−1 K−1; Earth–Sun distance = 1.50 × 1011 m; R☉ = 6.96 × 108 m; σ = 5.67 × 10−8 W m−2 K−4)

012345678910t / min2025303540455055606570θ / °C
Temperature of the disc against time (graph drawn to scale)
(a)
(i)

Explain why the initial rate of rise of temperature should be used, and determine it.

(2)
(b)
(i)

Determine the intensity of the sunlight absorbed by the disc.

(2)
(c)
(i)

Determine the surface temperature of the Sun implied by your answer to (b).

(2)
(d)
(i)

The accepted surface temperature of the Sun is 5.8 × 103 K. Suggest one reason for the difference between this value and your answer to (c).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
At t = 0 the disc is at air temperature, so it loses no energy to its surroundings: all the absorbed power raises its temperature✓ 1Later the gradient falls as the losses grow.
Tangent to the curve at t = 0: gradient = 7.5 K min−1✓ 1Accept 7.0–8.0 K min−1. A chord from 0 to 2 min (6.6 K min−1) is not a tangent: 0 for this mark.
Part (b)(i)
Power absorbed = mc Δθ/Δt = 0.0250 × 385 × 7.5/60 = 1.20 W✓ 1Allow ECF from (a).
I = 1.20/(π × 0.0200²) = 9.6 × 102 W m−2✓ 1Accept 9.0–10.2 × 102 W m−2.
Part (c)(i)
L = 4πd²I = 4π × (1.50 × 1011)² × 957 = 2.7 × 1026 W✓ 1Allow ECF from (b).
T = (L/4πR☉²σ)1/4 = 5.3 × 103 K✓ 1Accept 5.2–5.4 × 103 K. ALT: T = (Id²/σR☉²)1/4 in one step.
Part (d)(i)
The atmosphere absorbs and scatters part of the sunlight (or the disc is not a perfect absorber), so the intensity reaching the disc is less than the intensity at this distance from the Sun, giving a lower T✓ 1Allow ECF from (c). The reason must reduce the measured intensity, consistent with the candidate's comparison. Do not accept "heat losses" unless linked to underestimating the initial gradient.

Answers: (a)(i) 7.5 K min−1  ·  (b)(i) 9.6 × 102 W m−2  ·  (c)(i) 5.3 × 103 K (the remaining parts are explanations — see the table above)

58IBP2H-043-00-09Stellar spectra Paper 2Easy6 marks

An ultraviolet space telescope observes a star whose continuous spectrum peaks at a wavelength of 350 nm. The radius of the star is 2.1 times the radius of the Sun.
Radius of Sun = 6.96×108 m; luminosity of Sun = 3.83×1026 W; surface temperature of Sun = 5.8×103 K; Wien constant = 2.9×10−3 m K; Stefan–Boltzmann constant σ = 5.67×10−8 W m−2 K−4.

(a)

Calculate the surface temperature of the star.

(2)
(b)

Calculate the luminosity of the star.

(2)
(c)

Using your answers, outline where the star lies on a Hertzsprung–Russell diagram relative to the Sun.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses T = 2.9×10−3 / λmax = 2.9×10−3 / 350×10−9✓ 1
T = 8.3×103 K✓ 1
Part (b)
Uses L = 4πR2σT4 with R = 2.1 × 6.96×108 = 1.46×109 m✓ 1
L = 7.2×1027 W✓ 1
Allow ECF from the temperature calculated in the first part.—
Part (c)
L/LSun = 7.2×1027 / 3.83×1026 ≈ 19, so the star is plotted above the Sun (allow ECF from the luminosity calculated in the second part)✓ 1
Its temperature of 8.3×103 K is greater than 5.8×103 K, so it is plotted to the left of the Sun, because temperature increases to the left✓ 1

Exam tip: Convert nanometres to metres before using Wien's law. A wavelength left in nm gives a temperature that is wrong by a factor of 10⁹.

59IBP2H-043-01-09Stellar parallax Paper 2Easy4 marks

A space astrometry mission orbits close to Earth. It measures the parallax angle of a faint red dwarf as 4.2×10−3 arcseconds.
1 pc = 3.086×1016 m.

SunEarth (one position)Earth (six months later)star1 AUp = 4.2 × 10⁻³″distance d = ?
Diagram not to scale
(a)

Calculate the distance to the red dwarf in parsecs.

(1)
(b)

Calculate the distance to the red dwarf in metres.

(1)
(c)

Outline why parallax cannot be used to measure the distances to very distant stars.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = 1/p = 1/(4.2×10−3) = 238 pc (accept 2.4×102 pc)✓ 1
Part (b)
d = 238.1 × 3.086×1016 = 7.35×1018 m (accept 7.3×1018 m to 7.4×1018 m; allow ECF from (a))✓ 1
Part (c)
At large distances the parallax angle becomes extremely small✓ 1
The angle becomes comparable to, or smaller than, the angular resolution or measurement uncertainty of the instrument, so it cannot be measured reliably✓ 1

Exam tip: d = 1/p gives the distance in parsecs only when p is in arcseconds. Keep the unrounded value from (a) when you convert to metres.

60IBP2H-043-02-09Stellar parallax Paper 2Easy4 marks

A space-based astrometry telescope measures the stellar parallax angle of a nearby star as 0.0125 arc-seconds.

SunEarth (one position)Earth (six months later)star1 AUp = 0.0125″distance d = ?
Diagram not to scale
(a)

Define the parsec.

(1)
(b)

Calculate the distance to the star in parsecs.

(1)
(c)

Calculate this distance in metres. (1 pc = 3.09 × 1016 m)

(1)
(d)

Outline why the parallax method cannot be used for very distant stars.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The distance at which a baseline of 1 AU (the mean Earth–Sun distance) subtends an angle of 1 arc-second, i.e. the distance of a star whose parallax angle is 1 arc-second .✓ 1
Part (b)
d = 1/p = 1/0.0125 = 80 pc .✓ 1
Part (c)
d = 80 × 3.09 × 1016 = 2.47 × 1018 m (accept 2.5 × 1018 m) . ECF from (b).✓ 1
Part (d)
For distant stars the parallax angle becomes too small to measure precisely, so the percentage uncertainty in the distance becomes too large .✓ 1

Exam tip: The relation d = 1/p only works with d in parsecs and p in arc-seconds. Convert to metres only after using it.

61IBP3H-E5-20r22-09Stellar parallax Paper 2Easy5 marks

A robotic space telescope observes a star whose parallax angle is 0.125 arcseconds.

SunEarth (one position)Earth (six months later)star1 AUp = 0.125″distance d = ?
Diagram not to scale
(a)

Outline what is meant by the parallax angle of a star.

(2)
(b)

Calculate the distance to the star in parsecs and in light years, given that 1 pc = 3.26 ly.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
the apparent shift in position of the star against distant background stars when viewed from opposite sides of Earth's orbit, six months apart✓ 1
parallax angle is the angle subtended at the star by the Earth–Sun distance (1 AU), equivalent to half of the total angular shift✓ 1
Part (b)
M1 use \(d=\dfrac{1}{p}\)✓ 1
A1 8.00 pc (accept 8.0)✓ 1
A1 26.1 ly (8.00 × 3.26 = 26.08)✓ 1

Exam tip: Keep distance in parsecs when using the parallax angle in arcseconds, and convert units only at the end.

62E-2-04Stellar parallax Paper 2Medium10 marks

The diagram shows the Earth at two positions, E₁ and E₂, six months apart in its orbit of radius 1 AU about the Sun S. The parallax angle p of the nearby star Q, marked on the diagram, is 0.0400 arc-second.

(1 AU = 1.50 × 1011 m, 1 pc = 3.26 ly, L☉ = 3.85 × 1026 W)

SE₁E₂1 AUQpdistant starsnot to scale
Parallax of star Q (not to scale)
(a)
(i)

One parsec is the distance of a star whose parallax angle is 1 arc-second. Show, using the diagram, that 1 pc is about 3.1 × 1016 m.

(2)
(ii)

Determine the distance to Q in light-years.

(1)
(b)
(i)

The apparent brightness of Q is 2.9 × 10−10 W m−2. Determine the luminosity of Q in terms of L☉.

(2)
(ii)

The parallax angle of Q is measured to ±0.0010 arc-second. Determine the percentage uncertainty in the luminosity of Q.

(2)
(c)
(i)

The smallest parallax angle that can be measured with this telescope is 0.0050 arc-second. Star R has the same luminosity as Q and an apparent brightness of 1.0 × 10−12 W m−2. Deduce whether the distance to R can be measured by parallax with this telescope.

(2)
(ii)

Suggest why parallax angles measured with a telescope in space are more precise than those measured from the Earth's surface.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
1 arc-second = (1/3600)° = 4.85 × 10−6 rad; from the diagram tan p = (1 AU)/d✓ 1The conversion to radians (or use of tan with the angle in degrees) must be seen.
d = 1.50 × 1011/4.848 × 10−6 = 3.09 × 1016 m✓ 1Must see the full substitution or an answer to at least 3 s.f. (3.09 × 1016 m).
Part (a)(ii)
d = 1/p = 1/0.0400 = 25.0 pc = 25.0 × 3.26 = 81.5 ly✓ 1Unit required.
Part (b)(i)
d = 25.0 × 3.09 × 1016 = 7.7 × 1017 m; L = 4πd²b = 2.2 × 1027 W✓ 1Allow ECF from (a)(i) and (a)(ii).
L = 5.7 L☉✓ 1Accept 5.5–5.7 L☉.
Part (b)(ii)
Percentage uncertainty in p, and so in d = 0.0010/0.0400 = 2.5 %✓ 1
L ∝ d², so the percentage uncertainty in L = 5 %✓ 1Allow ECF from (b)(i). The uncertainty in b is ignored.
Part (c)(i)
Same L, so d ∝ 1/√b: dR = 25.0 × √(2.9 × 10−10/1.0 × 10−12) = 4.3 × 102 pc✓ 1ALT: d = √(L/4πb) = 1.3 × 1019 m = 4.3 × 102 pc. Allow ECF from (b)(i).
pR = 1/426 = 0.0023 arc-second, smaller than 0.0050 arc-second, so the distance cannot be measured by parallax✓ 1ALT: the greatest measurable distance is 1/0.0050 = 200 pc, less than 4.3 × 102 pc. The conclusion must agree with the candidate's own value.
Part (c)(ii)
There is no atmospheric turbulence (refraction) to blur and shift the star images✓ 1

Answers: (a)(i) 3.09 × 1016 m  ·  (a)(ii) 81.5 ly  ·  (b)(i) 5.7 L☉  ·  (b)(ii) 5 %  ·  (c)(i) No: p ≈ 0.0023″ (d ≈ 4.3 × 102 pc) (the remaining parts are explanations — see the table above)

63E-2-27The Sun as a red giant Paper 2Medium10 marks

The Hertzsprung–Russell diagram shows the predicted evolutionary track of the Sun. S marks the Sun today, P the top of its red-giant stage and W the white dwarf that it eventually becomes.

(L☉ = 3.85 × 1026 W, σ = 5.67 × 10−8 W m−2 K−4, 1 AU = 1.50 × 1011 m)

10⁻⁴10⁻³10⁻²10⁻¹10⁰10¹10²10³10⁴40000200001000050002500surface temperature / Kluminosity / L☉SPWmain sequence
Predicted evolutionary track of the Sun (both axes logarithmic)
(a)
(i)

Outline why the Sun will leave the main sequence and move towards P.

(2)
(ii)

State the luminosity of the Sun at P.

(1)
(b)
(i)

Determine the intensity of the Sun's radiation at the Earth's orbit when the Sun is at P.

(2)
(ii)

A planet of radius r orbits at distance d from a star of luminosity L. It absorbs a fraction (1 − α) of the radiation incident on it, where α is its albedo, and radiates as a black body at a uniform temperature T from its whole surface. Show that T⁴ = (1 − α)L/(16πσd²).

(2)
(iii)

Determine the equilibrium temperature of the Earth when the Sun is at P. Take α = 0.30.

(2)
(iv)

Deduce, without substituting values, the ratio (equilibrium temperature of Mars)/(equilibrium temperature of the Earth). Mars orbits at 1.52 AU and has the same albedo as the Earth.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
When the hydrogen in the core is used up, fusion in the core stops and the core contracts and heats up✓ 1
Hydrogen fusion continues in a shell around the core; the outer layers expand greatly and cool, so the luminosity rises while the surface temperature falls✓ 1
Part (a)(ii)
≈ 2.5 × 103 L☉✓ 1Accept 2.0–3.0 × 103 L☉.
Part (b)(i)
L = 2.5 × 103 × 3.85 × 1026 = 9.6 × 1029 W✓ 1Allow ECF from (a)(ii).
b = L/4π(1.50 × 1011)² = 3.4 × 106 W m−2✓ 1Accept 2.7–4.1 × 106 W m−2.
Part (b)(ii)
Power absorbed = (1 − α) × L/(4πd²) × πr² (the planet intercepts radiation on its cross-section)✓ 1
Power emitted = 4πr²σT⁴; equating, r cancels and T⁴ = (1 − α)L/(16πσd²)✓ 1
Part (b)(iii)
T⁴ = (1 − α)b/(4σ) = 0.70 × 3.40 × 106/(4 × 5.67 × 10−8) = 1.05 × 1013 K⁴✓ 1Allow ECF from (b)(i) and (b)(ii).
T = 1.8 × 103 K✓ 1Accept 1.7–1.9 × 103 K.
Part (b)(iv)
T ∝ d−1/2, so the ratio = (1/1.52)1/2 = 0.81✓ 1Allow ECF from (b)(ii).

Answers: (a)(ii) 2.5 × 103 L☉  ·  (b)(i) 3.4 × 106 W m−2  ·  (b)(iii) 1.8 × 103 K  ·  (b)(iv) 0.81 (the remaining parts are explanations — see the table above)

64E-2-28Nuclear fusion Paper 2Medium13 marks

When the hydrogen in the core of a star of two solar masses has been used up, the core contracts until helium fusion begins at a temperature of about 108 K. Hydrogen fusion needs only about 1.5 × 107 K. Helium fusion occurs in two steps:

3 42He → 126C     and then     126C + 42He → 168O

Atomic masses: 1H 1.007 825 u, 4He 4.002 602 u, 12C 12.000 000 u, 16O 15.994 915 u. 1 u = 931.5 MeV c−2 = 1.66 × 10−27 kg, 1 MeV = 1.60 × 10−13 J, L☉ = 3.85 × 1026 W, solar mass = 1.99 × 1030 kg, 1 year = 3.16 × 107 s.

(a)

Starting helium fusion.

(i)

When hydrogen fusion in the core stops, the core contracts. Explain, in terms of energy, why the temperature of the core rises as it contracts.

(2)
(b)

Energy released.

(i)

Show that the energy released when three helium nuclei form one carbon nucleus is about 7.3 MeV.

(2)
(ii)

Calculate the energy released when a carbon nucleus captures a helium nucleus to form oxygen.

(1)
(iii)

Hydrogen fusion releases 26.7 MeV for every four hydrogen atoms converted into one helium atom. Determine the ratio (energy released per kilogram of hydrogen fused to helium) / (energy released per kilogram of helium fused to carbon).

(2)
(c)

The helium-burning stage.

(i)

During this stage helium fusion to carbon supplies 30 L☉. Calculate the mass of helium fused per second.

(2)
(ii)

The core contains 0.30 solar masses of helium. Estimate the duration of the helium-burning stage, in years.

(2)
(iii)

Suggest two reasons why the helium-burning stage is much shorter than the star's main-sequence lifetime.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
As the core contracts, the gravitational potential energy of its matter decreases (the gravitational forces do work on the core material)✓ 1Do not accept an answer that only states that the pressure no longer balances gravity: the question asks about energy.
This energy is transferred to the random kinetic energy of the nuclei and electrons; since their average kinetic energy is proportional to the absolute temperature (Ek = (3/2)kBT), the temperature rises✓ 1The link between mean kinetic energy and temperature is needed for MP2. Accept "some of the energy is radiated away, the rest heats the core".
Part (b)(i)
Δm = 3 × 4.002 602 − 12.000 000 = 0.007 806 u✓ 1Atomic masses can be used: the electron masses cancel.
E = 0.007 806 × 931.5 = 7.27 MeV ≈ 7.3 MeV✓ 1
Part (b)(ii)
Δm = 12.000 000 + 4.002 602 − 15.994 915 = 0.007 687 u, E = 0.007 687 × 931.5 = 7.16 MeV✓ 1Accept 7.2 MeV.
Part (b)(iii)
Per unit mass of fuel: hydrogen 26.7 MeV/(4 × 1.0078 u) = 6.62 MeV u−1 (6.4 × 1014 J kg−1); helium 7.27 MeV/(3 × 4.0026 u) = 0.606 MeV u−1 (5.8 × 1013 J kg−1)✓ 1Both needed (or the two values in J kg−1).
Ratio ≈ 11✓ 1Accept 10.9–11.1. Allow ECF from (b)(i).
Part (c)(i)
Energy per kg of helium = 7.27 × 1.60 × 10−13/(3 × 4.0026 × 1.66 × 10−27) = 5.8 × 1013 J kg−1✓ 1Allow ECF from (b)(i).
Rate = 30 × 3.85 × 1026/5.84 × 1013 = 2.0 × 1014 kg s−1✓ 1Accept 1.9–2.0 × 1014.
Part (c)(ii)
Mass of helium = 0.30 × 1.99 × 1030 = 6.0 × 1029 kg; t = 6.0 × 1029/1.98 × 1014 = 3.0 × 1015 s✓ 1
≈ 1 × 108 years (9.5 × 107 years)✓ 1Accept 9–10 × 107 years. Allow ECF from (c)(i).
Part (c)(iii)
Helium fusion releases about 11 times less energy per kilogram of fuel than hydrogen fusion (from (b)(iii))✓ 1Any two of: (1) about 11 times less energy per kg (from (b)(iii)); (2) only the core helium, a small fraction of the star's mass, is available; (3) the star is more luminous than on the main sequence, so it uses fuel faster. [1] each, max 2. Allow ECF from (b)(iii).
Only the core helium (a small fraction of the star's mass) is available; the star is more luminous than on the main sequence, so it uses its fuel faster✓ 1

Answers: (b)(i) 7.27 MeV  ·  (b)(ii) 7.16 MeV  ·  (b)(iii) ≈ 11  ·  (c)(i) 2.0 × 1014 kg s−1  ·  (c)(ii) ≈ 1 × 108 years (the remaining parts are explanations — see the table above)

65E-2-29The Hertzsprung–Russell diagram Paper 2Medium11 marks

The Hertzsprung–Russell diagram shows the Sun and three other stars S, V and W. The dashed lines are lines of constant radius. Stars in the shaded band have a luminosity that varies periodically.

(R☉ = 6.96 × 108 m, L☉ = 3.85 × 1026 W, σ = 5.67 × 10−8 W m−2 K−4, 1 AU = 1.50 × 1011 m)

10−410−210010210410640000200001000050002500surface temperature / K (logarithmic scale)luminosity / L☉ (logarithmic scale)0.01 R☉1 R☉100 R☉1000 R☉bandSunSVW
HR diagram showing the Sun and stars S, V and W. Dashed lines: constant stellar radius. Shaded: a narrow band that crosses the main sequence. Both axes logarithmic.
(a)

Reading the diagram.

(i)

State the name of the shaded band.

(1)
(ii)

Identify the type of star represented by S and by W.

(2)
(iii)

Use the diagram to estimate the radius of W.

(1)
(b)

Star S has a surface temperature of 3600 K and a luminosity of 1.0 × 105 L☉.

(i)

Show that the radius of S is about 800 R☉.

(2)
(ii)

Express this radius in AU and comment on your answer.

(1)
(c)

Star V lies in the shaded band. Its luminosity varies between 60 L☉, when its surface temperature is 7200 K, and 40 L☉, when its surface temperature is 6200 K.

(i)

Determine the radius of V at maximum and at minimum luminosity, in terms of R☉.

(2)
(ii)

Explain, using your answers, why V is most luminous when it is smallest.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Instability strip✓ 1
Part (a)(ii)
S: (red) supergiant — very luminous and cool, at the top right✓ 1Accept red supergiant; do not accept red giant alone.
W: white dwarf — hot but very faint, at the lower left below the main sequence✓ 1
Part (a)(iii)
W lies just above the 0.01 R☉ line, so its radius is about 0.01 R☉✓ 1Accept 0.01–0.02 R☉.
Part (b)(i)
L = 4πR²σT⁴ gives R/R☉ = √(L/L☉) × (T☉/T)²✓ 1Or R = √(L/4πσT⁴) with L in W: 5.7 × 1011 m.
= √(1.0 × 105) × (5800/3600)² = 316 × 2.60 = 820 ≈ 800 R☉✓ 1Using SI values gives 815 R☉ — both acceptable.
Part (b)(ii)
R = 5.7 × 1011/1.50 × 1011 = 3.8 AU; S would extend well beyond the orbit of Mars (about 1.5 AU) — larger than the orbits of all the inner planets✓ 1Accept 3.7–3.9 AU with a sensible comment.
Part (c)(i)
Maximum: R = √60 × (5800/7200)² = 5.0 R☉✓ 1Accept 4.9–5.1 R☉; SI route acceptable.
Minimum: R = √40 × (5800/6200)² = 5.5 R☉✓ 1Accept 5.4–5.6 R☉.
Part (c)(ii)
L ∝ R²T⁴: at maximum light R² is smaller by a factor (5.0/5.5)² ≈ 0.83 but T⁴ is larger by (7200/6200)⁴ ≈ 1.8✓ 1
The temperature effect dominates, so the net luminosity is 0.83 × 1.8 ≈ 1.5 times larger (60/40) even though the star is smallest✓ 1

Answers: (a)(iii) ≈ 0.01 R☉  ·  (b)(i) ≈ 820 R☉  ·  (b)(ii) 3.8 AU  ·  (c)(i) 5.0 R☉ and 5.5 R☉ (the remaining parts are explanations — see the table above)

66E-2-30Stellar evolution Paper 2Medium10 marks

Two stars A and B formed at the same time in the same cluster. Star A has a mass of 20 solar masses and a main-sequence luminosity of 4.0 × 104 L☉. Star B has a mass of 1.0 solar mass and a luminosity of 1.0 L☉; its main-sequence lifetime is about 1.0 × 1010 years.

(Binding energy per nucleon: ²⁸Si 8.448 MeV, ⁵⁶Ni 8.643 MeV, ⁵⁶Fe 8.790 MeV — close to the maximum of the binding-energy curve.)

(a)

The main sequence.

(i)

Explain, in terms of the equilibrium of the star, why the core of A must be hotter than the core of B.

(2)
(ii)

Assume that each star fuses the same fraction of its mass while on the main sequence. Estimate the main-sequence lifetime of A.

(2)
(iii)

Explain why A has the shorter lifetime even though it has 20 times more fuel.

(1)
(b)

The end of fusion in A. In the final stage of core fusion in A, silicon fuses: 2 2814Si → 5628Ni.

(i)

Calculate the energy released in this reaction.

(2)
(ii)

The core of A eventually becomes iron. Explain why the core then collapses.

(2)
(c)

After the collapse, a supernova occurs and the remnant of A is a neutron star.

(i)

State the remnant that would be formed if the collapsing core had a much greater mass.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
A has much more mass, so the inward gravitational force on its layers (the weight of the overlying layers) is much greater✓ 1
To balance it a much greater outward pressure is needed in the core, which requires a higher core temperature (and fusion rate)✓ 1
Part (a)(ii)
Lifetime ∝ (fuel)/(rate of use) ∝ M/L: tA/tB = (20/1.0)/(4.0 × 104/1.0) = 5.0 × 10−4✓ 1
tA ≈ 5.0 × 10−4 × 1.0 × 1010 = 5 × 106 years✓ 1Accept 4–6 × 106 years. A ratio inverted (M/L the wrong way up) scores [1].
Part (a)(iii)
The fusion rate rises very steeply with core temperature, so A's luminosity (rate of fuel use) is 4 × 104 times B's while its fuel is only 20 times greater✓ 1Allow ECF from (a)(ii).
Part (b)(i)
Increase in total binding energy = 56 × 8.643 − 2 × 28 × 8.448 = 56 × 0.195✓ 1
= 10.9 MeV✓ 1Accept 10.9–11 MeV.
Part (b)(ii)
Iron is near the peak of the binding-energy curve, so fusing it would absorb rather than release energy: fusion in the core stops✓ 1
No energy is released to maintain the outward (radiation and gas) pressure, so gravity is no longer balanced and the core collapses✓ 1
Part (c)(i)
A black hole✓ 1

Answers: (a)(ii) ≈ 5 × 106 years  ·  (b)(i) 10.9 MeV (the remaining parts are explanations — see the table above)

67E-2-31Fusion in the Sun Paper 2Medium13 marks

In the nineteenth century physicists debated what could power the Sun. The Sun has mass 1.99 × 1030 kg, radius 6.96 × 108 m and luminosity 3.85 × 1026 W. Geological and radioactive dating show that it has been shining for about 4.6 × 109 years at roughly its present luminosity.

(G = 6.67 × 10−11 N m2 kg−2, c = 3.00 × 108 m s−1, 1 year = 3.16 × 107 s; atomic masses: 1H 1.007 825 u, 4He 4.002 602 u)

(a)

Energy needed.

(i)

Calculate the total energy radiated by the Sun so far.

(1)
(b)

Two rejected hypotheses.

(i)

The best chemical fuels release about 5 × 107 J per kilogram. Determine how long the Sun could shine at its present luminosity if it were made entirely of such fuel.

(2)
(ii)

The energy released by the gravitational contraction of the Sun to its present size is of order GM²/R. Determine how long this energy could supply the present luminosity, and comment on your answers to (b)(i) and (b)(ii).

(2)
(c)

Fusion.

(i)

The net result of hydrogen fusion in the Sun is that four hydrogen atoms become one helium atom. Show that about 0.7 % of the mass of the hydrogen is converted into energy.

(2)
(ii)

Determine the total mass of hydrogen that has been fused in the Sun so far.

(2)
(iii)

Only the hydrogen in the core, about 10 % of the Sun's mass, is ever hot and dense enough to fuse while the Sun is on the main sequence. Deduce how far the Sun is through its main-sequence life.

(2)
(iv)

Suggest why the Sun's luminosity has changed little over 4.6 × 109 years even though 0.44 of its core hydrogen has been used.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
E = 3.85 × 1026 × 4.6 × 109 × 3.16 × 107 = 5.6 × 1043 J✓ 1
Part (b)(i)
Chemical energy = 5 × 107 × 1.99 × 1030 = 1.0 × 1038 J✓ 1
t = 1.0 × 1038/3.85 × 1026 = 2.6 × 1011 s ≈ 8 × 103 years✓ 1
Part (b)(ii)
GM²/R = 6.67 × 10−11 × (1.99 × 1030)²/6.96 × 108 = 3.8 × 1041 J; t = 3.8 × 1041/3.85 × 1026 = 9.9 × 1014 s ≈ 3 × 107 years✓ 1
Both are far shorter than 4.6 × 109 years (by factors of about 6 × 105 and 150), so neither chemical reactions nor contraction can be the Sun's main energy source✓ 1
Part (c)(i)
Δm = 4 × 1.007 825 − 4.002 602 = 0.028 698 u✓ 1
Fraction = 0.028 698/(4 × 1.007 825) = 7.1 × 10−3 ≈ 0.7 %✓ 1
Part (c)(ii)
Mass converted to radiated energy = E/c² = 5.6 × 1043/(3.00 × 108)² = 6.2 × 1026 kg✓ 1
Mass of hydrogen fused = 6.2 × 1026/7.1 × 10−3 = 8.7 × 1028 kg✓ 1Accept 8.5–8.9 × 1028 kg; ecf from (a).
Part (c)(iii)
Hydrogen available in the core = 0.10 × 1.99 × 1030 = 2.0 × 1029 kg; fraction used = 8.7 × 1028/2.0 × 1029 = 0.44✓ 1
The Sun is a little less than halfway through its main-sequence life (total ≈ 4.6 × 109/0.44 ≈ 1 × 1010 years)✓ 1Accept "about halfway".
Part (c)(iv)
The fusion rate is self-regulated by the equilibrium between the inward gravitational force and the outward (radiation and gas) pressure✓ 1
Any small increase (or decrease) in core temperature or density makes the core expand (or contract), which returns the fusion rate to its equilibrium value — negative feedback, so the luminosity stays nearly constant✓ 1

Answers: (a)(i) 5.6 × 1043 J  ·  (b)(i) ≈ 8 × 103 years  ·  (b)(ii) ≈ 3 × 107 years  ·  (c)(i) 0.71 %  ·  (c)(ii) 8.7 × 1028 kg  ·  (c)(iii) ≈ 0.44 of the way (the remaining parts are explanations — see the table above)

68IBP2H-043-00-13Stellar evolution Paper 2Medium7 marks

A star of initial mass 20 solar masses is on the main sequence. It later evolves off the main sequence, and its life ends in a supernova that leaves a collapsed core of mass 1.7 solar masses.
Take the Oppenheimer–Volkoff limit to be about 2.5 solar masses.

(a)

Outline what is meant by the Chandrasekhar limit.

(2)
(b)

Identify the remnant left by this star and state what prevents its further collapse.

(2)
(c)

Describe the path of this star on a Hertzsprung–Russell diagram, from its position on the main sequence to just before the supernova.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
It is the maximum mass of a stable white dwarf (remnant core), about 1.4 solar masses✓ 1
Above this mass, electron degeneracy pressure cannot balance gravitational collapse✓ 1
Part (b)
Neutron star, because the core mass of 1.7 MSun is above the Chandrasekhar limit (1.4 MSun) but below the Oppenheimer–Volkoff limit (about 2.5 MSun)✓ 1
Neutron degeneracy pressure✓ 1
Part (c)
On the main sequence it lies in the upper left of the diagram, where stars are hot and very luminous✓ 1
When core hydrogen is exhausted, it moves to the right at roughly constant (or slightly increasing) luminosity, as its surface cools and its radius increases✓ 1
It becomes a red supergiant in the upper right of the diagram✓ 1

Exam tip: When identifying a stellar remnant, compare the core mass with both limits: the Chandrasekhar limit (about 1.4 solar masses) and the Oppenheimer–Volkoff limit.

69IBP2H-043-00-15The Hertzsprung–Russell diagram Paper 2Medium8 marks

The table gives data for four stars. The Sun has a surface temperature of 5800 K.
Star — L / LSun — Surface temperature / K
P — 0.010 — 12 000
Q — 1.0 — 5800
R — 1.0×104 — 3500
S — 3.0×104 — 30 000

(a)

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

(2)
(b)

Determine \(\frac{R_P}{R_Q}\).

(2)
(c)

Determine the radius of star R in units of RSun.

(2)
(d)

S and Q are both main-sequence stars. State and explain which has the longer main-sequence lifetime.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
P is the white dwarf (hot but very low luminosity)✓ 1
R is the red supergiant (cool but very high luminosity)✓ 1
Part (b)
R ∝ √L / T2, so RP/RQ = √0.010 × (5800/12 000)2✓ 1
RP/RQ = 0.023✓ 1
Part (c)
R/RSun = √(1.0×104) × (5800/3500)2✓ 1
R ≈ 2.7×102 RSun✓ 1
Part (d)
Q (the Sun-like star) has the longer lifetime✓ 1
S is far more massive and more luminous; since lifetime ∝ M/L, and L rises much faster than M, S uses its fuel far more quickly✓ 1

Exam tip: Working in solar units (L/LSun, T/TSun) avoids σ and 4π completely when only ratios are needed.

70IBP2H-043-00-16The Hertzsprung–Russell diagram Paper 2Medium8 marks

Models of the Sun's future predict that it will become a red giant with radius 150 RSun and surface temperature 3500 K. Later, it will become a white dwarf with radius 0.012 RSun and surface temperature 25 000 K. The Sun's present surface temperature is 5800 K.

(a)

Calculate the luminosity of the red giant in terms of the present solar luminosity LSun.

(2)
(b)

Calculate the luminosity of the white dwarf in terms of LSun.

(2)
(c)

Explain why the white dwarf's luminosity decreases over time.

(2)
(d)

Estimate the peak wavelength of the white dwarf's radiation and state the region of the electromagnetic spectrum it lies in.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
L/LSun = 1502 × (3500/5800)4✓ 1
L ≈ 3.0×103 LSun✓ 1
Part (b)
L/LSun = 0.0122 × (25 000/5800)4✓ 1
L ≈ 0.050 LSun✓ 1
Part (c)
No fusion occurs in a white dwarf, so it has no energy source✓ 1
It radiates stored thermal energy, so its temperature falls; with its radius fixed, L ∝ T4 falls✓ 1
Part (d)
λmax = 2.9×10−3 / 25 000 = 1.16×10−7 m ≈ 1.2×10−7 m✓ 1
Ultraviolet✓ 1

Common error: Forgetting to raise the temperature ratio to the fourth power is the most frequent slip in Stefan–Boltzmann ratio calculations.

71IBP2H-043-01-10Stellar spectra and luminosity Paper 2Medium6 marks

A medical thermal-imaging camera detects radiation from human skin at a temperature of 307 K. Astronomers use the same black-body physics to study a red supergiant. The supergiant's spectrum peaks at 828 nm and its luminosity is 1.0×105 L☉.
Wien constant = 2.9×10−3 m K; σ = 5.67×10−8 W m−2 K−4; L☉ = 3.83×1026 W.

(a)

Calculate the peak wavelength of the radiation emitted by the skin and state the region of the electromagnetic spectrum it lies in.

(2)
(b)

Show that the surface temperature of the supergiant is about 3500 K.

(1)
(c)

Determine the radius of the supergiant.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
λmax = 2.9×10−3/307 = 9.4×10−6 m✓ 1
Infrared✓ 1
Part (b)
T = 2.9×10−3/(828×10−9) = 3.50×103 K✓ 1
Part (c)
L = 1.0×105 × 3.83×1026 = 3.83×1031 W✓ 1
Rearranging L = 4πR2σT4 gives R = √(L/(4πσT4))✓ 1
R = √(3.83×1031/(4π × 5.67×10−8 × 35004)) = 6.0×1011 m; ECF from (b)✓ 1

Common error: Convert the luminosity to watts before substituting. Using L = 1.0×10⁵ directly gives a radius that is far too small.

72IBP2H-043-01-13Stellar parallax Paper 2Medium9 marks

The table gives data for two stars, P and Q.

Star P: parallax angle 0.0500 arcsec; apparent brightness 2.4×10−9 W m−2; peak wavelength 290 nm
Star Q: parallax angle 0.0125 arcsec; apparent brightness 1.5×10−10 W m−2; peak wavelength 725 nm

1 pc = 3.086×1016 m; Wien constant = 2.9×10−3 m K.

(a)

Show that Q is four times further from Earth than P.

(1)
(b)

Calculate the distance to P in metres.

(1)
(c)

Calculate the luminosity of P.

(2)
(d)

Deduce the ratio \(\frac{L_P}{L_Q}\).

(2)
(e)

Determine the ratio \(\frac{R_Q}{R_P}\).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
d ∝ 1/p and 0.0500/0.0125 = 4, so dQ = 4dP (equivalently 80 pc and 20 pc)✓ 1
Part (b)
dP = 20 pc = 20 × 3.086×1016 = 6.17×1017 m✓ 1
Part (c)
L = 4πd2b✓ 1
L = 4π × (6.17×1017)2 × 2.4×10−9 = 1.1×1028 W; ECF from (b)✓ 1
Part (d)
LP/LQ = (bP/bQ)(dP/dQ)2 = 16 × (1/4)2✓ 1
= 1, so the two stars have equal luminosity✓ 1
Part (e)
TP = 2.9×10−3/290×10−9 = 1.0×104 K and TQ = 4.0×103 K✓ 1
For equal luminosities R2T4 is the same for both stars, so RQ/RP = (TP/TQ)2✓ 1
RQ/RP = 2.52 = 6.25; ECF from (d)✓ 1

Exam tip: Work with ratios wherever possible. Many constants cancel, and the result here, equal luminosities, becomes obvious.

73IBP2H-043-01-15The Hertzsprung–Russell diagram Paper 2Medium8 marks

A white dwarf has a surface temperature of 2.5×104 K and a luminosity of 0.026 L☉.
Wien constant = 2.9×10−3 m K; σ = 5.67×10−8 W m−2 K−4; L☉ = 3.83×1026 W.

(a)

Calculate the peak wavelength of the white dwarf's spectrum and state the region of the electromagnetic spectrum it lies in.

(2)
(b)

Determine the radius of the white dwarf.

(2)
(c)

Explain, with reference to (a) and (b), where the white dwarf lies on an HR diagram relative to the main sequence.

(2)
(d)

Outline why the white dwarf does not collapse even though no fusion occurs in it.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
λmax = 2.9×10−3/2.5×104 = 1.2×10−7 m✓ 1
Ultraviolet✓ 1
Part (b)
L = 0.026 × 3.83×1026 = 9.96×1024 W, used in R = √(L/(4πσT4))✓ 1
R = √(9.96×1024/(4π × 5.67×10−8 × (2.5×104)4)) = 6.0×106 m✓ 1
Part (c)
Its high surface temperature places it on the left (hot) side of the diagram✓ 1
Its very small radius means its luminosity is far lower than that of main-sequence stars of the same temperature, so it lies well below the main sequence, in the lower left✓ 1
Part (d)
Electron degeneracy pressure acts outwards and balances the inward gravitational force✓ 1
The electrons are packed so closely that, by the Pauli exclusion principle, they cannot all occupy the same low-energy states; this resists further compression. Also accept: this is possible only while the mass is below the Chandrasekhar limit of about 1.4 M☉✓ 1

Exam tip: On an HR diagram, temperature increases to the left. Hot stars with a small radius therefore sit in the lower left, below the main sequence.

74IBP2H-043-02-10The Sun as a red giant Paper 2Medium6 marks

A spacecraft orbiting Mars is powered by solar panels. The panels have a total area of 15 m2 and an efficiency of 22%, and they always face the Sun. The luminosity of the Sun is 3.83 × 1026 W. The mean distance of Mars from the Sun is 2.28 × 1011 m.

(a)

Calculate the apparent brightness of the Sun at the orbit of Mars.

(2)
(b)

Calculate the electrical power output of the panels at Mars.

(1)
(c)

The spacecraft needs at least 1.2 kW of electrical power. Determine the greatest distance from the Sun at which these panels could still supply this power.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
b = L/(4πd2) = 3.83 × 1026/(4π × (2.28 × 1011)2)✓ 1
b = 586 W m−2✓ 1
Part (b)
P = 586 × 15 × 0.22 = 1.9 × 103 W . ECF from (a).✓ 1
Part (c)
Required brightness b = 1200/(15 × 0.22) = 364 W m−2✓ 1
d = √(L/(4πb)) = √(3.83 × 1026/(4π × 364))✓ 1
d = 2.9 × 1011 m✓ 1
Alternatively, scale the result from (b): d = 2.28 × 1011 × √(1935/1200). ECF from (b).—

Exam tip: Efficiency multiplies the incident power (b × A). Find the incident power first, then apply the efficiency.

75IBP2H-043-02-11Stellar parallax Paper 2Medium9 marks

A student analyses data for two stars from an online star catalogue.

Star P: peak wavelength 290 nm; apparent brightness 4.0 × 10−10 W m−2; parallax 0.020 arc-seconds.
Star Q: peak wavelength 725 nm; apparent brightness 9.0 × 10−10 W m−2; parallax 0.050 arc-seconds.

Use 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, L☉ = 3.83 × 1026 W and R☉ = 6.96 × 108 m.

(a)

Calculate the surface temperature of star P.

(1)
(b)

Determine the luminosity of star Q.

(3)
(c)

Determine the radius of star Q.

(3)
(d)

Using your answers, suggest the region of the Hertzsprung–Russell diagram in which star Q lies. Justify your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
T = 2.9 × 10−3/(290 × 10−9) = 1.0 × 104 K✓ 1
Part (b)
d = 1/0.050 = 20 pc = 6.18 × 1017 m✓ 1
L = 4πd2b = 4π × (6.18 × 1017)2 × 9.0 × 10−10✓ 1
L = 4.3 × 1027 W✓ 1
Part (c)
TQ = 2.9 × 10−3/(725 × 10−9) = 4000 K✓ 1
R = √(L/(4πσT4)) = √(4.3 × 1027/(4π × 5.67 × 10−8 × 40004))✓ 1
R = 4.9 × 109 m . ECF from (b).✓ 1
Part (d)
Q is cool (T ≈ 4000 K) yet has L ≈ 11 L☉, or equivalently R ≈ 7 R☉✓ 1
A main-sequence star at 4000 K would be smaller and less luminous than the Sun, so Q lies above the main sequence, in the red giant region . ECF from (b) and (c).✓ 1

Common error: Convert the peak wavelength from nm to m before using Wien's law. A wavelength left in nm gives temperatures that are 10⁹ times too small.

76IBP2H-043-02-12An eclipsing binary star Paper 2Medium7 marks

A university's robotic telescope records the spectrum of a star. The spectrum shows a continuous background crossed by dark lines. One strong dark line, at 656 nm, is due to hydrogen. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)

(a)

Explain how the dark lines in a stellar spectrum are formed.

(3)
(b)

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

(2)
(c)

This line is absorbed by hydrogen atoms with electrons in the n = 2 level. Explain why the line is much stronger in a star with surface temperature 9500 K than in the Sun (5800 K).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Continuous radiation from the hot, dense interior passes through the cooler outer atmosphere of the star✓ 1
Atoms there absorb only photons whose energies equal differences between their energy levels, so only specific wavelengths are absorbed✓ 1
The excited atoms re-emit in all directions, so less radiation reaches the observer at those wavelengths and dark lines appear✓ 1
Part (b)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108/(656 × 10−9) = 3.03 × 10−19 J✓ 1
E = 3.032 × 10−19/1.60 × 10−19 = 1.90 eV (accept 1.89 eV from a rounded intermediate value)✓ 1
Part (c)
In the Sun's cooler atmosphere, most hydrogen atoms have electrons in the ground state, n = 1, so few can absorb this line✓ 1
At 9500 K, collisions excite a much larger fraction of atoms to n = 2, so many more atoms can absorb 656 nm photons (n = 2 → n = 3)✓ 1

Exam tip: Line strength depends on how many atoms are in the right energy level. That depends on temperature much more than on how much of the element is present. Keep unrounded values in your calculator until the final step.

77IBP2H-043-02-13Apparent brightness Paper 2Medium7 marks

Sirius is the brightest star in the night sky. Its apparent brightness is 1.2 × 10−7 W m−2, its distance from Earth is 2.64 pc and its surface temperature is 9900 K. (1 pc = 3.09 × 1016 m, σ = 5.67 × 10−8 W m−2 K−4)

(a)

Calculate the luminosity of Sirius.

(2)
(b)

Determine the radius of Sirius.

(2)
(c)

A sports-stadium lamp radiates 500 W uniformly in all directions. Estimate the distance from the lamp at which its apparent brightness equals that of Sirius.

(2)
(d)

Outline one reason why the estimate in (c) is unlikely to match what an observer actually sees.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = 2.64 × 3.09 × 1016 = 8.16 × 1016 m✓ 1
L = 4πd2b = 4π × (8.16 × 1016)2 × 1.2 × 10−7 = 1.0 × 1028 W✓ 1
Part (b)
R = √(L/(4πσT4)) = √(1.0 × 1028/(4π × 5.67 × 10−8 × 99004))✓ 1
R = 1.2 × 109 m . ECF from (a).✓ 1
Part (c)
d = √(P/(4πb)) = √(500/(4π × 1.2 × 10−7))✓ 1
d ≈ 1.8 × 104 m (about 18 km)✓ 1
Part (d)
Any one valid reason :✓ 1
• the lamp does not radiate uniformly, since floodlights are directional—
• much of the lamp's output is not visible light, and its spectrum differs from that of Sirius—
• light from the lamp passes horizontally through much more absorbing and scattering air—

Exam tip: The relation b = L/(4πd²) applies to any isotropic source, not just stars. An everyday comparison like this is a good check on scale.

78IBP2S-042-04-13Stellar evolution Paper 2Medium9 marks

An astrobiology team is deciding whether to search for life on planets orbiting a main-sequence star of mass 12 solar masses. The Sun's main-sequence lifetime is 1.0×1010 years. Assume L ∝ M3.5 for main-sequence stars. Assume also that every main-sequence star fuses the same fraction of its mass before leaving the main sequence.

(a)

Show that the luminosity of the star is about 6×103 times that of the Sun.

(2)
(b)

Estimate the main-sequence lifetime of the star.

(3)
(c)

Outline the evolution of the star after it leaves the main sequence.

(3)
(d)

Life on Earth took about 109 years to develop. Comment on the team's plan.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
L/L☉ = (M/M☉)3.5 = 123.5✓ 1
= 5.99×103 ≈ 6×103✓ 1
Part (b)
Lifetime ∝ fuel available ÷ rate of use ∝ M/L (or ∝ M−2.5)✓ 1
T = 1.0×1010 × 12/(6.0×103), or T = 1.0×1010 × 12−2.5✓ 1
T ≈ 2.0×107 years. ECF from (a).✓ 1
Part (c)
It becomes a red supergiant, fusing successively heavier elements up to iron✓ 1
Its core collapses and the star explodes as a supernova✓ 1
It leaves a neutron star, or a black hole if the remnant core mass exceeds the Oppenheimer–Volkoff limit✓ 1
Part (d)
The plan is unlikely to succeed, because the star's lifetime (about 2×107 years) is much shorter than 109 years. Both the conclusion and the lifetime comparison are needed for the mark; a mention of the planet being exposed to the supernova is credit-worthy extra detail but is not required. ECF from (b).✓ 1

Common error: Do not divide the Sun's lifetime by L alone. The star also has 12 times more fuel, so the lifetime scales as M/L = M⁻².⁵.

79IBP2S-042-04-15Stellar evolution Paper 2Medium8 marks

Astronomers estimate the age of the Galaxy from the coolest white dwarfs they can find. A newly formed white dwarf has a surface temperature of 25 000 K. Its radius stays constant as it cools.

(a)

State what prevents a white dwarf from collapsing under gravity.

(1)
(b)

State the maximum possible mass of a white dwarf.

(1)
(c)

Calculate the peak wavelength of the newly formed white dwarf and state the region of the spectrum in which it lies.

(2)
(d)

Explain why very old white dwarfs are difficult to detect with optical telescopes.

(2)
(e)

Determine the factor by which the luminosity has decreased when the surface temperature reaches 6000 K.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Electron degeneracy pressure✓ 1
Part (b)
The Chandrasekhar limit, about 1.4 solar masses✓ 1
Part (c)
λmax = 2.9×10−3/25 000 = 1.2×10−7 m✓ 1
Ultraviolet✓ 1
Part (d)
Their luminosity is very low, because they are small and their luminosity falls as T4 while they cool✓ 1
Their peak emission shifts to longer wavelengths, towards the infrared, so only a small fraction of their output is visible light✓ 1
Part (e)
R is constant, so L ∝ T4✓ 1
(25 000/6000)4 = 3.0×102✓ 1

Common error: In (e), the radius cancels only because it is constant. Say so explicitly; the full Stefan–Boltzmann law still applies.

80IBP3H-E5-20r22-10Stellar spectra Paper 2Medium6 marks

The spectrum of a star has its peak intensity at a wavelength of 483 nm. Use Wien's constant 2.90×10−3 m K, σ = 5.67×10−8 W m−2 K−4, solar luminosity 3.83×1026 W and solar radius 6.96×108 m.

(a)

Calculate the surface temperature of the star, giving your answer to 3 significant figures.

(2)
(b)

The star has a radius of 1.8 times the solar radius. Using \(T=6.00\times10^{3}\) K, calculate the luminosity of the star.

(3)
(c)

Determine the luminosity of the star as a multiple of the solar luminosity, to 2 significant figures.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(T=\dfrac{2.90\times10^{-3}}{483\times10^{-9}}\)✓ 1
A1 \(6.00\times10^{3}\) K✓ 1
Part (b)
M1 \(L=4\pi\sigma R^2T^4\)✓ 1
M1 \(R=1.8\times6.96\times10^{8}=1.25\times10^{9}\) m substituted✓ 1
A1 \(1.45\times10^{27}\) W✓ 1
Part (c)
A1 3.8 (accept 3.78 to 3.79)✓ 1

Exam tip: State clearly which temperature value you carry into later parts; here the question tells you to use the rounded value.

81IBP3H-E5-20r22-12Binding energy and fusion Paper 2Medium6 marks

A proposed fusion reaction in a stellar core is deuterium + deuterium → helium-3 + neutron. The nuclear masses are: deuterium 2.013553 u, helium-3 3.014932 u, neutron 1.008665 u. Use 1 u = 931.5 MeV c−2.

(a)

Define the binding energy of a nucleus.

(1)
(b)

Calculate the energy released in one reaction, in MeV.

(3)
(c)

Explain why fusion releases energy for light nuclei but not for nuclei heavier than iron-56.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the minimum energy required to separate a nucleus completely into its individual protons and neutrons✓ 1
Part (b)
M1 mass of reactants minus mass of products \(=2(2.013553)-(3.014932+1.008665)\)✓ 1
A1 mass difference = 0.003509 u✓ 1
A1 \(E=0.003509\times931.5=3.27\) MeV✓ 1
Part (c)
binding energy per nucleon increases with nucleon number up to a maximum near iron-56✓ 1
fusing light nuclei produces a nucleus with higher binding energy per nucleon, so energy is released; fusing nuclei beyond iron would give lower binding energy per nucleon, so energy would have to be supplied✓ 1

Exam tip: Use nuclear masses throughout and keep every digit of the mass difference until the final multiplication.

82IBP3H-E5-20r22-13The Hertzsprung–Russell diagram Paper 2Medium7 marks

A red supergiant has surface temperature 3500 K and luminosity 1.20×1031 W. Use σ = 5.67×10−8 W m−2 K−4 and solar radius 6.96×108 m.

(a)

Determine the radius of the star.

(3)
(b)

Calculate the radius as a multiple of the radius of the Sun, to 3 significant figures.

(1)
(c)

State the region of the Hertzsprung–Russell diagram in which this star lies.

(1)
(d)

Explain, using the luminosity equation, why a star that is cooler than the Sun but much more luminous must have a much larger radius.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 rearrange \(R=\sqrt{\dfrac{L}{4\pi\sigma T^4}}\)✓ 1
M1 correct substitution of \(L\), \(\sigma\) and \(T\)✓ 1
A1 \(3.35\times10^{11}\) m✓ 1
Part (b)
A1 481 (using \(3.35\times10^{11}/6.96\times10^{8}\); accept 480 to 481)✓ 1
Part (c)
A1 upper right (cool and very luminous)✓ 1
Part (d)
\(L\propto R^2T^4\), so a lower \(T\) reduces the power radiated per unit area✓ 1
to give a larger total luminosity the surface area must be much greater, so the radius is large✓ 1

Common error: Temperature enters as T to the fourth power, but the radius enters only as R squared. A cool star can be very luminous only by being very large.

83IBP3H-E5-20r22-14Stellar parallax Paper 2Medium6 marks

A fishing-fleet navigation survey records a star with a parallax angle of 0.050 arcseconds and an apparent brightness of 3.4×10−10 W m−2. Use 1 pc = 3.09×1016 m, solar luminosity 3.83×1026 W.

(a)

Calculate the distance to the star in metres.

(2)
(b)

Determine the luminosity of the star.

(2)
(c)

Calculate the luminosity of the star as a multiple of solar luminosity, to 3 significant figures, and deduce its radius relative to the Sun if its surface temperature equals that of the Sun.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 \(d=\dfrac{1}{0.050}=20\) pc, then multiply by \(3.09\times10^{16}\)✓ 1
A1 \(6.18\times10^{17}\) m✓ 1
Part (b)
M1 \(L=4\pi d^2b\)✓ 1
A1 \(1.63\times10^{27}\) W✓ 1
Part (c)
A1 \(L/L_\odot=4.26\) (using \(1.6318\times10^{27}\) W; accept 4.26 to 4.27)✓ 1
A1 radius = \(\sqrt{4.26}=2.06\) times the Sun's radius, since \(L\propto R^2\) at equal temperature✓ 1

Exam tip: Use the exact distance from part (a) in later parts, not a rounded value, when the question gives no instruction otherwise.

84E-2-08Conditions for fusion Paper 2Hard9 marks

In a proposed fusion power station, deuterium and tritium nuclei fuse: ²₁H + ³₁H → ⁴₂He + ¹₀n + 17.6 MeV.

Model each nucleus as a sphere of radius R = R0A1/3, with R0 = 1.20 × 10−15 m, and assume that fusion requires the two nuclei to touch. (k = 8.99 × 109 N m² C−2, e = 1.60 × 10−19 C, kB = 1.38 × 10−23 J K−1, 1 MeV = 1.60 × 10−13 J, 1 u = 1.661 × 10−27 kg)

(a)
(i)

Show that the electric potential energy of a deuterium nucleus and a tritium nucleus that are just touching is about 0.44 MeV.

(2)
(ii)

Estimate the temperature at which the average kinetic energy of the nuclei would equal this potential energy.

(2)
(iii)

The plasma in the power station is heated to about 1.5 × 108 K. Explain why a useful rate of fusion is nevertheless obtained, and why the plasma must also have a high density.

(2)
(b)
(i)

Tritium has a half-life of 12.3 years. Suggest why the tritium fuel must be manufactured.

(1)
(ii)

Determine the mass of tritium used per day by a power station in which the fusion power is 500 MW.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Separation of centres = sum of radii = 1.20 × 10−15 × (21/3 + 31/3) = 3.24 × 10−15 m✓ 1Using one radius only scores 0 for MP1.
Ep = ke²/r = 7.10 × 10−14 J = 0.444 MeV✓ 1Must see 0.444 MeV or more figures.
Part (a)(ii)
(3/2)kBT = 7.10 × 10−14 J✓ 1Allow ECF from (a)(i).
T = 3.4 × 109 K✓ 1Accept 3.3–3.5 × 109 K.
Part (a)(iii)
The nuclei have a spread of kinetic energies: a small fraction have energies many times the average, enough to overcome the repulsion and reach the separation at which the strong force acts✓ 1Allow ECF from (a)(ii). Do not accept answers that ignore the spread of kinetic energies.
A high density gives a high collision rate, so these rare high-energy collisions happen often enough to release a useful power✓ 1
Part (b)(i)
Its half-life is very short compared with the age of the Earth, so any tritium formed long ago has decayed: almost none exists naturally✓ 1
Part (b)(ii)
Reactions per second = 500 × 106/(17.6 × 1.60 × 10−13) = 1.78 × 1020 s−1✓ 1
Mass per day = 1.78 × 1020 × 86 400 × 3 × 1.661 × 10−27 = 0.076 kg✓ 1Accept 0.076–0.077 kg (using 3.016 u gives 0.077 kg). Allow ECF from MP1.

Answers: (a)(ii) 3.4 × 109 K  ·  (b)(ii) 0.076 kg (the remaining parts are explanations — see the table above)

85E-2-11Neutron stars and nuclear density Paper 2Hard10 marks

After a supernova, the collapsed core of a massive star forms a neutron star of mass 1.4 M☉. Model the neutron star as a uniform sphere of neutrons packed together at the density of nuclear matter.

Nuclear radii are given by R = R0A1/3 with R0 = 1.20 × 10−15 m. (Mass of a nucleon mn = 1.675 × 10−27 kg, M☉ = 1.99 × 1030 kg, G = 6.67 × 10−11 N m² kg−2, c = 3.00 × 108 m s−1)

(a)
(i)

Show that the density of nuclear matter is ρ = 3mn/(4πR0³), independent of A.

(2)
(ii)

Calculate the density of nuclear matter.

(1)
(b)
(i)

Estimate the radius of the neutron star.

(2)
(ii)

Suggest why the actual radius of a neutron star of this mass is likely to be smaller than your estimate.

(1)
(c)
(i)

Calculate the escape speed from the surface of the neutron star as a fraction of the speed of light.

(2)
(ii)

A collapsing core of greater mass forms a black hole. Deduce the radius to which this neutron star would have to be compressed for the escape speed from its surface to equal c, and comment on your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Mass of a nucleus ≈ Amn; volume = (4/3)πR0³A✓ 1
ρ = Amn/((4/3)πR0³A) = 3mn/(4πR0³): A cancels✓ 1
Part (a)(ii)
ρ = 3 × 1.675 × 10−27/(4π × (1.20 × 10−15)³) = 2.3 × 1017 kg m−3✓ 1Allow ECF from (a)(i).
Part (b)(i)
V = 1.4 × 1.99 × 1030/2.31 × 1017 = 1.2 × 1013 m³✓ 1Allow ECF from (a)(ii).
R = (3V/4π)1/3 = 1.4 × 104 m✓ 1Accept 14 km.
Part (b)(ii)
The enormous gravitational forces compress the matter to more than nuclear density (especially near the centre)✓ 1
Part (c)(i)
vesc = √(2 × 6.67 × 10−11 × 2.79 × 1030/1.42 × 104) = 1.6 × 108 m s−1✓ 1Allow ECF from (b)(i).
= 0.54c✓ 1Accept 0.5–0.6c.
Part (c)(ii)
c = √(2GM/R) gives R = 2GM/c² = 4.1 × 103 m✓ 1
The neutron star (radius ≈ 14 km) is about 3 times larger than this, so it is not a black hole✓ 1Allow ECF from (b)(i).

Answers: (a)(ii) 2.3 × 1017 kg m−3  ·  (b)(i) 1.4 × 104 m  ·  (c)(i) 0.54c  ·  (c)(ii) 4.1 × 103 m (the remaining parts are explanations — see the table above)

86E-2-14The cooling of a white dwarf Paper 2Hard11 marks

A white dwarf has mass 0.60 M☉, radius 8.7 × 106 m and surface temperature 1.2 × 104 K. It consists mainly of carbon-12 and no fusion takes place in it. Its interior temperature is about 1.0 × 107 K.

Model the internal energy of the white dwarf as the random kinetic energy of its carbon nuclei, treated as the particles of an ideal gas. (M☉ = 1.99 × 1030 kg, L☉ = 3.85 × 1026 W, 1 u = 1.661 × 10−27 kg, kB = 1.38 × 10−23 J K−1, σ = 5.67 × 10−8 W m−2 K−4, 1 year = 3.16 × 107 s)

(a)
(i)

Show that the white dwarf contains about 6 × 1055 carbon nuclei.

(1)
(ii)

Determine the internal energy of the white dwarf.

(2)
(b)
(i)

Show that the luminosity of the white dwarf is about 3 × 10−3 L☉.

(2)
(ii)

Estimate the time, in years, for the white dwarf to radiate its internal energy.

(2)
(iii)

State one assumption made in your estimate in (b)(ii) and explain whether the real cooling time is longer or shorter than your estimate.

(2)
(c)
(i)

Describe and explain the path of the white dwarf on a Hertzsprung–Russell diagram as it cools.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
N = 0.60 × 1.99 × 1030/(12 × 1.661 × 10−27) = 5.99 × 1055✓ 1Must see 5.99 × 1055 or the full substitution.
Part (a)(ii)
Average kinetic energy per nucleus = (3/2)kBT = 1.5 × 1.38 × 10−23 × 1.0 × 107 = 2.07 × 10−16 J✓ 1
U = 5.99 × 1055 × 2.07 × 10−16 = 1.2 × 1040 J✓ 1Allow ECF from (a)(i).
Part (b)(i)
L = 4π(8.7 × 106)² × 5.67 × 10−8 × (1.2 × 104)⁴ = 1.12 × 1024 W✓ 1
= 2.9 × 10−3 L☉✓ 1Must see 2.9 × 10−3 or more figures.
Part (b)(ii)
t = U/L = 1.24 × 1040/1.12 × 1024 = 1.1 × 1016 s✓ 1Allow ECF from (a)(ii) and (b)(i).
= 3.5 × 108 years✓ 1Accept 3–4 × 108 years.
Part (b)(iii)
Assumption: the luminosity (surface temperature) stays constant while the energy is radiated✓ 1Accept also: no energy is gained, e.g. from contraction.
In reality the surface temperature, and so L ∝ T⁴, falls as the star cools, so energy is radiated ever more slowly: the real time is longer✓ 1MP2 depends on a relevant assumption.
Part (c)(i)
No fusion and no contraction: the radius stays (almost) constant while the temperature falls, so L falls as T⁴✓ 1
It moves down and to the right, along a line of constant radius, in the region below the main sequence✓ 1

Answers: (a)(ii) 1.2 × 1040 J  ·  (b)(ii) 3.5 × 108 years (the remaining parts are explanations — see the table above)

87E-2-16An eclipsing binary star Paper 2Hard12 marks

Stars A and B, both on the main sequence, orbit their common centre of mass in circular orbits. The Earth lies in the plane of the orbits, so each star passes in turn in front of the other. B has a smaller radius than A and is completely hidden when it passes behind A.

The graph shows the total apparent brightness b of the pair, as a fraction of its normal value, over 6 days. Treat the disc of each star as uniformly bright. (Wien constant = 2.9 × 10−3 m K, c = 3.00 × 108 m s−1)

0123456t / days0.650.700.750.800.850.900.951.001.05b / normal brightness
Brightness of the binary star against time (graph drawn to scale)
(a)
(i)

Determine the orbital period of the pair.

(1)
(ii)

Explain why the two eclipses produce different decreases in brightness.

(2)
(iii)

Show that (decrease in brightness in the deep dip)/(decrease in brightness in the shallow dip) = (TA/TB)⁴.

(2)
(iv)

The spectrum of A peaks at a wavelength of 387 nm. Determine the surface temperature of B.

(2)
(b)
(i)

The hydrogen line of laboratory wavelength 656.280 nm in the spectrum of A varies periodically between 656.103 nm and 656.457 nm. Determine the orbital speed of A.

(2)
(ii)

The same line in the spectrum of B is shifted by up to ±0.301 nm. Deduce the ratio of the masses MA/MB.

(2)
(iii)

Explain why your answer to (b)(ii) is consistent with your answer to (a)(iv).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Time between successive deep (or shallow) dips = 2.5 days✓ 1Accept 2.4–2.6 days. Half this value scores 0.
Part (a)(ii)
Deep dip: the smaller star B passes in front of A and hides part of A's disc; shallow dip: B is completely hidden behind A✓ 1
In both cases the hidden area is the same (πRB²), so the different losses mean that the two stars emit different powers per unit area (different surface temperatures)✓ 1
Part (a)(iii)
In each eclipse an area πRB² of a stellar surface is hidden; the power per unit area emitted is σT⁴✓ 1Allow ECF from (a)(ii).
Deep: Δb ∝ πRB²σTA⁴; shallow: Δb ∝ πRB²σTB⁴; the ratio is (TA/TB)⁴✓ 1
Part (a)(iv)
TA = 2.9 × 10−3/387 × 10−9 = 7.5 × 103 K✓ 1
From the graph the decreases are 0.28 and 0.08: TB = 7.5 × 103 × (0.08/0.28)1/4 = 5.5 × 103 K✓ 1Allow ECF from (a)(iii). Accept 5.3–5.7 × 103 K. Using the minimum brightness values (0.72, 0.92) instead of the decreases scores [1].
Part (b)(i)
Maximum shift Δλ = 0.177 nm (half the range)✓ 1Using the whole range (0.354 nm) scores [1] max.
vA = cΔλ/λ = 3.00 × 108 × 0.177/656.280 = 8.1 × 104 m s−1✓ 1
Part (b)(ii)
The total momentum about the centre of mass is zero (the stars are always on opposite sides with the same period), so MAvA = MBvB✓ 1
MA/MB = vB/vA = 0.301/0.177 = 1.70✓ 1Allow ECF from (b)(i). Accept 1.7. An inverted ratio (0.59) scores [1].
Part (b)(iii)
On the main sequence a more massive star is hotter (and more luminous); A is both more massive and hotter than B✓ 1Allow ECF from (a)(iv) and (b)(ii).

Answers: (a)(i) 2.5 days  ·  (a)(iv) 5.5 × 103 K  ·  (b)(i) 8.1 × 104 m s−1  ·  (b)(ii) 1.70 (the remaining parts are explanations — see the table above)

88E-2-37A transiting exoplanet Paper 2Hard19 marks

The main-sequence star K has a mass of 1.00 M☉. A planet, of much smaller mass, moves around K in a circular orbit of period 3.52 days. The orbit is seen edge-on, so once in every orbit the planet passes in front of K (a transit).

The graph shows the apparent brightness b of K, as a fraction of its normal value, during one transit. Take the duration of the transit to be the time between the two points at which the brightness is half-way down the dip. Treat the disc of K as uniformly bright.

(G = 6.67 × 10−11 N m² kg−2, M☉ = 1.99 × 1030 kg, R☉ = 6.96 × 108 m, radius of the Earth = 6.37 × 106 m)

-3-2-10123t / h0.9800.9840.9880.9920.9961.0001.004b / normal brightness
Brightness of star K during a transit (graph drawn to scale)
(a)
(i)

Show that the radius of the planet's orbit is about 7 × 109 m.

(2)
(ii)

Calculate the orbital speed of the planet.

(1)
(b)
(i)

Determine, using the graph, the radius of K in terms of R☉.

(2)
(ii)

State one assumption made in your answer to (b)(i).

(1)
(iii)

Show that the radius of the planet is r = R√(Δb/b), where Δb/b is the fractional decrease in brightness during the transit and R is the radius of K.

(2)
(iv)

Determine the radius of the planet, giving your answer in units of the Earth's radius.

(2)
(v)

The brightness takes 19 minutes to fall from its normal value to its minimum value. Show that this is consistent with your answer to (b)(iv).

(2)
(c)
(i)

Measurements of the motion of K show that the mass of the planet is 1.3 × 1027 kg. Determine the mean density of the planet.

(2)
(ii)

The mean densities of rocky planets lie between about 3000 kg m−3 and 5500 kg m−3. Suggest what your answer to (c)(i) indicates about the planet.

(1)
(d)
(i)

The fractional decrease in brightness is (0.0165 ± 0.0005) and the duration of the transit is (2.50 ± 0.05) h. The mass of K and the period are known precisely. Determine the percentage uncertainty in the radius of the planet.

(2)
(ii)

Suggest two reasons why most of the planets discovered by the transit method are large planets in small orbits.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
The gravitational force provides the centripetal force: GMm/a² = m4π²a/T², so a³ = GMT²/4π²✓ 1
a = (6.67 × 10−11 × 1.99 × 1030 × (3.52 × 86 400)²/4π²)1/3 = 6.78 × 109 m✓ 1Must see the full substitution or 6.78 × 109 m. The period must be in seconds.
Part (a)(ii)
v = 2πa/T = 2π × 6.78 × 109/(3.52 × 86 400) = 1.40 × 105 m s−1✓ 1Allow ECF from (a)(i).
Part (b)(i)
Duration of the transit from the graph = 2.5 h = 9.0 × 103 s✓ 1Accept 2.4–2.6 h.
The planet moves a distance 2R in this time: R = vt/2 = 1.40 × 105 × 9.0 × 103/2 = 6.3 × 108 m = 0.91 R☉✓ 1Allow ECF from (a)(ii). Accept 0.87–0.94 R☉. The answer must be expressed in R☉ for MP2.
Part (b)(ii)
The planet crosses along a diameter of the disc of K (or: the planet moves in a straight line at constant speed across the disc / the motion of K itself is negligible)✓ 1Any one.
Part (b)(iii)
The planet hides an area πr² of the disc of K, of area πR², which is uniformly bright✓ 1
So the fraction of the light blocked is Δb/b = πr²/πR², giving r = R√(Δb/b)✓ 1
Part (b)(iv)
From the graph Δb/b = 0.0165✓ 1Accept 0.016–0.017.
r = 6.3 × 108 × √0.0165 = 8.1 × 107 m = 13 Earth radii✓ 1Allow ECF from (b)(i) and (b)(iii). Accept 12–13.5 Earth radii; the answer must be expressed in Earth radii for MP2.
Part (b)(v)
While the brightness falls, the planet moves from first touching the edge of the disc to lying wholly inside it: a distance equal to its own diameter 2r✓ 1
2r = 1.40 × 105 × 19 × 60 = 1.6 × 108 m, so r = 8.0 × 107 m, in agreement with (b)(iv)✓ 1Allow ECF from (a)(ii) and (b)(iv). A comparison with the answer to (b)(iv) is needed.
Part (c)(i)
Volume = (4/3)πr³ = (4/3)π × (8.1 × 107)³ = 2.2 × 1024 m³✓ 1Allow ECF from (b)(iv).
ρ = 1.3 × 1027/2.2 × 1024 = 5.8 × 102 kg m−3✓ 1Accept 5.0–6.8 × 102 kg m−3.
Part (c)(ii)
Its density is far below that of rock (and even below that of water), so it is a gas giant made mainly of hydrogen and helium, not a rocky planet✓ 1Allow ECF from (c)(i).
Part (d)(i)
With M and T exact, R ∝ t: 0.05/2.50 = 2.0 %; the fractional dip contributes ½ × (0.0005/0.0165) = ½ × 3.0 % = 1.5 %✓ 1
Percentage uncertainty in r = 2.0 % + 1.5 % = 3.5 %✓ 1Allow ECF from (b). Adding 3.0 % without halving (5.0 %) scores [1].
Part (d)(ii)
The dip Δb/b ∝ r², so only a large planet blocks enough light to be detected above the scatter of the measurements (an Earth-sized planet would give a dip of only about 1 × 10−4 here)✓ 1Any two of: (1) the r² dependence of the dip; (2) a short period means many transits can be observed and confirmed in a given observing time; (3) a transit is seen only if the orbit is close to edge-on, and the range of orientations for which this happens is larger when the orbit is small compared with the star. [1] each, [2 max].
A planet in a small orbit has a short period, so its transits repeat often and can be confirmed quickly; also its transits are seen over a wider range of orbit orientations✓ 1

Answers: (a)(ii) 1.40 × 105 m s−1  ·  (b)(i) 0.91 R☉  ·  (b)(iv) 13 Earth radii  ·  (c)(i) 5.8 × 102 kg m−3  ·  (d)(i) 3.5 % (the remaining parts are explanations — see the table above)

89IBP2H-043-00-11The Sun as a red giant Paper 2Hard9 marks

An astronomy outreach team collects the following data for a bright star X.
Property — Value
Parallax angle — 8.0 milliarcseconds
Apparent brightness — 2.0×10−9 W m−2
Peak wavelength of spectrum — 1.0 μm
1 pc = 3.09×1016 m; Wien constant = 2.9×10−3 m K; σ = 5.67×10−8 W m−2 K−4; RSun = 6.96×108 m.

(a)

Calculate the distance to X in metres.

(2)
(b)

Determine the luminosity of X.

(2)
(c)

Calculate the surface temperature of X.

(1)
(d)

Determine the radius of X.

(2)
(e)

Identify the type of star X is and outline its likely future if its mass is similar to that of the Sun.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
d = 1/p = 1/0.0080 = 125 pc✓ 1
d = 125 × 3.09×1016 = 3.86×1018 m ≈ 3.9×1018 m✓ 1
Part (b)
L = 4πd2b = 4π(3.86×1018)2 × 2.0×10−9✓ 1
L ≈ 3.7×1029 W; accept 3.7–3.8×1029 W, since using the rounded d = 3.9×1018 m gives 3.8×1029 W✓ 1
ECF from (a)—
Part (c)
T = 2.9×10−3 / 1.0×10−6 = 2.9×103 K✓ 1
Part (d)
R = √(L / 4πσT4), with values substituted correctly✓ 1
R ≈ 8.6×1010 m, which is about 124 RSun (accept 8.6–8.7×1010 m)✓ 1
ECF from (b) and (c)—
Part (e)
Red giant, because it has a low surface temperature, a large radius and a high luminosity✓ 1
It will eject its outer layers as a planetary nebula, leaving its core as a white dwarf✓ 1

Exam tip: Convert parallax from milliarcseconds to arcseconds before using d = 1/p: 8.0 mas is 0.0080 arcsec. Carry unrounded values through later parts to avoid rounding drift.

90IBP2H-043-00-12Nuclear fusion Paper 2Hard8 marks

In the core of a main-sequence star, the net effect of hydrogen fusion is that four hydrogen nuclei form one helium-4 nucleus, two positrons and two neutrinos. The positrons annihilate with electrons.
Atomic mass of hydrogen-1 = 1.007825 u; atomic mass of helium-4 = 4.002603 u; 1 u = 931.5 MeV c−2 = 1.661×10−27 kg; e = 1.60×10−19 C.
The star has a mass of 3.98×1030 kg, of which 70% is hydrogen, and a constant luminosity of 4.3×1027 W. Assume 10% of this hydrogen is fused during the main-sequence stage. 1 year = 3.15×107 s.

(a)

Show that the energy released per helium-4 nucleus formed, including positron annihilation, is about 27 MeV.

(2)
(b)

Determine the energy released per kilogram of hydrogen fused.

(2)
(c)

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

(3)
(d)

State one assumption that makes this an estimate.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Mass defect = 4 × 1.007825 − 4.002603 = 0.028697 u (atomic masses include the annihilating electrons)✓ 1
E = 0.028697 × 931.5 = 26.7 MeV✓ 1
Part (b)
26.7 MeV = 4.28×10−12 J, and the mass of four H atoms = 4.0313 × 1.661×10−27 = 6.70×10−27 kg✓ 1
Energy per kg = 4.28×10−12 / 6.70×10−27 = 6.4×1014 J kg−1✓ 1
Allow ECF from the previous part.—
Part (c)
Mass fused = 0.10 × 0.70 × 3.98×1030 = 2.79×1029 kg✓ 1
Total energy = 2.79×1029 × 6.4×1014 = 1.78×1044 J✓ 1
t = 1.78×1044 / 4.3×1027 = 4.1×1016 s ≈ 1.3×109 years✓ 1
Allow ECF from the previous part.—
Part (d)
Any one of the following: the luminosity is constant over the whole main sequence; exactly 10% of the hydrogen (that in the core) is fused; all the fusion energy leaves as luminosity, ignoring the energy carried away by neutrinos✓ 1

Common error: If nuclear masses are used instead of atomic masses, the positron annihilation energy must be added separately. Using atomic masses includes it automatically.

91IBP2H-043-00-14The Sun as a red giant Paper 2Hard7 marks

A space agency is studying a rocky exoplanet that orbits a red dwarf star. The star has a surface temperature of 3200 K and a radius of 0.20 RSun. The planet's orbital radius is 0.030 AU.
RSun = 6.96×108 m; 1 AU = 1.50×1011 m; σ = 5.67×10−8 W m−2 K−4.

(a)

Calculate the luminosity of the red dwarf.

(2)
(b)

Calculate the intensity of the star's radiation at the planet.

(2)
(c)

The planet absorbs all the radiation incident on it and radiates as a black body from its whole surface. Determine the planet's equilibrium temperature.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
L = 4π(0.20 × 6.96×108)2 × 5.67×10−8 × 32004✓ 1
L = 1.4×1024 W (accept 1.45×1024 W)✓ 1
Part (b)
I = L/(4πd2) with d = 0.030 × 1.50×1011 = 4.5×109 m✓ 1
I = 5.7×103 W m−2 (accept 5.5×103 W m−2 from L = 1.4×1024 W; allow ECF from (a))✓ 1
Part (c)
Power absorbed = πr2I✓ 1
Power emitted = 4πr2σT4, so T = (I/4σ)1/4✓ 1
T ≈ 4.0×102 K (accept 395–398 K, i.e. 3.9×102 K from rounded values; allow ECF from (b))✓ 1

Common error: The planet intercepts radiation over its cross-section πr² but emits over its whole surface 4πr². Using the same area for both overestimates T.

92IBP2H-043-01-11Conditions for fusion Paper 2Hard10 marks

A prototype fusion power plant is designed to use the deuterium–tritium reaction
\(^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0\text{n}\)
Atomic masses: 2H = 2.014102 u, 3H = 3.016049 u, 4He = 4.002603 u. Neutron mass = 1.008665 u.
1 u = 931.5 MeV c−2 = 1.661×10−27 kg; 1 MeV = 1.60×10−13 J.

(a)

Calculate the energy released in one reaction, in MeV.

(2)
(b)

The initial kinetic energy of the reactants is negligible. Determine the kinetic energy of the neutron.

(3)
(c)

The plant produces a fusion power of 500 MW. Calculate the mass of deuterium consumed in one day.

(3)
(d)

Explain why the fuel must be at a temperature of order 108 K.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Δm = (2.014102 + 3.016049) − (4.002603 + 1.008665) = 0.018883 u✓ 1
E = 0.018883 × 931.5 = 17.6 MeV✓ 1
Part (b)
The total momentum is zero, so the helium nucleus and the neutron have momenta of equal magnitude✓ 1
Ek = p2/2m, so Ek,n/Ek,He = mHe/mn✓ 1
Ek,n = 17.6 × 4.0026/(4.0026 + 1.0087) = 14.1 MeV; ECF from (a)✓ 1
Part (c)
Energy per reaction = 17.59 × 1.60×10−13 = 2.81×10−12 J, so the reaction rate = 5.00×108/2.81×10−12 = 1.78×1020 s−1✓ 1
Reactions per day = 1.78×1020 × 86400 = 1.53×1025✓ 1
Mass = 1.53×1025 × 2.014102 × 1.661×10−27 = 0.051 kg; ECF from (a)✓ 1
Part (d)
Both nuclei are positively charged and repel each other electrostatically✓ 1
A very high temperature gives the nuclei enough kinetic energy to approach closely enough for the strong nuclear force to act and fusion to occur✓ 1

Exam tip: In a two-body decay from rest, the lighter particle takes the larger share of the kinetic energy, in the inverse ratio of the masses.

93IBP2H-043-01-12The cooling of a white dwarf Paper 2Hard8 marks

A cold molecular cloud consists of hydrogen molecules (H2). It has a radius of 2.0×1016 m and a temperature of 15 K. Model the cloud as a uniform sphere of mass M containing N molecules. Its gravitational potential energy is \(-\frac{3GM^2}{5R}\) and the total random kinetic energy of its molecules is \(\frac{3}{2}NkT\).
Mass of an H2 molecule, m = 3.34×10−27 kg; G = 6.67×10−11 N m2 kg−2; k = 1.38×10−23 J K−1.

(a)

Outline what is meant by the Jeans criterion.

(2)
(b)

Show that the minimum mass for collapse is \(M_J = \frac{5RkT}{2Gm}\).

(2)
(c)

Calculate MJ for this cloud.

(2)
(d)

The actual mass of the cloud is 3.0×1031 kg. Determine the maximum temperature at which this cloud, with the same radius, would collapse.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A gas cloud will begin to collapse under gravity✓ 1
when the magnitude of its gravitational potential energy exceeds the total random kinetic energy of its particles, which means the cloud mass exceeds the Jeans mass✓ 1
Part (b)
Set 3GM2/(5R) = (3/2)NkT and substitute N = M/m✓ 1
3GM2/(5R) = 3MkT/(2m), which rearranges to M = 5RkT/(2Gm)✓ 1
Part (c)
MJ = (5 × 2.0×1016 × 1.38×10−23 × 15)/(2 × 6.67×10−11 × 3.34×10−27)✓ 1
= 4.6×1031 kg (about 23 solar masses)✓ 1
Part (d)
3.0×1031 kg < MJ, so the cloud does not collapse at 15 K. Since MJ ∝ T, the cloud must cool until MJ = 3.0×1031 kg✓ 1
T = 15 × 3.0×1031/4.6×1031 ≈ 9.7–9.8 K (9.7 K with the unrounded MJ; 9.8 K with 4.6×1031 kg), so collapse occurs below about 9.7 K. Allow ECF from (c)✓ 1

Common error: N is the number of molecules, not the number of moles. Use N = M/m, with m the mass of one H₂ molecule.

94IBP2H-043-01-14Nuclear fusion Paper 2Hard9 marks

A main-sequence star has a mass of 6.0 M☉. For main-sequence stars, L ∝ M3.5. In the star's core, the net effect of hydrogen fusion is that four protons form one helium-4 nucleus and release 26.7 MeV.
L☉ = 3.83×1026 W; M☉ = 1.99×1030 kg; proton mass = 1.67×10−27 kg; 1 MeV = 1.60×10−13 J; 1 year = 3.15×107 s.

(a)

Show that the luminosity of the star is about 2×1029 W.

(2)
(b)

Calculate the mass of hydrogen converted to helium each second.

(3)
(c)

Assume that 10% of the star's mass is fused during its time on the main sequence. Estimate its main-sequence lifetime in years.

(2)
(d)

Explain why this star spends much less time on the main sequence than the Sun.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
L = 6.03.5 L☉ = 529 L☉✓ 1
L = 529 × 3.83×1026 = 2.03×1029 W✓ 1
Part (b)
Energy per helium nucleus formed = 26.7 × 1.60×10−13 = 4.27×10−12 J✓ 1
Helium nuclei formed per second = 2.03×1029/4.27×10−12 = 4.74×1040 s−1✓ 1
Mass of hydrogen used per second = 4.74×1040 × 4 × 1.67×10−27 = 3.17×1014 ≈ 3.2×1014 kg s−1; ECF from (a)✓ 1
Part (c)
Mass fused = 0.10 × 6.0 × 1.99×1030 = 1.19×1030 kg✓ 1
Lifetime = 1.19×1030/3.17×1014 = 3.77×1015 s ≈ 1.2×108 years (3.7×1015 s from the rounded 3.2×1014 kg s−1 is also accepted); ECF from (b)✓ 1
Part (d)
The star has 6 times the Sun's fuel, but its luminosity, and so its rate of fuel use, is about 500 times greater✓ 1
The lifetime therefore scales as M/L ∝ M−2.5, which is about 1/90 of the Sun's lifetime✓ 1

Common error: Each helium nucleus formed uses four protons. Forgetting the factor of 4 underestimates the hydrogen consumed.

95IBP2H-043-02-14Nuclear fusion Paper 2Hard8 marks

Large underground detectors measure the flux of neutrinos from the Sun. In the proton–proton chain, the overall reaction is
4 1H → 4He + 2e+ + 2νe
Nuclear masses: proton = 1.007276 u, helium-4 nucleus = 4.001506 u, positron = 0.000549 u. 1 u = 931.5 MeV c−2. L☉ = 3.83 × 1026 W, Earth–Sun distance = 1.50 × 1011 m, e = 1.60 × 10−19 C.

(a)

Explain why fusion in the Sun's core requires both a very high temperature and a very high density.

(2)
(b)

Show that the energy released in the overall reaction, not including positron annihilation, is about 25 MeV.

(2)
(c)

Including the annihilation of the positrons, each overall reaction releases 26.7 MeV. Estimate the number of helium nuclei produced in the Sun each second.

(2)
(d)

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

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A high temperature gives protons enough kinetic energy to overcome, or tunnel through, the electrostatic repulsion✓ 1
A high density makes collisions frequent enough to give a significant reaction rate✓ 1
Part (b)
Δm = 4(1.007276) − 4.001506 − 2(0.000549) = 0.026500 u✓ 1
E = 0.026500 × 931.5 = 24.7 MeV ≈ 25 MeV✓ 1
Part (c)
Energy per reaction = 26.7 × 106 × 1.60 × 10−19 = 4.27 × 10−12 J✓ 1
N = 3.83 × 1026/4.27 × 10−12 = 9.0 × 1037 s−1✓ 1
Part (d)
Neutrinos emitted per second = 2 × 9.0 × 1037 = 1.8 × 1038 s−1✓ 1
Flux = 1.8 × 1038/(4π × (1.50 × 1011)2) ≈ 6.3 × 1014 m−2 s−1. Accept 6.3–6.4 × 1014 m−2 s−1 . Allow error carried forward from (c).✓ 1

Common error: When nuclear masses are used, the mass of the two positrons must be subtracted. Leaving them out overestimates the energy released.

96IBP2H-043-02-16Stellar evolution Paper 2Hard8 marks

A nearby white dwarf has a surface temperature of 25 000 K and a radius of 7.0 × 106 m. It is 2.6 pc from Earth. (σ = 5.67 × 10−8 W m−2 K−4, 1 pc = 3.09 × 1016 m, L☉ = 3.83 × 1026 W)

(a)

Outline why a white dwarf does not collapse under gravity even though no fusion takes place in it.

(2)
(b)

State the Chandrasekhar limit and outline its significance for the evolution of a stellar core.

(2)
(c)

Calculate the luminosity of the white dwarf.

(2)
(d)

Determine the apparent brightness of the white dwarf at Earth.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The electrons are packed so closely that they form a degenerate gas✓ 1
The Pauli exclusion principle produces electron degeneracy pressure, which balances gravity✓ 1
Part (b)
It is the maximum mass of a white dwarf, about 1.4 M☉✓ 1
A remnant core more massive than this cannot be supported by electron degeneracy pressure, so it collapses further to a neutron star or black hole✓ 1
Part (c)
L = 4πR2σT4 = 4π × (7.0 × 106)2 × 5.67 × 10−8 × 25 0004✓ 1
L = 1.4 × 1025 W (about 0.036 L☉)✓ 1
Part (d)
d = 2.6 × 3.09 × 1016 = 8.03 × 1016 m✓ 1
b = L/(4πd2) = 1.36 × 1025/(4π × (8.03 × 1016)2) = 1.7 × 10−10 W m−2 . ECF from (c).✓ 1

Common error: White dwarfs are hot but dim because their surface area is tiny. A high temperature does not by itself mean a high luminosity.

97IBP2S-042-04-16Stellar evolution Paper 2Hard9 marks

In a star cluster, all the stars formed at the same time. Today, the most massive star still on the main sequence has a mass of 1.6 solar masses. The Sun's main-sequence lifetime is 1.0×1010 years, and L ∝ M3.5 for main-sequence stars.

(a)

Outline the condition for a main-sequence star to be in equilibrium.

(2)
(b)

Show that the main-sequence lifetime T of a star of mass M is proportional to M−2.5.

(2)
(c)

Determine the age of the cluster.

(2)
(d)

The cluster originally contained a star of mass 20 solar masses. Explain what this star is likely to be now.

(3)
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Marking pointMarkNotes
Part (a)
The outward pressure (thermal gas pressure and radiation pressure) is maintained by energy released in core fusion✓ 1
This outward pressure balances the inward gravitational pull at every layer of the star✓ 1
Part (b)
T ∝ energy available ÷ rate of energy output ∝ M/L✓ 1
L ∝ M3.5, so T ∝ M/M3.5 = M−2.5✓ 1
Part (c)
Age = main-sequence lifetime of a 1.6 M☉ star = 1.0×1010 × 1.6−2.5✓ 1
Age = 3.1×109 years✓ 1
Part (d)
Its lifetime is 1.0×1010 × 20−2.5 = 5.6×106 years✓ 1
This is much shorter than the cluster age, so the star left the main sequence long ago, became a red supergiant and exploded as a supernova. ECF from (c).✓ 1
It is now a neutron star or a black hole✓ 1

Exam tip: The heaviest star still on the main sequence acts as a clock. Its main-sequence lifetime equals the age of the cluster, because all the stars formed together.

98IBP2S-042-00-15Stellar evolution Paper 2Hard8 marks

Mission planners are judging whether planets around a main-sequence star of mass 10 M☉ are worth searching for life. For main-sequence stars, L ∝ M3.5. Assume that the mass available for fusion is the same fraction of the total mass for every main-sequence star. The Sun's main-sequence lifetime is 1.0×1010 years. 1 AU is the Earth–Sun distance.

(a)

Calculate the luminosity of the star in terms of the Sun's luminosity L☉.

(1)
(b)

Determine the distance from the star, in AU, at which a planet would receive the same intensity of radiation as Earth receives from the Sun.

(2)
(c)

Calculate the main-sequence lifetime of the star.

(2)
(d)

Suggest, with a reason, whether complex life is likely to develop on a planet orbiting this star.

(1)
(e)

Outline the likely evolution of this star after it leaves the main sequence.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
L = 103.5 L☉ ≈ 3.2×103 L☉✓ 1
Part (b)
Same intensity requires L/(4πd2) = L☉/(4π(1 AU)2), so d = √(L/L☉) AU✓ 1
d = √3162 ≈ 56 AU✓ 1
Allow ECF from (a).—
Part (c)
T ∝ M/L ∝ M−2.5, so T = 1.0×1010 × 10−2.5✓ 1
T ≈ 3.2×107 years✓ 1
Part (d)
Unlikely: the star stays on the main sequence for only ~107 years, far shorter than the ~109 years life took to become complex on Earth. Accept any equivalent valid time comparison.✓ 1
Part (e)
It becomes a red supergiant, fusing successively heavier elements up to iron✓ 1
It ends in a supernova, leaving a neutron star (or possibly a black hole)✓ 1

Common error: Do not use L ∝ M³.⁵ directly for the lifetime. The lifetime is the fuel supply (∝ M) divided by the luminosity, which gives T ∝ M⁻².⁵.

99IBP3H-E5-20r22-15Nuclear fusion Paper 2Hard8 marks

In the proton–proton chain, four hydrogen-1 nuclei fuse to form one helium-4 nucleus and release 26.7 MeV. The Sun's luminosity is 3.83×1026 W. Use 1 eV = 1.60×10−19 J, proton mass 1.67×10−27 kg, \(c=3.00\times10^{8}\) m s−1. Use the unrounded value from each part in the next.

(a)

Calculate the number of helium nuclei produced each second in the Sun.

(3)
(b)

Calculate the mass of hydrogen fused each second.

(2)
(c)

Calculate the mass converted into energy each second, and hence the percentage of the fused hydrogen mass that is converted into energy.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 energy per reaction \(=26.7\times10^{6}\times1.60\times10^{-19}=4.27\times10^{-12}\) J✓ 1
M1 number per second \(=\dfrac{3.83\times10^{26}}{4.27\times10^{-12}}\)✓ 1
A1 \(8.97\times10^{37}\) s−1✓ 1
Part (b)
M1 mass \(=4\times1.67\times10^{-27}\times\) number per second✓ 1
A1 \(5.99\times10^{11}\) kg s−1✓ 1
Part (c)
M1 \(\Delta m=\dfrac{L}{c^2}=\dfrac{3.83\times10^{26}}{(3.00\times10^{8})^2}\)✓ 1
A1 \(4.26\times10^{9}\) kg s−1✓ 1
A1 \(\dfrac{4.26\times10^{9}}{5.99\times10^{11}}\times100=0.711\%\)✓ 1

Exam tip: Work in SI units throughout: convert MeV to joules before dividing the power by the energy per reaction.