Lecture 11
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Transcript of Lecture 11
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Lecture 11Lecture 11
Energy transport
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Review: Nuclear energyReview: Nuclear energy
• If each reaction releases an energy the amount of energy released per unit mass is just
TXXr xiixix 0
24 rdr
dLr
• The sum over all reactions gives the nuclear reaction contribution to in our fifth fundamental equation:
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Proton-proton chain (PPI)Proton-proton chain (PPI)
HHeHeHe
HeHH
eHHH e
11
42
32
32
32
11
21
21
11
11
2
The net reaction is: 2224 42
11
eeHeH
But each of the above reactions occurs at its own rate. The first step is the slowest because it requires a proton to change into a neutron:
eenp Energy
This occurs via the weak force. The rate of this reaction determines the rate of Helium production
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Proton-proton chain (PPII and PPIII)Proton-proton chain (PPII and PPIII)
HeHLi
LieBe
BeHeHe
e
42
11
73
73
74
74
42
32
2
Alternatively, helium-3 can react with helium-4 directly:
HeBe
eBeB
BHBe
e
42
84
84
85
85
11
74
2
Yet another route is via the collision between a proton and the beryllium-7 nucleus
This reaction only occurs 0.3% of the time in the Sun.
In the Sun, this reaction occurs 31% of the time; PPI occurs 69% of the time.
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The PP chainThe PP chain
The nuclear energy generation rate for the PP chain, including all three branches:
kgWeTX Tpp /1038.2
3/1680.333/2
62
54
KTT 66 10/
Near T~1.5x107 K (i.e. the central temperature of the Sun):
W/kg1007.1 46
25
7 TXpp
355 /10 mkg
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ExampleExample
W/kg1007.1 46
25
7 TXpp
If we imagine a core containing 10% of the Sun’s mass, composed entirely of hydrogen (X=1), calculate the total energy produced by the PP reaction.
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The CNO cycleThe CNO cycle
There is a second, independent cycle in which carbon, nitrogen and oxygen act as catalysts. The main branch (accounting for 99.6% of CNO reactions) is:
HeCHN
eNO
OHN
NHC
eCN
NHC
e
e
42
126
11
157
157
158
158
11
147
147
11
136
136
137
137
11
126
kgWeTXX TCNOCNO /1067.8
3/1628.1523/2
6525
W/kg1024.8 9.1965
27 TXXCNOCNO at T~1.5x107 K
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Helium collisionsHelium collisions
Recall that the temperature at which quantum tunneling becomes possible is:
2
422
21
20 3
4
4
1
kh
eZZT
• As hydrogen is converted into helium, the mean molecular weight increases.
• To keep the star in approximate pressure equilibrium, the density and temperature of the core must rise
Hm
kTP
As H burning progresses, the temperature increases and eventually He burning becomes possible
K 109.1 22
21
7 ZZmH
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The triple-alpha processThe triple-alpha process
The burning of helium occurs via the triple alpha process:
CHeBe
BeHeHe126
42
84
84
42
42
The intermediate product 8-beryllium is very unstable, and will decay if not immediately struck by another Helium. Thus, this is almost a 3-body interaction
kgWeTY T /1009.51
8027.4438
325
113
W/kg1085.3 0.418
325
83 TY
Note the very strong temperature dependence. A 10% increase in T increases the energy generation by a factor 50.
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NucleosynthesisNucleosynthesis
At the temperatures conducive to helium burning, other reactions can take place by the capturing of -particles (He atoms).
NeHeO
OHeC2010
42
168
168
42
126
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NucleosynthesisNucleosynthesis
The binding energy per nucleon describes the stability of a nucleus. It is easier to break up a nucleus with a low binding energy.
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BreakBreak
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SummarySummary
We have now established four important equations:
2r
GM
dr
dP r
24 rdr
dM r
Hm
kTP
Hydrostatic equilibrium:
Mass conservation:
Equation of state:
There are 5 variables (P,,Mr, T and Lr) and 4 equations. To solve the stellar structure we will need to know something about the energy transportation.
24 rdr
dLr Energy production
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Energy transportEnergy transport
Radiation: the photons carry the energy as they move through the star, and are absorbed at a rate that depends on the opacity.
Convection: buoyant, hot mass will rise
Conduction: collisions between particles transfer kinetic energy of particles. This is usually not important because gas densities are too low.
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Radiation transportRadiation transport
When we considered the properties of radiation, we found an equation relating the pressure gradient to the radiative flux:
radrad F
cdr
dP
From this we can derive an expression for the temperature gradient, assuming a blackbody.
In regions of high opacity, or high radiative flux, the temperature gradient must be steep to transport the energy outward.
3264
3
Tr
L
dr
dT r