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25.3Stellar evolution extension

The creation of the elements

The hydrogen in the universe and most of the helium (and some lithium) were produced at the very earliest times in the life of the universe. The rest of the elements were produced in a process known as nucleosynthesis the formation of the elements in stars, in the process of stellar evolution.

We have seen that for low mass stars, the dominant nuclear fusion reaction is the proton-proton chain which results in the fusion of four hydrogen nuclei into one nucleus of helium. For stars more massive than about 1.5 solar masses there is a second way to fuse hydrogen into helium. This is the CNO (carbon, nitrogen, oxygen) cycle and is described by the series of fusion reactions:

H 1 1+C 6 12→N 7 13+γ N 7 13→C 6 13+e+1 0+ν

H 1 1+C 6 13→N 7 14+γ

H 1 1+N 7 14→O 8 15+γ O 8 15→N 7 15+e+1 0+ν

H 1 1+N 7 15→C 6 12+He 2 4

Notice that the net effect is just to turn four hydrogen nuclei into one of helium just like the proton-proton chain. The heavier elements produced in intermediate stages are all used up. Carbon has 6 units of positive charge so the Coulomb barrier that needs to be overcome for carbon to fuse is much higher. This requires higher temperatures which can be found in the more massive stars.

The proton-proton chain and the CNO cycle are both main sequence star processes. What happens beyond the main sequence? Almost everything that happens to a star from now on depends on the mass of the star.

For low mass stars, the helium produced in the p-p chain is collecting in the core of the star surrounded by a thin shell of hydrogen and a bigger hydrogen envelope, Figure 25.12.

Figure 25.12: a The structure of a low mass star after it leaves the main sequence. b After helium begins to fuse, carbon collects in the core.

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In more massive stars, carbon will replace helium in the core once carbon starts getting produced in the triple alpha process:

He 2 4+He 2 4→Be 4 8+γ

He 2 4+Be 4 8→C 6 12+γ

Only hydrogen in the thin shell is undergoing nuclear fusion. The temperature and pressure of helium build up and eventually helium begins to fuse through the triple alpha process. This called the ‘helium flash’. This only lasts a few minutes but during this time enormous amounts of energy are produced. Carbon collects in the core, with helium now fusing in a thin shell around carbon. The hydrogen in the thin shell is still fusing so the star now has nuclear fusion in two shells, the H and He shells. This release of energy blows away the outer layers of the star in an explosion called a planetary nebula; mass is thrown into space leaving behind a helium or carbon core.

For high mass stars (those with more than 8 solar masses) helium fuses with carbon to produce oxygen:

He 2 4+C 6 12→O 8 16+γ

For even more massive stars neon, sodium and magnesium are produced:

C 6 12+C 6 12→Ne 10 20+He 2 4

C 6 12+C 6 12→Na 11 23+H 1 1

C 6 12+C 6 12→Mg 12 23+n 0 1

Silicon is then produced through fusion of oxygen:

O 8 16+O 8 16→Si 14 28+He 2 4

The process continues until iron is formed. This creates the onion like layered structure of the star with progressively heavier elements as we move in towards the centre. Fusion cannot produce elements heavier than iron, since the binding energy per nucleon peaks near iron and further fusion is not energetically possible. Thus, a massive star ends its cycle of nuclear reactions with iron at its core surrounded by progressively lighter elements, as shown in Figure 25.13. Elements heavier than iron are produced during supernova explosions through neutron absorption.

Figure 25.13: The central core of fully evolved massive star consists of iron with layers of successively lighter elements surrounding it.

In summary:

●A star with an initial mass M < 0.25M⊙ ends up as a white dwarf with a helium core.

●A star with an initial mass 0.25M⊙ < M < 8M⊙ ends up as a white dwarf with a carbon core.

●A star with an initial mass 8M⊙ < M < 12M⊙ ends up as a white dwarf with an oxygen/neon/magnesium core.

●A star with an initial mass 12M⊙ < M < 40M⊙ ends up as a neutron star.

●A star with an initial mass M > 40M⊙ ends up as a black hole.

Links

The study of stars requires knowledge from very many different areas of physics including gravitation, hydrodynamics, atomic physics, plasma physics, electrodynanics, magnetohydrodynamics, condensed matter physics, relativity and of course nuclear physics. Treating the material in a star as an ideal gas gives approximately correct results for some main sequence stars but fails completely for white dwarf and neutron stars where the ideal gas approximation has to be replaced by other more exotic states of matter.

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NATURE OF SCIENCE

The supreme problem solver and the death of a star

Figure 25.14: Hans Bethe (1906–2005) and Subrahmanyan Chandrasekhar (1910–1995).

Much about the nature of nuclear fusion in stars was discovered by Hans Bethe. Freeman Dyson, one of Bethe’s greatest students, called him the ‘supreme problem solver of the twentieth century’. Bethe was active as a physicist well into his nineties! After providing the details of the proton–proton chain, he wrote a paper on what was to become known as the carbon–nitrogen–oxygen cycle (the CNO cycle), a series of reactions taking place in heavier stars as they evolve. Bethe found out that the New York Academy of Sciences was offering a $500 prize for the best paper on energy production in stars. Originally planning publication in another journal, Bethe entered the competition, won it and with the prize money managed to do a great favour for his mother: relocate her furniture from Germany to the United States. The family, fearing persecution, had left Germany in 1939. For his work on the production of energy in stars, Bethe received the 1967 Nobel Prize in Physics. Bethe is also famous for a paper he did not write: in a paper written by G. Gamow and R. Alpher, Gamow added the name of Bethe so the paper is now known as the alpha, beta, gamma paper. He was in charge of the theoretical division of the Manhattan project that developed the atomic bomb during World War II. He later became a critic of the nuclear arms race and warned of the dangers of nuclear weapons.

Subrahmanyan Chandrasekhar was an Indian, and later American, astrophysicist. His work on collapsed stars was started on the long boat ride from India to England where he was going to study at the University of Cambridge.

His work was presented in 1935 at a meeting of the Royal Astronomical Society in England. His ideas were publicly ridiculed by Sir Arthur Eddington, his colleague and mentor at Cambridge. Undeterred, Chandraseshkar continued his work, and seven years later he moved to the University of Chicago where he remained for the rest of his academic career, becoming one of the greatest and most influential astrophysicists of the twentieth century. A long overdue Nobel Prize in Physics was awarded to Chandrasekhar in 1983. A moving article on Chandrasekhar’s work, in Nature 438, 1086 (2005), by Freeman Dyson, is aptly entitled ‘The death of a star’.

TEST YOUR UNDERSTANDING

6 aState what is meant by an HR diagram.

bDescribe the main features of the HR diagram.

7 aEstimate the ratio of radii of two stars, A and B, of the same luminosity and temperatures T A = 3.2 × 104 K and T B = 2.9 × 103 K.

bEstimate the ratio of radii of two stars, C and D, of the same temperature and luminosities L C = 5.0 × 1028 W and L D = 2.4 × 1022 W.

cPlace the stars A, B, C and D on a copy of the HR diagram (Figure 25.15).

Figure 25.15: For question 7.

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8A main sequence star has temperature 3000 K. Use the HR diagram in the text to estimate its radius in terms of the solar radius.

9The parallax of a star is 0.0125 arc seconds. What is its distance in metres?

10The parallax of a star is 0.022 arc seconds. The apparent brightness of the star is 5.0 × 10−10 W m−2 and its surface temperature is 4500 K. What is the radius of this star?

11Lunar occultation is a method for finding the radius of a star. The idea is to measure how long it takes for the moon to cover the star as the moon passes in front of the star.

aThe moon makes a full revolution around the earth in 27 days. Show that the angular speed of the moon is 2.7 × 10−6 rad s−1.

bA star is covered by the moon in 17 ms. What is the angular diameter of the star?

cThe parallax of the star is 0.167 arcsec. What is the radius of the star?

12Describe the formation of a red giant star.

13 aDescribe what is meant by a planetary nebula.

bSuggest why most photographs show planetary nebulae as rings—doesn’t the gas surround the core from all directions?

cTo eject mass from a star in a planetary nebula, the core of the star must exert a force on the outer layers of the star. What force is then exerted on the core and what does this imply?

14Verify that at each stage of the proton–proton chain the indicated energies are released.

1 1 H+1 1 H→1 2 H+1 0 e++0 0 ν⏟0.42 MeV,1 1 H+1 2 H→2 3 He+0 0 γ⏟5.49 MeV,2 3 He+2 3 He→2 4 He+2 1 1 H⏟12.86 MeV

Hence determine the total energy produced when four protons fuse to make one helium nucleus. (You will need the off-syllabus information that when one positron collides with one electron an amount of energy equal to 1.02 MeV is produced. The positron in stage 1 will certainly do that with a surrounding electron.)

Atomic masses: H 1 1=1.007825 u, H 1 2=2.014103 u, He 2 3=3.016029 u, He 2 4=4.002603 u.

15Describe the evolution of a main sequence star of mass:

a2 solar masses

b20 solar masses.

16 aDescribe the formation of a white dwarf star.

bState two properties of a white dwarf.

17Describe two differences between a main sequence star and a white dwarf.

18A white dwarf of mass half that of the sun and radius equal to one earth radius is formed. Estimate the density of this white dwarf. (M⊙ = 2.0 × 1030 kg, R E = 6.4 × 106 m.)

19Explain what is meant by the terms

ainstability region

bsupernova.

20State and explain two changes in the appearance of a main sequence star as it moves into the red giant phase.

21Describe two differences between a main sequence star and a neutron star.

22 aDescribe the formation of a neutron star.

bState two properties of a neutron star.

23 aThe triple alpha process is the series of reactions

He 2 4+He 2 4→Be 4 8+γ

He 2 4+Be 4 8→C 6 12+γ

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Show that the net effect of the triple alpha process is to fuse three helium nuclei into one of carbon-12.

bShow that the energy released in the triple alpha process is 7.27 MeV. (Helium mass = 4.0026 u.)

cAssume that all the mass of a star is made entirely of helium. The mass is 3.0 × 1032 kg and the luminosity is 1.6 × 1034 W. How long would it take this star to convert all the helium to carbon?

24The CNO cycle is a series of nuclear fusion reactions taking place in massive stars. It consists of the reactions

H 1 1+C 6 12→N 7 13+γ N 7 13→C 6 13+e+1 0+ν

H 1 1+C 6 13→N 7 14+γ

H 1 1+N 7 14→O 8 15+γ O 8 15→N 7 15+e+1 0+ν

H 1 1+N 7 15→C 6 12+He 2 4

aExplain why the CNO cycle happens only in massive stars.

bExplain why the net effect of the CNO cycle is to turn four hydrogen nuclei into one of helium.