E.5 Fusion and stars – IB Physics exam questions with mark scheme

1Hard4 marks

The Sun has a luminosity of 3.83 × 10²⁶ W. Its energy comes from fusion, in which four protons form one helium nucleus, releasing 26.7 MeV.

(a) Calculate the number of helium nuclei formed each second. [2]

(b) Calculate the mass converted to energy each second. [2]

(a) Energy per fusion in joules: 26.7 × 10⁶ × 1.60 × 10⁻¹⁹ = 4.27 × 10⁻¹² J(1)
(a) Rate = L ÷ E = 8.97 × 10³⁷ s⁻¹ (allow 8.95 × 10³⁷ to 9.0 × 10³⁷)(1)
(b) Use ΔE = Δm c², so Δm/Δt = L/c²(1)
(b) 4.26 × 10⁹ kg s⁻¹ (allow 4.3 × 10⁹)(1)
Answer: (a) 8.97 × 10³⁷ s⁻¹; (b) 4.26 × 10⁹ kg s⁻¹
Examiner tip: Convert MeV to J (×1.60 × 10⁻¹³ per MeV) before dividing into the power.

2Hard5 marks

A star has a surface temperature of 3600 K and a luminosity of 4.0 × 10³⁰ W. Assume it radiates as a black body.

(a) Calculate the peak wavelength of its emission and identify the region of the spectrum. [2]

(b) Calculate the radius of the star. [2]

(c) Suggest what type of star this is, justifying your answer using its temperature and luminosity compared with the Sun (L = 3.83 × 10²⁶ W, T = 5800 K). [1]

(a) λ_max = b/T = 2.90 × 10⁻³ ÷ 3600 = 8.06 × 10⁻⁷ m(1)
(a) Infrared (just beyond the visible red end)(1)
(b) L = σAT⁴ with A = 4πR², so R = √(L/4πσT⁴)(1)
(b) R = 1.8 × 10¹¹ m(1)
(c) Red supergiant/giant – cooler than the Sun (red) yet ~10⁴ times more luminous, so it must have a far larger surface area(1)
Answer: (a) 8.06 × 10⁻⁷ m, infrared; (b) 1.8 × 10¹¹ m; (c) red giant/supergiant
Examiner tip: Use T in kelvin and raise it to the fourth power; the area of a sphere is 4πR².

3Core3 marks

A main-sequence star remains the same size for millions of years.

(a) Explain, in terms of the forces acting on the outer layers of the star, why the star is stable. [2]

(b) Hydrogen fusion in the core of the star requires a high temperature. State why a high density of nuclei is also required for fusion to sustain the star. [1]

(a) gravitational force/pressure acts inwards (towards the centre) on the outer layers; allow 'gravity pulls the layers in'(1)
(a) outward (radiation pressure and gas/thermal) pressure from the hot core balances it / the forces are equal and opposite, so there is no resultant force (hydrostatic equilibrium); do not accept 'pressures cancel' without saying one is outward and one inward(1)
(b) so that collisions/fusion reactions are frequent enough (high reaction rate) to release energy at the rate needed / to supply the luminosity; ignore references to temperature alone(1)
Answer: (a) inward gravity is balanced by outward radiation and gas pressure, so there is no resultant force; (b) high density gives a high enough rate of fusion reactions
Examiner tip: Name both directions: an answer that says only 'the pressures are balanced' without saying gravity acts inwards and radiation pressure outwards rarely gets both marks.

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4Core4 marks

The parallax angle of a nearby star, measured from Earth over six months, is 0.0125 arc-seconds.

(a) Calculate the distance to the star in parsecs. [1]

(b) Calculate the distance in metres. [1]

(c) Explain why the stellar parallax method cannot be used to measure the distance to stars that are very far away. [2]

(a) $d = \dfrac{1}{0.0125} = 80$ pc(1)
(b) $80 \times 3.09 \times 10^{16} = 2.5 \times 10^{18}$ m; allow ECF from (a); accept $2.47 \times 10^{18}$ m(1)
(c) the further away the star, the smaller the parallax angle (angle inversely proportional to distance)(1)
(c) for distant stars the angle is too small to be measured (accurately) / is smaller than the uncertainty/resolution of the measurement; allow 'blurring by the atmosphere limits the smallest measurable angle'(1)
Answer: (a) 80 pc; (b) $2.5 \times 10^{18}$ m; (c) angle becomes too small to measure
Examiner tip: In (c), 'the star is too far away' is not enough: you must say the parallax angle becomes too small to measure.

5Standard4 marks

The final step of the proton–proton chain in the Sun is

$$^{3}_{2}\text{He} + {}^{3}_{2}\text{He} \rightarrow {}^{4}_{2}\text{He} + 2\,{}^{1}_{1}\text{H}$$

Atomic masses: $^{3}$He = 3.016029 u, $^{4}$He = 4.002602 u, $^{1}$H = 1.007825 u.

(a) Calculate, in MeV, the energy released in this reaction. [2]

(b) Explain why this reaction can only take place at temperatures of millions of kelvin. [2]

(a) mass defect $= 2(3.016029) - 4.002602 - 2(1.007825) = 0.013806$ u(1)
(a) $0.013806 \times 931.5 = 12.9$ MeV; accept 12.86 MeV; allow use of $\Delta m c^2$ with conversion to MeV(1)
(b) both nuclei are positively charged so there is (strong) electrostatic/Coulomb repulsion between them(1)
(b) a high temperature gives the nuclei enough kinetic energy/speed to come close enough (overcome the repulsion) for the strong nuclear force to act and fuse them; do not accept 'to get energy for the reaction' alone(1)
Answer: (a) 12.9 MeV; (b) nuclei need very large kinetic energy to overcome Coulomb repulsion
Examiner tip: Watch the factor of 2 on both sides of the equation: forgetting either the second He-3 or the second proton gives a wildly wrong mass defect.

6Standard4 marks

The Hertzsprung–Russell (HR) diagram shows six stars labelled O, M, S, G, X and W. The shaded band is the main sequence.

(a) State the letter of the star that is fusing hydrogen in its core and has the greatest mass. [1]

(b) State the region of the HR diagram in which star W lies. [1]

(c) Stars M and W have the same surface temperature. Use the diagram to deduce the ratio $\dfrac{\text{radius of M}}{\text{radius of W}}$. [2]

(a) O(1)
(b) white dwarf (region); do not accept 'dwarf' alone(1)
(c) $L = \sigma 4\pi R^2 T^4$ with the same $T$ so $R \propto \sqrt{L}$ / $\dfrac{R_M}{R_W} = \sqrt{\dfrac{L_M}{L_W}}$; luminosity ratio read as $\dfrac{10}{10^{-3}} = 10^4$(1)
(c) ratio $= 100$; award [2] for a bald correct answer(1)
Answer: (a) O; (b) white dwarf; (c) 100
Examiner tip: Read the vertical axis carefully: it is logarithmic, so M (10 solar luminosities) and W (10⁻³) differ by a factor of 10⁴, not 4.

7Hard5 marks

Observations of a star give the following data.

parallax angle = 0.0040 arc-seconds
apparent brightness = 2.1 × 10⁻⁹ W m⁻²
wavelength at which its spectrum peaks = 290 nm

(a) Show that the luminosity of the star is about 1.6 × 10³⁰ W. [2]

(b) Calculate the radius of the star. [2]

(c) The radius of the Sun is 7.0 × 10⁸ m. Using your answer to (b), suggest whether this star is on the main sequence. [1]

(a) $d = \dfrac{1}{0.0040} = 250$ pc $= 250 \times 3.09 \times 10^{16} = 7.7 \times 10^{18}$ m(1)
(a) $L = 4\pi d^2 b = 4\pi (7.7 \times 10^{18})^2 (2.1 \times 10^{-9}) = 1.57 \times 10^{30}$ W; at least 3 s.f. needed for a 'show that'(1)
(b) $T = \dfrac{2.9 \times 10^{-3}}{290 \times 10^{-9}} = 1.0 \times 10^4$ K(1)
(b) $R = \sqrt{\dfrac{L}{4\pi\sigma T^4}} = 1.5 \times 10^{10}$ m; allow ECF from wrong $T$; accept $1.4\text{–}1.5 \times 10^{10}$ m (1.6 × 10³⁰ W gives 1.5 × 10¹⁰ m)(1)
(c) radius is about 20 times that of the Sun, too large for a main-sequence star of this temperature, so it is (probably) a giant / not on the main sequence; allow a consistent conclusion from an ECF radius if justified by comparison with the Sun(1)
Answer: (a) $L \approx 1.57 \times 10^{30}$ W; (b) $R \approx 1.5 \times 10^{10}$ m; (c) about 21 solar radii, so a giant, not main sequence
Examiner tip: Convert parsecs to metres before using $L = 4\pi d^2 b$; leaving $d$ in parsecs is the most common error and loses every later mark except by ECF.

8Hard6 marks

Star P has a mass equal to the mass of the Sun. Star Q has a mass of 20 solar masses. Both stars are on the main sequence.

Compare and contrast the evolution of star P and star Q from the time they leave the main sequence to the end of their lives. In your answer, refer to the fusion processes, the path of each star on the HR diagram and the final state of each star. [6]

both stars leave the main sequence when hydrogen in the core is used up (similarity)(1)
core contracts and heats; outer layers expand and cool, so both move up and to the right on the HR diagram (similarity)(1)
P becomes a red giant; Q becomes a red supergiant (larger, more luminous)(1)
P fuses helium to carbon (and oxygen) only; Q, with its greater core mass and temperature, fuses successively heavier elements up to iron(1)
fusion stops at iron because iron has the highest binding energy per nucleon, so further fusion releases no energy(1)
P ejects its outer layers as a planetary nebula; its core becomes a white dwarf (lower left of HR diagram) that gradually cools(1)
Q's iron core collapses, producing a supernova explosion(1)
Q's remnant is a neutron star or (if the remnant is massive enough) a black hole(1)
allow reference to the Chandrasekhar limit (about 1.4 solar masses) for the white dwarf and the Oppenheimer–Volkoff limit (about 2–3 solar masses) for the neutron star, but these are not required(1)

Level	Marks	What it looks like
Level 3	5–6	Gives a full, well-linked comparison: states at least one clear similarity and explains the differences in fusion stages, HR-diagram path and end state for both stars, linking the differences to mass (e.g. heavier elements and core collapse because of the greater core mass). Terminology is correct throughout.
Level 2	3–4	Describes the evolution of both stars with mostly correct end states and at least one other difference (fusion or HR path), but the comparison is partly implicit, a stage is missing, or the link to mass is not made.
Level 1	1–2	Makes some relevant points (e.g. P becomes a white dwarf, Q becomes a neutron star or black hole) with little description of the process or comparison.
Level 0	0	No relevant content.

Answer: P: red giant → planetary nebula → white dwarf (He → C, O only). Q: red supergiant → fusion to Fe → supernova → neutron star or black hole.
Examiner tip: 'Compare and contrast' needs explicit similarities as well as differences; answers that describe P and then Q separately with no link are capped in the middle level.

9Standard4 marks

The graph shows how the binding energy per nucleon varies with nucleon number. Values for three nuclides are marked.

In the core of a red giant, three helium-4 nuclei fuse to form one carbon-12 nucleus.

(a) Using the graph, calculate the energy released in this reaction. [2]

(b) Massive stars fuse elements in their cores until the core is mainly iron. Explain, with reference to the graph, why fusion in the core stops at iron. [2]

(a) binding energies: carbon-12 $= 12 \times 7.68 = 92.2$ MeV and $3 \times$ helium-4 $= 12 \times 7.07 = 84.8$ MeV(1)
(a) energy released $= 92.16 - 84.84 = 7.3$ MeV; accept 7.32 MeV; award [1 max] for $7.68 - 7.07 = 0.61$ MeV(1)
(b) iron(-56) has the (maximum) highest binding energy per nucleon / is at the peak of the curve(1)
(b) fusing iron into heavier nuclei would give products with lower binding energy per nucleon, so the reaction would absorb energy rather than release it / no energy is released to support the core (against gravity)(1)
Answer: (a) 7.3 MeV; (b) iron is at the peak of the curve, so fusion beyond it would need energy input
Examiner tip: Multiply binding energy per nucleon by the number of nucleons before subtracting: 7.68 − 7.07 is a per-nucleon difference, not the energy released.

10Stretch4 marks

Stars A and B form a binary system, so they are the same distance from Earth. The spectra of the stars peak at the wavelengths shown.

star A: peak wavelength 290 nm
star B: peak wavelength 116 nm

The apparent brightness of star A is 450 times the apparent brightness of star B.

(a) Determine the ratio $\dfrac{\text{radius of A}}{\text{radius of B}}$. [3]

(b) Star B is not undergoing fusion. Identify the type of star B is, with a reason based on your answer to (a). [1]

(a) $\dfrac{T_B}{T_A} = \dfrac{290}{116} = 2.5$ (from Wien's law, $T \propto \dfrac{1}{\lambda_{max}}$); allow calculation of both temperatures: $T_A = 1.0 \times 10^4$ K, $T_B = 2.5 \times 10^4$ K(1)
(a) same distance so $\dfrac{L_A}{L_B} = \dfrac{b_A}{b_B} = 450$ and $\dfrac{R_A}{R_B} = \sqrt{\dfrac{L_A}{L_B}} \times \left(\dfrac{T_B}{T_A}\right)^2$(1)
(a) $\sqrt{450} \times 2.5^2 = 133$; accept 130; award [1 max] for 3.4 (temperature ratio inverted)(1)
(b) white dwarf: B is hotter than A but much smaller (radius about 1/130 of A) / very small and hot, with no fusion, it is a cooling stellar remnant(1)
Answer: (a) $R_A/R_B \approx 130$; (b) white dwarf (hot but very small)
Examiner tip: Students often invert the temperature ratio: the star with the shorter peak wavelength is the hotter one, so B is hotter and must be smaller to be fainter.

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