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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, px ~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 px ~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)