E.5 Fusion and stars flashcards: IB Physics SL and HL

All 75 E.5 Fusion and stars cards

1. DefinitionSL & HL

   Define nuclear fusion.

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   Nuclear fusion is the joining together of two light nuclei to form a heavier nucleus, with the release of energy. Exam tip: the mark scheme wants "two light NUCLEI" + "form a single heavier nucleus" + "energy released" — writing "atoms" or "particles" is not credited, and omitting "energy released" costs a mark. The energy comes from an increase in binding energy per nucleon (valid only for products lighter than iron-56), so the total rest mass falls and E = Δmc². Energy is a scalar, measured in J or MeV.

2. DefinitionSL & HL

   Define the Coulomb barrier and state the conditions needed to overcome it.

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   The Coulomb barrier is the electrostatic potential energy barrier that two positively charged nuclei must surmount, arising from their mutual repulsion, before they come close enough (≈ 10⁻¹⁵ m) for the attractive strong nuclear force to act and fusion to occur. Conditions: very high temperature (≈ 10⁷ K in the Sun's core) so nuclei have large random kinetic energy, and very high density/pressure so the collision rate is large enough. Exam tip: state BOTH temperature and density — "it must be hot" alone is one mark of two. Quantum tunnelling and the high-energy tail of the Maxwell distribution allow fusion below the classical barrier energy.

3. DefinitionSL & HL

   Define plasma and explain why controlled fusion on Earth is not yet commercially viable.

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   A plasma is an ionised gas of free nuclei and electrons, electrically neutral overall, formed when a gas is heated so strongly that electrons are stripped from atoms. Viability problem: the plasma must be held at ≈ 10⁸ K at sufficient density for long enough (confinement time), but no solid container can touch it, so magnetic confinement (tokamak) or inertial confinement is used; energy input to heat and confine currently exceeds energy output. Exam tip: "it is too hot" is not enough — the mark scheme wants the confinement issue plus the statement that break-even/net energy gain has not been sustained.

4. DefinitionSL & HL

   Define stellar (hydrostatic) equilibrium.

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   Stellar equilibrium is the state in which the inward gravitational force on each layer of a star is exactly balanced by the outward radiation pressure and gas (thermal) pressure produced by fusion in the core, so the star's radius stays constant. Exam tip: the two-mark answer is "gravitational pressure/force inwards" AND "radiation and gas pressure from fusion outwards", stated as balanced. A very common loss is writing "gravity balances the explosion". The equilibrium is stable and self-regulating: extra fusion raises T and pressure, the star expands and cools, and the fusion rate falls again.

5. DefinitionSL & HL

   Define the luminosity of a star.

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   The luminosity L of a star is the total power radiated by the star over all wavelengths, i.e. the total energy emitted per unit time from its entire surface. SI unit: watt (W); it is a scalar. Exam tip: luminosity is an intrinsic property of the star and does NOT depend on how far away the observer is — the mark is lost if the definition mentions the observer or "brightness at Earth". Often quoted relative to the Sun, L_sun = 3.85 × 10²⁶ W. Determined from surface area and surface temperature via L = σAT⁴.

6. DefinitionSL & HL

   Define the apparent brightness of a star.

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   Apparent brightness b is the power received from the star per unit area at the observer, measured perpendicular to the direction of the star. SI unit: W m⁻² ; it is a scalar. Exam tip: the mark scheme insists on "per unit area" and "at the observer/at Earth" — "how bright the star looks" scores zero. Apparent brightness depends on both the star's luminosity and its distance, b = L/(4πd²), which is why a nearby dim star can appear brighter than a distant luminous one. Interstellar dust absorption makes measured b slightly too small.

7. DefinitionSL & HL

   Define a black body and state how a star's surface temperature is obtained from its spectrum.

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   A black body is a perfect emitter and perfect absorber: it absorbs all radiation incident on it at every wavelength and emits the maximum possible power at every wavelength for its temperature. A star is modelled as a black body, so its continuous spectrum has a characteristic peak; measuring the wavelength λ_max of that peak and applying Wien's displacement law λ_max T = 2.9 × 10⁻³ m K gives the surface temperature T in kelvin. Exam tip: T is the SURFACE (photosphere) temperature, not the core temperature — a frequent and costly slip.

8. DefinitionSL & HL

   Define stellar parallax and the parallax angle.

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   Stellar parallax is the apparent shift in position of a nearby star against the background of very distant stars, observed as the Earth moves around the Sun. The parallax angle p is HALF the total angular shift over six months — equivalently the angle subtended at the star by a baseline of 1 AU (the Earth–Sun radius). It is measured in arcseconds (1″ = 1/3600 of a degree). Exam tip: forgetting the factor of two (using the full six-month shift as p) is the classic error, and it halves the calculated distance.

9. DefinitionSL & HL

   Define the parsec.

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   One parsec is the distance at which a star has a parallax angle of exactly one arcsecond, i.e. the distance at which 1 AU subtends an angle of 1″. 1 pc = 3.09 × 10¹⁶ m = 3.26 ly. Exam tip: the definition must reference BOTH 1 AU (or the Earth's orbital radius) and 1 arcsecond; "a big astronomical distance" earns nothing. The parsec exists purely so that d/pc = 1/(p/arcsec), which is why the equation only works in those two units. Nearest star Proxima Centauri: p ≈ 0.77″, d ≈ 1.3 pc.

10. DefinitionSL & HL

    Define the astronomical unit and the light year.

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    The astronomical unit (AU) is the mean distance between the Earth and the Sun, 1 AU = 1.50 × 10¹¹ m. The light year (ly) is the distance travelled by light in vacuum in one year, 1 ly = c × 1 yr = 3.00 × 10⁸ × 3.15 × 10⁷ ≈ 9.46 × 10¹⁵ m. Exam tip: both are distances, not times — describing a light year as "how long light takes" scores zero. Useful ordering for estimates: 1 AU < 1 ly < 1 pc, with 1 pc = 206 265 AU = 3.26 ly.

11. DefinitionSL & HL

    Define a main-sequence star.

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    A main-sequence star is a star that is in stable equilibrium and is fusing hydrogen into helium in its core, principally by the proton–proton chain. Main-sequence stars occupy the broad diagonal band running from hot, luminous stars at the top left of the Hertzsprung–Russell diagram to cool, dim stars at the bottom right. Exam tip: the required mark is "hydrogen fusing to helium in the core" — "a normal star" is not credited. Mass determines position: more massive means hotter, more luminous, larger radius and a much shorter main-sequence lifetime. Our Sun is a G2 main-sequence star.

12. DefinitionSL & HL

    Define a red giant and a red supergiant.

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    A red giant is a cool (≈ 3000–5000 K), very large, luminous star formed when a low- or intermediate-mass star (≲ 8 M_sun) exhausts core hydrogen: the core contracts and heats while the outer envelope expands and cools, and helium fusion begins in or around the core. A red supergiant is the corresponding, far larger and more luminous, evolved stage of a massive star (≳ 8 M_sun), which can fuse successively heavier elements up to iron in onion-like shells. Exam tip: both are red because their SURFACE is cool; they are luminous only because L = σAT⁴ with an enormous surface area A.

13. DefinitionSL & HL

    Define a white dwarf and state the Chandrasekhar limit.

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    A white dwarf is the hot, dense, Earth-sized remnant core left after a low-mass star ejects its outer layers as a planetary nebula. No fusion occurs in it; it is supported against gravitational collapse by electron degeneracy pressure and slowly cools and fades. The Chandrasekhar limit is the maximum mass a white dwarf can have, approximately 1.4 solar masses; above this, electron degeneracy pressure cannot support the star. Exam tip: white dwarfs sit at the bottom LEFT of the HR diagram — high surface temperature but very low luminosity because their radius is tiny.

14. DefinitionSL & HL

    Define a neutron star and a black hole, and state the Oppenheimer–Volkoff limit.

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    A neutron star is the extremely dense remnant core (radius ≈ 10 km) left after a supernova, composed almost entirely of neutrons and supported against collapse by neutron degeneracy pressure. The Oppenheimer–Volkoff limit is the maximum mass of a neutron star, approximately 2–3 solar masses. If the remnant core exceeds this limit, no known pressure can halt the collapse and a black hole forms — a region whose gravitational field is so strong that not even light can escape. Exam tip: these limits apply to the mass of the REMNANT core, not the mass of the original main-sequence star.

15. DefinitionSL & HL

    Define nucleosynthesis and state where elements heavier than iron are formed.

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    Nucleosynthesis is the formation of heavier nuclei from lighter ones by nuclear fusion. In main-sequence and evolved stars, successive fusion stages build elements up to iron-56, which lies at the peak of the binding-energy-per-nucleon curve. Beyond iron, fusion is endothermic (it absorbs energy), so heavier elements cannot be produced in normal stellar burning; they are formed by rapid neutron capture and subsequent β⁻ decay in the extreme conditions of a supernova explosion. Exam tip: the mark scheme requires the reason — "binding energy per nucleon decreases beyond Fe-56, so energy must be supplied".

16. DefinitionSL & HL

    State the stellar spectral classes in order and what determines the class.

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    The spectral classes in order of DECREASING surface temperature are O, B, A, F, G, K, M (mnemonic: "Oh Be A Fine Girl/Guy, Kiss Me"). Approximate surface temperatures: O ≈ 30 000 K and above (blue), B ≈ 20 000 K (blue-white), A ≈ 9000 K (white), F ≈ 7000 K (yellow-white), G ≈ 5500 K (yellow, the Sun), K ≈ 4500 K (orange), M ≈ 3000 K (red). Class is determined by surface temperature, revealed by the colour/λ_max and by which absorption lines appear in the spectrum, since temperature controls which atoms are ionised or excited. Class fixes T but NOT L: a K-class red giant and a K-class main-sequence star share a surface temperature but differ enormously in radius and hence luminosity. Exam tip: the sequence runs hot → cool, the same direction as the reversed temperature axis of an HR diagram.

17. EquationSL & HLData booklet: Yes

    State the Stefan–Boltzmann law for the luminosity of a star and define every symbol.

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    L = σAT⁴, and for a spherical star A = 4πR² so L = 4πR²σT⁴. L = luminosity, total power radiated (W); σ = Stefan–Boltzmann constant = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ ; A = surface area (m²); R = radius (m); T = ABSOLUTE surface temperature (K). Valid for a black body (emissivity e = 1); for a general body L = eσAT⁴ with 0 ≤ e ≤ 1. Data-booklet status: printed. Common misuse: using T in °C, or forgetting the fourth power — doubling T multiplies L by 16. Check: R = 7.0 × 10⁸ m, T = 5800 K gives L ≈ 3.9 × 10²⁶ W, the Sun.

18. EquationSL & HLData booklet: Yes

    State the equation relating apparent brightness, luminosity and distance.

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    b = L/(4πd²). b = apparent brightness (W m⁻²); L = luminosity (W); d = distance from star to observer (m). Valid for a point source radiating isotropically into empty space with negligible absorption. Data-booklet status: printed. The 4πd² is the surface area of a sphere of radius d over which the power spreads. Common misuse: substituting d in parsecs or light years — it must be in metres for b in W m⁻². Rearranged forms you must know: L = 4πd²b and d = √(L/4πb). Check: the Sun, L = 3.85 × 10²⁶ W at d = 1.5 × 10¹¹ m gives b ≈ 1.4 × 10³ W m⁻², the solar constant.

19. EquationSL & HLData booklet: Yes

    State Wien's displacement law and define every symbol.

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    λ_max T = 2.9 × 10⁻³ m K. λ_max = the wavelength at which the emitted power per unit wavelength is a maximum (m); T = absolute surface temperature of the black body (K); the constant 2.9 × 10⁻³ m K is Wien's constant. Valid for black-body (continuous thermal) radiation. Data-booklet status: printed. Common misuse: leaving λ_max in nanometres — convert to metres first, or the temperature is out by 10⁹. Note λ_max ∝ 1/T: hotter stars peak at shorter wavelengths (bluer). Check: the Sun at 5800 K gives λ_max = 2.9 × 10⁻³/5800 = 5.0 × 10⁻⁷ m = 500 nm, in the green-yellow.

20. EquationSL & HLData booklet: Yes

    State the stellar parallax equation and the units it demands.

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    d = 1/p. d = distance to the star measured in PARSECS; p = parallax angle measured in ARCSECONDS. Valid only in these units, and only for relatively nearby stars (p greater than about 0.001″, i.e. d less than about 1000 pc) because for smaller angles the measurement uncertainty swamps the value. Data-booklet status: printed. Common misuse: putting p in degrees or radians, or using the full six-month angular shift instead of half of it. Check: p = 0.10″ gives d = 10 pc = 32.6 ly = 3.09 × 10¹⁷ m. Conversions to memorise: 1 pc = 3.26 ly = 3.09 × 10¹⁶ m.

21. EquationSL & HLData booklet: No – memorise

    State the mass–luminosity relation for main-sequence stars.

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    L ∝ M³·⁵, usually used as L/L_sun = (M/M_sun)³·⁵. L = luminosity (W or in solar units); M = stellar mass (kg or in solar units); the index 3.5 is an empirical average (values 3 to 4 occur). Valid ONLY for main-sequence stars — never apply it to red giants, white dwarfs or neutron stars. Data-booklet status: not printed in the 2025 booklet — memorise it. Common misuse: forgetting to work in solar units, so a proportionality is turned into a wrong absolute value. Check: M = 10 M_sun gives L = 10³·⁵ ≈ 3.2 × 10³ L_sun.

22. EquationSL & HLData booklet: No – derive

    Define the main-sequence lifetime of a star, state how it depends on stellar mass and derive the index.

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    The main-sequence lifetime t is the time a star spends fusing hydrogen to helium in its core in stable equilibrium. t ∝ M/L, since fuel available ∝ M and rate of consumption ∝ L. Substituting L ∝ M³·⁵ gives t ∝ M/M³·⁵ = M⁻²·⁵. In solar units t/t_sun = (M/M_sun)⁻²·⁵ with t_sun ≈ 10¹⁰ years. t = main-sequence lifetime (s or years); M = mass (kg or solar masses); L = luminosity. Use t, not T, so as not to clash with temperature. Valid for main-sequence hydrogen burning only. Data-booklet status: not printed — derive it from L ∝ M³·⁵ in the answer, which itself gains credit. Common misuse: writing t ∝ M³·⁵, i.e. assuming more fuel means a longer life. Check: M = 25 M_sun gives t ≈ 10¹⁰ × 25⁻²·⁵ ≈ 3 × 10⁶ years.

23. EquationSL & HLData booklet: No – memorise

    State the overall reaction of the proton–proton chain and the energy it releases.

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    4 ¹₁H → ⁴₂He + 2e⁺ + 2ν_e + energy, releasing about 26.7 MeV per helium nucleus formed (including the annihilation of the two positrons). ¹₁H = proton; ⁴₂He = helium-4 nucleus (alpha particle); e⁺ = positron; ν_e = electron neutrino; gamma photons carry away much of the energy. Valid for core temperatures of order 10⁷ K in low- and intermediate-mass stars. Data-booklet status: not printed — memorise. Common misuse: omitting the neutrinos or writing electrons instead of positrons; charge must balance (4(+1) = +2 + 2(+1)). Check: Δm = 4(1.007825) − 4.002603 = 0.0287 u; 0.0287 × 931.5 ≈ 26.7 MeV.

24. EquationSL & HLData booklet: Yes

    State the mass–energy equation used for fusion energy calculations and the conversion factor for u.

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    E = Δmc² (booklet, given as E = mc²). E = energy released (J); Δm = mass defect, the difference between the total rest mass of the reactants and of the products (kg); c = 3.00 × 10⁸ m s⁻¹. Shortcut used in every IB nuclear question: 1 u = 1.66 × 10⁻²⁷ kg is equivalent to 931.5 MeV, so E(MeV) = Δm(u) × 931.5. Valid for any nuclear reaction. Common misuse: leaving Δm in u while using c² in SI, giving an answer 10²⁷ times too large. Check: Δm = 0.0287 u → 26.7 MeV = 26.7 × 1.60 × 10⁻¹³ = 4.28 × 10⁻¹² J.

25. EquationSL & HLData booklet: No – derive

    Show how a star's radius is obtained from its luminosity and surface temperature.

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    From L = 4πR²σT⁴, rearrange to R = √(L/(4πσT⁴)), or in solar units R/R_sun = (T_sun/T)²√(L/L_sun). R = radius (m); L = luminosity (W); T = surface temperature (K); σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. Valid for a spherical black-body star. Data-booklet status: L = σAT⁴ is printed; this rearrangement must be derived. Common misuse: taking the square root of the whole bracket but forgetting that T is raised to the fourth power inside it. Check: a red giant with L = 10⁴ L_sun and T = 3000 K has R ≈ (5800/3000)² × 10² ≈ 370 R_sun.

26. EquationSL & HLData booklet: No – derive

    Show how the distance to a star is found from its luminosity and apparent brightness.

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    From b = L/(4πd²), rearrange to d = √(L/(4πb)). d = distance (m); L = luminosity (W), obtained from a standard candle such as a Cepheid or from the star's HR-diagram position; b = apparent brightness measured at Earth (W m⁻²). Valid for isotropic emission with no interstellar absorption. Data-booklet status: derived from the printed b = L/4πd². Common misuse: forgetting the square root, or quoting the answer in parsecs without dividing by 3.09 × 10¹⁶. Check: L = 3.85 × 10²⁸ W, b = 1.0 × 10⁻⁹ W m⁻² gives d = √(3.85 × 10²⁸/1.26 × 10⁻⁸) = √(3.06 × 10³⁶) = 1.8 × 10¹⁸ m ≈ 57 pc.

27. EquationSL & HLData booklet: Yes

    State the equation used to estimate the temperature needed for fusion from average kinetic energy.

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    Average random kinetic energy per particle E_k = (3/2)k_B T. E_k = average translational kinetic energy (J); k_B = Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹ ; T = absolute temperature (K). Valid for an ideal gas / fully ionised plasma of point particles. Data-booklet status: printed. Common misuse: assuming every proton has this energy — it is an AVERAGE, and fusion in the Sun proceeds via the small high-energy tail of the distribution plus quantum tunnelling. Check: to reach the 2.3 × 10⁻¹³ J Coulomb barrier classically needs T ≈ 1 × 10¹⁰ K, yet the Sun's core is only 1.5 × 10⁷ K.

28. EquationSL & HLData booklet: Yes

    State the equation for the Coulomb potential energy barrier between two protons.

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    E_p = kq₁q₂/r. E_p = electric potential energy (J); k = 8.99 × 10⁹ N m² C⁻² ; q₁, q₂ = charges (C), each +1.60 × 10⁻¹⁹ C for protons; r = separation of centres (m). Valid for point charges in vacuum; take r ≈ 1.0 × 10⁻¹⁵ m, the range of the strong nuclear force. Data-booklet status: printed. Common misuse: using kq₁q₂/r² — that is the FORCE, not the energy. Check: E_p = 8.99 × 10⁹ × (1.60 × 10⁻¹⁹)²/1.0 × 10⁻¹⁵ = 2.3 × 10⁻¹³ J = 1.4 MeV, the energy two protons must have to touch.

29. EquationSL & HLData booklet: Yes

    State the red-shift equation used for receding galaxies.

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    z = Δλ/λ₀ ≈ v/c. z = red shift (no unit); Δλ = λ_observed − λ_emitted, the increase in wavelength (m); λ₀ = wavelength emitted in the source's rest frame (m); v = recession speed of the galaxy (m s⁻¹); c = 3.00 × 10⁸ m s⁻¹. Valid only for v much less than c (non-relativistic approximation). Data-booklet status: printed. Common misuse: using the observed wavelength as λ₀ in the denominator, or forgetting that a positive z means the galaxy is moving AWAY. Check: a 656 nm hydrogen line observed at 663 nm gives z = 7/656 = 0.0107, so v ≈ 3.2 × 10⁶ m s⁻¹.

30. EquationSL & HLData booklet: Yes

    State the astronomical distance conversions required in stellar calculations.

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    1 AU = 1.50 × 10¹¹ m; 1 ly = 9.46 × 10¹⁵ m; 1 pc = 3.09 × 10¹⁶ m = 3.26 ly = 2.06 × 10⁵ AU. Symbols: AU = astronomical unit (mean Earth–Sun distance); ly = light year (distance light travels in vacuum in one year, c × 3.15 × 10⁷ s); pc = parsec (distance giving a parallax angle of 1″). All are distances, unit metre, and all are scalars. Data-booklet status: printed — these values are listed in the data booklet's astronomical data, but the relationships between them are worth memorising. Common misuse: substituting parsecs into b = L/4πd². Check: 3.00 × 10⁸ × 3.15 × 10⁷ = 9.46 × 10¹⁵ m per light year.

31. Graph/diagramSL & HL

    Describe the axes and main regions of a Hertzsprung–Russell diagram.

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    Axes: luminosity (usually L/L_sun) on the vertical axis, logarithmic, increasing upwards over roughly 10⁻⁴ to 10⁶; surface temperature in kelvin on the horizontal axis, logarithmic and DECREASING to the right (about 40 000 K on the left to 2500 K on the right), equivalently spectral class O B A F G K M. Regions: the main sequence, a BAND (not a line) containing about 90% of all stars, runs diagonally from top left (hot, luminous, massive) to bottom right (cool, dim, low mass); red giants sit upper right; supergiants across the very top; white dwarfs bottom left. The Sun lies on the main sequence at 5800 K, L = 1 L_sun. Exam tip: label the reversed temperature axis or you lose the mark.

32. Graph/diagramSL & HL

    Explain how lines of constant radius appear on an HR diagram and how to use them.

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    Because L = 4πR²σT⁴, a fixed R gives L ∝ T⁴, which on log–log axes is a straight line of gradient 4 — but since the temperature axis is reversed, constant-radius lines run diagonally from top left down to bottom right, roughly parallel to the main sequence, with larger R towards the top right. To extract a radius: read L and T for the star, then use R = √(L/(4πσT⁴)), or interpolate between the drawn constant-R lines. Changing a parameter: at fixed T, a star ten times more luminous has √10 ≈ 3.2 times the radius; at fixed L, R ∝ T⁻², so halving T multiplies R by four. Exam tip: white dwarfs lie below the R = R_sun line, giants far above it.

33. Graph/diagramSL & HL

    Describe the black-body spectrum curves used to compare stars.

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    Axes: intensity (power emitted per unit area per unit wavelength, W m⁻² m⁻¹ or arbitrary units) against wavelength λ in metres or nanometres. Shape: a continuous asymmetric curve rising steeply from zero at short λ to a single peak at λ_max, then falling more gradually towards long λ, never crossing another curve. Area under the curve = total power per unit area = σT⁴, so it grows as the fourth power of T. Extracting a quantity: read λ_max off the peak and use λ_max T = 2.9 × 10⁻³ m K. Changing T: a hotter star's curve peaks at shorter λ AND lies entirely above the cooler star's curve at every wavelength.

34. Graph/diagramSL & HL

    Describe the linearised graph used to test the mass–luminosity relation.

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    Plot log(L/L_sun) on the vertical axis against log(M/M_sun) on the horizontal axis, both dimensionless. Taking logs of L = kM^n gives log L = n log M + log k, so the graph is a straight line for main-sequence stars. The gradient equals the index n (≈ 3.5) and the vertical intercept at log M = 0 gives log k, which is 0 in solar units since the Sun lies at the origin. Exam tip: use the plotted best-fit gradient, not one data point; if the axes are plotted in SI units the intercept is no longer zero. Non-main-sequence stars scatter off the line, which is the evidence that the relation applies only to the main sequence.

35. Graph/diagramSL & HL

    Describe the binding energy per nucleon curve and how it explains fusion energy release.

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    Axes: binding energy per nucleon in MeV (vertical, 0 to about 9) against nucleon number A (horizontal, 0 to about 240). Shape: a steep rise from ¹H (zero, no binding) through ²H (1.1 MeV) and ⁴He (7.1 MeV, a local peak) and on through the light nuclei, a maximum of about 8.8 MeV per nucleon at iron-56, then a slow decline to about 7.6 MeV for uranium-238. Fusion of nuclei to the LEFT of the peak moves the product up the curve, so binding energy per nucleon increases, mass is lost and energy is released; beyond iron this is impossible, which is why heavier elements need a supernova. Extract energy released = (BE per nucleon of products × A) − (that of reactants).

36. Graph/diagramSL & HL

    Describe the light curve of a Cepheid variable and the period–luminosity graph.

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    Light curve: apparent brightness (W m⁻²) or magnitude on the vertical axis against time in days on the horizontal axis; the shape is a repeating asymmetric saw-tooth — a rapid rise to maximum followed by a slower decline — and the period is read as the time between successive maxima. Period–luminosity graph: average luminosity (L/L_sun, logarithmic) against period in days (logarithmic); it is a straight line of positive gradient, so longer period means greater luminosity. Method: read the period from the light curve, use the period–luminosity line to get L, measure average b, then d = √(L/4πb). Exam tip: measure the period peak-to-peak, not peak-to-trough.

37. Graph/diagramSL & HL

    Describe the evolutionary track of a Sun-like star on the HR diagram.

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    Starting on the main sequence at about 5800 K and 1 L_sun, the track moves up and to the RIGHT as core hydrogen is exhausted: luminosity rises by a factor of 10³ while surface temperature falls to about 3000 K, taking the star into the red-giant region (top right). After the planetary nebula is ejected, the exposed core moves sharply LEFT at almost constant luminosity, then drops DOWN and slightly right as it becomes a white dwarf, ending at the bottom left (hot, T ≈ 10⁴ K, but L ≈ 10⁻³ L_sun). A massive star instead moves right to the supergiant region and ends abruptly as a supernova, leaving a neutron star or black hole.

38. Graph/diagramSL & HL

    Describe how a graph is used to confirm the inverse-square dependence of apparent brightness on distance.

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    Direct plot: b (W m⁻²) against d (m) gives a decreasing curve that never reaches either axis, from which nothing can be read reliably. Linearisation: plot b on the vertical axis against 1/d² (m⁻²) on the horizontal axis. Since b = L/(4πd²), the result is a straight line through the origin of gradient L/4π, so the source's luminosity is L = 4π × gradient. Alternatively plot log b against log d: a straight line of gradient −2 confirms the inverse-square law, with intercept log(L/4π). Exam tip: a line through the origin is the evidence for direct proportionality — quote both the straightness and the origin.

39. Concept/explainSL & HL

    Explain why extremely high temperatures are required for hydrogen fusion to occur in the core of a star.

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    • Fusing nuclei must approach to within the range of the strong nuclear force, about 10⁻¹⁵ m
    • Both nuclei are positively charged, so they repel electrostatically; the potential energy rises as they approach, forming a Coulomb barrier of order 1 MeV for two protons
    • The only source of the energy needed to climb this barrier is the random thermal kinetic energy of the particles, and average KE ∝ T (E_k ≈ (3/2)k_BT)
    • Hence a temperature of order 10⁷ K is needed so that a significant fraction of nuclei in the high-energy tail of the Maxwell distribution can approach closely enough
    • Quantum tunnelling allows fusion at temperatures below the classical barrier estimate, which is why the Sun's core at 1.5 × 10⁷ K fuses at all. Exam tip: incomplete answers say "heat makes atoms join"; the mark is for linking temperature to kinetic energy and to overcoming electrostatic repulsion.
40. Concept/explainSL & HL

    Outline the proton–proton chain and explain the origin of the energy it releases.

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    • The overall reaction is 4 ¹H → ⁴He + 2e⁺ + 2ν_e + energy, releasing about 26.7 MeV per helium nucleus formed
    • It proceeds in steps: two protons fuse to deuterium with emission of a positron and a neutrino, deuterium captures a proton to give ³He, and two ³He nuclei combine to give ⁴He plus two protons
    • The mass of the products is less than the mass of the reactants; this mass defect Δm appears as energy through E = Δmc²
    • Equivalently, ⁴He has a much greater binding energy per nucleon (≈ 7.07 MeV) than ¹H (zero), and energy is released whenever binding energy per nucleon increases
    • The positrons annihilate with electrons, adding to the released energy; the neutrinos escape carrying a small fraction away. Exam tip: students lose marks by writing "mass is destroyed" — the correct statement is that mass–energy is conserved and rest mass is converted to other forms.
41. Concept/explainSL & HL

    Explain how a main-sequence star maintains a stable size, referring to the forces and pressures involved.

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    • Gravitation pulls every layer of the star inwards towards the centre, tending to collapse it
    • Fusion in the core releases energy, keeping the gas hot; the resulting outward gas (thermal) pressure and radiation pressure act against gravity
    • The star is in hydrostatic equilibrium: at every radius the outward pressure gradient exactly balances the weight of the material above
    • The equilibrium is self-regulating — a small contraction raises core temperature and density, increasing the fusion rate and pressure, which pushes the star back out; a small expansion cools the core and slows fusion
    • This balance holds for as long as core hydrogen lasts, which is why main-sequence stars have nearly constant L and T. Exam tip: answers must name both the inward agent (gravity) and the outward agent (thermal/radiation pressure from fusion); "the forces are equal" alone is not enough without saying which forces.
42. Concept/explainSL & HL

    Explain why controlled nuclear fusion has not yet been achieved as a commercial energy source on Earth.

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    • The fuel must be heated to over 10⁸ K so that nuclei have enough kinetic energy to overcome the Coulomb barrier — hotter than a stellar core because terrestrial densities are far lower
    • At that temperature the fuel is a fully ionised plasma that would melt and be cooled by any material container, so it must be confined magnetically (tokamak) or by inertial confinement (laser compression)
    • Stars confine their plasma by their own enormous gravity, which cannot be reproduced on Earth
    • Maintaining the plasma at high density for long enough (the Lawson criterion) is difficult because of instabilities and energy losses
    • So far the energy input to heat and confine the plasma has been comparable to or greater than the energy output, so no sustained net gain has been commercially delivered. Exam tip: state the two distinct problems — reaching the temperature and confining the plasma — rather than only saying "it is too hot".
43. Concept/explainSL & HL

    Explain why a red giant can be thousands of times more luminous than the Sun even though its surface is cooler.

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    • Luminosity depends on both surface temperature and surface area: L = σAT⁴ = σ4πR²T⁴
    • Lowering T from 5800 K to about 3000 K reduces the power emitted per unit area by a factor of roughly (5800/3000)⁴ ≈ 14
    • However the radius of a red giant is of order 100 times the Sun's, increasing the emitting area by about 10⁴
    • The area increase overwhelms the temperature decrease, giving a net luminosity of order 10³ L_Sun
    • On the HR diagram this places red giants in the upper right, above the lines of constant radius that pass through the main sequence
    • The colour is red because Wien's law shifts λ_max to longer wavelengths at lower T. Exam tip: a bare statement "it is bigger" scores little; quantify using L ∝ R²T⁴ and compare the two competing factors.
44. Concept/explainSL & HLData booklet: No – memorise

    Explain why massive main-sequence stars have much shorter lifetimes than low-mass stars.

    Show answer

    • The fuel available is essentially the hydrogen in the core, so the total energy a star can release is proportional to its mass M
    • The rate at which it releases energy is its luminosity L, so the main-sequence lifetime t ∝ M/L
    • Observations give the mass–luminosity relation L ∝ M^3.5 for main-sequence stars
    • Combining, t ∝ M/M^3.5 = M^−2.5, so lifetime falls steeply with mass
    • A 10 M_Sun star therefore lives about 10^−2.5 ≈ 1/320 as long as the Sun: a few tens of millions of years rather than about 10¹⁰ years
    • Physically, greater mass means stronger gravity, higher core temperature and a far faster fusion rate, so the larger fuel supply is consumed disproportionately quickly. Exam tip: students often argue "more mass means more fuel so it lasts longer" — the mark is for recognising that luminosity rises much faster than mass.
45. Concept/explainSL & HL

    Outline the evolution of a Sun-like star (about 1 solar mass) after it leaves the main sequence.

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    • When core hydrogen is exhausted the core, now mostly helium, contracts under gravity and heats up while hydrogen fusion continues in a shell around it
    • The outer layers expand and cool, so the star becomes a red giant: luminosity rises and surface temperature falls, moving it to the upper right of the HR diagram
    • Core temperature eventually reaches about 10⁸ K and helium fuses to carbon and oxygen (the helium flash and subsequent core helium burning)
    • The star is not massive enough to fuse carbon; instabilities eject the outer layers, which form a planetary nebula ionised by the hot exposed core
    • The remnant core, of mass below the Chandrasekhar limit of about 1.4 M_Sun, becomes a white dwarf supported by electron degeneracy pressure, which cools and fades along the lower left of the HR diagram. Exam tip: name the stages in order and state the Chandrasekhar limit as the condition for the white-dwarf outcome.
46. Concept/explainSL & HL

    Outline the possible end states of a star much more massive than the Sun, and the mass limits that decide between them.

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    • A massive star swells to a red supergiant and fuses successively heavier elements in onion-like shells, up to iron in the core
    • Iron has the greatest binding energy per nucleon, so further fusion absorbs rather than releases energy; the core can no longer support itself and collapses catastrophically
    • The rebound and neutrino burst blow off the outer layers in a Type II supernova
    • If the remnant core mass exceeds the Chandrasekhar limit (≈ 1.4 M_Sun) electron degeneracy pressure fails and electrons combine with protons to give neutrons: a neutron star supported by neutron degeneracy pressure
    • If the remnant exceeds the Oppenheimer–Volkoff limit (≈ 2–3 M_Sun) no known pressure can halt collapse and a black hole forms. Exam tip: these limits apply to the mass of the REMNANT core, not the original main-sequence mass — a very common confusion in mark schemes.
47. Concept/explainSL & HL

    Explain, in terms of binding energy per nucleon, why elements up to iron are produced in stars but heavier elements are not.

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    • The binding energy per nucleon curve rises steeply from hydrogen, peaks near ⁵⁶Fe at about 8.8 MeV per nucleon, then falls slowly for heavier nuclei
    • Fusion releases energy only when the products have a higher binding energy per nucleon than the reactants, i.e. only up to iron
    • Successive fusion stages in a massive star — hydrogen, helium, carbon, oxygen, silicon — each need higher temperatures and each build up to iron in the core
    • Fusing beyond iron would absorb energy, removing the pressure support and triggering core collapse
    • Nuclei heavier than iron are formed by rapid neutron capture in the extreme neutron flux of a supernova (and neutron-star mergers), where the required energy is supplied by the explosion
    • Nature of Science link: the elements in our bodies were made in earlier generations of stars. Exam tip: say "binding energy per nucleon", not "binding energy" — the total binding energy keeps rising with A.
48. Concept/explainSL & HLData booklet: Yes

    Explain how the method of stellar parallax is used to determine the distance to a nearby star, and why it fails for distant stars.

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    • As the Earth orbits the Sun, a nearby star appears to shift its position against the background of very distant stars
    • The parallax angle p is defined as HALF the total angular shift observed over six months, i.e. the angle subtended at the star by the Earth–Sun distance of 1 AU
    • For small angles d = 1/p, with p in arcseconds and d in parsecs; 1 pc = 3.26 ly = 3.09 × 10¹⁶ m
    • A parsec is by definition the distance at which 1 AU subtends 1 arcsecond
    • As d increases p decreases, so the shift becomes comparable with the uncertainty in measuring the angle; beyond a few hundred parsecs from the ground the fractional uncertainty becomes unacceptable
    • More distant objects therefore require standard candles such as Cepheid variables. Exam tip: the definition of p as the half-angle over one AU is regularly dropped; also state that the background stars are assumed fixed.
49. Concept/explainSL & HLData booklet: Yes

    Explain how the surface temperature and the radius of a distant star can be determined from measurements made on Earth.

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    • Record the star's black-body spectrum and find the wavelength λ_max at which the emitted intensity peaks
    • Wien's displacement law λ_max T = 2.9 × 10⁻³ m K then gives the surface temperature T; the spectral class (OBAFGKM) and colour give the same information more coarsely
    • Measure the apparent brightness b, the power received per unit area at Earth, with a calibrated detector
    • Obtain the distance d independently, by parallax for nearby stars or a standard candle for distant ones
    • Then the luminosity follows from b = L/(4πd²), giving L = 4πd²b
    • Finally L = σ4πR²T⁴ is rearranged for R, so the radius is deduced without ever resolving the star as a disc. Exam tip: the chain b → L requires the distance; students who quote L = σAT⁴ without first finding L from b and d cannot complete the deduction.
50. Concept/explainSL & HLData booklet: Yes

    Explain how Cepheid variables are used as standard candles to measure the distance to a galaxy.

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    • A Cepheid variable is a star whose outer layers pulsate periodically, so its luminosity varies with a regular, easily measured period of days to weeks
    • There is a well-established period–luminosity relation: the longer the period, the greater the average luminosity
    • The relation is calibrated using nearby Cepheids whose distances are known independently, for example from parallax
    • For a Cepheid in a distant galaxy, the light curve gives the period, the period–luminosity graph gives L, and the measured average apparent brightness b then gives the distance from d = √(L/(4πb))
    • Because Cepheids are very luminous they can be resolved in other galaxies, so the distance to the whole galaxy is obtained
    • Assumptions: negligible absorption of light by intervening dust, and that all Cepheids obey the same calibrated relation. Exam tip: state clearly which quantity is measured (period, b) and which is deduced (L, then d).
51. Concept/explainSL & HLData booklet: Yes

    Explain what is meant by the red shift of light from distant galaxies and what it indicates.

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    • Spectral absorption lines from a distant galaxy are identified by their pattern but are observed at longer wavelengths than the same lines measured in a laboratory
    • This shift towards the red end of the spectrum is described by z = Δλ/λ₀, and for speeds much less than c, z ≈ v/c, so the galaxy is receding with speed v
    • Almost all galaxies show red shift, and more distant galaxies show greater red shift — the observational basis of an expanding universe
    • Strictly the effect is cosmological: space itself expands and stretches the wavelength in transit, rather than the galaxy moving through space, although at low z the Doppler treatment gives the same numbers
    • Nature of Science link: Hubble's combination of Cepheid distances and red shifts overturned the static-universe model. Exam tip: students must compare with a LABORATORY wavelength; "the light looks red" is not evidence of red shift.
52. Worked problemSL & HLData booklet: Yes

    The proton–proton chain has the overall form 4 ¹H → ⁴He + 2e⁺ + 2ν_e. Given atomic masses ¹H = 1.007825 u and ⁴He = 4.002603 u, determine the energy released per helium nucleus formed.

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    Principle: energy released = Δm c², with 1 u = 931.5 MeV c⁻². Mass of reactants = 4 × 1.007825 u = 4.031300 u. Mass defect Δm = 4.031300 − 4.002603 = 0.028697 u. Energy = 0.028697 × 931.5 MeV = 26.7 MeV (3 s.f.). In joules: 26.7 × 1.60 × 10⁻¹³ = 4.28 × 10⁻¹² J. Check/Trap: using ATOMIC masses automatically includes the four electrons of the hydrogen atoms; two of these annihilate with the two positrons, so the 26.7 MeV already contains the annihilation energy — do not subtract 2 × 0.511 MeV. If nuclear masses were quoted instead, the positron rest masses would have to be added to the product side. A small part of this energy (about 0.5 MeV) escapes as neutrinos and never heats the star.

53. Worked problemSL & HLData booklet: Yes

    The Sun radiates with luminosity 3.85 × 10²⁶ W. Estimate the mass of hydrogen consumed per second, given that each proton–proton chain releases 26.7 MeV and consumes four protons. Take m_p = 1.67 × 10⁻²⁷ kg.

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    Principle: number of reactions per second = L ÷ energy per reaction. Energy per reaction = 26.7 × 1.60 × 10⁻¹³ = 4.27 × 10⁻¹² J. Rate of reactions = 3.85 × 10²⁶ / 4.27 × 10⁻¹² = 9.01 × 10³⁷ s⁻¹. Protons used per second = 4 × 9.01 × 10³⁷ = 3.61 × 10³⁸ s⁻¹. Mass of hydrogen per second = 3.61 × 10³⁸ × 1.67 × 10⁻²⁷ = 6.0 × 10¹¹ kg s⁻¹ (2 s.f.). Check/Trap: this is the mass of hydrogen PROCESSED, not the mass lost. The mass actually converted to energy is L/c² = 3.85 × 10²⁶ / (3.00 × 10⁸)² = 4.3 × 10⁹ kg s⁻¹, about 0.7% of the hydrogen throughput — consistent with the mass defect fraction 0.0287/4.031.

54. Worked problemSL & HLData booklet: Yes

    A star has radius 8.4 × 10⁸ m and surface temperature 9200 K. Calculate its luminosity, treating it as a black body.

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    Principle: L = σAT⁴ with A = 4πR² for a sphere. A = 4π(8.4 × 10⁸)² = 4π × 7.056 × 10¹⁷ = 8.87 × 10¹⁸ m². T⁴ = (9200)⁴ = 7.16 × 10¹⁵ K⁴. L = 5.67 × 10⁻⁸ × 8.87 × 10¹⁸ × 7.16 × 10¹⁵. σA = 5.03 × 10¹¹, so L = 5.03 × 10¹¹ × 7.16 × 10¹⁵ = 3.6 × 10²⁷ W (2 s.f.). Check/Trap: A is the total SURFACE area 4πR², not πR² and not the diameter squared; forgetting the factor 4π is the single most common error. Sanity check: this is about 9 L_Sun, reasonable for a star roughly 20% larger and 60% hotter than the Sun since L ∝ R²T⁴ gives 1.2² × 1.59⁴ ≈ 9.

55. Worked problemSL & HLData booklet: Yes

    The peak of a star's black-body spectrum occurs at 380 nm. Determine its surface temperature and state its approximate spectral class.

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    Principle: Wien's displacement law λ_max T = 2.9 × 10⁻³ m K. T = 2.9 × 10⁻³ / (380 × 10⁻⁹) = 2.9 × 10⁻³ / 3.80 × 10⁻⁷ = 7.6 × 10³ K (2 s.f.). A surface temperature of about 7600 K corresponds to spectral class A (white), just hotter than class F. Check/Trap: convert nanometres to metres before dividing — an answer of 7.6 × 10⁻⁶ K or 7.6 × 10¹² K signals a power-of-ten slip. Quick sanity check: the Sun at 5800 K peaks at 500 nm, and this star peaks at a shorter wavelength, so it must be hotter — consistent. Note λ_max ∝ 1/T, so halving λ_max doubles T.

56. Worked problemSL & HLData booklet: Yes

    A star has apparent brightness 4.2 × 10⁻¹² W m⁻² and is 42 ly from Earth. Determine its luminosity. (1 ly = 9.46 × 10¹⁵ m.)

    Show answer

    Principle: b = L/(4πd²), so L = 4πd²b. Convert: d = 42 × 9.46 × 10¹⁵ = 3.97 × 10¹⁷ m. d² = 1.58 × 10³⁵ m². 4πd² = 1.98 × 10³⁶ m². L = 1.98 × 10³⁶ × 4.2 × 10⁻¹² = 8.3 × 10²⁴ W (2 s.f.). Check/Trap: the light year is a DISTANCE, not a time — a common slip is to multiply by 3.15 × 10⁷ s. Also square the distance after converting, never before. Sanity check: 8.3 × 10²⁴ W is about 0.02 L_Sun, so this is a dim red dwarf, entirely plausible for a nearby star that is faint despite being close.

57. Worked problemSL & HLData booklet: Yes

    A red supergiant has luminosity 1.2 × 10⁵ L_Sun and surface temperature 3400 K. Determine its radius as a multiple of the solar radius. Take T_Sun = 5800 K.

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    Principle: L = σ4πR²T⁴, so for a ratio R/R_Sun = (L/L_Sun)^½ × (T_Sun/T)². (L/L_Sun)^½ = (1.2 × 10⁵)^½ = 346. (T_Sun/T)² = (5800/3400)² = (1.706)² = 2.91. R/R_Sun = 346 × 2.91 = 1.0 × 10³ (2 s.f.). Check/Trap: the ratio method removes σ and 4π, so never substitute the constants; and remember T enters as T⁴ but as T² after taking the square root, with the SUN's temperature on top because a cooler star must be larger for the same L. Sanity check: R ≈ 1000 R_Sun ≈ 7 × 10¹¹ m, about 4.6 AU — so such a star placed at the Sun would swallow the orbit of Mars — typical of a supergiant such as Betelgeuse.

58. Worked problemSL & HLData booklet: Yes

    The parallax angle of a star is measured as 0.038 arcseconds. Determine its distance in parsecs, light years and metres.

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    Principle: d/pc = 1/(p/arcsec). d = 1/0.038 = 26 pc (2 s.f.). In light years: 26 × 3.26 = 86 ly. In metres: 26 × 3.09 × 10¹⁶ = 8.1 × 10¹⁷ m. Check/Trap: the equation works ONLY with p in arcseconds and d in parsecs; converting the angle to radians or degrees first is the classic error. Remember p is half the total angular shift observed six months apart, so if a question quotes a total shift of 0.076 arcsec you must halve it. Sanity check: a smaller angle must give a larger distance — the nearest star, Proxima Centauri, has p = 0.77 arcsec and d = 1.3 pc, so 0.038 arcsec correctly places this star far further away.

59. Worked problemSL & HLData booklet: No – memorise

    A main-sequence star has mass 12 M_Sun. Using L ∝ M^3.5 and a solar main-sequence lifetime of 1.0 × 10¹⁰ years, estimate its main-sequence lifetime.

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    Principle: lifetime t ∝ (fuel)/(rate of use) ∝ M/L ∝ M/M^3.5 = M^−2.5. So t/t_Sun = (M/M_Sun)^−2.5 = 12^−2.5. 12^2.5 = 12² × √12 = 144 × 3.464 = 499. t = 1.0 × 10¹⁰ / 499 = 2.0 × 10⁷ years (2 s.f.). Check/Trap: the exponent is −2.5, not −3.5; students frequently forget the factor M for the increased fuel supply. Also note the answer is an order-of-magnitude estimate: only the core hydrogen, roughly 10%, is actually available, but this affects the Sun and the star similarly and cancels in the ratio. Sanity check: massive stars live tens of millions of years, so 2 × 10⁷ yr is the expected order.

60. Worked problemSL & HLData booklet: Yes

    Two stars in a binary system have the same surface temperature, but star A is 64 times more luminous than star B. Determine the ratio of their radii. (Paper 1 style.)

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    Principle: L = σ4πR²T⁴; with T identical, L ∝ R². So L_A/L_B = (R_A/R_B)². R_A/R_B = √64 = 8. Star A has 8 times the radius of star B. Check/Trap: the temperature dependence cancels ONLY because T is stated to be equal — if the stars had different temperatures the T⁴ factor would dominate. Do not take the fourth root of 64 (a common distractor answer of 2.83); the fourth power applies to T, the square to R. Same-temperature stars also share colour and spectral class, so on an HR diagram they lie on the same vertical line with A above B.

61. Worked problemSL & HLData booklet: Yes

    A hydrogen absorption line with laboratory wavelength 656.3 nm is observed at 672.1 nm in the spectrum of a distant galaxy. Determine the red shift and the recession speed of the galaxy.

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    Principle: z = Δλ/λ₀ and, for v ≪ c, z ≈ v/c. Δλ = 672.1 − 656.3 = 15.8 nm. z = 15.8/656.3 = 0.02407 ≈ 0.0241. v = zc = 0.0241 × 3.00 × 10⁸ = 7.2 × 10⁶ m s⁻¹ (2 s.f.), i.e. about 2.4% of the speed of light, directed away from Earth. Check/Trap: divide Δλ by the LABORATORY wavelength, not the observed one; and since both are in nanometres there is no need to convert — the ratio is dimensionless. The non-relativistic approximation is safe here because z ≪ 1; at z of order 1 the relativistic formula would be required.

62. Worked problemSL & HLData booklet: Yes

    Estimate the temperature at which two protons would classically have enough kinetic energy to reach a separation of 2.0 × 10⁻¹⁵ m, and comment on the Sun's core temperature of 1.5 × 10⁷ K.

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    Principle: electrostatic PE at closest approach = ke²/r, supplied by thermal KE ≈ (3/2)k_BT per particle. E_p = 8.99 × 10⁹ × (1.60 × 10⁻¹⁹)²/(2.0 × 10⁻¹⁵) = 8.99 × 10⁹ × 2.56 × 10⁻³⁸/2.0 × 10⁻¹⁵ = 1.15 × 10⁻¹³ J (≈ 0.72 MeV). Setting (3/2)k_BT = 1.15 × 10⁻¹³ gives T = 2 × 1.15 × 10⁻¹³/(3 × 1.38 × 10⁻²³) = 5.6 × 10⁹ K. Comment: this is about 400 times the Sun's core temperature, so classical physics forbids fusion there. Check/Trap: fusion still occurs because the Maxwell–Boltzmann distribution has a high-energy tail and because protons can quantum-tunnel through the Coulomb barrier — an explanation the mark scheme expects in the comment mark. Both equations used, E_p = kq₁q₂/r and E_K = (3/2)k_BT, are printed in the data booklet.

63. Worked problemSL & HLData booklet: Yes

    A Cepheid variable in a distant galaxy has a pulsation period of 10 days. The period–luminosity graph gives L = 1.1 × 10³⁰ W for this period. Its average apparent brightness is 2.4 × 10⁻¹⁴ W m⁻². Determine the distance to the galaxy in parsecs.

    Show answer

    Principle: b = L/(4πd²) so d = √(L/(4πb)). 4πb = 4π × 2.4 × 10⁻¹⁴ = 3.02 × 10⁻¹³. L/(4πb) = 1.1 × 10³⁰/3.02 × 10⁻¹³ = 3.64 × 10⁴². d = √(3.64 × 10⁴²) = 1.91 × 10²¹ m. In parsecs: 1.91 × 10²¹/3.09 × 10¹⁶ = 6.2 × 10⁴ pc, i.e. about 62 kpc. Check/Trap: take the square root of the full quantity including the power of ten — halve the exponent only for an even exponent, so write 36.4 × 10⁴¹ as 3.64 × 10⁴² first. Assumption: no absorption of light by interstellar dust, which would make the galaxy appear dimmer and so falsely more distant.

64. Worked problemSL & HLData booklet: Yes

    A main-sequence star has mass 0.40 M_Sun. Estimate its luminosity as a fraction of the Sun's, and hence its apparent brightness if it lies 3.2 pc from Earth. (L_Sun = 3.85 × 10²⁶ W.)

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    Principle: mass–luminosity relation L ∝ M^3.5. L/L_Sun = 0.40^3.5 = 10^(3.5 log 0.40) = 10^(3.5 × (−0.398)) = 10^−1.393 = 0.0405. So L = 0.0405 × 3.85 × 10²⁶ = 1.56 × 10²⁵ W. Distance d = 3.2 × 3.09 × 10¹⁶ = 9.89 × 10¹⁶ m. b = L/(4πd²) = 1.56 × 10²⁵/(4π × 9.78 × 10³³) = 1.56 × 10²⁵/1.23 × 10³⁵ = 1.3 × 10⁻¹⁰ W m⁻² (2 s.f.). Check/Trap: the exponent 3.5 applies to the mass RATIO, so never substitute a mass in kilograms. Sanity check: L is only 4% of the Sun's, as expected for a red dwarf, and b ≈ 10⁻¹⁰ W m⁻² is right at the naked-eye limit (about 10⁻¹⁰ W m⁻², apparent magnitude 6), so such a star is at best marginally visible without a telescope.

65. Worked problemSL & HLData booklet: Yes

    The solar constant at the Earth's orbit is 1.36 × 10³ W m⁻² and the Earth–Sun distance is 1.50 × 10¹¹ m. Determine the luminosity of the Sun and hence its surface temperature, given a solar radius of 6.96 × 10⁸ m.

    Show answer

    Principle: L = 4πd²b, then T = (L/(σ4πR²))^¼. 4πd² = 4π × 2.25 × 10²² = 2.83 × 10²³ m². L = 2.83 × 10²³ × 1.36 × 10³ = 3.85 × 10²⁶ W. Surface area A = 4π(6.96 × 10⁸)² = 6.09 × 10¹⁸ m². T⁴ = L/(σA) = 3.85 × 10²⁶/(5.67 × 10⁻⁸ × 6.09 × 10¹⁸) = 3.85 × 10²⁶/3.45 × 10¹¹ = 1.12 × 10¹⁵. T = (1.12 × 10¹⁵)^¼ = 5.8 × 10³ K (2 s.f.). Check/Trap: use the ORBITAL radius d in the brightness step and the SOLAR radius R in the Stefan–Boltzmann step — mixing them is the standard error. Cross-check with Wien: λ_max = 2.9 × 10⁻³/5800 = 500 nm, in the green, as observed.

66. Exam technique/trapSL & HL

    Explain the distinction between the luminosity and the apparent brightness of a star, and the trap this creates in exam questions.

    Show answer

    The trap: students treat "brighter" as meaning "more luminous". Luminosity L is the total power radiated by the star in all directions, measured in watts, and is an intrinsic property independent of the observer. Apparent brightness b is the power received per unit area at the detector on Earth, in W m⁻², and depends on distance through b = L/(4πd²). Why students fall for it: everyday language uses "bright" for both. Correct approach: read the units in the question stem — W means L, W m⁻² means b. A dim-looking star may be hugely luminous but far away, which is exactly why two of the three quantities L, b and d must be known to find the third. Command-term note: "distinguish" requires a statement about BOTH quantities with the difference made explicit, not two separate definitions left side by side.

67. Exam technique/trapSL & HL

    State the most common errors made when reading or sketching a Hertzsprung–Russell diagram, and how to avoid them.

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    The traps: (1) plotting temperature increasing to the right — on an HR diagram T DECREASES to the right, so hot blue O stars are on the left; (2) treating the axes as linear when both are logarithmic, so equal divisions represent equal factors of ten and interpolation must be done logarithmically; (3) placing white dwarfs on the main sequence — they lie bottom left, hot but very dim; (4) drawing the main sequence as a thin line rather than a band. Why students fall for it: every other graph they meet has the independent variable increasing rightwards. Correct approach: label the left end "hot/blue" and the right end "cool/red" before plotting anything, and mark the Sun at (5800 K, 1 L_Sun) as an anchor point. Command-term note: "sketch" still requires labelled axes with units and correctly placed named regions.

68. Exam technique/trapSL & HL

    Give guidance on the command terms used in stellar astrophysics questions, with an example of what each demands.

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    "State" — a bare answer, no reasoning: state the Chandrasekhar limit as 1.4 M_Sun. "Outline" — a brief account of the main points: outline the proton–proton chain in two or three steps. "Describe" — say WHAT happens without needing the cause: describe the path of a 1 M_Sun star on the HR diagram. "Explain" — give reasons or causes; every marking point must contain a "because": explain why massive stars are short-lived. "Determine"/"Calculate" — a numerical answer with working, unit and sensible significant figures. "Deduce" — reach a conclusion from given data and state the reasoning chain. "Suggest" — propose a plausible explanation not necessarily taught directly. "Compare" — statements that treat both items together, using words such as "whereas" or "both". Why marks are lost: writing a description when an explanation is demanded is the single largest source of lost marks in this topic.

69. Exam technique/trapSL & HL

    Identify the unit and conversion traps in astrophysics calculations and how to avoid them.

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    The traps: (1) parallax must be in ARCSECONDS for d = 1/p to give parsecs — converting to radians or degrees first gives nonsense; (2) 1 pc = 3.26 ly = 3.09 × 10¹⁶ m, 1 ly = 9.46 × 10¹⁵ m, 1 AU = 1.50 × 10¹¹ m — mixing these is common; (3) a light year is a distance, never a time; (4) wavelengths given in nm must become metres before use in Wien's law; (5) masses expressed as multiples of M_Sun must stay as ratios in L ∝ M^3.5 and must be converted to kilograms only if a value in kilograms is wanted. Why students fall for it: the constants are all supplied, so students substitute without checking which system the equation assumes. Correct approach: write each quantity with its unit in the substitution line; a missing or wrong unit forfeits the final answer mark even when the number is right.

70. Exam technique/trapSL & HL

    Explain the misconceptions about mass, energy and binding energy that are most common in fusion questions.

    Show answer

    The traps: (1) writing that "mass is converted into energy and so is destroyed" — the correct statement is that rest mass is converted to other forms of energy while mass–energy is conserved; (2) saying fusion releases energy because the products have greater BINDING ENERGY — it is binding energy per NUCLEON that must increase, since total binding energy rises with A for fission products too; (3) claiming fusion works for any nuclei — beyond iron, fusion absorbs energy; (4) confusing the Coulomb barrier with the strong force: repulsion acts at all separations, attraction only within about 10⁻¹⁵ m. Why students fall for it: the binding-energy curve is memorised as a shape without its axis label. Correct approach: always quote the peak at ⁵⁶Fe, about 8.8 MeV per nucleon, and argue "products lie closer to the peak, so energy is released".

71. Exam technique/trapSL & HL

    Describe an experiment to determine the luminosity of the Sun from a measurement of the solar constant, including variables, limitations and improvements.

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    Apparatus: a shallow blackened metal can of known mass and specific heat capacity containing a known mass of water, a thermometer or temperature probe, a stopwatch, and a card of measured area. Method: place the blackened surface perpendicular to the Sun's rays, record temperature against time for several minutes, and find the initial gradient dθ/dt. Then incident power P = (m_w c_w + m_c c_c)(dθ/dt) and intensity b = P/A; L = 4πd²b with d = 1.50 × 10¹¹ m. Independent variable: time; dependent: temperature; controlled: area exposed, angle of incidence, surface emissivity, ambient temperature. Limitations: atmospheric absorption and scattering reduce b to roughly 1000 W m⁻² at the ground, heat losses to surroundings, imperfect blackness. Improvements: use the INITIAL gradient to minimise heat loss, insulate the sides, apply a cooling correction, keep the surface normal to the rays, and repeat.

72. Exam technique/trapSL & HL

    Describe how to verify the inverse-square law b ∝ 1/d² using a laboratory lamp, and explain how the data should be analysed and its uncertainties treated.

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    Apparatus: small filament lamp on a constant, monitored supply as a point source, a light sensor or LDR-based meter on a metre rule, darkened room. Method: record sensor reading b at measured distances d from 0.10 m to 1.00 m, repeating each reading. Controlled variables: lamp power and supply voltage, sensor orientation and gain, background light (subtract a zero-reading). Analysis: linearise by plotting b against 1/d², which should be a straight line through the origin; alternatively plot ln b against ln d and check for a gradient of −2. Uncertainties: convert absolute uncertainties in b and d to error bars, note that the fractional uncertainty in 1/d² is twice that in d, and find the uncertainty in the gradient from maximum and minimum gradient lines through the error bars. Limitations: the filament is not a point (add a systematic offset d₀), reflections from walls and bench raise readings at large d.

73. Exam technique/trapSL & HL

    Explain how uncertainty in a measured parallax angle propagates into the distance, and how this limits the method.

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    The trap: students halve or double the uncertainty at random. Since d = 1/p, the quantities are inversely proportional, so the FRACTIONAL uncertainty in d equals the fractional uncertainty in p: Δd/d = Δp/p. Worked case: p = 0.020 ± 0.002 arcsec gives d = 50 pc with Δd/d = 10%, so d = 50 ± 5 pc. For p = 0.0020 ± 0.002 arcsec the fractional uncertainty is 100% and the distance is meaningless. Why students fall for it: they add absolute uncertainties instead of combining fractional ones for a product or quotient. Correct approach: use fractional uncertainties for multiplication, division and powers (multiplying by the power), and absolute uncertainties only for sums and differences. This is precisely why parallax fails at large distances and standard candles must take over — the uncertainty, not the equation, sets the limit.

74. Exam technique/trapSL & HL

    Explain the difference between random and systematic errors in stellar brightness measurements, and how each affects a deduced distance.

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    Random errors scatter repeated readings of apparent brightness about the true value; they arise from detector noise, fluctuating atmospheric transparency and reading the scale. They are reduced by repeating and averaging, and they show up as scatter of points about a best-fit line. Systematic errors shift every reading the same way: an uncorrected zero offset on the light sensor, a mis-calibrated detector, or — most importantly in astronomy — absorption of starlight by interstellar dust. Effect on distance: since d = √(L/(4πb)), an underestimate of b caused by dust makes the star appear further away than it is; repetition cannot remove this. Precision versus accuracy: many consistent readings of a dust-dimmed star are precise but inaccurate. Correct approach in exams: name the specific source, classify it, state its DIRECTION of effect on the final answer, and give a targeted remedy such as correcting for extinction.

75. Exam technique/trapSL & HL

    Explain the significant-figure and estimation conventions expected in astrophysics answers.

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    The trap: copying a full calculator display, or rounding to one figure too early. Rule: quote the final answer to the same number of significant figures as the LEAST precise piece of given data, typically 2 or 3 in this topic, and keep extra figures in intermediate values. In multi-step chains such as b → L → R, rounding at each stage can shift the radius by several per cent because of the fourth-power dependence on T. For order-of-magnitude questions ("estimate"), one significant figure and a correct power of ten earn full marks, and a stated assumption — a black body, a spherical star, no absorbing dust — is often worth a mark on its own. Also give every answer with a unit and, where relevant, express it usefully: R as a multiple of R_Sun, t in years, d in parsecs. Command-term note: "estimate" invites approximation, but the method must still be shown.

Practise this topic with exam-style questions: E.5 Fusion and stars questions (SL) · E.5 Fusion and stars questions (HL) · all flashcards

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