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Answer to Question Number 1 (a) Parsec: a distance unit such that the parallax subtended by an object due to the orbit of the Earth about the sun is one second of arc at a distance of 1 parsec. Apparent magnitude: The magnitude a star appears in the sky. Absolute magnitude: The magnitude a star would appear to be at a distance of 10 pc. (3) Consider the ratio of apparent brightness of a star at a distance d and at 10pc. Then, Ld / L10 = 100/d2 from inverse square law. Taking log10 gives: log10(Ld / L10 ) = 2 - 2log10d 1 Recall definition of stellar magnitudes: 1 magnitude corresponds to a factor of 1001/5 with smaller magnitudes brighter. Then, Ld / L10 =100(M - m)/5, hence log10(Ld / L10 ) = 2/5 log10 (M - m) 2 Combining 1 and 2, (m - M) = 5log10d -5 (3) (b) Class Colour Temp. (x103 K) Spectral lines (7) O Blue-violet 28 – 50 Ionised atoms G Yellow 5–6 Can+, other ionised and neutral metals M Red-orange 2.5 – 3.5 TiO and Ca (3) H Balmer absorption: Consider both fraction ionised (Saha equation) and fraction excited to n=2 level (Boltzmann) Hot, O stars: practically 100% ionisation, hence small neutral population for absorption, hence very weak Balmer absorption Cooler, G stars, temperature insufficient for high degree of ionisation of H, but sufficient to cause some excitation of n=2, hence strong Balmer absorption. Cool M stars, temperature insufficient for high degree of ionisation of H or excitation of n=2, hence weak Balmer absorption. (3) Nearby faint vs distant luminous star of same spectral class: Consider width of spectral lines. Faint star, small diameter (eg a red dwarf) hence high density hence lines will be pressure broadened. Luminous star, large diameter (eg, a red giant) hence low density hence little line broadening. (2) (c) (6) Answer to Question Number 2 (a) Observable properties of a white dwarf: Temperature ~ 10-50000K, diameter ~same as Earth, mass <= 1.4 solar masses, luminosity <1% sun, composition – rich in carbon and oxygen. Stage of evolution: Corresponds to final phase of a sun-like star. No energy generation, hence radiation is purely from cooling. Young white dwarfs often seen in association with planetary nebulae. (4) (b) Principle source of pressure in a white dwarf – electron degeneracy pressure. Electrons form a Fermi gas and the pressure can be understood to arise from the Heisenberg Uncertainty Principle, px ~h (2) Pressure-density relationship For a non-relativistic gas, the pressure is given by P = n/3<p.v> where n is the electron number density, p is the momentum and v is the velocity of the electrons. Heisenberg tells us px ~h. Also, mean spacing x ~ n-1/3 and p~p Hence, p~h/n-1/3 and v~p/m where m is the mass of an electron Hence P = 1/3 nvp n5/3 (5) Mass-radius relationship We require an estimate of the pressure required to support the star against gravity. Consider a column of star material with unit area and length = R, the star’s radius Gravitational acceleration acting on column ~ GM/R 2 Unit area Weight of column of material = P R Mass of column ~ R ~ M/R 2 Hence P ~ GM 2/R4 Equating this pressure with that derived from degeneracy considerations we find: h 2 5 3 GM 2 P~ n ~ 4 m R Noting that n ~ /mp and M ~ R3 we obtain the proportionality: R M 13 (5) (c) The results in (b) break down when the density of the white dwarf rises sufficiently that the Heisenberg Uncertainty Principle demands the degenerate electrons become relativistic. (i.e. the electron mean spacing is reduced, requiring the mean momentum to increase). The pressure now only rises as n4/3, insufficiently fast to support a massive white dwarf against gravity. An upper limit is therefore placed on the mass of a white dwarf (the Chandrasekhar mass, ~1.4 solar masses) above which gravitational collapse resumes leading to the formation of a neutron star or black hole. (4) Answer to Question Number 3 (a) The probability of barrier penetration is given by: E 12 Ppenetration exp G E Where EG is the Gamow energy and E is the collieion energy. Note that the probability of barrier penetration increases with collision energy. However, the probability that a pair of nuclei will have a collision energy E with the palsma at some temperature, T is governed by Maxwell-Boltzmann statistics, i.e. Which is a decreasing function of T. Pcollision exp E kT Hence, the probability of a fusion event occuring can be approximated by: 1 E G 2 E P exp kT E Which will clearly be low for both high and low impact energies, and peaks at some impact energy defined by the Gamow energy and T: 1 E G ( kT ) 2 3 E0 4 (note: quoting this final result is not essential to obtain full marks) (b) Proton-proton cycle: First step: fuse p + p to give d (+ e+ + ) (6) Second step: fuse p + d to give 3He (+ ) Third step: fuse 3He + 3He to give 4He + 2p This chain liberates 26.2 MeV and accounts for 85% of the sun's energy production. The first step is mediated by the weak nuclear force, and is therefore slow. Indeed, in the sun, the mean time for step 1 to occur is about 1010 years. The remaining steps are comparatively fast. The remaining 15% of the sun's energy production is mainly via two branches of the p-p chain, involving fusion with 7Be and 7Li. The CNO cycle operates in higher mass stars, and involves again fusion of protons into helium, this time mediated by nuclear reactions with C, N and O. p+ 12C + 13N 13N p+ 13C 14N + p+ 14N 15O + 15O p+ 15N 12C 13C 15N + e+ + e + e+ + e + 4He Total energy release: 23.8 MeV In the CNO cycle, the p + mean lifetime of 14 14 N step is the slowest, hence this sets the reaction rate. The N in a sun-like star (where CNO is a minor process) is about 5x108 years, hence the CNO cycle can proceed faster than p-p. The reason p-p is dominant in low mass stars, whereas CNO dominates in high mass stars is due to the temperature dependence of the reactions. The rate of p-p is proportional to T4, whilst CNO goes as about T16. Hence, a modest increase in core temperature (higher mass star cores are hotter than lower mass) leads to a massive increase in the CNO rate, but only a modest rise in the p-p rate. (8) (c) Nuclei heavier than helium form in the latter stages of a star's life, where power production via proton fusion drops due to the drop in proton concentration. At this point, the stellar core contracts and heats up, triggering helium fusion to the triple- process (3 4He 12 12 C via C). This occurs via an intermediate step producing the short-lived species 8Be, which may fuse ith a third 4He giving an excited state of 12C. Fusion with further 4He nuclei gives 16 O and 20 Ne. At this stage in its evolution, a sun- like star has become a red giant. In a sun-like star, fusion stops at this point once the 4He has been exhausted, leaving a high temperature, degenerate core as a white dwarf star. The timescale of helium fusion in a sun-like star is about 10% of the hydrogen burning phase, for the sun about 109 years. In high mass (> about 8 solar masses), fusion proceeds further, producing most elements of the periodic table up to 57Fe. At this point, fusion cannot liberate energy, and the core collapses rapidly under gravity liberating sufficient energy to disrupt the star in a supernova explosion, leaving a high density neutron star or, if the core exceeds about 4 solar masses, a black hole. The timescales for these final stages are very short. For example, carbon burnuing in a 25 solar mass star takes about 600 years, and the final stage, silicon burning to iron is complete in less than a day. (6)