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IB Physics E.5: Fusion and stars worked solutions

Worked solutions to every practice question in E.5: Fusion and stars of Photon Physics: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question B1

Type B1 · F1 · balance a fusion equation[1 mark]

In one fusion reaction a deuterium nucleus joins a helium-3 nucleus: {}^{2}_{1}\mathrm{H} + {}^{3}_{2}\mathrm{He} \to {}^{4}_{2}\mathrm{He} + X. What is X?

A. a neutron

B. a proton

C. a positron

D. a deuterium nucleus

Show solutionHide solution

Nucleon number: 2 + 3 = 4 + 1, so X has one nucleon. Charge: 1 + 2 = 2 + 1, so X carries one positive charge. One nucleon with charge +1 is a proton. A fails the charge count, C the nucleon count, D both.

Final answer

Question B2

Type B2 · S3 · conditions for fusion[1 mark]

Fusion in the core of a star needs both a very high temperature and a very high density. Which row gives the reason for each?

A. temperature: electrons are removed from the nuclei; density: the strong force becomes longer in range

B. temperature: the gas becomes dense enough; density: nuclei move fast enough

C. temperature: the star's surface must be hot to radiate; density: gravity is weaker

D. temperature: nuclei move fast enough to come close despite their electric repulsion; density: close approaches happen often enough

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Nuclei are positive and repel. Only very fast nuclei get close enough for the strong force to act, and a high temperature means high kinetic energies. A high density makes such encounters frequent, so enough reactions happen each second. C confuses the core with the surface; A gives the strong force a property it does not have.

Final answer

Question B3

Type B3 · S1 · why a star is stable[1 mark]

Why does a main-sequence star keep a constant radius?

A. Fusion takes place only in the core, so the outer layers do not move.

B. The inward pull of gravity on each layer is balanced by the outward pressure of hot gas and radiation, kept up by energy from fusion.

C. The cool outer layers form a rigid shell.

D. The surrounding space pushes inward on the star.

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Equilibrium needs two opposing effects of equal size. Gravity pulls every layer inward; the pressure of the hot gas and of radiation pushes outward, and fusion keeps supplying the energy that maintains that pressure. Space is close to a vacuum (D) and a star has no solid shell (C).

Final answer

Question B4

Type B4 · F2 · mass and binding energy after fusion[1 mark]

Two light nuclei fuse and energy is released. Compared with the two original nuclei, what is true of the products?

A. smaller total mass; greater total binding energy

B. smaller total mass; smaller total binding energy

C. greater total mass; greater total binding energy

D. the same total mass; greater total binding energy

Show solutionHide solution

Released energy has come from rest mass, so the products are lighter. The same nucleons are now more tightly bound, so the total binding energy has risen by exactly the energy released. D would release no energy.

Final answer

Question B5

Type B5 · D2 · parallax to distance[1 mark]

A star has a parallax angle of 0.025 arcsecond. What is its distance from Earth?

A. 12\,\mathrm{ly}

B. 40\,\mathrm{ly}

C. 130\,\mathrm{ly}

D. 0.082\,\mathrm{ly}

Show solutionHide solution

d = 1/0.025 = 40\,\mathrm{pc} and 1\,\mathrm{pc} = 3.26\,\mathrm{ly}, so d = 40\times3.26 = 130\,\mathrm{ly}. B stops at parsecs; A divides by 3.26; D multiplies the angle by 3.26.

Final answer

Question B6

Type B6 · D1 · parallax method and angle[1 mark]

Observations made six months apart show that a nearby star has moved through a total angle of 0.080 arcsecond against the distant background stars. What is the distance to the star?

A. 12.5\,\mathrm{pc}

B. 25\,\mathrm{pc}

C. 50\,\mathrm{pc}

D. 0.040\,\mathrm{pc}

Show solutionHide solution

The two views are 2 AU apart, so the total shift is twice the parallax angle: p = 0.040 arcsecond and d = 1/0.040 = 25\,\mathrm{pc}. A uses the whole shift as p; D quotes the angle itself.

Final answer

Question B7

Type B7 · D3 · inverse-square brightness[1 mark]

Star X is nine times as luminous as star Y and three times as far from Earth. What is \dfrac{\text{apparent brightness of X}}{\text{apparent brightness of Y}}?

A. \tfrac13

B. 1

C. 3

D. 27

Show solutionHide solution

From b = L/4\pi d^2 the ratio is (L_X/L_Y)(d_Y/d_X)^2 = 9\times(1/3)^2 = 1. A forgets to square the distance ratio and inverts; C divides by the distance ratio only once; D multiplies by it.

Final answer

Question B8

Type B8 · R1 · temperature and composition[1 mark]

The continuous spectrum of a star has its peak at a wavelength of 480\,\mathrm{nm}. What is the surface temperature of the star?

A. 6.0\,\mathrm{K}

B. 6.0\times10^{3}\,\mathrm{K}

C. 6.0\times10^{6}\,\mathrm{K}

D. 1.4\times10^{-9}\,\mathrm{K}

Show solutionHide solution

Wien's law: T = 2.9\times10^{-3}/(480\times10^{-9}) = 6.0\times10^{3}\,\mathrm{K}. A treats nm as 10^{-6}\,\mathrm{m}, C as 10^{-12}\,\mathrm{m}, and D multiplies instead of dividing.

Final answer

Question B9

Type B9 · R2 · radius and luminosity ratios[1 mark]

Star P has twice the surface temperature of star Q and 64 times its luminosity. What is \dfrac{\text{radius of P}}{\text{radius of Q}}?

A. 2

B. 4

C. 8

D. 16

Show solutionHide solution

L = 4\pi R^2\sigma T^4 gives R \propto \sqrt{L}/T^2, so R_P/R_Q = \sqrt{64}/2^2 = 8/4 = 2. C ignores the temperature; B divides by T instead of T^2; D forgets the square root.

Final answer

Question B10

Type B10 · H1 · HR regions[1 mark]

A star has a surface temperature of 20\,000\,\mathrm{K} and a luminosity of about one hundredth of the Sun's. In which region of the HR diagram does it lie?

A. red giants

B. main sequence

C. white dwarfs

D. supergiants

Show solutionHide solution

Hot and dim places it at the lower left, below the main sequence: a white dwarf. A main-sequence star this hot is thousands of times more luminous than the Sun; giants and supergiants are luminous and mostly cool.

Final answer

Question B11

Type B11 · H2 · radii on the HR diagram[1 mark]

Two stars lie on the same horizontal line of an HR diagram. Star A has a surface temperature of 3000\,\mathrm{K} and star B of 12\,000\,\mathrm{K}. What is \dfrac{\text{radius of A}}{\text{radius of B}}?

A. \tfrac{1}{16}

B. 4

C. 16

D. 256

Show solutionHide solution

Same height means the same luminosity, so R^2T^4 is the same for both: R_A/R_B = (T_B/T_A)^2 = 4^2 = 16. The cooler star must be the bigger one; A inverts the ratio, B forgets to square, D uses the fourth power.

Final answer

Question B12

Type B12 · H3 · evolution on the HR diagram[1 mark]

Which sequence describes the life of a star with about the same mass as the Sun, from the time it starts fusing hydrogen?

A. main sequence \to red supergiant \to supernova \to neutron star

B. red giant \to main sequence \to white dwarf

C. main sequence \to white dwarf \to red giant

D. main sequence \to red giant \to white dwarf

Show solutionHide solution

Core hydrogen fusion is the main-sequence stage. When the core hydrogen runs out the star swells into a red giant, throws off its envelope, which forms a planetary nebula, and what is left of the core is a white dwarf. A is the path of a star many times more massive than the Sun.

Final answer

Tutorial practice

Question T1

Type T1 · F3 · energy from particle masses + F1[5 marks]

Two deuterium nuclei can fuse: {}^{2}_{1}\mathrm{H} + {}^{2}_{1}\mathrm{H} \to {}^{3}_{2}\mathrm{He} + X. Nuclear masses: deuterium 2.013553\,\mathrm{u}, helium-3 3.014932\,\mathrm{u}.

(a) Identify X. [1]

(b) Show that the mass that disappears in one reaction is about 3.5\times10^{-3}\,\mathrm{u}. [2]

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(a) Nucleon number 4 = 3 + 1 and charge 2 = 2 + 0: X is {}^{1}_{0}\mathrm{n} (A1). (b) Mass before = 2\times2.013553 = 4.027106\,\mathrm{u}; mass after = 3.014932 + 1.008665 = 4.023597\,\mathrm{u}, with the neutron mass from the data booklet (M1); difference = 3.509\times10^{-3}\,\mathrm{u} (A1). (c) E = 3.509\times10^{-3}\times931.5 = 3.27\,\mathrm{MeV} (A1); 3.27\times1.60\times10^{-13} = 5.2\times10^{-13}\,\mathrm{J} (A1).

Final answer

(a) a neutron. (b) 3.509\times10^{-3}\,\mathrm{u}. (c) 3.27\,\mathrm{MeV}; 5.2\times10^{-13}\,\mathrm{J}.

Question T2

Type T2 · F4 · energy from binding energies + H5[5 marks]

In the core of a red giant, carbon-12 can capture a helium-4 nucleus: {}^{12}_{6}\mathrm{C} + {}^{4}_{2}\mathrm{He} \to {}^{16}_{8}\mathrm{O} + \gamma. Binding energy per nucleon: helium-4 7.074\,\mathrm{MeV}, carbon-12 7.680\,\mathrm{MeV}, oxygen-16 7.976\,\mathrm{MeV}.

(a) Calculate the energy released in one reaction. [3]

(b) Explain, with reference to binding energy per nucleon, why fusion in a star's core cannot release energy once the core is made of iron. [2]

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(a) Total binding energies: oxygen 16\times7.976 = 127.616\,\mathrm{MeV}; carbon 12\times7.680 = 92.160\,\mathrm{MeV}; helium 4\times7.074 = 28.296\,\mathrm{MeV} (M1). Energy released = binding energy after minus binding energy before (M1) = 127.616 - 120.456 = 7.16\,\mathrm{MeV} (A1). (b) Binding energy per nucleon rises up to the region of iron and then falls (A1). Fusing iron nuclei would make nuclei with a smaller binding energy per nucleon, so the total binding energy would fall and energy would be absorbed, not released (A1).

Final answer

(a) 7.16\,\mathrm{MeV}. (b) Iron lies close to maximum binding energy per nucleon. Further fusion reduces binding per nucleon, requiring an external energy input.

Question T3

Type T3 · F6 · reaction rate, fuel use, lifetime + F5[6 marks]

A main-sequence star has a mass of 4.0\times10^{30}\,\mathrm{kg} and a constant luminosity of 3.0\times10^{27}\,\mathrm{W}. Each complete conversion of four protons into one helium-4 nucleus releases 26.7\,\mathrm{MeV}. Assume that 10% of the star's mass is hydrogen that will be fused in its core.

(a) Show that fusing 1.0\,\mathrm{kg} of hydrogen releases about 6\times10^{14}\,\mathrm{J}. [2]

(b) Calculate the total energy available from the hydrogen that will be fused. [1]

(c) Estimate the time the star spends on the main sequence, in years (1\,\mathrm{year} = 3.16\times10^{7}\,\mathrm{s}). [2]

(d) State one assumption made in (c). [1]

Show solutionHide solution

(a) Energy per chain = 26.7\times1.60\times10^{-13} = 4.27\times10^{-12}\,\mathrm{J}; fuel per chain = 4\times1.673\times10^{-27} = 6.69\times10^{-27}\,\mathrm{kg} (M1); 4.27\times10^{-12}/6.69\times10^{-27} = 6.4\times10^{14}\,\mathrm{J\,kg^{-1}} (A1). (b) 0.10\times4.0\times10^{30}\times6.4\times10^{14} = 2.6\times10^{44}\,\mathrm{J} (A1). (c) t = 2.6\times10^{44}/3.0\times10^{27} = 8.5\times10^{16}\,\mathrm{s} (M1) = 2.7\times10^{9} years (A1). (d) The luminosity stays at its present value for the whole time (A1).

Final answer

(a) 6.4\times10^{14}\,\mathrm{J}. (b) 2.6\times10^{44}\,\mathrm{J}. (c) 2.7\times10^{9} years. (d) The luminosity stays constant (or: all the energy released leaves as radiation).

Question T4

Type T4 · D4 · parallax, brightness, luminosity + D2[7 marks]

A star has a parallax angle of 0.0125 arcsecond and an apparent brightness of 3.0\times10^{-10}\,\mathrm{W\,m^{-2}}.

(a) Show that its distance is about 2.5\times10^{18}\,\mathrm{m}. [2]

(b) Calculate its luminosity. [2]

(d) A second star has the same luminosity but one quarter of the apparent brightness. Determine its parallax angle. [2]

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(a) d = 1/0.0125 = 80\,\mathrm{pc} (M1); 80\times3.26\times9.46\times10^{15} = 2.47\times10^{18}\,\mathrm{m} (A1). (b) L = 4\pi bd^2 (M1) = 4\pi\times3.0\times10^{-10}\times(2.47\times10^{18})^2 = 2.3\times10^{28}\,\mathrm{W} (A1). (c) 2.29\times10^{28}/3.83\times10^{26} = 60 (A1). (d) Equal luminosity, so d \propto 1/\sqrt b: the star is twice as far (M1) and its parallax is half as large, 6.3\times10^{-3} arcsecond (A1).

Final answer

(a) 80\,\mathrm{pc} = 2.47\times10^{18}\,\mathrm{m}. (b) 2.3\times10^{28}\,\mathrm{W}. (c) 60L_\odot. (d) 6.3\times10^{-3} arcsecond.

Question T5

Type T5 · R3 · radius from measurements + R1, H1[6 marks]

The spectrum of a star peaks at a wavelength of 725\,\mathrm{nm}. The star's luminosity is 2.0\times10^{29}\,\mathrm{W}. The radius of the Sun is 6.96\times10^{8}\,\mathrm{m}.

(a) Calculate the surface temperature of the star. [1]

(b) Determine the radius of the star as a multiple of the Sun's radius. [3]

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(a) T = 2.9\times10^{-3}/725\times10^{-9} = 4.0\times10^{3}\,\mathrm{K} (A1). (b) Rearranging L = 4\pi R^2\sigma T^4 for the radius gives R = \sqrt{L/4\pi\sigma T^4} (M1) = \sqrt{2.0\times10^{29}/(4\pi\times5.67\times10^{-8}\times4000^4)} = 3.3\times10^{10}\,\mathrm{m} (A1); 3.31\times10^{10}/6.96\times10^{8} = 48 (A1). (c) Red giant (A1): its surface is cooler than the Sun's but its luminosity is about 500 times greater, which needs a radius tens of times larger (A1).

Final answer

(a) 4.0\times10^{3}\,\mathrm{K}. (b) 3.3\times10^{10}\,\mathrm{m}, about 48R_\odot. (c) A red giant: cool (about 4000\,\mathrm{K}) yet very luminous, with a radius tens of times the Sun's.

Question T6

Type T6 · H4 · final remnant from mass + H3[5 marks]

Star A has about the same mass as the Sun. Star B has about twenty times the Sun's mass.

(a) Describe the final stage of each star after it leaves the main sequence. [2]

(b) Explain what stops the remnant of star A from collapsing, and why this support works only below a certain mass. [2]

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(a) A becomes a red giant, throws off its outer layers as a planetary nebula, and its core remains as a white dwarf (A1). B becomes a red supergiant, its core collapses and it explodes as a supernova, leaving a neutron star or a black hole (A1). (b) Electron degeneracy pressure holds up the white dwarf; no fusion is needed (A1). Above the Chandrasekhar limit the inward pull of gravity is too strong for this pressure and the core collapses further (A1). (c) Neutron degeneracy pressure can no longer resist gravity, and the core collapses to a black hole (A1).

Final answer

(a) A: red giant, planetary nebula, white dwarf. B: red supergiant, supernova, neutron star or black hole. (b) Electron degeneracy pressure; above the Chandrasekhar limit gravity overcomes it. (c) It collapses to a black hole.

Question T7

Type T7 · R2 · radius and luminosity ratios + H2, H1[5 marks]

Star X lies on the main sequence. Its surface temperature is 9000\,\mathrm{K} and its luminosity is 25 times that of the Sun. The Sun's surface temperature is 5800\,\mathrm{K}.

(a) Calculate \dfrac{\text{radius of X}}{\text{radius of the Sun}}. [2]

(b) Star W has the same surface temperature as X and one hundredth of its radius. Calculate the luminosity of W as a fraction of the Sun's. [2]

Show solutionHide solution

(a) R \propto \sqrt{L}/T^2 (M1): R_X/R_\odot = \sqrt{25}\times(5800/9000)^2 = 2.1 (A1). (b) At equal temperature L \propto R^2 (M1): L_W = 25\times(1/100)^2 = 2.5\times10^{-3} of the Sun's (A1). (c) Hot yet very dim, so W is a white dwarf, lying under the main sequence on its hot side (A1).

Final answer

(a) 2.1. (b) 2.5\times10^{-3}. (c) White dwarfs.

Question T8

Type T8 · D1 · parallax method and angle + D2[5 marks]

Parallax is used to find the distances of nearby stars.

(a) Outline the observations needed to find a star's parallax angle, and how the angle is obtained from them. [3]

(b) A star is 250\,\mathrm{pc} from Earth. Calculate its parallax angle. [1]

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(a) The position of the star is measured against much more distant stars, which appear fixed (A1). The measurements are made six months apart, from opposite ends of Earth's orbit, a baseline of 2 AU (A1). The parallax angle is half the total angular shift (A1). (b) p = 1/250 = 4.0\times10^{-3} arcsecond (A1). (c) For distant stars p becomes so small that it is comparable with the uncertainty in measuring it (A1).

Final answer

(a) Measure the star's position against very distant background stars from two points of Earth's orbit six months apart; half the total angular shift is the parallax angle. (b) 4.0\times10^{-3} arcsecond. (c) The angle becomes too small to measure accurately.

Advanced practice

Question A1

Type A1 · S2 · how balance is restored + S1[6 marks]

A main-sequence star is in equilibrium. A small disturbance makes its core contract slightly. In this model, the rate of fusion in the core rises steeply when the core temperature rises.

(a) Explain how the star returns to equilibrium. [4]

(b) Late in its life the core has no hydrogen left. Explain why the same process can no longer keep the core stable. [2]

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(a) As the core contracts, gravitational potential energy is transferred to the particles, so the core's temperature and density rise (A1). The fusion rate increases strongly (A1). More energy is released, so the gas and radiation pressure rise and push the layers outward (A1). The core expands and cools until the fusion rate and the pressure fall back to the values that balance gravity: equilibrium is restored (A1). (b) The response depends on hydrogen fusion speeding up; with no core hydrogen there is no fuel to respond (A1). The core keeps contracting and heating until a new fuel, helium, starts to fuse (A1).

Final answer

(a) Contraction heats and compresses the core, fusion speeds up, the outward pressure rises above what gravity needs and the core expands back. (b) With no hydrogen there is no core fusion to speed up, so the pressure cannot rise to stop the contraction.

Question A2

Type A2 · F5 · energy per kilogram of fuel + F3[7 marks]

Deuterium (hydrogen-2) can fuse with tritium (hydrogen-3) to give helium-4 and a neutron. Atomic masses: deuterium 2.014102\,\mathrm{u}, tritium 3.016049\,\mathrm{u}, helium-4 4.002603\,\mathrm{u}.

(a) Explain why atomic masses can be used here without any correction for electrons, and calculate the energy released in one reaction in MeV. [2]

(b) Calculate the energy released per kilogram of a deuterium–tritium mixture with one deuterium atom for each tritium atom. [2]

(d) The fission of one uranium-235 nucleus releases about 200\,\mathrm{MeV}. Determine \dfrac{\text{energy per kilogram of the D–T mixture}}{\text{energy per kilogram of uranium-235}}. [2]

Show solutionHide solution

(a) There are two electrons on each side (one in each hydrogen atom, two in the helium atom), so their masses cancel (A1). \Delta m = 5.030151 - 5.011268 = 0.018883\,\mathrm{u} and E = 0.018883\times931.5 = 17.6\,\mathrm{MeV} (A1). (b) Fuel mass per reaction = 5.030151\times1.661\times10^{-27} = 8.355\times10^{-27}\,\mathrm{kg} (M1); 17.59\times1.60\times10^{-13}/8.355\times10^{-27} = 3.4\times10^{14}\,\mathrm{J\,kg^{-1}} (A1). (c) 0.018883/5.030151 = 3.8\times10^{-3}, under half a percent (A1). (d) Uranium: 200\times1.60\times10^{-13}/(235\times1.661\times10^{-27}) = 8.2\times10^{13}\,\mathrm{J\,kg^{-1}} (M1); ratio = 3.37\times10^{14}/8.20\times10^{13} = 4.1 (A1). One fission releases far more energy than one fusion, but per kilogram fusion wins.

Final answer

(a) Two electrons on each side; 17.6\,\mathrm{MeV}. (b) 3.4\times10^{14}\,\mathrm{J\,kg^{-1}}. (c) 3.8\times10^{-3}. (d) about 4.1.

Question A3

Type A3 · H5 · heavier elements and iron + F4[7 marks]

Late in the life of a very massive star, its core fuses successively heavier nuclei. Binding energy per nucleon: silicon-28 8.448\,\mathrm{MeV}, nickel-56 8.643\,\mathrm{MeV}, iron-56 8.790\,\mathrm{MeV}, tellurium-112 8.40\,\mathrm{MeV}.

(a) Explain why the core must be hotter to fuse silicon nuclei than to fuse hydrogen nuclei. [2]

(b) In a simple model, two silicon-28 nuclei fuse to form nickel-56. Calculate the energy released. [2]

(d) Outline what happens to the star once its core is mostly iron. [1]

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(a) Each silicon nucleus has charge +14e, so the electric repulsion between two of them is 14^2 = 196 times that between two protons at the same separation (A1). To come close enough for the strong force to act they need much larger kinetic energies, which means a much higher core temperature (A1). (b) 56\times8.643 - 56\times8.448 (M1) = 10.9\,\mathrm{MeV} released (A1). (c) 112\times8.40 - 112\times8.790 = -43.7\,\mathrm{MeV} (A1): the products are less tightly bound, so this reaction absorbs energy and cannot supply the pressure that holds the core up (A1). (d) With no energy source the core collapses in a fraction of a second and the star explodes as a supernova (A1).

Final answer

(a) Silicon nuclei carry 14 times the charge of a proton, so their repulsion is far greater and they must be much faster. (b) 10.9\,\mathrm{MeV}. (c) -43.7\,\mathrm{MeV}: energy would have to be supplied, so iron cannot fuel the core. (d) The core collapses and the star explodes as a supernova.

Question A4

Type A4 · D5 · two distances disagree + D4[6 marks]

A star has a measured parallax of 2.50 milliarcseconds and an apparent brightness of 2.0\times10^{-10}\,\mathrm{W\,m^{-2}}. From its spectrum, its luminosity is estimated as 5.0\times10^{29}\,\mathrm{W}.

(a) Calculate the distance to the star from its parallax, in pc. [1]

(b) Calculate the distance from its luminosity and apparent brightness, in pc. [3]

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(a) p = 2.50\times10^{-3} arcsecond, so d = 1/2.50\times10^{-3} = 400\,\mathrm{pc} (A1). (b) d = \sqrt{L/4\pi b} (M1) = \sqrt{5.0\times10^{29}/(4\pi\times2.0\times10^{-10})} = 1.41\times10^{19}\,\mathrm{m} (A1); 1.41\times10^{19}/3.084\times10^{16} = 457\,\mathrm{pc} (A1). (c) The brightness distance is about 14% larger, more than a rounding difference (A1). Dust absorbs some of the light, so b is smaller than it would be and the brightness distance comes out too large; or the parallax of a few milliarcseconds and the estimated luminosity each have a large uncertainty (A1).

Final answer

(a) 400\,\mathrm{pc}. (b) 1.41\times10^{19}\,\mathrm{m} = 457\,\mathrm{pc}. (c) About 14% larger; dust between us and the star dims it (larger luminosity distance), or the tiny parallax and the luminosity estimate both carry large uncertainties.

Question A5

Type A5 · H2 · radii on the HR diagram + R2, H4[6 marks]

A white dwarf has a radius of 7.0\times10^{6}\,\mathrm{m} and a surface temperature of 20\,000\,\mathrm{K}. Over a long time it cools to 10\,000\,\mathrm{K} without changing its radius. The Sun's luminosity is 3.83\times10^{26}\,\mathrm{W}.

(a) Calculate the luminosity of the white dwarf at 20\,000\,\mathrm{K}, as a fraction of the Sun's. [2]

(b) State the factor by which its luminosity falls as it cools to 10\,000\,\mathrm{K}. [1]

(c) On an HR diagram with logarithmic scales, explain why its path is a straight line, and state the direction in which it moves. [2]

(d) Explain why the white dwarf cools at all. [1]

Show solutionHide solution

(a) L = 4\pi R^2\sigma T^4 = 4\pi\times(7.0\times10^{6})^2\times5.67\times10^{-8}\times(2.0\times10^{4})^4 (M1) = 5.6\times10^{24}\,\mathrm{W} = 0.015L_\odot (A1). (b) 2^4 = 16 (A1). (c) With R fixed, \log L = 4\log T + \text{constant}, a straight line on log axes, the line of constant radius (A1). Both T and L fall, so it moves down and to the right, towards lower temperature (A1). (d) No fusion takes place in a white dwarf; it radiates the thermal energy it already has, so its temperature falls (A1).

Final answer

(a) 5.6\times10^{24}\,\mathrm{W}, about 0.015L_\odot. (b) 16. (c) L \propto T^4 at constant radius, so \log L is a linear function of \log T; it moves down and to the right. (d) It has no fusion, so it radiates away stored thermal energy.

Connected questions

Question X1

Type X1 · H4 · final remnant from mass + S1; Connected · E.5 + D.1 + A.2[8 marks]

A neutron star has a mass of 2.8\times10^{30}\,\mathrm{kg} and a radius of 12\,\mathrm{km}. It rotates once every 0.033\,\mathrm{s}.

(a) Calculate the gravitational field strength at its surface. [2]

(b) Calculate the speed of a point on its equator. [1]

(d) Calculate the mean density of the star. [1]

(e) Explain what holds the neutron star up, and what would happen if its mass were greater than the Oppenheimer–Volkoff limit. [2]

Show solutionHide solution

(a) D.1: g = GM/R^2 (M1) = 6.67\times10^{-11}\times2.8\times10^{30}/(1.2\times10^{4})^2 = 1.3\times10^{12}\,\mathrm{N\,kg^{-1}} (A1). (b) A.2: v = 2\pi R/T = 2\pi\times1.2\times10^{4}/0.033 = 2.3\times10^{6}\,\mathrm{m\,s^{-1}} (A1). (c) a = v^2/R = 4.4\times10^{8}\,\mathrm{m\,s^{-2}} (A1); the gravitational force per kilogram is about 3000 times larger than the centripetal force needed, so surface matter stays bound (A1). (d) \rho = M/\tfrac43\pi R^3 = 3.9\times10^{17}\,\mathrm{kg\,m^{-3}} (A1). (e) Neutron degeneracy pressure balances gravity (A1). Above the Oppenheimer–Volkoff limit it cannot, and the star collapses to a black hole (A1).

Final answer

(a) 1.3\times10^{12}\,\mathrm{N\,kg^{-1}}. (b) 2.3\times10^{6}\,\mathrm{m\,s^{-1}}. (c) 4.4\times10^{8}\,\mathrm{m\,s^{-2}}, far less than g, so gravity easily provides the centripetal force. (d) 3.9\times10^{17}\,\mathrm{kg\,m^{-3}}. (e) Pressure from degenerate neutrons supports the remnant. Exceeding the limiting mass allows gravity to collapse it into a black hole.

Question X2

Type X2 · R3 · radius from measurements + R1, F6; Connected · E.5 + B.1 + E.3[8 marks]

The intensity of sunlight just above Earth's atmosphere is the solar constant, 1.36\times10^{3}\,\mathrm{W\,m^{-2}}. Earth is 1.00\,\mathrm{AU} from the Sun. The Sun's spectrum peaks at 500\,\mathrm{nm} and its mass is 2.0\times10^{30}\,\mathrm{kg}.

(a) Show that the Sun radiates a total power of about 3.8\times10^{26}\,\mathrm{W}. [2]

(b) Calculate the Sun's surface temperature. [1]

(d) Calculate the rate at which the Sun loses mass. [1]

(e) Estimate the fraction of its mass that the Sun loses this way in 1.0\times10^{10} years. [2]

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(a) B.1: L = 4\pi d^2b (M1) = 4\pi\times(1.50\times10^{11})^2\times1.36\times10^{3} = 3.85\times10^{26}\,\mathrm{W} (A1). (b) T = 2.9\times10^{-3}/500\times10^{-9} = 5.8\times10^{3}\,\mathrm{K} (A1). (c) R = \sqrt{L/4\pi\sigma T^4} (M1) = \sqrt{3.85\times10^{26}/(4\pi\times5.67\times10^{-8}\times5800^4)} = 6.9\times10^{8}\,\mathrm{m} (A1). (d) E.3: \Delta m/\Delta t = L/c^2 = 3.85\times10^{26}/(3.00\times10^{8})^2 = 4.3\times10^{9}\,\mathrm{kg\,s^{-1}} (A1). (e) 4.27\times10^{9}\times1.0\times10^{10}\times3.16\times10^{7} = 1.35\times10^{27}\,\mathrm{kg} (M1); 1.35\times10^{27}/2.0\times10^{30} = 7\times10^{-4} (A1).

Final answer

(a) 3.85\times10^{26}\,\mathrm{W}. (b) 5.8\times10^{3}\,\mathrm{K}. (c) 6.9\times10^{8}\,\mathrm{m}. (d) 4.3\times10^{9}\,\mathrm{kg\,s^{-1}}. (e) about 7\times10^{-4}.

Question X3

Type X3 · R1 · temperature and composition; Connected · E.5 + E.1 + C.5[8 marks]

The spectrum of a star shows a continuous band of colours crossed by dark lines. One dark line, caused by hydrogen, is at 656.3\,\mathrm{nm} in the laboratory but at 656.8\,\mathrm{nm} in the star's spectrum. The continuous spectrum peaks at 414\,\mathrm{nm}.

(a) Explain how the dark lines are formed. [2]

(b) Determine the photon energy, in eV, that corresponds to the 656.3\,\mathrm{nm} line. [2]

(c) Determine the speed of the star along the line of sight, and state whether it is moving towards or away from Earth. [2]

(d) Calculate the surface temperature of the star. [1]

(e) State how the dark lines show which elements are present in the star's outer layers. [1]

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(a) E.1: light from the hot surface passes through cooler gas in the star's outer layers (A1). Atoms absorb photons whose energy matches the difference between two of their energy levels and re-emit them in all directions, so less light of those wavelengths reaches us (A1). (b) E = hc/\lambda (M1) = 6.63\times10^{-34}\times3.00\times10^{8}/656.3\times10^{-9} = 3.03\times10^{-19}\,\mathrm{J} = 1.89\,\mathrm{eV} (A1). (c) C.5: v = c\,\Delta\lambda/\lambda = 3.00\times10^{8}\times0.5/656.3 = 2.3\times10^{5}\,\mathrm{m\,s^{-1}} (A1); the wavelength is longer, so the star is moving away (A1). (d) T = 2.9\times10^{-3}/414\times10^{-9} = 7.0\times10^{3}\,\mathrm{K} (A1). (e) The pattern of line wavelengths is matched with the laboratory spectra of known elements (A1).

Final answer

(a) Cooler gas in the outer layers absorbs photons whose energies equal differences between its energy levels and re-emits them in all directions. (b) 1.89\,\mathrm{eV}. (c) 2.3\times10^{5}\,\mathrm{m\,s^{-1}}, away. (d) 7.0\times10^{3}\,\mathrm{K}. (e) Each element has its own set of line wavelengths, matched with laboratory spectra.

Question X4

Type X4 · S3 · conditions for fusion + F3; Connected · E.5 + B.1 + B.3[8 marks]

The core of a red giant is fully ionised helium at a temperature of 1.0\times10^{8}\,\mathrm{K} and a density of 1.0\times10^{8}\,\mathrm{kg\,m^{-3}}. Model the core as an ideal gas of helium nuclei and free electrons. The mass of a helium nucleus is 6.64\times10^{-27}\,\mathrm{kg}.

(a) Calculate the mean kinetic energy of a helium nucleus in the core, in keV. [1]

(b) Explain why helium needs a higher temperature to fuse than hydrogen. [2]

(d) Three helium-4 atoms fuse to form a carbon-12 atom. The atomic mass of helium-4 is 4.002603\,\mathrm{u} and that of carbon-12 is 12.000000\,\mathrm{u}. Calculate the energy released. [2]

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(a) B.1: \bar E_k = \tfrac32k_BT = 1.5\times1.38\times10^{-23}\times1.0\times10^{8} = 2.07\times10^{-15}\,\mathrm{J} = 13\,\mathrm{keV} (A1). (b) A helium nucleus has charge +2e, so two of them repel four times as strongly as two protons at the same separation (A1); they need greater kinetic energies to get close enough for the strong force to act, so a higher temperature (A1). (c) Helium nuclei per cubic metre = 1.0\times10^{8}/6.64\times10^{-27} = 1.51\times10^{34} (M1); each nucleus brings two electrons, so particles = 4.52\times10^{34}\,\mathrm{m^{-3}} (M1); B.3: P = nk_BT = 4.52\times10^{34}\times1.38\times10^{-23}\times1.0\times10^{8} = 6.2\times10^{19}\,\mathrm{Pa} (A1). (d) \Delta m = 3\times4.002603 - 12.000000 = 0.007809\,\mathrm{u} (M1); 0.007809\times931.5 = 7.27\,\mathrm{MeV} (A1).

Final answer

(a) 13\,\mathrm{keV}. (b) Helium nuclei carry twice the charge of protons, so they repel four times as strongly at the same separation and must move faster. (c) 6.2\times10^{19}\,\mathrm{Pa}. (d) 7.27\,\mathrm{MeV}.

From the book

Photon Physics: Study Guide for the IB Diploma (SL & HL)

Fusion and stars from the core ideas to full exam questions, with question types, worked examples and graded practice.

About the book