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E.5

Fusion and stars

Theme E · Nuclear and quantum physics · SL and HL

This is all that survives of the old Option D, Astrophysics, and it is much smaller than that option was. Stellar structure, the mass–luminosity relationship and the Hertzsprung–Russell diagram remain. Cosmology — Hubble’s law, the expanding universe, dark matter and the fate of the universe — is no longer in the syllabus. It remains excellent Extended Essay territory, but it will not be examined.
Everything in this topic is deduced from light that arrived here. Nobody has visited a star, taken its temperature or weighed it — and yet we know the Sun’s core temperature, Betelgeuse’s radius and how long either has left. This is how.

🎯What you need to be able to do

  • Explain what keeps a main-sequence star stable, and why that equilibrium is self-correcting.
  • Describe the proton–proton chain and calculate the energy it releases.
  • Distinguish luminosity from apparent brightness, and use \( b = L/4\pi d^2 \).
  • Use stellar parallax to find a distance, and state the limits of the method.
  • Use Wien’s displacement law and the Stefan–Boltzmann law, and find a star’s radius from them.
  • Read a Hertzsprung–Russell diagram and identify the main sequence, red giants and white dwarfs.
  • Describe the types of star and what supports each against gravity.
  • Outline how a star forms, and describe stellar evolution as a function of mass.
  • HLState the Jeans criterion, the CNO cycle, and the Chandrasekhar and Oppenheimer–Volkoff limits.

⚖️What a star actually is

A star is a self-gravitating ball of plasma in hydrostatic equilibrium: at every depth, the inward pull of gravity on a layer is exactly balanced by the outward pressure of the gas and radiation beneath it. That balance is the definition of the main sequence, and a star holds it for about 90% of its life.

On the left, a star drawn as a glowing disc with a hot core at 1.6 times ten to the seventh kelvin. Arrows at eight points around it show the two competing forces: blue arrows pointing inwards for gravity, and amber arrows pointing outwards for radiation and gas pressure. On the right, the feedback loop that keeps the balance stable, drawn as a cycle: suppose the core contracts a little, then the temperature and density rise, so the fusion rate rises steeply, so the outward pressure rises, and the core expands back. A caption notes that this is a negative feedback loop, which is exactly why a star is stable.
The balance is not delicate — it is self-correcting. Squeeze the core and it fuses faster, pushes back and re-expands; let it expand and fusion slows and it settles again. That is why stars are not perched on a knife edge.

The equilibrium ends when the hydrogen in the core is used up. Everything after that happens comparatively quickly, and what happens is decided almost entirely by the star’s mass.

☀️Where the energy comes from

In a star like the Sun, the energy source is the proton–proton chain: four protons are converted, in three steps, into one helium-4 nucleus.

The proton-proton chain in three numbered steps. Step one: two protons fuse to give a deuterium nucleus, a positron and a neutrino; this is by far the slowest step because a proton must turn into a neutron. Step two: deuterium plus a proton gives helium-3 and a gamma ray. Step three: two helium-3 nuclei meet to give helium-4 and two protons, which are handed back to start again. The net result is four hydrogen-1 nuclei giving helium-4, two positrons, two neutrinos and 26.7 mega-electronvolts, worked from the masses as four times 1.007825 minus 4.002603 equals 0.028697 unified mass units, which is 26.73 MeV; only 0.7 per cent of the mass becomes energy. A small graph compares the pp chain, whose rate goes roughly as the fourth power of temperature, with the CNO cycle, which goes as the twentieth power and takes over above 1.8 times ten to the seventh kelvin.
The first step is the bottleneck: it needs a proton to turn into a neutron, which is a weak-interaction process and therefore extraordinarily slow. A given proton in the Sun’s core waits billions of years for it — which is precisely why the Sun burns slowly enough to be useful.
\[ 4\,^{1}_{1}\text{H} \rightarrow\ ^{4}_{2}\text{He} + 2e^{+} + 2\nu + 26.7\ \text{MeV} \]

HLHotter stars use a different route to the same product. The CNO cycle uses carbon as a catalyst, and because its rate goes roughly as \(T^{20}\) against the pp chain’s \(T^{4}\), it takes over completely above about \( 1.8 \times 10^{7} \) K. The Sun’s core, at \( 1.57 \times 10^{7} \) K, sits just below the crossover.

✏️Worked example 1 — the energy from one helium nucleus, and what the Sun loses

Atomic masses: hydrogen-1, 1.007825 u; helium-4, 4.002603 u. (a) Find the energy released when four hydrogen nuclei fuse to one helium nucleus. (b) The Sun’s luminosity is \( 3.83 \times 10^{26} \) W. At what rate is it losing mass?

(a) Mass difference first.

\[ \Delta m = 4(1.007825) - 4.002603 = 4.031300 - 4.002603 = 0.028697\ \text{u} \]
\[ E = 0.028697 \times 931.5 = 26.7\ \text{MeV} \]

(b) Energy leaving means mass leaving. Rearranging \( E = mc^2 \) as a rate:

\[ \frac{\Delta m}{\Delta t} = \frac{L}{c^{2}} = \frac{3.83 \times 10^{26}}{(3.00 \times 10^{8})^{2}} = 4.3 \times 10^{9}\ \text{kg s}^{-1} \]
Sanity check. 4.3 million tonnes a second sounds unsustainable, and it is not. Over a year that is \( 1.3 \times 10^{17} \) kg, against a solar mass of \( 2.0 \times 10^{30} \) kg — a fraction of \( 7 \times 10^{-14} \) per year. In 4.6 billion years the Sun has shed well under a hundredth of one per cent of itself this way. Note also that only 0.7% of the hydrogen’s mass becomes energy, so the hydrogen consumed is about \( 6 \times 10^{11} \) kg per second — a much larger number, and a different question.

💡Luminosity, apparent brightness, and the difference

Two quantities get called “brightness” and they are not the same thing. Luminosity \(L\) is the total power the star radiates, in watts — a property of the star alone. Apparent brightness \(b\) is the power per square metre arriving here, in W m−2 — a property of the star and how far away it is.

\[ b = \frac{L}{4\pi d^{2}} \]

The \( 4\pi d^2 \) is the surface area of a sphere of radius \(d\): the star’s output spreads over an ever-larger sphere as it travels, so the power crossing each square metre falls as \( 1/d^2 \).

At the top, a cone of light from a star passes through screens at distances of one, two and three units. The screens have areas A, 4A and 9A, so the power per square metre falls to b, b over 4 and b over 9, giving b equals L over four pi d squared. Below, the geometry of stellar parallax: the Earth is shown at two positions six months apart, in January and July, at the ends of a 2 AU baseline across its orbit around the Sun. Sight lines from each position to a nearby star are extended past it to a field of distant background stars, against which the near star appears to shift over the year. The angle p at the star is marked, with a note that it is hugely exaggerated in the drawing and is always under one arcsecond. The relation is d in parsecs equals one over p in arcseconds, with Proxima Centauri at p equals 0.769 arcseconds giving 1.30 parsecs or 4.2 light years.
Apparent brightness alone cannot tell you whether a star is modest and near or monstrous and far. You need the distance too — which is why parallax has to come first.

Stellar parallax supplies that distance for nearby stars. As the Earth moves from one side of its orbit to the other, a nearby star appears to shift against the far more distant background. Half of that annual shift is the parallax angle \(p\), and the distance follows from simple geometry:

\[ d\,/\,\text{parsec} = \frac{1}{p\,/\,\text{arcsecond}} \]

A parsec is defined as exactly the distance at which \( p = 1'' \); it works out at \( 3.09 \times 10^{16} \) m, or 3.26 light years. The method runs out when \(p\) shrinks below the measurement uncertainty, which for ground-based work is a few hundred parsecs — a tiny neighbourhood by galactic standards.

✏️Worked example 2 — from a parallax angle to a luminosity

A star has a parallax angle of \( 0.067'' \) and an apparent brightness of \( 1.5 \times 10^{-9} \) W m−2. Find its distance in metres and its luminosity. Take 1 parsec \( = 3.09 \times 10^{16} \) m.

Distance from the parallax.

\[ d = \frac{1}{0.067} = 14.9\ \text{pc} = 14.9 \times 3.09 \times 10^{16} = 4.6 \times 10^{17}\ \text{m} \]

Luminosity from the inverse-square law. Rearranging \( b = L/4\pi d^2 \):

\[ L = 4\pi d^{2}b = 4\pi (4.6 \times 10^{17})^{2}(1.5 \times 10^{-9}) = 4.0 \times 10^{28}\ \text{W} \]
Sanity check. That is about 100 times the Sun’s \( 3.83 \times 10^{26} \) W, which is an entirely ordinary value for a star a little more massive than the Sun. Note the order of operations: the parallax gives the distance, and only then can the brightness be turned into a luminosity. Doing it the other way round is impossible, and questions are often set to check that you know why.

🌈Colour, temperature and size

A star radiates very nearly as a black body, and that single fact is what makes it measurable. Two laws follow from the black-body spectrum, and between them they give the temperature and the radius.

Wien’s displacement law\( \lambda_{\text{max}}T = 2.90 \times 10^{-3}\ \text{m K} \)
Stefan–Boltzmann law\( L = 4\pi R^{2}\sigma T^{4} \)
Stefan’s constant\( \sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\,\text{K}^{-4} \)
A graph of intensity per unit wavelength against wavelength in nanometres, showing black-body curves at 4000, 5000 and 6000 kelvin, with the visible band from 380 to 700 nanometres shaded. Each curve has its peak marked, and a dashed line joining the three peaks shows that the peak moves to shorter wavelengths as the temperature rises, while the area under the curve grows rapidly. Alongside, the two laws are set out: Wien's displacement law, lambda-max times T equals 2.90 times ten to the minus three metre kelvin, which turns a colour into a surface temperature and gives 502 nanometres for the Sun at 5772 kelvin; and the Stefan-Boltzmann law, L equals four pi R squared sigma T to the fourth, which turns a known luminosity and temperature into a radius. A note points out the fourth power: double the temperature and the same star radiates sixteen times as much.
Both effects are visible at once: as the temperature rises the peak moves left (Wien) and the whole curve lifts dramatically (Stefan). It is the second that carries the fourth power, and that is the one people forget.

Put the pieces together and a point of light becomes a measured object. Parallax gives the distance; distance with apparent brightness gives the luminosity; the colour gives the temperature; and luminosity with temperature gives the radius. Four numbers about something nobody will ever visit, every one of them extracted from arriving light.

✏️Worked example 3 — how big is it?

The star from Worked example 2 has a luminosity of \( 4.0 \times 10^{28} \) W, and its spectrum peaks at 322 nm. Find its surface temperature and its radius. The Sun’s radius is \( 7.0 \times 10^{8} \) m.

Temperature, from Wien.

\[ T = \frac{2.90 \times 10^{-3}}{\lambda_{\text{max}}} = \frac{2.90 \times 10^{-3}}{322 \times 10^{-9}} = 9.0 \times 10^{3}\ \text{K} \]

Radius, from Stefan–Boltzmann. Rearrange \( L = 4\pi R^2\sigma T^4 \):

\[ R = \sqrt{\frac{L}{4\pi\sigma T^{4}}} = \sqrt{\frac{4.0 \times 10^{28}}{4\pi (5.67 \times 10^{-8})(9000)^{4}}} \]
\[ R = \sqrt{8.56 \times 10^{18}} = 2.9 \times 10^{9}\ \text{m} \approx 4.2\,R_{\odot} \]
Sanity check. Hotter than the Sun and a hundred times more luminous, so it must be bigger — and four solar radii is a reasonable size for an A-type main-sequence star. The fourth power is where this goes wrong: \( (9000)^4 = 6.6 \times 10^{15} \), and dropping a factor of ten there shifts the radius by a factor of three.
The trap: a cool star can be far more luminous than a hot one. Betelgeuse has a surface temperature of only 3600 K, well below the Sun’s 5772 K, so each square metre of it radiates less than a fifth as much. Yet it is 126 000 times more luminous, because \( L = 4\pi R^2\sigma T^4 \) has an \(R^2\) in it as well, and Betelgeuse’s radius is about 900 solar radii. Temperature alone never tells you luminosity, and luminosity alone never tells you temperature — which is exactly why the H–R diagram needs two axes.

📊The Hertzsprung–Russell diagram

Plot luminosity against surface temperature for a large sample of stars and they do not scatter evenly: they fall into a few well-defined groups. That plot is the Hertzsprung–Russell diagram, and it is the single most useful chart in astrophysics.

A Hertzsprung-Russell diagram plotting luminosity relative to the Sun, on a logarithmic scale from ten to the minus four to ten to the sixth, against surface temperature running backwards from 40000 kelvin on the left to 2000 kelvin on the right, with the spectral classes O B A F G K M shown along the top. Four populated regions are shaded: the main sequence running diagonally from hot and bright at the top left to cool and faint at the bottom right; red giants in a band at upper right; supergiants across the top; and white dwarfs at lower left. Named stars are marked: the Sun, Sirius A, Sirius B, Betelgeuse, Rigel, Proxima Centauri, Aldebaran and Arcturus. A panel explains how to read each region, noting that about 90 per cent of stars lie on the main sequence with hydrogen fusing in the core, that red giants must be enormous because they are cool yet luminous, and that white dwarfs must be tiny because they are hot yet faint.
The temperature axis runs backwards. That is a historical accident — Hertzsprung and Russell plotted spectral class O B A F G K M, which only later turned out to be a sequence of falling temperature — but every H–R diagram since has kept it.

The reasoning behind each region is worth being able to reproduce, because it is Stefan–Boltzmann applied twice:

main sequenceabout 90% of stars; hydrogen fusing in the core, ordered along the band by MASS
red giantscool, so little power per square metre — yet very luminous, so they must be huge
white dwarfshot, so a great deal per square metre — yet very faint, so they must be tiny
The trap: the main sequence is not a path a star travels along. A star does not slide down the main sequence as it ages. It arrives at one point on the band — the point fixed by the mass it was born with — stays put for most of its life, and then leaves the band altogether when the core hydrogen runs out. The main sequence is a place stars sit, not a route they follow, and describing it as an evolutionary track is a standard way to lose marks.

⭐The kinds of star

A logarithmic scale of stellar radius in solar radii, spanning ten decades. A red supergiant such as Betelgeuse is 900 solar radii, about four astronomical units across; a red giant such as Aldebaran is about 40; the Sun is 1, which is 7.0 times ten to the eighth metres; a white dwarf such as Sirius B is 0.013 solar radii, about 1.4 Earth radii, despite having 0.98 solar masses; and a neutron star is about ten to the minus five solar radii, some twelve kilometres across and heavier than the Sun. A second panel lists what holds each up: gas and radiation pressure from fusion in a main-sequence star at 1400 kilograms per cubic metre; fusion in a shell around a spent core in a red giant; electron degeneracy pressure in a white dwarf at ten to the ninth; neutron degeneracy pressure in a neutron star at ten to the seventeenth, where a teaspoon would weigh a billion tonnes; and nothing at all in a black hole.
Sirius B has 98% of the Sun’s mass packed into a ball the size of the Earth. What stops it collapsing further is not heat — it has no fusion at all — but electron degeneracy pressure, a purely quantum effect that persists even at zero temperature.

Degeneracy pressure is worth a moment because it is the reason dead stars exist at all. The Pauli exclusion principle forbids two electrons from occupying the same quantum state, so compressing a gas of electrons forces them into higher-momentum states, and that costs energy. The resistance that follows does not go away as the star cools. It is what holds up a white dwarf, and the same argument applied to neutrons holds up a neutron star.

🌌How a star begins

Stars form from cold, thin clouds of molecular gas — typically around 20 K and light years across. Gravity pulls such a cloud inwards; the random thermal motion of its molecules pushes outwards. Which wins is decided by the cloud’s mass.

Four stages of star formation drawn as shrinking circles. First a molecular cloud, cold at about 20 kelvin and very thin but enormous, light years across. Then it contracts, as gravity wins over gas pressure and the cloud fragments as it falls. Then a protostar, in which gravitational energy becomes heat so the centre warms as it shrinks. Finally fusion ignites at about ten to the seventh kelvin, when the proton-proton chain starts and the contraction halts. The whole sequence takes a few tens of millions of years for a star like the Sun. Two panels follow: the Jeans criterion, that the cloud collapses if its mass exceeds the Jeans mass, so collapse is favoured by high mass, high density and low temperature; and a worked figure showing that a cloud at 20 kelvin with ten to the tenth molecules per cubic metre has a Jeans mass of about twenty solar masses, so anything more massive must collapse, fragmenting as it does, which is why stars are born in clusters.
Nothing is fusing during the collapse. A protostar shines on gravitational potential energy alone, converted to heat as material falls inwards — and it becomes a star only at the moment fusion starts and the contraction stops.

HLThe condition is the Jeans criterion: a cloud collapses if its mass exceeds the Jeans mass \( M_J \). Since gravity grows with mass and density while the opposing pressure grows with temperature, collapse is favoured by high mass, high density and low temperature. A cloud at 20 K holding \( 10^{10} \) molecules per cubic metre has a Jeans mass of roughly 20 solar masses — and since it fragments as it falls, it produces a cluster of stars rather than one.

🕐How long a star lasts, and how it ends

Main-sequence lifetime falls sharply with mass, which is counter-intuitive until you see why: a more massive star has more fuel, but it burns it very much faster. Luminosity goes roughly as \( M^{3.5} \), so the lifetime, which is fuel divided by burn rate, goes as \( M/M^{3.5} = M^{-2.5} \).

✏️Worked example 4 — how long the Sun has

The Sun’s mass is \( 2.0 \times 10^{30} \) kg and its luminosity is \( 3.83 \times 10^{26} \) W. Only the core, about 10% of the mass, is hot enough to fuse, and 0.7% of the mass that fuses is released as energy. Estimate the Sun’s main-sequence lifetime.

How much fuel, and how much energy it holds.

\[ m_{\text{fuel}} = 0.10 \times 2.0 \times 10^{30} = 2.0 \times 10^{29}\ \text{kg} \]
\[ E = 0.007\,m_{\text{fuel}}c^{2} = 0.007 (2.0 \times 10^{29})(3.00 \times 10^{8})^{2} = 1.3 \times 10^{44}\ \text{J} \]

Divide by the rate it is spent.

\[ t = \frac{E}{L} = \frac{1.3 \times 10^{44}}{3.83 \times 10^{26}} = 3.3 \times 10^{17}\ \text{s} \approx 1.0 \times 10^{10}\ \text{years} \]
Sanity check. Ten billion years, against an accepted figure of about ten billion — and the Sun is 4.6 billion years old, so it is a little under halfway through. The estimate is only as good as the 10% assumption, which is where all the uncertainty lives; the physics either side of it is exact. A star of ten solar masses, by the \( M^{-2.5} \) scaling, lasts about \( 10^{-2.5} \) of that — some 30 million years.

What happens after the main sequence is decided by mass, and the branch point is at about eight solar masses.

Two evolutionary paths. A star below about eight solar masses goes from main sequence, with hydrogen fusing in the core, to red giant, when the core hydrogen is spent and a shell fuses instead, then to planetary nebula as the outer layers drift away, leaving a white dwarf: the bare core, cooling forever. A star above about eight solar masses goes from main sequence, burning hot and fast for millions rather than billions of years, to red supergiant, fusing heavier and heavier elements in shells right up to iron, and then to supernova, because the iron core cannot fuse further so it collapses and rebounds; the remnant is either a neutron star, for a core between 1.4 and about 3 solar masses, or a black hole above that. A scale below shows the two mass limits: the Chandrasekhar limit at 1.4 solar masses between white dwarf and neutron star, and the Oppenheimer-Volkoff limit at about 3 solar masses between neutron star and black hole, both applying to the remnant core rather than the original star.
Fusion in stars stops at iron, and E.3 already explained why: iron-56 is at the peak of the binding-energy-per-nucleon curve, so fusing it would absorb energy. The core loses its pressure support the instant it turns to iron.

Two mass limits set the ending, and both apply to the remnant core, not to the star it came from — a distinction questions test directly, because a star loses a great deal of mass before it dies:

Chandrasekhar limit1.4 \(M_{\odot}\) — above it, electron degeneracy pressure fails
Oppenheimer–Volkoff limitabout 3 \(M_{\odot}\) — above it, neutron degeneracy fails too

Below 1.4 solar masses the remnant is a white dwarf; between 1.4 and about 3 it is a neutron star; above that nothing known can halt the collapse and it becomes a black hole. The Sun, at one solar mass and losing more before the end, will finish as a white dwarf — which is also why every element in you heavier than iron had to be made somewhere else: in the supernova of a star far larger than the Sun.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Explain what keeps a main-sequence star in equilibrium, and why that equilibrium is stable rather than precarious.
At every depth the inward gravitational force on a layer is balanced by the outward pressure of the gas and radiation beneath it — hydrostatic equilibrium.
It is stable because the balance self-corrects. If the core contracts slightly, its temperature and density rise; the fusion rate is very sensitive to temperature, so it rises steeply; the outward pressure therefore rises and pushes the core back out. If the core expands, the reverse happens. That is negative feedback, so a star sits in the balance rather than being perched on it.
2. Write the overall reaction of the proton–proton chain and find the energy released, given atomic masses of 1.007825 u for hydrogen-1 and 4.002603 u for helium-4.
\[ 4\,^{1}_{1}\text{H} \rightarrow\ ^{4}_{2}\text{He} + 2e^{+} + 2\nu \] Mass difference: \( 4(1.007825) - 4.002603 = 0.028697 \) u.
Energy: \( 0.028697 \times 931.5 = 26.7 \) MeV.
Only about 0.7% of the mass involved is converted, which is what makes hydrogen fusion a slow enough process to power a star for billions of years rather than seconds.
3. A star has a parallax angle of \( 0.25'' \) and an apparent brightness of \( 6.4 \times 10^{-9} \) W m−2. Find its distance and its luminosity. Take 1 pc \( = 3.09 \times 10^{16} \) m.
\( d = 1/p = 1/0.25 = 4.0 \) pc \( = 4.0 \times 3.09 \times 10^{16} = 1.24 \times 10^{17} \) m.
\( L = 4\pi d^{2}b = 4\pi (1.24 \times 10^{17})^{2}(6.4 \times 10^{-9}) = 1.2 \times 10^{27} \) W.
That is about three solar luminosities. Note that the parallax must be converted to a distance in metres before it goes anywhere near the inverse-square law — mixing parsecs and metres is the usual slip here.
4. A star’s spectrum peaks at 580 nm and its luminosity is \( 1.0 \times 10^{30} \) W. Find its surface temperature and radius, and say what kind of star it is.
Wien: \( T = 2.90 \times 10^{-3}/580 \times 10^{-9} = 5.0 \times 10^{3} \) K.
Stefan–Boltzmann, rearranged: \[ R = \sqrt{\frac{L}{4\pi\sigma T^{4}}} = \sqrt{\frac{1.0 \times 10^{30}}{4\pi(5.67 \times 10^{-8})(5000)^{4}}} = 4.7 \times 10^{10}\ \text{m} \] That is about 68 solar radii. A star cooler than the Sun but 2600 times more luminous, and 68 times its radius, is a red giant — it can only be so luminous at that temperature by being enormous.
5. On an H–R diagram, explain how you can tell that white dwarfs must be very small and red giants very large, without being told either radius.
Use \( L = 4\pi R^{2}\sigma T^{4} \) twice.
White dwarfs lie at high temperature but very low luminosity. High \(T\) means each square metre radiates a great deal, so to have a small total \(L\) there must be very little surface area — a small \(R\). They turn out to be roughly Earth-sized.
Red giants lie at low temperature but very high luminosity. Low \(T\) means each square metre radiates little, so to have a huge total \(L\) there must be an enormous surface area — a very large \(R\). They are tens to hundreds of solar radii.
6. Two stars have the same radius, but one has twice the surface temperature of the other. Compare their luminosities, and compare the wavelengths at which they peak.
Luminosity: \( L \propto T^{4} \) at fixed \(R\), so the hotter star is \( 2^{4} = 16 \) times more luminous.
Peak wavelength: \( \lambda_{\text{max}} \propto 1/T \), so the hotter star peaks at half the wavelength — it looks bluer.
Both results come from the same black-body spectrum: raising the temperature lifts the whole curve steeply and shifts its peak to the left.
7. Explain why a star of 25 solar masses spends far less time on the main sequence than the Sun, even though it starts with far more fuel.
It has 25 times the fuel, but its luminosity goes roughly as \( M^{3.5} \), so it burns that fuel about \( 25^{3.5} \approx 7.8 \times 10^{4} \) times faster. Lifetime is fuel divided by burn rate, so \[ \frac{t}{t_{\odot}} \approx \frac{M}{M^{3.5}} = M^{-2.5} = 25^{-2.5} = 3.2 \times 10^{-4} \] Against the Sun’s \( 10^{10} \) years that is about 3 million years. Massive stars are spectacularly luminous and spectacularly short-lived, which is why the O and B stars you can see are all young.
8. Describe what happens to a star of one solar mass after it leaves the main sequence, and name the remnant.
The core hydrogen runs out, so the core contracts under gravity while a shell of hydrogen around it begins to fuse. The extra shell luminosity inflates the outer layers, which expand and cool: the star becomes a red giant. Helium fusion begins in the core. Eventually the outer layers are shed as a planetary nebula, leaving the bare, hot core behind. With no fusion left it simply cools: it is a white dwarf, supported by electron degeneracy pressure. Because its mass is below the Chandrasekhar limit of 1.4 solar masses, that pressure is enough to hold it up indefinitely.
9. State the Chandrasekhar and Oppenheimer–Volkoff limits, and explain what is being weighed in each case.
Chandrasekhar limit, 1.4 solar masses: the greatest mass that electron degeneracy pressure can support. Above it, a white dwarf cannot exist and the collapse continues.
Oppenheimer–Volkoff limit, about 3 solar masses: the greatest mass that neutron degeneracy pressure can support. Above it, nothing known halts the collapse and a black hole forms.
Both limits apply to the mass of the remnant core, not to the star’s original mass. A star loses a great deal of mass as a red giant or in a supernova, so a star that began with, say, 20 solar masses may still leave a core light enough to become a neutron star.
10. HLState the Jeans criterion and use it to explain why stars form in cold, dense clouds and are born in clusters.
A cloud collapses under its own gravity if its mass exceeds the Jeans mass, \( M > M_J \). Gravity is the collapsing agent, and the thermal motion of the gas is what resists it, so collapse is favoured by high mass, high density and low temperature — which is exactly the condition of a cold molecular cloud at around 20 K.
As the cloud collapses its density rises, which lowers the Jeans mass, so sub-regions that were previously stable now exceed their own Jeans mass and begin to collapse independently. The cloud therefore fragments as it falls, and a single cloud of a few thousand solar masses produces a whole cluster of stars rather than one very large one.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • HyperPhysics — stellar structure, fusion and the HR diagram
  • NASA — stellar evolution and the life cycles of stars