Summary to follow. 7 syllabus statements · 23 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 7 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Stellar equilibrium
Stellar equilibrium
A star is stable when, at every depth, the inward gravitational force on its gas is balanced by the outward force from the pressure beneath it; the guide describes this as an equilibrium between outward radiation pressure and inward gravitational forces. The energy released by fusion in the core maintains the outward pressure. If the outward pressure falls, gravity compresses the star; if it rises, the star expands until the balance is restored. Equilibrium means that the forces balance, not that no forces act.
Radiation pressure (in a star)
The outward pressure that supports a star against gravity, maintained by the energy flowing out from fusion in the core. Strictly it is the combined pressure of the radiation and of the hot gas that the radiation keeps hot; in a star like the Sun most of it is gas pressure, but both depend on the energy released by fusion. SI unit: pascal (Pa).
Students often think A star in equilibrium has no forces acting on it, because it is neither expanding nor contracting. In fact No. Very large forces act: gravity pulls every part of the star inwards and the outward pressure pushes it out. Equilibrium means these forces balance, so the resultant force on each part is zero.
Students often think A star's rotation produces an outward centrifugal force that balances its gravity. In fact No. The support comes from the outward pressure maintained by the energy released in fusion. Rotation is not needed: the Sun rotates far too slowly for its rotation to matter to its overall balance.
Nuclear fusion
Nuclear fusion
The joining of two light nuclei to form a heavier nucleus. For nuclei lighter than iron the product has a greater binding energy per nucleon, so the total rest mass decreases and energy is released, E = Δmc². In main-sequence stars hydrogen nuclei (protons) are fused into helium-4 in the core; this is the source of the energy they radiate.
Energy released in a fusion reaction
The energy released is E = Δmc², where Δm is the decrease in total rest mass from reactants to products. With masses in u, Δm × 931.5 gives the energy in MeV; with Δm in kg (1 u = 1.661 × 10⁻²⁷ kg), multiplying by c² gives joules. The number in front of each particle in the equation must be applied: for ³He + ³He → ⁴He + 2 ¹H, Δm = 2m(³He) − m(⁴He) − 2m(¹H) = 0.013805 u, about 12.9 MeV or 2.06 × 10⁻¹² J. Unit: J (often given in MeV).
Hydrogen fusion in the Sun
Through a sequence of reactions (the proton–proton chain), four protons become one helium-4 nucleus, two positrons and two neutrinos. Using atomic masses, 4 × 1.007825 u − 4.002603 u = 0.028697 u, so about 26.7 MeV is released for each helium-4 nucleus formed (this includes the energy from annihilation of the positrons; a small part is carried away by the neutrinos). About 0.7% of the mass of the hydrogen fused is converted to energy, so the Sun's mass decreases by about 4 × 10⁹ kg every second.
Students often think Joining light nuclei into heavier ones increases the total mass, so a star gets heavier as it makes helium. In fact No. A helium-4 nucleus has less mass than the four protons from which it forms; the missing mass (about 0.7%) is released as energy, so fusion slowly decreases the star's mass.
Students often think Stars are balls of burning gas: their energy comes from chemical combustion, like a fire or a gas flame, which must first be ignited. In fact No. Burning is a chemical reaction with oxygen, and chemical reactions release far too little energy per kilogram to power a star for billions of years. Stars are powered by nuclear fusion.
Temperature condition for fusion
Temperature condition for fusion
Nuclei are positively charged and repel one another electrically. To fuse they must come within about 10⁻¹⁵ m, where the short-range, attractive strong nuclear force can bind them, and this needs collisions with very large kinetic energies. Because the average kinetic energy of particles is proportional to the kelvin temperature, fusion needs core temperatures of the order of 10⁷ K (about 1.5 × 10⁷ K at the centre of the Sun); even then only the small fraction of collisions between the fastest nuclei lead to fusion. The matter there is fully ionized.
Density condition for fusion
The rate of fusion depends on how often nuclei collide as well as on how energetic the collisions are. A very high density (about 1.5 × 10⁵ kg m⁻³ at the centre of the Sun) means many nuclei per cubic metre, so collisions are frequent enough for fusion to release energy at the rate a star radiates it. Gravitational contraction of a star's core raises both its density and its temperature, which is how the conditions for fusion are reached.
Students often think Because fusion needs a very high temperature, it must take in energy: the thermal energy of the core is used up by the reactions. In fact No. The high temperature is needed only so that nuclei get close enough to fuse. Each fusion reaction then releases far more energy than the kinetic energy the nuclei needed, so fusion is a net source of energy.
Students often think Nuclei need a high kinetic energy to overcome the strong nuclear force, which pushes them apart. In fact No. The barrier is the electric repulsion between the positively charged nuclei. The strong nuclear force is attractive and short-range: once the nuclei are close enough, it binds them together.
Main-sequence lifetime and stellar mass
Main-sequence lifetime and stellar mass
A more massive main-sequence star has more hydrogen, but its core is hotter and denser, so it fuses hydrogen far faster: its luminosity is greater by a much larger factor than its mass. It therefore spends much less time on the main sequence. The Sun will spend about 10¹⁰ years there; a star of 10 M⊙, several thousand times as luminous, only a few times 10⁷ years.
Evolution of a low-mass star
For a star of low initial mass such as the Sun: when the hydrogen in the core is used up, the core contracts and heats, hydrogen fusion continues in a shell around it, and the outer layers expand and cool, so the star becomes a red giant. Helium in the core then fuses to carbon (and oxygen). The core never becomes hot enough for further fusion stages; the outer layers are ejected as a planetary nebula and the exposed core remains as a white dwarf, in which no fusion takes place.
Evolution of a high-mass star
For a star of much greater initial mass (more than about 8 M⊙): after the main sequence it becomes a red supergiant, and its core becomes hot enough for successive stages of fusion that build heavier elements, up to iron. Iron is at the peak of the binding energy curve, so fusing it releases no energy; the iron core collapses and the star explodes as a supernova, leaving a neutron star or, for the most massive stars, a black hole. Mass therefore decides both how long a star lives and how it ends.
Students often think A more massive star has more fuel, so it lasts longer, just as a larger fuel tank lasts longer. In fact No. It has more hydrogen, but its luminosity is greater by a much larger factor, so it uses its fuel far faster and has a much shorter main-sequence lifetime.
Students often think Small stars have less energy stored and, like small hot objects, cool down and fade sooner than large ones. In fact No. Low-mass stars fuse hydrogen slowly and have the longest main-sequence lifetimes, much longer than the Sun's.
Hertzsprung–Russell (HR) diagram
Hertzsprung–Russell (HR) diagram
A graph of the luminosity of stars (vertical axis, usually a logarithmic scale in units of L⊙) against their surface temperature (horizontal axis, logarithmic, with temperature DECREASING to the right). Hot stars are on the left and cool stars on the right; luminous stars at the top and dim stars at the bottom. Stars are not scattered at random but fall into distinct regions: the main sequence, the red giants, the supergiants and the white dwarfs.
Main sequence
The diagonal band running from hot, luminous stars at the upper left to cool, dim stars at the lower right; about 90% of stars lie on it. Main-sequence stars are fusing hydrogen into helium in their cores. A star stays at nearly the same place on the band for this whole stage, and its position is set mainly by its mass: the more massive the star, the hotter, larger and more luminous it is, and the higher up the band it lies.
Red giants and supergiants
Red giants lie above the main sequence on the right-hand side: cool (a few thousand kelvin) but luminous, typically tens to a few thousand times the Sun's luminosity, because they are very large (tens of times the Sun's radius). Supergiants lie in a band across the whole top of the diagram, from hot blue supergiants at the upper left to cool red supergiants at the upper right, with luminosities of roughly 10⁴ to 10⁶ L⊙; red supergiants have radii of hundreds of times the Sun's. All are stars that have left the main sequence.
White dwarfs
Stars at the lower left of the HR diagram: hot (typically around 10⁴ K) but of low luminosity (well below 1 L⊙) because they are tiny, about the size of the Earth, roughly 1% of the Sun's radius. A white dwarf is the exposed core of a low-mass star after the red-giant stage; no fusion takes place in it, and it slowly cools, moving down and to the right on the diagram.
Instability strip
A narrow, nearly vertical band of the HR diagram that extends upward from the main sequence through the giant and supergiant regions, at intermediate surface temperatures. Stars in it pulsate: they expand and contract in a regular cycle, so their luminosity varies periodically.
Lines of constant radius
From L = σ4πR²T⁴, stars of the same radius have L ∝ T⁴. On an HR diagram with logarithmic scales these are straight diagonal lines running from the upper left to the lower right. Lines towards the upper right correspond to larger radii and lines towards the lower left to smaller radii, which shows at a glance that supergiants are the largest stars and white dwarfs the smallest.
Stellar spectrum: temperature and composition
The continuous part of a star's spectrum is close to a black-body spectrum, so its peak wavelength gives the surface temperature through Wien's displacement law, λ_max T = 2.9 × 10⁻³ m K. Superimposed on it are dark absorption lines, at wavelengths absorbed by atoms and ions in the star's cooler outer layers. Matching these with the lines of elements measured in the laboratory shows which elements are present in the star's outer layers.
Students often think The main sequence is the path that every star follows during its life, from hot and luminous at the upper left to cool and dim at the lower right. In fact No. A star stays at nearly the same point on the main sequence throughout its hydrogen-fusing life; its position is set by its mass. When the hydrogen in its core runs out, it leaves the main sequence.
Students often think Main-sequence stars are all about the size of the Sun, because they form a single narrow band on the HR diagram. In fact No. Main-sequence radii range from about a tenth of the Sun's radius for cool, dim stars to several times it for hot, luminous ones.
Astronomical unit (AU)
Astronomical unit (AU)
The mean distance from the Earth to the Sun: 1 AU = 1.50 × 10¹¹ m. Used for distances within the Solar System and as the baseline in the definition of stellar parallax.
Light-year (ly)
The distance travelled by light in a vacuum in one year: 1 ly = 9.46 × 10¹⁵ m, about 6.3 × 10⁴ AU. It is a unit of distance, not of time.
Stellar parallax and the parsec
As the Earth orbits the Sun, a nearby star appears to shift against much more distant background stars. The parallax angle p is HALF the angular shift between observations made six months apart: the angle subtended at the star by 1 AU. With p in arc-seconds, the distance in parsecs is d = 1/p; one parsec is the distance at which 1 AU subtends an angle of 1 arc-second. 1 pc = 3.09 × 10¹⁶ m = 3.26 ly ≈ 2.06 × 10⁵ AU. Because p is smaller for more distant stars, the method works only for relatively nearby stars.
Students often think The parallax angle is the total angle through which the star appears to move against the background between two observations six months apart. In fact No. The parallax angle p is HALF the total angular shift between observations made six months apart: it is the angle subtended at the star by 1 AU, not by the 2 AU diameter of the Earth's orbit.
Students often think The parsec and the light-year are the same unit, or interchangeable, so d = 1/p gives light-years. In fact No. With p in arc-seconds, d = 1/p gives the distance in parsecs; 1 pc = 3.26 ly.
Determining a stellar radius
Determining a stellar radius
Rearranging the Stefan–Boltzmann law L = σ4πR²T⁴ gives R = √(L/(4πσT⁴)), with σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. The luminosity is found from the apparent brightness and the distance (L = 4πd²b), and the surface temperature from the peak of the spectrum (Wien's law). For two stars, R₁/R₂ = √(L₁/L₂) × (T₂/T₁)². SI unit: m (often expressed as a multiple of the Sun's radius).
Students often think A more luminous star is simply a bigger star, so the radius ratio can be found from the luminosities alone, as √(L₁/L₂), with temperature ignored. In fact No. L = σ4πR²T⁴ depends on temperature as well as radius, so R ∝ √L/T². A cool star must be much larger than a hot one to have the same luminosity.
Students often think Rearranging L = σ4πR²T⁴ for R gives R ∝ √L/T: the square root of T⁴ is taken to be T. In fact No. The square root of T⁴ is T², so R = √(L/(4πσ))/T², that is R ∝ √L/T².
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 A main-sequence star stays the same size for millions of years. Which statement explains why it does not collapse?
Answer and reasoning
Its rotation produces an outward centrifugal force on its gas that balances the inward pull of gravity. — A student who treats centrifugal force as a real outward force picks this. A star does not need to rotate to be stable, and the Sun rotates far too slowly for this to matter; the outward support comes from pressure maintained by fusion.
No forces act on its gas at all, which is why the star is neither expanding nor contracting. — A student who thinks equilibrium means 'no forces' picks this. Enormous forces act: gravity pulls every layer inwards. The star keeps its size because the outward pressure balances gravity, so the resultant force is zero.
The inward gravitational forces on its gas are balanced by the outward radiation pressure maintained by fusion. — In a stable star the inward gravitational force on each layer is balanced by the outward pressure maintained by the energy released in fusion in the core. The forces are large; it is their balance that keeps the size constant.
Gravity is negligible in space, so the hot gas of the star has no weight that needs supporting. — A student who believes there is no gravity in space picks this. A star's own mass produces a huge gravitational field; without the outward pressure maintained by fusion the star would collapse.
2 What is the source of the energy radiated by a main-sequence star such as the Sun?
Answer and reasoning
Nuclear fusion of hydrogen into helium in its core — In the core, hydrogen nuclei fuse to form helium-4. The products have less total rest mass than the reactants, and the difference is released as energy, E = Δmc².
Chemical burning of hydrogen gas in oxygen, like a flame — A student who pictures the Sun as a ball of fire picks this. Combustion releases a few eV per reaction, far too little to keep a star shining for billions of years; fusion releases millions of eV per reaction.
Nuclear fission of heavy nuclei such as uranium and thorium — A student who equates 'nuclear energy' with fission reactors picks this. Stars are mostly hydrogen and helium; they release energy by joining light nuclei, not by splitting heavy ones.
Heat left over from its formation, lost slowly as it cools — A student who thinks a star glows only because it is hot picks this. A body radiating with no energy source would cool; the Sun has radiated at a similar rate for billions of years because fusion continually replaces the energy.
3 Hydrogen fusion in a star needs a core temperature of the order of 10⁷ K. Why is such a high temperature needed?
Answer and reasoning
Fusion absorbs energy, and the thermal energy of the core supplies it. — A student who assumes a reaction that needs heat must absorb energy picks this. Each fusion reaction releases far more energy than the colliding nuclei needed; the high temperature only gets them close enough to fuse.
Hydrogen must reach its ignition temperature before it can burn. — A student who thinks stars burn like a fire picks this. Nothing burns in a star's core: there is no combustion, and chemical reactions could not supply a star's energy. The temperature is needed for nuclear fusion.
Nuclei need enough energy to overcome the strong nuclear force. — A student who thinks the strong force is the barrier picks this. The strong nuclear force is attractive and short-range; it is what binds the nuclei once they are close. The barrier is their electric repulsion.
Nuclei must move fast enough to overcome electric repulsion. — Nuclei are positively charged and repel. Only at very high temperatures do enough nuclei move fast enough to approach within about 10⁻¹⁵ m, where the attractive strong nuclear force can bind them.
4 Which sequence of stages describes the evolution of a star of 1 M⊙, such as the Sun, after it forms?
Answer and reasoning
Main sequence → red giant → supernova explosion — A student who thinks every star explodes when its fuel runs out picks this. Only stars of high initial mass (more than about 8 M⊙) end as supernovae. A star of 1 M⊙ becomes a red giant, sheds its outer layers gently as a planetary nebula and leaves a white dwarf.
Main sequence → red giant → planetary nebula → white dwarf — When core hydrogen is used up, a 1 M⊙ star swells into a red giant. Its core never gets hot enough for fusion beyond carbon and oxygen; the outer layers are shed as a planetary nebula and the exposed core becomes a white dwarf.
Main sequence → red giant → planetary nebula → black hole — A student who thinks all stars end as black holes picks this. Only the cores of the most massive stars become black holes, after a supernova; a 1 M⊙ star sheds a planetary nebula and leaves a white dwarf.
Main sequence → white dwarf → red giant → red supergiant — A student who thinks stars grow as they age, from dwarfs into giants, picks this. The white dwarf is the LAST stage: it is the exposed core left after the red-giant stage, and no fusion takes place in it. A star of 1 M⊙ never becomes a supergiant.
5 An HR diagram is drawn with luminosity on the vertical axis (increasing upwards) and surface temperature on the horizontal axis (decreasing to the right). The main sequence is a band running from the upper left to the lower right. Which statement about main-sequence stars is correct?
Answer and reasoning
Their position on the band is set mainly by mass: more massive stars lie higher up. — All main-sequence stars fuse hydrogen in their cores. The more massive a star, the hotter, larger and more luminous it is, so it lies further towards the upper left.
Each star moves steadily along the band, from upper left to lower right, as it ages. — A student who reads the main sequence as a track picks this. A star stays at nearly the same point on the band while it fuses hydrogen in its core; the band is made of many stars of different masses.
They all have about the same radius as the Sun, as they form a single narrow band. — A student who takes the Sun as typical of every main-sequence star picks this. The band crosses many lines of constant radius: from about a tenth of the Sun's radius at the lower right to several times it at the upper left.
Those at the upper left last longest, since they have the most hydrogen to fuse. — A student who reasons 'more fuel, longer life' picks this. Stars at the upper left are the most massive, but they are so luminous that they use their hydrogen fastest and have the shortest main-sequence lifetimes.
6 Two measurements of the position of a nearby star against very distant background stars, made six months apart, differ by an angle of 0.0700 arc-second. Take 1 pc = 3.09 × 10¹⁶ m, 1 ly = 9.46 × 10¹⁵ m and 1 AU = 1.50 × 10¹¹ m. What is the distance to the star in AU?
Answer and reasoning
2.94 × 10⁶ AU — A student who uses the whole 0.0700 arc-second shift as the parallax angle gets d = 14.3 pc = 2.94 × 10⁶ AU, half the true distance. The parallax angle is half the shift between observations six months apart.
1.80 × 10⁶ AU — A student who treats d = 1/p as giving light-years takes 28.6 ly = 2.70 × 10¹⁷ m = 1.80 × 10⁶ AU. With p in arc-seconds, 1/p gives parsecs, and a parsec is 3.26 ly.
7.21 × 10³ AU — A student who takes the distance as proportional to the parallax angle uses d = 0.0350 pc and gets 7.21 × 10³ AU. Distance is the reciprocal: d = 1/p = 28.6 pc.
5.89 × 10⁶ AU — p = 0.0700/2 = 0.0350 arc-second, so d = 1/0.0350 = 28.6 pc = 28.6 × 3.09 × 10¹⁶ m = 8.83 × 10¹⁷ m. Dividing by 1.50 × 10¹¹ m gives 5.89 × 10⁶ AU.
Working Parallax angle p = ½ × 0.0700″ = 0.0350″. d = 1/p = 1/0.0350 = 28.57 pc. In metres: 28.57 × 3.09 × 10¹⁶ m = 8.829 × 10¹⁷ m. In AU: 8.829 × 10¹⁷ m/(1.50 × 10¹¹ m) = 5.89 × 10⁶ AU (equivalently 93.3 ly).
7 An HR diagram is drawn with luminosity on the vertical axis (increasing upwards) and surface temperature on the horizontal axis (decreasing to the right). A red giant on the diagram has a luminosity of 100 L⊙ and a surface temperature of 4000 K. A white dwarf has a luminosity of 0.010 L⊙ and a surface temperature of 16 000 K. Both are modelled as black bodies. What is (radius of the red giant)/(radius of the white dwarf)?
Answer and reasoning
1.0 × 10² — A student who uses the luminosities alone gets √(100/0.010) = 100. The white dwarf is four times hotter, so for a given luminosity it needs 4² = 16 times less radius; the ratio is 100 × 16 = 1.6 × 10³.
1.6 × 10³ — R ∝ √L/T², so R_RG/R_WD = √(100/0.010) × (16 000/4000)² = 100 × 16 = 1.6 × 10³. The red giant is cooler yet far more luminous because it is vastly larger.
4.0 × 10² — A student who takes √(T⁴) as T gets 100 × (16 000/4000) = 4.0 × 10². Rearranging L = σ4πR²T⁴ gives R ∝ √L/T², so the temperature ratio must be squared.
2.6 × 10⁶ — A student who forgets the square root, as if L were proportional to R, gets (100/0.010) × 4⁴ = 2.6 × 10⁶. That is the ratio of R², not of R; its square root is 1.6 × 10³.
Working L = σ4πR²T⁴ ⇒ R = √(L/(4πσ))/T², so R_RG/R_WD = √(L_RG/L_WD) × (T_WD/T_RG)² = √(100/0.010) × (16 000 K/4000 K)² = 100 × 16 = 1.6 × 10³.
8 In the core of a star, hydrogen fusion stops when the hydrogen there has been used up. What happens to the core next, and why?
Answer and reasoning
It stays the same size and simply cools down, like a fire that has gone out once its fuel is used up. — A student who pictures a star as a fire running out of fuel picks this. Gravity has not gone away: without the pressure maintained by fusion, nothing balances it, so the core contracts (and heats) rather than simply cooling at constant size.
It explodes at once as a supernova, because a star blows itself apart as soon as its fuel has run out. — A student who thinks every star explodes when its fuel runs out picks this. Losing the energy source removes OUTWARD support, so the core is pulled inwards. Only massive stars end in a supernova, and only after further stages of fusion.
It contracts, because the helium made by fusion has made the core more massive, so its gravity is now stronger than before. — A student who thinks fusion adds mass picks this. The core does contract, but not for this reason: fusing hydrogen into helium slightly DECREASES the total mass, because the mass difference is released as energy. The core contracts because the outward pressure maintained by fusion has gone, so gravity is no longer balanced.
It contracts, because without fusion the outward radiation pressure no longer balances the inward gravitational force. — Fusion maintained the outward pressure that balanced gravity. When it stops, the pressure falls, the resultant force is inwards and the core contracts. The contraction heats the core, which leads to the next stage of the star's evolution.
9 One reaction in the Sun's core is ³He + ³He → ⁴He + 2 ¹H. The atomic masses are 3.016029 u for helium-3, 4.002603 u for helium-4 and 1.007825 u for hydrogen-1 (the electron masses balance). Take 1 u = 1.661 × 10⁻²⁷ kg and c = 3.00 × 10⁸ m s⁻¹. How much energy is released in one reaction?
Answer and reasoning
6.88 × 10⁻²¹ J — A student who multiplies the mass defect by c instead of c² gets 2.293 × 10⁻²⁹ × 3.00 × 10⁸ = 6.88 × 10⁻²¹. That has the unit kg m s⁻¹, not J; E = Δmc² needs c² = 9.00 × 10¹⁶ m² s⁻².
5.98 × 10⁻¹⁰ J — A student who takes the whole mass of the helium-4 atom formed as the mass converted gets 4.002603 u × c² = 5.98 × 10⁻¹⁰ J. The helium nucleus is not destroyed; only the decrease in total mass, 0.013805 u, is released as energy.
1.53 × 10⁻¹⁰ J — A student who counts each particle once, ignoring the 2 in front of ¹H, gets Δm = 2(3.016029) − 4.002603 − 1.007825 = 1.021630 u and 1.53 × 10⁻¹⁰ J. Two ¹H atoms are produced, so 2 × 1.007825 u must be subtracted.
Working Δm = 2m(³He) − [m(⁴He) + 2m(¹H)] = 2(3.016029 u) − [4.002603 u + 2(1.007825 u)] = 6.032058 u − 6.018253 u = 0.013805 u. In kg: 0.013805 × 1.661 × 10⁻²⁷ kg = 2.293 × 10⁻²⁹ kg. E = Δmc² = 2.293 × 10⁻²⁹ kg × (3.00 × 10⁸ m s⁻¹)² = 2.06 × 10⁻¹² J (equivalently 0.013805 × 931.5 MeV = 12.9 MeV).
10 At the centre of the Sun the temperature is about 1.5 × 10⁷ K and the density about 1.5 × 10⁵ kg m⁻³. Fusion takes place only in this central core. How, if at all, does the very high density affect fusion in the core?
Answer and reasoning
It makes collisions between nuclei frequent enough for fusion to release energy at a high rate. — High temperature makes individual collisions energetic enough; high density means many nuclei per cubic metre, so collisions, and hence fusion reactions, happen often enough to release energy at the rate the Sun radiates it.
It brings nuclei close enough together for the gravity between them to pull them into each other. — A student who transfers the role of gravity from the whole star to pairs of nuclei picks this. Between two nuclei, gravity is about 10³⁶ times weaker than their electric repulsion. Gravity acts on the star as a whole, compressing and heating the core.
It has no effect on fusion; only the high temperature of the core affects the rate of fusion. — A student who thinks temperature is the only condition picks this. The rate of fusion also depends on how often nuclei collide; a hot but thin gas has too few collisions to release energy at a significant rate.
It gives each nucleus more kinetic energy, which it needs to overcome electric repulsion. — A student who runs density and temperature together picks this. The average kinetic energy of the nuclei depends on the temperature; density sets how many nuclei there are per unit volume and so how often they collide.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
13 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Star X has a mass of 10 M⊙ and star Y a mass of 1 M⊙; both are main-sequence stars. X has about ten times as much hydrogen available for fusion as Y, but it is about 3000 times as luminous. How does the main-sequence lifetime of X compare with that of Y?
Answer and reasoning
About 10 times as long as Y's — A student who reasons 'more fuel, longer life' uses the fuel ratio alone and gets 10. X has ten times the fuel but uses it about 3000 times as fast, so its lifetime is 10/3000 ≈ 1/300 of Y's.
About the same length as Y's — A student who assumes every star lives about as long as the Sun gets a ratio of 1. The same process runs at very different rates: lifetime ∝ fuel/L = 10/3000, so X lasts only about 1/300 as long as Y.
About 1/300 as long as Y's — Lifetime ∝ (fuel available)/(rate of use) ∝ fuel/L. For X relative to Y this is 10/3000 = 3.3 × 10⁻³ ≈ 1/300: X has more hydrogen, but it uses it far faster.
About 1/3000 as long as Y's — A student who uses the luminosity ratio alone gets 1/3000. X also has ten times as much hydrogen to fuse, so its lifetime is 10/3000 ≈ 1/300 of Y's, ten times longer than 1/3000.
Working Main-sequence lifetime ≈ (hydrogen available for fusion) ÷ (rate at which it is used), and the rate of use is proportional to the luminosity, so lifetime ∝ fuel/L. (lifetime of X)/(lifetime of Y) = (10/1) ÷ (3000/1) = 10/3000 = 3.3 × 10⁻³ ≈ 1/300: X spends about 1/300 as long on the main sequence as Y.
2 An HR diagram is drawn with luminosity on the vertical axis (increasing upwards) and surface temperature on the horizontal axis (decreasing to the right). Stars A and B lie on the same line of constant radius. A has a surface temperature of 9000 K and B a surface temperature of 6000 K. Both are modelled as black bodies. What is the ratio (luminosity of A)/(luminosity of B)?
Answer and reasoning
1.5 — A student who treats luminosity as proportional to T gets 9000/6000 = 1.5. At a fixed radius L ∝ T⁴, so the ratio is 1.5⁴ = 5.1.
1.0 — A student who reads the line as a line of constant luminosity gets 1.0. The stars share the same RADIUS; the line slopes diagonally, and the hotter star, higher up, is more luminous.
0.2 — A student who believes cooler stars are more luminous inverts the ratio: (6000/9000)⁴ = 0.2. At equal radius the hotter star is always the more luminous; red giants are luminous because they are large, not because they are cool.
5.1 — On a line of constant radius, L = σ4πR²T⁴ with R the same, so L_A/L_B = (T_A/T_B)⁴ = (9000/6000)⁴ = 1.5⁴ = 5.1. The hotter star lies higher up the line, to the left.
Working Same radius, so L_A/L_B = (4πR²σT_A⁴)/(4πR²σT_B⁴) = (T_A/T_B)⁴ = (9000 K/6000 K)⁴ = 1.5⁴ = 5.06 ≈ 5.1.
3 An HR diagram is drawn with luminosity on the vertical axis (increasing upwards) and surface temperature on the horizontal axis (decreasing to the right). The instability strip is a narrow, nearly vertical band that extends upwards from the main sequence at intermediate surface temperatures. What is characteristic of the stars in it?
Answer and reasoning
They are massive stars that are about to explode as supernovae at the end of their lives. — A student who reads 'instability' in its everyday sense picks this. The stars in the strip are pulsating, and their luminosity rises and falls regularly; the term does not mean they are about to explode.
Their luminosity varies periodically, because they pulsate, expanding and contracting. — Stars in the instability strip pulsate: they expand and contract in a regular cycle, and their luminosity varies periodically as they do so.
Their light seems to flicker because their light output changes many times a second. — A student who thinks twinkling comes from the star picks this. Rapid flickering (twinkling) is caused by the Earth's atmosphere and affects all stars; stars in the strip vary in a regular cycle lasting hours to months.
They are collapsing steadily, because their gravity exceeds the outward radiation pressure. — A student who thinks any change means a star has lost its equilibrium picks this. A pulsating star oscillates about equilibrium, returning to the same size and luminosity each cycle; it is not steadily collapsing.
4 The spectrum of a star is a continuous spectrum crossed by dark lines. Two of the dark lines are at wavelengths that match lines in the emission spectrum of sodium measured in the laboratory. What do these two dark lines show?
Answer and reasoning
Sodium is absent from the star, so light at those two wavelengths is missing. — A student who reads a dark line as a sign that an element is missing picks this. The light is missing BECAUSE sodium is there: sodium atoms absorb exactly the wavelengths they would emit.
The lines are produced by sodium atoms in the star emitting light at those wavelengths. — A student who thinks every spectral line is emitted light picks this. Emission lines would be bright; dark lines on a continuous spectrum are absorption lines.
Sodium in the star's outer layers is absorbing light of those two wavelengths. — Continuous radiation from deeper, hotter layers passes through the cooler outer layers. Sodium atoms there absorb their own characteristic wavelengths, so dark lines at the sodium wavelengths show that sodium is present in the outer layers.
Sodium is being made by nuclear fusion in the core of the star itself. — A student who thinks the spectrum shows the core picks this. The light we receive comes from the outer layers, so the lines show what is present there, not what fusion is producing in the core.
5 Proxima Centauri, the nearest star to the Sun, has a parallax angle of 0.768 arc-second. Take 1 pc = 3.09 × 10¹⁶ m and 1 ly = 9.46 × 10¹⁵ m. What is its distance from the Earth in light-years?
Answer and reasoning
2.51 ly — A student who takes the distance as proportional to the parallax angle uses d = 0.768 pc and gets 2.51 ly. The distance is the reciprocal of p: d = 1/0.768 = 1.302 pc.
1.30 ly — A student who thinks d = 1/p gives light-years stops at 1.30. That value is in parsecs. A parsec is about 3.26 light-years, so the number of light-years must be about 3.26 times the number of parsecs; converting through metres with the values given gives 4.25 ly.
4.25 ly — d = 1/p = 1/0.768 = 1.302 pc = 1.302 × 3.09 × 10¹⁶ m = 4.02 × 10¹⁶ m. Dividing by 9.46 × 10¹⁵ m gives 4.25 ly.
0.40 ly — A student who converts in the wrong direction divides 1.302 by 3.26 and gets 0.40. A parsec is LONGER than a light-year (1 pc ≈ 3.26 ly), so the distance in light-years must be the larger number, 4.25 ly.
Working d = 1/p = 1/0.768 = 1.302 pc. In metres: 1.302 × 3.09 × 10¹⁶ m = 4.023 × 10¹⁶ m. In light-years: 4.023 × 10¹⁶ m/(9.46 × 10¹⁵ m) = 4.25 ly.
6 An HR diagram is drawn with luminosity on the vertical axis (increasing upwards) and surface temperature on the horizontal axis (decreasing to the right). A star in the upper right of the diagram has a luminosity of 5.0 × 10⁴ L⊙. Its spectrum peaks at a wavelength of 725 nm. Take L⊙ = 3.83 × 10²⁶ W, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ and Wien's displacement law λ_max T = 2.9 × 10⁻³ m K. The star is modelled as a black body. What is its radius?
Answer and reasoning
3.2 × 10¹¹ m — T = 2.9 × 10⁻³/725 × 10⁻⁹ = 4000 K; L = 5.0 × 10⁴ × 3.83 × 10²⁶ = 1.92 × 10³¹ W. R = √(L/(4πσT⁴)) = 3.2 × 10¹¹ m, about 470 times the Sun's radius: a red supergiant.
1.0 × 10²³ m — A student who forgets to take the square root gets L/(4πσT⁴) = 1.05 × 10²³. That quantity is R², in m²; its square root is 3.2 × 10¹¹ m.
6.5 × 10¹¹ m — A student who takes the surface area as πR² gets R = √(L/(πσT⁴)) = 6.5 × 10¹¹ m, twice the true value. The star radiates from its whole spherical surface, 4πR².
8.2 × 10¹⁶ m — A student who treats luminosity as proportional to T uses L = σ4πR²T and gets 8.2 × 10¹⁶ m, larger than a light-year. The Stefan–Boltzmann law has T⁴.
Working T = (2.9 × 10⁻³ m K)/(725 × 10⁻⁹ m) = 4.0 × 10³ K. L = 5.0 × 10⁴ × 3.83 × 10²⁶ W = 1.915 × 10³¹ W. R = √(L/(4πσT⁴)) = √(1.915 × 10³¹/(4π × 5.67 × 10⁻⁸ × (4.0 × 10³)⁴)) = √(1.050 × 10²³ m²) = 3.2 × 10¹¹ m (about 470 solar radii, consistent with its position among the red supergiants).
7 An HR diagram is drawn with luminosity on the vertical axis (increasing upwards) and surface temperature on the horizontal axis, running from 40 000 K at the left-hand end to 2500 K at the right-hand end. Star P is plotted at the lower left of the diagram, well below the main sequence. Which description of star P is correct?
Answer and reasoning
A white dwarf: hot, but tiny, so its luminosity is low. — The left of the axis is hot, and the bottom is dim. From L = σ4πR²T⁴, a hot star can have a low luminosity only if its surface area is tiny, about the size of the Earth. This is the white-dwarf region: the exposed core of a low-mass star after the red-giant stage, in which no fusion takes place.
A cool red star, dim because its surface temperature is low. — A student who assumes temperature increases to the right, as on most graphs, reads the lower left as cool. The axis runs from 40 000 K at the left to 2500 K at the right, so the lower left is HOT; a hot star with a low luminosity must be tiny: a white dwarf.
A young dwarf that will later swell into a red giant. — A student who thinks stars grow as they age, starting as dwarfs, picks this. A white dwarf is the LAST stage of a low-mass star: the exposed core left after the red-giant stage. No fusion takes place in it, and it never swells into a giant; it simply cools.
A Sun-sized star that is dim because it has nearly run out of fuel. — A student who pictures a star as a fire that dims as its fuel runs out picks this. A star as hot as P and as large as the Sun would be more luminous than the Sun (L = σ4πR²T⁴), so it could not lie well below the main sequence. P is dim because it is tiny: a white dwarf.
8 The HR diagram shows the main sequence, the Sun and a star X. Which statement about star X is correct?
Answer and reasoning
It is very hot, so each square metre of its surface radiates a lot of power. — A student who reads the temperature axis as increasing to the right, as on an ordinary graph, takes X to be a very hot star. The axis is marked 40 000 K at the left and 2500 K at the right: temperature decreases to the right on an HR diagram, and X at about 3500 K is one of the coolest stars shown.
It is cool, and cool red stars are much more luminous than hot blue ones. — A student who takes red giants and supergiants as showing that cool stars are the luminous ones picks this. Coolness lowers, not raises, the power emitted per square metre (L = σ4πR²T⁴). The lower right of the main sequence is full of cool stars that are very dim; X is luminous in spite of being cool, because of its huge surface area.
It is near to Earth, which is why its luminosity appears to be so high. — A student who thinks of brightness as depending mainly on distance, as with a nearby street light, picks this. The vertical axis is luminosity, the total power the star radiates, which is a property of the star and does not depend on its distance from Earth. Star X emits about 10⁵ times the power of the Sun wherever it is.
It is cool, but its surface area is so large that its power output is high. — X is plotted at about 3500 K, far to the right where temperature is low, yet at about 10⁵ L⊙. A cool surface radiates little power per square metre, so a luminosity that high needs an enormous surface area: X is a red supergiant, and its position on the diagram (cool and very luminous) can only mean a very large radius.
9 The HR diagram shows the evolutionary track of a star of 1 M⊙, from position 1 on the main sequence to position 2 and then to position 3. Which statement is consistent with the change from position 2 to position 3 shown on the diagram?
Answer and reasoning
It throws off its outer layers as a planetary nebula, leaving only its small, hot core. — From 2 to 3 the star's luminosity falls by about five orders of magnitude while its surface temperature rises from about 3500 K to about 20 000 K, so what is left is hot but tiny: a white dwarf. A 1 M⊙ red giant ejects its cool outer layers as a planetary nebula, exposing the hot, dense core, which is exactly the hot, dim object plotted at 3.
It explodes as a supernova, and what is now plotted is the remnant of the explosion. — A student who believes every star ends in a supernova picks this. A supernova needs a star of well over 8 M⊙; a 1 M⊙ star never reaches the core conditions for one. Its outer layers drift away gently as a planetary nebula, and the hot core left behind is a white dwarf, which is what position 3 (hot, dim) shows.
Its core collapses into a black hole, which is plotted so low because it is so dim. — A student who thinks every star eventually collapses into a black hole picks this. Only the most massive stars leave black holes. Position 3 is a star with a surface temperature of about 20 000 K radiating 10⁻² L⊙: a hot, small, still-shining object, a white dwarf, not a black hole, which emits no light of its own and could not be plotted this way.
Its fuel is used up, so it cools and fades, keeping the giant size it had before. — A student who pictures the star fading like an ember, keeping its size, picks this. The diagram shows the opposite of cooling: from 2 to 3 the star moves far to the LEFT, so its surface temperature rises from about 3500 K to about 20 000 K. A star that kept the size of a red giant and got hotter would become more luminous, not 10⁵ times dimmer; the fall in luminosity shows the radius has shrunk enormously.
10 The HR diagram shows two stars, A and B, with dotted guide lines to the axes. Both stars are modelled as black bodies. Using the luminosities and surface temperatures read from the diagram, what is (radius of B)/(radius of A)?
Answer and reasoning
4.0 × 10¹ — R_B/R_A = √(L_B/L_A) × (T_A/T_B)². B is 100 times as luminous, giving a factor √100 = 10, and A is twice as hot, giving (10 000/5000)² = 4. The ratio is 10 × 4 = 40: B, a giant above the main sequence in the upper right, is about 40 times the radius of A.
1.0 × 10¹ — √(L_B/L_A) = √100 = 10 uses the luminosities alone. Luminosity depends on temperature as well as radius (L = σ4πR²T⁴): A is twice as hot as B, so each square metre of A radiates 2⁴ = 16 times as much. To be 100 times as luminous while cooler, B must be larger than the luminosity ratio alone suggests.
2.0 × 10¹ — √100 × (10 000/5000) = 10 × 2 = 20 takes the square root of T⁴ to be T. Rearranging L = σ4πR²T⁴ gives R² = L/(σ4πT⁴), so R ∝ √L/T², and the temperature ratio must be squared: 2² = 4, giving 40.
1.6 × 10³ — (L_B/L_A) × (T_A/T_B)⁴ = 100 × 16 = 1600 comes from R = L/(σ4πT⁴), with no square root. Luminosity is proportional to R², the surface area, not to R, so R² = L/(σ4πT⁴) and the radius ratio is √(L_B/L_A) × (T_A/T_B)² = 10 × 4 = 40.
Working From the diagram: L_A = 10² L⊙ at T_A = 10 000 K; L_B = 10⁴ L⊙ at T_B = 5000 K. L = σ4πR²T⁴, so R = √(L/(σ4π))/T² and R_B/R_A = √(L_B/L_A) × (T_A/T_B)² = √(10⁴/10²) × (10 000/5000)² = 10 × 2² = 10 × 4 = 40 = 4.0 × 10¹.
11 The diagram (not to scale) shows the Earth at two positions, E₁ and E₂, six months apart in its orbit around the Sun, and the two lines of sight from these positions to a nearby star X. The marked angle is the angle between these two lines of sight. Take 1 pc = 3.26 ly. What is the distance from the Sun to X in light-years?
Answer and reasoning
65.2 ly — The marked angle is subtended at X by the whole diameter of the orbit (E₁ to E₂, 2 AU). The parallax angle p is half of it, 0.0500 arc-second, subtended by 1 AU. d = 1/p = 20.0 pc, and 1 pc is 3.26 ly, so d = 20.0 × 3.26 = 65.2 ly.
32.6 ly — 1/0.100 = 10.0 pc, then 10.0 × 3.26 = 32.6 ly, uses the marked angle as p. The diagram shows that the marked angle is subtended by the full E₁–E₂ baseline of 2 AU; p is defined for a baseline of 1 AU (Sun to Earth), so p is half the marked angle, 0.0500 arc-second, and the star is twice as far: 20.0 pc = 65.2 ly.
20.0 ly — 1/0.0500 = 20.0 is the distance in parsecs, not light-years. d = 1/p gives parsecs, by definition of the parsec (the distance at which 1 AU subtends 1 arc-second), and 1 pc = 3.26 ly, so the distance is 20.0 × 3.26 = 65.2 ly.
6.13 ly — 20.0/3.26 = 6.13 divides by the conversion factor instead of multiplying. A parsec is the larger unit (1 pc = 3.26 ly), so a distance in light-years is a bigger number than the same distance in parsecs: 20.0 pc = 65.2 ly, not 6.13 ly.
Working The parallax angle p is the angle subtended at the star by the radius of the Earth's orbit (1 AU), i.e. the angle between the Sun–X line and the E₁–X line. The marked angle, between the E₁–X and E₂–X lines, is subtended by the full diameter (2 AU), so p = 0.100/2 = 0.0500 arc-second. d = 1/p = 1/0.0500 = 20.0 pc. Converting: 20.0 pc × 3.26 ly pc⁻¹ = 65.2 ly.
12 The flow chart shows the stages in the evolution of stars of two different masses after they leave the main sequence. One box, marked X, has been left empty. What belongs in box X?
Answer and reasoning
Red dwarf — A student who takes 'dwarf' to mean the shrunken remains of the giant, and keeps the giant's colour, picks this. A red dwarf is a small main-sequence star still fusing hydrogen, not a remnant. The core exposed when the planetary nebula is ejected is hot, at tens of thousands of kelvin, so it is a white dwarf.
New star — A student who reads 'nebula' as a star-forming cloud picks this. A planetary nebula is the thin, expanding shell of gas a low-mass star ejects at the end of its life; it disperses rather than collapsing. The chart's next box is the hot core left at its centre, a white dwarf.
White dwarf — Box X is the end of the branch for stars of less than about 8 M⊙: red giant, then planetary nebula, then the remnant. When such a star ejects its outer layers as a planetary nebula, the hot core left behind is a white dwarf, which then cools slowly over billions of years. Supernovae, neutron stars and black holes belong only to the more massive branch.
Neutron star — A student who takes white dwarf, neutron star and black hole to be successive stages of any dying star picks the denser remnant here. They are alternatives, chosen by the star's mass, not a sequence. A neutron star forms from the collapsed core of a massive star after a supernova; the remnant of a star below about 8 M⊙, left after the planetary nebula, is a white dwarf and stays one.
13 The diagram shows the spectrum of a star, a continuous spectrum crossed by absorption lines (drawn as gaps in the band), together with the laboratory emission spectra of hydrogen and of sodium drawn to the same wavelength scale. What can be deduced from the diagram?
Answer and reasoning
The star's outer layers contain sodium; hydrogen is not detected in them. — A student who reads an absorption line as meaning 'this element is absent, so its light is missing' reaches this. An absorption line is produced BY the element: hydrogen atoms in the outer layers absorb light at their own wavelengths, so the absorption lines at hydrogen's wavelengths show that hydrogen is present. The absence of a line at 589 nm means sodium is not detected, not that it is present.
Hydrogen atoms in the outer layers are emitting light at those four wavelengths. — A student who thinks every spectral line is an emission line picks this, but the star's lines are ABSORPTION lines, gaps in the continuous spectrum: less light reaches us at those wavelengths, not more. Hydrogen atoms in the cooler outer layers absorb those wavelengths from the continuous spectrum of the hotter layers beneath. The bright lines are in the laboratory hydrogen spectrum, where a hot gas emits.
The star's core is fusing hydrogen, and that fusion produces the four lines. — A student who takes spectral lines to show what the core is fusing picks this. The lines are formed in the star's outer layers, where atoms absorb light passing outwards; radiation from the core is scattered and re-emitted countless times before it leaves the star and carries no direct record of the fusion reactions. The lines show what the outer layers contain, and nothing about the core.
The star's outer layers contain hydrogen; no sodium is detected in them. — The star's four absorption lines fall at exactly the wavelengths of the four hydrogen emission lines (410, 434, 486 and 656 nm), so hydrogen in the star's cooler outer layers is absorbing those wavelengths from the continuous spectrum below. There is no absorption line at the sodium wavelength, 589 nm, so the spectrum gives no evidence of sodium in the outer layers.
That was your twenty minutes. Real practice on E.5 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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