Key Idea: This topic is the physics of stars, from the reaction that powers them to how they live, shine and die. It ties together six ideas: fusion releases energy from a mass defect (E = mc²); the outward pressure it creates balances gravity to keep a star stable; a star has a finite lifetime set by its fuel and its luminosity; we measure stars by their brightness and distance; we read their temperature, size and type from their light (Wien, Stefan-Boltzmann, the H-R diagram); and they end their lives — and forge the heavier elements — according to their mass. It is examined on both papers. Paper 1A is quick multiple-choice — which conditions allow fusion, how brightness changes with distance, a parallax distance, where a star sits on the H-R diagram, the order of a life cycle. Paper 2 is longer structured work — calculate the energy released in MeV, 'show that' a lifetime is about N years, estimate a mass lost or a stellar radius, find a radius ratio from L and T, or outline how an element is confirmed in a star.
📋 Key formulas
Five of these carry the data-booklet badge (look for it). The lifetime relation t = E/L is not printed separately — you build it from 'luminosity = energy used each second', so you remember that one. The mass lost Δm = E/c² and the radius ratio are just E = mc² and Stefan-Boltzmann rearranged.
- energy released by the reaction (J, or MeV)
- mass that disappears — the mass defect Δm (kg, or u)
- speed of light, 3.00 × 10⁸ m s⁻¹ (given constant)
- main-sequence lifetime — how long the star keeps fusing hydrogen (s)
- total energy the fusible hydrogen can release (J)
- luminosity — the energy the star radiates each second (W = J s⁻¹)
- apparent brightness — power received per unit area at the observer (W m⁻²)
- luminosity — total power the star radiates in all directions (W)
- distance from the star to the observer (m)
- surface area of the sphere the light has spread over (m²)
- distance to the star, in parsecs (pc)
- parallax angle — half the star's apparent yearly shift, in arc-seconds (″)
- peak wavelength — where the star's radiation is most intense (m)
- surface (absolute) temperature of the star (K)
- Wien's constant, 2.9 × 10⁻³ m K (given)
- luminosity — total power the star radiates (W)
- Stefan-Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (given)
- surface area of the star; for a sphere A = 4πR² (m²)
- surface (absolute) temperature of the star (K)
🌟 The six ideas at a glance
💡 Luminosity vs apparent brightness — the distinction that recurs
Luminosity is the total power the star pours out (fixed). Apparent brightness is what reaches us per m² (it drops with distance). The H-R diagram and Stefan-Boltzmann use luminosity; a detector reads apparent brightness — read which one the question gives you. And the temperature axis runs backwards on the H-R diagram: hot stars are on the LEFT, cool ones on the right.
✏️ Worked exam-style questions
IB-style question — energy released by a fusion reaction
Deep in a star's core, light nuclei fuse into a single heavier nucleus. The combined mass of the original nuclei exceeds the mass of the product by a mass defect Δm = 0.022300 u. Calculate the energy released by this reaction, in MeV. (1 u = 931.5 MeV c⁻².)
Solution:
The released energy comes from the mass defect. Write the given $E = mc²$ first; with the 931.5 MeV-per-u shortcut the c² is built in, so multiply Δm (in u) by 931.5:
Substitute the mass defect Δm = 0.022300 u:
Work it out — keep the unit:
E = 0.022300 × 931.5 ≈ 20.8 MeV ≈ 21 MeV. The tiny missing mass became the released energy. (For an answer in joules instead, convert Δm to kilograms first, then multiply by c².)
IB-style question — show that a star lives for about 8 billion years
A main-sequence star, Vega-B, has mass M = 2.0 × 10³⁰ kg. About 12% of its mass is hydrogen in the core that can fuse, and 0.70% of that fusible mass is released as energy. Its luminosity is L = 6.0 × 10²⁶ W. Show that its main-sequence lifetime is about 8 × 10⁹ years. (c = 3.00 × 10⁸ m s⁻¹; 1 year ≈ 3.16 × 10⁷ s.)
Solution:
Fusible-hydrogen mass — take 12% of the star's mass:
Energy released — write the given $E = mc²$; only 0.70% of that fuel becomes energy, so multiply by 0.0070:
Lifetime — use the relation $t = E/L$:
Convert seconds to years (÷ 3.16 × 10⁷):
The fuel is worth E ≈ 1.5 × 10⁴⁴ J, and at L = 6.0 × 10²⁶ W that lasts t = E/L ≈ 2.5 × 10¹⁷ s ≈ 8 × 10⁹ years. Cut the fuel down TWICE (core fraction, then ~0.7% conversion) before E = mc², and convert seconds to years at the end.
IB-style question — brightness, distance and parallax
A star has luminosity L = 3.2 × 10²⁸ W and is at a distance d = 5.0 × 10¹⁸ m from Earth. (a) Calculate its apparent brightness as seen from Earth. (b) A second, fainter star has a measured parallax angle of p = 0.020 arc-seconds. Find its distance, in parsecs.
Solution:
(a) Write the given inverse-square law — the brightness is the luminosity spread over the sphere of area 4π d²:
(a) Work out the sphere's area first, then divide:
(b) Use the given parallax formula — distance in parsecs is one over the angle in arc-seconds:
(b) Evaluate — the answer is in parsecs:
(a) b ≈ 1.0 × 10⁻¹⁰ W m⁻² — tiny, because the star's huge power is spread over an enormous sphere. (b) d = 1 ÷ 0.020 = 50 parsec; a smaller parallax angle would mean an even farther star. Keep luminosity (fixed) and apparent brightness (falls as 1/d²) separate.
IB-style question — temperature from the peak, then a radius ratio
Star X has a black-body peak at λₘₐₓ = 725 nm and Star Y has a peak at λₘₐₓ = 290 nm. (a) Find the surface temperature of each. (b) Star Y is also 9 times as luminous as Star X — find the ratio of Star Y's radius to Star X's radius. (Wien's constant = 2.9 × 10⁻³ m K.)
Solution:
(a) Use the given Wien's law, rearranged for temperature, with the peaks in metres:
(a) Star X: 725 nm = 725 × 10⁻⁹ m; Star Y: 290 nm = 290 × 10⁻⁹ m:
(b) Use the given $L = \sigma A T⁴$ with A = 4πR², so L ∝ R²T⁴. Write the ratio (σ and 4π cancel) and rearrange for the radius ratio:
(b) Substitute LY/LX = 9 and TY/TX = 10000 ÷ 4000 = 2.5:
(a) TX ≈ 4000 K, TY ≈ 10000 K — the bluer (shorter) peak is the hotter star. (b) RY/RX = √9 ÷ 2.5² ≈ 0.48, so Star Y is under half the radius of Star X yet far more luminous, because of its much higher temperature. The temperature ratio is raised to the FOURTH power before you square-root, so it appears squared in the radius formula.
IB-style question — read the H-R diagram and the life cycle
On an H-R diagram, Star P sits at the bottom-left (hot, very dim) and Star Q sits at the top-right (cool, very bright). (a) State the type of each star and which has the larger radius, with a reason. (b) Star Q has a mass of about fifteen times the Sun's. Identify, in order, the stages it will pass through to the end of its life.
Solution:
(a) Read the positions. Bottom-left = hot but dim → a white dwarf (Star P); top-right = cool but bright → a red giant / supergiant (Star Q).
(a) Compare the radii with L ∝ r²T⁴: Star Q is far more luminous yet cooler, so its radius must be much larger to radiate that much power — Star Q is bigger.
(b) Decide the path from the mass. Fifteen solar masses is a massive star, so it follows the supernova path, not the white-dwarf path.
(b) Write the stages in order, ending with the remnant:<br>main sequence → red supergiant → supernova → neutron star (or a black hole if the core is heavy enough).
(a) P = white dwarf, Q = red giant/supergiant; Q has the larger radius because it is more luminous but cooler (L ∝ r²T⁴). (b) main sequence → red supergiant → supernova → neutron star or black hole. Only MASSIVE stars go supernova — a Sun-like star ends gently as a white dwarf.
🧠 Quick self-check
Tap each card to reveal the answer.
🎯 Exam tips
Exam Tips
- Energy released: find the mass defect FIRST (the product is lighter), then E = mc². In MeV, just multiply Δm (in u) by 931.5; for joules, convert Δm to kg then × c².
- Equilibrium: gravity pulls IN, the PRESSURE from fusion's radiation + hot gas pushes OUT. Say it is the pressure (not the reactions themselves) that balances gravity, and that the balance is self-correcting.
- Lifetime = fuel's energy ÷ luminosity: t = E ÷ L. Cut the fuel down twice (core fraction, then ~0.7% conversion) before E = mc², and convert seconds to years (÷ 3.16 × 10⁷). Mass lost = E ÷ c².
- Keep luminosity (total power out, fixed) and apparent brightness (received per m², falls as 1/d²) apart. Inverse-square: twice as far → a quarter as bright. Equal brightness ⇒ d ∝ √L.
- Parallax: d(parsec) = 1/p, with p in ARC-SECONDS and d straight out in PARSECS. A smaller angle means a farther star.
- Wien: T = 2.9 × 10⁻³ ÷ λₘₐₓ, with λₘₐₓ in METRES (nm → ×10⁻⁹). Stefan-Boltzmann: L = σ(4πR²)T⁴, so L ∝ R²T⁴. For two stars take the RATIO so σ and 4π cancel: RB/RA = √(LB/LA) ÷ (TB/TA)².
- H-R diagram: temperature axis runs BACKWARDS (hot on the LEFT), luminosity UP. Read the type from the position; a cool star can still be bright if it is huge (red giant), a hot one dim if it is tiny (white dwarf).
- Evolution: MASS decides the path — only massive stars go supernova (the Sun ends as a white dwarf). To confirm an element in a star: split its light, find the dark ABSORPTION lines, and match the pattern to that element in the lab.