DP Physics HL E.5 Fusion and Stars

DP Physics HL E.5 Fusion and Stars - Full notes

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DP Physics HL | E.5 Fusion and Stars

DP PHYSICS · THEME E: NUCLEAR AND QUANTUM PHYSICS

E.5

Fusion and Stars

Full notes for HL · IB-style

Contents

1	Nuclear fusion
2	Stars and their energy
3	Measuring the stars
4	The Hertzsprung–Russell diagram
5	Stellar evolution
6	End points of stars
7	Exam technique and further practice
8	Common misconceptions and links
9	Exam-style question with markscheme
10	Check your understanding
11	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
explain fusion energy release using binding energy and the Coulomb barrier
calculate energy from fusion reactions
describe stellar equilibrium and star formation
use Stefan–Boltzmann and Wien's laws for stars
use parallax, luminosity and apparent brightness to find stellar properties
interpret the HR diagram and describe stellar evolution
describe stellar end points with the Chandrasekhar and Oppenheimer–Volkoff limits
describe successive fusion stages in massive stars

Sections marked (HL) are Additional Higher Level content: stellar end points and mass limits. Every other section is common to SL and HL.

Guiding questions

How do stars produce energy? How can we determine the properties of stars? How do stars evolve, and what are their end points?

How this topic is examined

E.5 appears in Papers 1 and 2. Expect fusion energy calculations, stellar radiation laws, parallax, brightness and luminosity calculations, and HR diagram interpretation; HL adds stellar end points.

Always label HR diagram axes correctly: temperature decreases to the right.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Nuclear fusion

Joining light nuclei.

Think about it

The Sun has been shining for 4.6 billion years. What fuel could possibly last that long?

Stars are powered by nuclear fusion, the joining of light nuclei. This section explains why fusion releases energy and why it needs the extreme conditions found in stellar cores.

1.1 What is fusion?

In nuclear fusion, two light nuclei join to form a heavier nucleus. Because the product has a higher binding energy per nucleon than the reactants (on the steep left side of the binding energy curve), energy is released. Fusion releases more energy per kilogram of fuel than fission.

Figure 1. Binding energy per nucleon for light nuclei: fusing them to helium-4 releases energy.

WORKED EXAMPLE 1 D–D fusion (exam level)

Two deuterium nuclei (binding energy 2.22 MeV each) fuse to form helium-4 (28.3 MeV). Estimate the energy released.

Solution

Energy = BE after − BE before = 28.3 − 2 × 2.22M1

= 23.9 MeVA1

1.2 The D–T reaction

The reaction of deuterium and tritium is the easiest to achieve on Earth: ²H + ³H → ⁴He + n, releasing 17.6 MeV. Most of the energy is carried by the neutron (14.1 MeV).

Figure 2. Deuterium–tritium fusion.

WORKED EXAMPLE 2 Energy from masses (exam level)

Masses: ²H 2.014102 u, ³H 3.016049 u, ⁴He 4.002603 u, n 1.008665 u. Find the energy released.

Solution

Δm = 5.030151 − 5.011268 = 0.018883 uM1

E = 0.018883 × 931.5 = 17.6 MeVA1

WORKED EXAMPLE 3 Energy sharing (exam level)

Using momentum conservation, show that the neutron takes about 4/5 of the energy.

Solution

Equal and opposite momenta: Ek = p²/2m, so E ∝ 1/mM1

En/Etotal = 4/(4 + 1) = 0.8 (14.1 MeV)A1

1.3 Overcoming the Coulomb barrier

Nuclei are positively charged and repel. To fuse, they must get within range of the strong force (about 1–3 fm), requiring huge kinetic energies. In practice this needs temperatures of about 10⁷ K in stars (and over 10⁸ K on Earth), plus high densities so collisions are frequent.

Figure 3. Potential energy of two nuclei against separation (schematic).

Key point

Stars fuse hydrogen at about 1.5 × 10⁷ K, far below this estimate. Fusion still occurs because some particles are much faster than average and quantum tunnelling lets them pass the barrier.

WORKED EXAMPLE 4 Barrier height (exam level)

Estimate the electric potential energy of two protons 2.0 fm apart.

Solution

E = ke²/r = 8.99 × 10⁹ × (1.6 × 10⁻¹⁹)²/2.0 × 10⁻¹⁵M1

= 1.15 × 10⁻¹³ J = 0.72 MeVA1

WORKED EXAMPLE 5 Temperature (exam level)

Estimate the temperature at which the mean kinetic energy (3/2)kT equals 0.72 MeV.

Solution

T = 2E/3k = 2 × 1.15 × 10⁻¹³/(3 × 1.38 × 10⁻²³)M1

≈ 5.6 × 10⁹ KA1

1.4 The proton–proton chain

In the Sun, four protons are converted into a helium-4 nucleus in a series of steps called the proton–proton chain. Overall 26.7 MeV is released per helium nucleus formed, partly carried away by neutrinos.

Figure 4. The main proton–proton chain in the Sun.

WORKED EXAMPLE 6 Net energy (exam level)

Show that converting four protons into helium-4 releases about 26.7 MeV (mp = 1.007276 u; ⁴He nucleus 4.001506 u; positron 0.000549 u).

Solution

Δm = 4(1.007276) − 4.001506 − 2(0.000549) = 0.026500 uM1

E = 24.7 MeV, plus 2 × 1.022 MeV from positron annihilation = 26.7 MeVA1

1.5 Magnetic confinement: the tokamak

No material container can hold a plasma at over 10⁸ K. A tokamak uses strong magnetic fields to keep the charged plasma particles spiralling away from the walls in a doughnut-shaped vessel. Neutrons escape the field and deposit their energy in a surrounding blanket, which can also breed tritium from lithium.

Figure 5. A tokamak confines hot plasma with magnetic fields (schematic).

Quick check

1. Why can magnetic fields confine the plasma but not the neutrons?

Answers: 1. neutrons are uncharged, so magnetic forces do not act on them

1.6 Solar neutrinos

The proton–proton chain produces neutrinos that escape the Sun directly, carrying information from its core. About 6.5 × 10¹⁴ solar neutrinos pass through every square metre on Earth each second. Detecting them confirmed that fusion powers the Sun.

WORKED EXAMPLE 7 Neutrino flux (exam level)

Each helium-4 produced releases 26.7 MeV and two neutrinos. Estimate the number of neutrinos the Sun emits per second.

Solution

He nuclei per second = 3.83 × 10²⁶/(26.7 × 1.6 × 10⁻¹³) = 9.0 × 10³⁷M1

Neutrinos = 1.8 × 10³⁸ s⁻¹A1

1.7 Fusion power on Earth

Fusion reactors aim to confine a plasma of deuterium and tritium at over 10⁸ K using strong magnetic fields (tokamaks such as ITER) or by compressing fuel pellets with lasers (inertial confinement). Fuel is abundant (deuterium from seawater; tritium bred from lithium), there is no long-lived high-level waste and no runaway chain reaction, but sustaining net energy gain remains very difficult.

Feature	Fission	Fusion
fuel	uranium, plutonium	deuterium, tritium
fuel supply	limited	practically unlimited
energy per nucleon	about 0.9 MeV	about 3.5 MeV (D–T)
waste	long-lived fission products	activated structure, short-lived
status	commercial since 1950s	experimental

In the real world

In 2022 the US National Ignition Facility produced more fusion energy than the laser energy delivered to its fuel pellet for the first time.

2 Stars and their energy

Stellar equilibrium and radiation.

2.1 Stellar equilibrium

A main-sequence star is stable because the inward pull of gravity is balanced by the outward pressure of hot gas and radiation from fusion in the core. If fusion slows, the core contracts and heats up, increasing fusion again: a self-regulating balance.

Figure 6. Hydrostatic equilibrium in a star.

Quick check

1. What happens to a star's core when it runs out of hydrogen?

Answers: 1. fusion falls, gravity wins and the core contracts and heats up

2.2 Formation of stars

Stars form when clouds of gas and dust (nebulae) collapse under gravity. Gravitational potential energy becomes thermal energy and the core heats up; when it reaches about 10⁷ K, hydrogen fusion begins and a main-sequence star is born.

Quick check

1. What energy source heats a protostar before fusion begins?

Answers: 1. gravitational potential energy released by contraction

2.3 Structure of the Sun

Energy released in the core travels outwards slowly by radiation through the radiative zone (taking tens of thousands of years), then by convection to the visible surface, the photosphere, from which it is radiated into space.

Figure 7. Internal structure of the Sun (not to scale).

Quick check

1. Why do we see the photosphere rather than the core?

Answers: 1. the gas above the photosphere is transparent; below it, light is repeatedly absorbed and re-emitted

2.4 Stellar spectra and composition

Absorption lines in a star's spectrum reveal the elements in its outer layers (E.1), while the overall shape of the continuous spectrum gives its temperature. Most stars are about 73% hydrogen and 25% helium by mass.

Quick check

1. How can we tell a star's temperature and composition from its light?

Answers: 1. peak wavelength gives temperature (Wien); absorption lines identify elements

2.5 Luminosity and black bodies

Luminosity L is the total power radiated by a star. Treating a star as a black body, the Stefan–Boltzmann law gives L = σAT⁴ = 4πR²σT⁴. Wien's law gives the peak wavelength: λmaxT = 2.9 × 10⁻³ m K, so hotter stars are bluer.

Figure 8. Black-body spectra for stars of different temperatures.

WORKED EXAMPLE 8 Wien (standard)

The Sun's spectrum peaks at 500 nm. Find its surface temperature.

Solution

T = 2.9 × 10⁻³/5.0 × 10⁻⁷M1

= 5800 KA1

WORKED EXAMPLE 9 Luminosity (exam level)

Calculate the Sun's luminosity (R = 6.96 × 10⁸ m, T = 5800 K).

Solution

L = 4π(6.96 × 10⁸)² × 5.67 × 10⁻⁸ × 5800⁴M1

= 3.9 × 10²⁶ WA1

2.6 The Sun's spectrum

The Sun's spectrum is close to that of a 5800 K black body, peaking near 500 nm in the visible range. About 40% of its output is visible light, about half is infrared and the rest is ultraviolet.

Figure 9. Black-body spectrum at 5800 K, approximating sunlight.

Quick check

1. Why does the Sun appear white from space rather than green, despite peaking at 500 nm?

Answers: 1. it emits strongly across all visible wavelengths, which combine to white

2.7 Multi-step problem: a red giant

WORKED EXAMPLE 10 Betelgeuse (exam level)

Betelgeuse has T = 3600 K and L = 1.0 × 10⁵ L⊙. Find (a) its peak wavelength; (b) its radius in metres and in AU.

Solution

(a) λ = 2.9 × 10⁻³/3600 = 810 nmA1

(b) R = R⊙ × √(10⁵) × (5800/3600)² = 821 R⊙ = 5.7 × 10¹¹ mM1 A1

= 3.8 AU: it would engulf Mars's orbitA1

2.8 Comparing stars

Ratios avoid repeated calculation: L ∝ R²T⁴.

WORKED EXAMPLE 11 Radius (exam level)

A star has luminosity 100 L⊙ and temperature 2900 K (Sun 5800 K). Find its radius in solar radii.

Solution

R² = (L/L⊙)(T⊙/T)⁴ = 100 × 16M1

R = 40 R⊙A1

WORKED EXAMPLE 12 Temperature (exam level)

A white dwarf has radius 0.010 R⊙ and luminosity 0.0010 L⊙. Find its temperature.

Solution

T⁴ = T⊙⁴ × (L/L⊙)/(R/R⊙)² = T⊙⁴ × 10M1

T = 5800 × 100.25 = 10 300 KA1

2.9 Mass lost by the Sun

Fusion converts mass into energy: the Sun loses mass at L/c², about 4 million tonnes per second, yet this is tiny compared with its total mass.

WORKED EXAMPLE 13 Mass loss (exam level)

Find the rate at which the Sun converts mass into energy.

Solution

Δm/Δt = L/c² = 3.83 × 10²⁶/9.0 × 10¹⁶M1

= 4.3 × 10⁹ kg s⁻¹A1

3 Measuring the stars

Distance and brightness.

Think about it

We cannot visit a star. How can we find out how far away it is and how much power it emits?

3.1 Stellar parallax

As Earth orbits the Sun, a nearby star appears to shift against distant background stars. Half the total angular shift is the parallax angle p. The distance in parsecs is d = 1/p with p in arcseconds. 1 pc = 3.26 ly = 3.09 × 10¹⁶ m. Parallax works up to a few thousand parsecs with space telescopes.

Figure 10. Stellar parallax: the star's apparent position shifts as Earth orbits the Sun.

WORKED EXAMPLE 14 Parallax (standard)

A star has parallax 0.25 arcseconds. Find its distance in pc, ly and m.

Solution

d = 1/0.25 = 4.0 pcA1

= 13 ly = 1.2 × 10¹⁷ mA1

Quick check

1. Why can parallax not be used for very distant stars?

Answers: 1. the angle becomes too small to measure

3.2 Units of distance

Unit	Definition	Size
astronomical unit (AU)	mean Earth–Sun distance	1.50 × 10¹¹ m
light-year (ly)	distance light travels in a year	9.46 × 10¹⁵ m
parsec (pc)	distance at which 1 AU subtends 1 arcsecond	3.09 × 10¹⁶ m = 3.26 ly

WORKED EXAMPLE 15 Light-year (basic)

Show that 1 ly ≈ 9.5 × 10¹⁵ m.

Solution

d = 3.0 × 10⁸ × 3.16 × 10⁷M1

= 9.5 × 10¹⁵ mA1

3.3 Apparent brightness

Apparent brightness b is the power received per unit area at Earth: b = L/4πd² (W m⁻²). It falls with the square of distance.

Figure 11. Apparent brightness falls as the inverse square of distance.

WORKED EXAMPLE 16 Solar constant (standard)

Calculate the Sun's apparent brightness at Earth (d = 1.50 × 10¹¹ m).

Solution

b = 3.83 × 10²⁶/(4π × (1.50 × 10¹¹)²)M1

= 1.35 × 10³ W m⁻²A1

WORKED EXAMPLE 17 Distance from brightness (exam level)

A star has L = 4.0 × 10²⁸ W and b = 2.0 × 10⁻⁹ W m⁻². Find its distance in pc.

Solution

d = √(L/4πb) = 1.26 × 10¹⁸ mM1

= 41 pcA1

3.4 Brightness and luminosity are different

The brightest stars in the night sky are not necessarily the most luminous: Sirius looks brightest because it is close, while Rigel is about 5000 times more luminous but much further away.

Figure 12. Apparent brightness of some well-known stars (log scale).

WORKED EXAMPLE 18 Which is brighter? (exam level)

Star P has L = 100 L⊙ at 50 pc; star Q has L = 1 L⊙ at 5 pc. Compare their apparent brightnesses.

Solution

b ∝ L/d²: P: 100/2500 = 0.04; Q: 1/25 = 0.04M1

They appear equally brightA1

3.5 Combining the laws

Measuring b, d (parallax) and λmax lets us find L, T and then R for a star: a powerful chain of reasoning used in every exam.

WORKED EXAMPLE 19 Full analysis (exam level)

A star 20 pc away has b = 5.0 × 10⁻¹⁰ W m⁻² and peak wavelength 380 nm. Find L, T and R.

Solution

d = 6.18 × 10¹⁷ m; L = 4πd²b = 2.4 × 10²⁷ WM1 A1

T = 2.9 × 10⁻³/3.8 × 10⁻⁷ = 7600 KA1

R = √(L/4πσT⁴) = 1.0 × 10⁹ m (about 1.4 R⊙)A1

4 The Hertzsprung–Russell diagram

Classifying stars.

4.1 Reading the HR diagram

The HR diagram plots luminosity (vertical, logarithmic) against surface temperature (horizontal, decreasing to the right). Most stars lie on the main sequence. Red giants and supergiants are cool but very luminous (so very large); white dwarfs are hot but dim (so very small).

Figure 13. A Hertzsprung–Russell diagram.

WORKED EXAMPLE 20 Size from position (exam level)

Explain why stars at the top right of the HR diagram must be large.

Solution

High luminosity but low temperatureR1

L = 4πR²σT⁴, so R must be largeR1

4.2 Lines of constant radius

Because L ∝ R²T⁴, lines of constant radius on a log–log HR diagram are straight diagonal lines. Stars on the same line have the same size; moving up to a higher line means a larger star.

Figure 14. HR diagram with lines of constant radius (dashed) and the main-sequence band.

Quick check

1. A star moves straight up the HR diagram at constant temperature. What happens to its radius?

Answers: 1. it increases

4.3 Spectral classes

Stars are classified by their spectra into classes O, B, A, F, G, K, M, from hottest (blue) to coolest (red). The Sun is a G star.

Class	Temperature / K	Colour	Example
O	> 30 000	blue	Zeta Puppis
B	10 000–30 000	blue-white	Rigel
A	7500–10 000	white	Sirius A
F	6000–7500	yellow-white	Procyon
G	5200–6000	yellow	Sun
K	3700–5200	orange	Arcturus
M	< 3700	red	Betelgeuse

4.4 Main-sequence properties

Along the main sequence, more massive stars are hotter and much more luminous: L ∝ M3.5 approximately. Massive stars burn their fuel far faster, so their lifetimes are much shorter.

Figure 15. Mass–luminosity relation for main-sequence stars.

Figure 16. Main-sequence lifetime against mass.

WORKED EXAMPLE 21 Luminosity (exam level)

A main-sequence star has mass 4 M⊙. Estimate its luminosity.

Solution

L = 43.5 L⊙M1

= 128 L⊙A1

WORKED EXAMPLE 22 Lifetime (exam level)

Estimate the lifetime of the star in the previous example (Sun: 10¹⁰ years).

Solution

t ∝ M/L = 4/128M1

≈ 3 × 10⁸ yearsA1

4.5 Star clusters

All stars in a cluster formed at about the same time and distance. Their HR diagram shows a main sequence that bends off towards the giants at a 'turn-off point'. Stars above the turn-off have already left the main sequence, so the turn-off temperature reveals the cluster's age: the lower the turn-off, the older the cluster.

Figure 17. HR diagram of a star cluster showing the main-sequence turn-off.

Quick check

1. A cluster's turn-off is at about 1 L⊙. Estimate its age.

Answers: 1. about the Sun's lifetime, 10¹⁰ years

5 Stellar evolution

Life cycles of stars.

5.1 Evolution of a Sun-like star

When hydrogen in the core runs out, the core contracts and heats while the outer layers expand and cool: the star becomes a red giant. Helium fusion begins in the core (forming carbon and oxygen). Eventually the outer layers drift off as a planetary nebula, leaving a hot, dense white dwarf that slowly cools.

Figure 18. Evolutionary paths for low-mass and high-mass stars.

Quick check

1. What is left after a Sun-like star dies?

Answers: 1. a white dwarf (with a planetary nebula)

5.2 Evolution of massive stars

Stars more than about 8 solar masses become red supergiants and fuse successively heavier elements up to iron in shells. Iron cannot release energy by fusion, so the core collapses catastrophically, triggering a supernova. The remnant is a neutron star or, for the most massive cores, a black hole.

Figure 19. Shell structure of a massive star just before a supernova (not to scale).

WORKED EXAMPLE 23 Iron (exam level)

Explain why fusion in stars stops at iron.

Solution

Iron-56 has the highest binding energy per nucleonR1

Fusing iron would absorb energy rather than release itR1

In the real world

The gold, uranium and iodine on Earth were made in supernovae and neutron-star mergers billions of years ago.

5.3 Tracks on the HR diagram

As a star evolves it moves across the HR diagram: from the main sequence up and right to the red giant region, then (for Sun-like stars) left across the top as the outer layers are lost, and down to the white dwarf region.

Figure 20. Approximate evolutionary track of the Sun on an HR diagram.

Quick check

1. In which direction does a white dwarf move on the HR diagram as it cools?

Answers: 1. down and to the right

5.4 Supernovae

A supernova can briefly outshine its entire galaxy, releasing about 10⁴⁴ J of light and far more in neutrinos. The explosion scatters heavy elements into space, and its shock waves can trigger new star formation.

Figure 21. Typical light curve of a core-collapse supernova (schematic).

In the real world

Supernova 1987A in a neighbouring galaxy was the first to be observed by neutrino detectors on Earth, confirming the core-collapse model.

Quick check

1. Why does a supernova fade over months?

Answers: 1. its light comes increasingly from radioactive decay of newly made nuclei such as nickel-56 and cobalt-56

5.5 Nucleosynthesis

Hydrogen and helium formed in the Big Bang. Stars make heavier elements by fusion up to iron; elements heavier than iron are made by neutron capture in red giants and in supernovae and neutron-star mergers, then spread into space to form new stars and planets.

Quick check

1. Where were the carbon atoms in your body made?

Answers: 1. inside earlier stars, by helium fusion

6 End points of stars

White dwarfs, neutron stars and black holes.

6.1 Mass limits (HL)

A white dwarf is supported against gravity by electron degeneracy pressure, which can only support a core of up to about 1.4 M⊙ (the Chandrasekhar limit). More massive collapsing cores form neutron stars, supported by neutron degeneracy pressure up to about 2–3 M⊙ (the Oppenheimer–Volkoff limit). Beyond that, nothing can stop collapse to a black hole.

Remnant core mass	End point	Supported by
< 1.4 M⊙	white dwarf	electron degeneracy pressure
1.4–3 M⊙	neutron star	neutron degeneracy pressure
> about 3 M⊙	black hole	nothing

Quick check

1. What is the Chandrasekhar limit?

Answers: 1. maximum mass of a white dwarf, about 1.4 M⊙

6.2 Density of remnants (HL)

White dwarfs are about the size of Earth; neutron stars are only about 20 km across, with densities similar to atomic nuclei.

WORKED EXAMPLE 24 Neutron star (exam level)

A neutron star has mass 1.5 M⊙ (M⊙ = 2.0 × 10³⁰ kg) and radius 10 km. Find its density.

Solution

ρ = 3.0 × 10³⁰/(⁴⁄₃π(1.0 × 10⁴)³)M1

= 7.2 × 10¹⁷ kg m⁻³: like nuclear matterA1

WORKED EXAMPLE 25 Escape speed (exam level)

Find the escape speed from the surface of this neutron star.

Solution

v = √(2GM/R) = √(2 × 6.67 × 10⁻¹¹ × 3.0 × 10³⁰/1.0 × 10⁴)M1

= 2.0 × 10⁸ m s⁻¹: two-thirds of the speed of lightA1

6.3 Stages of fusion in massive stars (HL)

Each successive stage of core fusion requires a higher temperature and lasts a much shorter time, because heavier nuclei have larger Coulomb barriers and each stage releases less energy per nucleon.

Fuel	Main products	Approximate duration (25 M⊙ star)
hydrogen	helium	7 million years
helium	carbon, oxygen	0.5 million years
carbon	neon, magnesium	600 years
neon	oxygen, magnesium	1 year
oxygen	silicon, sulfur	6 months
silicon	iron	1 day

Quick check

1. Why does each stage need a higher temperature?

Answers: 1. heavier nuclei have more charge, so a larger Coulomb barrier

7 Exam technique and further practice

Applying the ideas.

7.1 Method checklist

•Convert parsecs to metres (× 3.09 × 10¹⁶) before using b = L/4πd².
•Use ratios with L ∝ R²T⁴ when comparing stars.
•Read the HR diagram's temperature axis carefully: it increases to the left.
•Explain fusion energy using binding energy per nucleon.
•State the balance of forces when explaining stellar stability.

7.2 Multi-step problem: Sirius

WORKED EXAMPLE 26 Sirius A (exam level)

Sirius A has parallax 0.379 arcsec, apparent brightness 1.2 × 10⁻⁷ W m⁻² and surface temperature 9900 K. Find (a) its distance in m; (b) its luminosity; (c) its radius in solar radii.

Solution

(a) d = 2.64 pc = 8.2 × 10¹⁶ mA1

(b) L = 4πd²b = 1.0 × 10²⁸ W (about 26 L⊙)M1 A1

7.3 Graded practice

WORKED EXAMPLE 27 Wien (standard)

A red giant has λmax = 850 nm. Find its temperature.

Solution

T = 2.9 × 10⁻³/8.5 × 10⁻⁷M1

= 3400 KA1

WORKED EXAMPLE 28 Brightness ratio (standard)

Two identical stars are at 10 pc and 40 pc. Compare their apparent brightnesses.

Solution

b ∝ 1/d²M1

Nearer star is 16 times brighterA1

WORKED EXAMPLE 29 Fusion energy per kg (exam level)

Find the energy released per kilogram of D–T fuel (17.6 MeV per reaction; 5 u per reaction).

Solution

Reactions per kg = 1/(5 × 1.661 × 10⁻²⁷) = 1.20 × 10²⁶M1

E = 1.20 × 10²⁶ × 2.82 × 10⁻¹² = 3.4 × 10¹⁴ J kg⁻¹A1

7.4 Data: properties of stars

Use the data to find unknown properties of each star.

Star	L / L⊙	T / K	Radius / R⊙
Sun	1	5800	1
Vega	40	9600	?
Aldebaran	440	3900	?
Sirius B	0.025	25 000	?

WORKED EXAMPLE 30 Vega (exam level)

Find Vega's radius.

Solution

R = √(L/L⊙) × (T⊙/T)²M1

= √40 × (5800/9600)² = 2.3 R⊙A1

WORKED EXAMPLE 31 Aldebaran and Sirius B (exam level)

Find the radii of Aldebaran and Sirius B and classify each.

Solution

Aldebaran: √440 × (5800/3900)² = 46 R⊙: red giantA1

Sirius B: √0.025 × (5800/25000)² = 0.0085 R⊙: white dwarfA1

7.5 Graded practice: fusion and stars

WORKED EXAMPLE 32 Sun's lifetime (exam level)

The Sun will convert about 10% of its 1.5 × 10³⁰ kg of hydrogen into helium during its main-sequence life, releasing 0.7% of that mass as energy. Estimate its main-sequence lifetime.

Solution

E = 0.007 × 1.5 × 10²⁹ × 9.0 × 10¹⁶ = 9.5 × 10⁴³ JM1

t = E/L = 2.5 × 10¹⁷ s ≈ 8 × 10⁹ yearsA1

WORKED EXAMPLE 33 Parallax limit (standard)

Gaia measures parallaxes down to about 2 × 10⁻⁵ arcsec. Find the corresponding distance.

Solution

d = 1/pM1

= 50 000 pcA1

WORKED EXAMPLE 34 Colour (standard)

Explain why a 3000 K star appears red and a 20 000 K star appears blue-white.

Solution

Wien: λmax ≈ 970 nm (infrared/red end) for 3000 KR1

λmax ≈ 145 nm (UV, so the visible part is bluest) for 20 000 KR1

7.6 Graded practice: distances and brightness

WORKED EXAMPLE 35 Light-years (standard)

Proxima Centauri is 1.30 pc away. Find its distance in light-years and the time its light takes to reach us.

Solution

1.30 × 3.26 = 4.24 lyA1

Light takes 4.24 yearsA1

WORKED EXAMPLE 36 Luminosity from data (exam level)

A star 100 pc away has apparent brightness 1.0 × 10⁻¹¹ W m⁻². Find its luminosity in L⊙.

Solution

L = 4π(3.09 × 10¹⁸)² × 1.0 × 10⁻¹¹ = 1.2 × 10²⁷ WM1

= 3.1 L⊙A1

WORKED EXAMPLE 37 Doubling temperature (standard)

A star's temperature doubles at constant radius. By what factor does its luminosity change?

Solution

L ∝ T⁴M1

16 timesA1

7.7 Graded practice: explanations

WORKED EXAMPLE 38 Stability of the Sun (exam level)

Explain why the Sun neither collapses nor expands during its main-sequence life.

Solution

Gravity pulls inwards on all layersR1

Pressure from hot gas and radiation, maintained by core fusion, pushes outwards with equal effectR1

WORKED EXAMPLE 39 Massive stars die young (exam level)

Explain why massive stars have short lives even though they have more fuel.

Solution

Luminosity rises steeply with mass (L ∝ M³·⁵)R1

Fuel is used up much faster, so lifetime ∝ M/L fallsR1

WORKED EXAMPLE 40 Fusion on Earth (standard)

Give two reasons why fusion power is difficult to achieve.

Solution

Extremely high temperatures (> 10⁸ K) are needed to overcome repulsionR1

The plasma must be confined long enough at high density without touching the wallsR1

7.8 Self-test 2

Term	Meaning
fusion	joining of light nuclei, releasing energy
luminosity	total power emitted by a star (W)
apparent brightness	power received per unit area at Earth (W m⁻²)
parsec	distance at which 1 AU subtends 1 arcsecond
main sequence	stars fusing hydrogen in their cores
red giant	large, cool, luminous evolved star
white dwarf	small, hot, dense remnant of a low-mass star
Chandrasekhar limit (HL)	maximum white dwarf mass, 1.4 M⊙

Students often think…	Actually…
stars burn hydrogen like fire	they fuse nuclei
brighter-looking stars are always more luminous	brightness also depends on distance
the Sun will become a black hole	it will become a white dwarf
fusion is easy because nuclei attract	they repel; high temperatures are needed
HR diagram temperature increases to the right	it decreases to the right

Topic	Connection
E.3 Radioactive decay	binding energy
E.4 Fission	comparison of nuclear energy sources
B.1 Thermal energy transfers	black-body radiation, Stefan–Boltzmann, Wien
D.1 Gravitational fields	gravitational collapse

Idea	Remember
Fusion	light nuclei join; binding energy rises
Conditions	≈10⁷ K, high density
Radiation laws	L = 4πR²σT⁴; λmaxT = 2.9 × 10⁻³
Distance	parallax d = 1/p; b = L/4πd²
HR diagram	main sequence, giants, white dwarfs
Evolution	mass decides the end point

Part	Answer	Marks
(a)	d = 1/0.050 = 20 pc	M1
	6.2 × 10¹⁷ m	A1
(b)	L = 4πd²b	M1
	1.4 × 10²⁵ W	A1
(c)	T = 2.9 × 10⁻³/2.9 × 10⁻⁷ = 10 000 K	A1
(d)	R = √(L/4πσT⁴)	M1
	1.4 × 10⁸ m	A1
(e)	inward gravity balanced by outward gas and radiation pressure	R1
	fusion in core provides energy for pressure	R1
(f)	below the main sequence near white dwarfs (hot, small, low luminosity)	A1
(g)	1.4 solar masses (Chandrasekhar limit)	A1
	electron degeneracy pressure	A1