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Reminder

Please return Assignment 1 to the
School Office by 13:00 Weds. 11th
February (tomorrow!)
– The assignment questions will be reviewed
in next week’s workshop
Stellar Power Sources

Possibilities
– Chemical burning

only a few thousand years
– Gravitational contraction

a few million years
– Nuclear fusion

~1010 years for a sun-like star
Physics of Fusion

The Nuclear Potential
– Repulsive at large distances due to Coulomb
– Attractive at short range due to Strong Nuclear
Force

(1fermi = 10-15 m)
0
1
2
r (fermis)
3
4
Physics of Fusion
– Classically, for nuclei to fuse, a barrier of
height Ec has to be overcome:
Ec 
1.44Z A Z B
rN
Ec in MeV, ZA, ZB nuclear charges and rN the range of the
strong force in fermis (typically 1 fermi)
Physics of Fusion
– Typical stellar models give typical core
temperatures of ~ 107 K (kT ~ 1keV)
– Fraction of nuclei with sufficient energy to
overcome EC given by Boltzmann:

f E C   exp  1.44
1  10
Classically, fusion shouldn’t happen!
3
  10
 625
Physics of Fusion

Solution:
– Quantum Mechanics
– Protons described by Schroedinger
Equation
– Quantum mechanical tunnelling possible
Physics of Fusion
– Probability of barrier penetration given by:
P  exp  (rc  rN )
2
Where rc is the classical closest distance of approach of
the nuclei and  is defined by:
 r    r  and
2
2
2
2
 
E 
 Ec
2m r
Physics of Fusion
– The probability of barrier penetration can
be recast in terms of energy:
1
 E
2

P  exp   G E  


Where EG is the Gamow energy define by:
E G  Z A Z B   2mr c
2
2
 = fine structure constant (~1/137)
mr = reduced mass of nuclei
Physics of Fusion
– For two protons, EG =493 keV
– In a typical stellar core, kT ~ 1 keV
– Hence the probability of barrier penetration
is:

 493
P  exp 
1


1
2

10
  exp 22  2.8  10

Still slow, but there are a lot of protons, and we have a lot
of time!
Fusion Cross Sections

Consider a proton moving in a medium with n
protons per unit volume
– Probability of fusion occuring within a distance Dx
= ns Dx

s = reaction cross section
– Mean distance between collisions = mean free
path, l = 1/ns
– Mean time between collisions,
t = l/<v> = 1/ n<sv>

note s may depend on v
Fusion Cross Sections

The fusion cross section (Units, barns = 10-28
m2) as a function of energy is given by:
1

2
S E 
E


G

sE  
exp   E  
E


S(E) is a slowly varying function determined by the
nuclear physics of the reaction
1/E introduced to account for low energy behaviour
Fusion Reaction Rates

Consider the reaction rate between two
nuclei, A and B, travelling with relative speed,
v, with concentrations nA and nB, with cross
section s
– Mean time for an A nucleus to fuse with a B is:
tA = 1/ nB<sv>
– Hence, the total fusion rate per unit volume is:
RAB = nAnB <sv>
Fusion Reaction Rates
– To obtain <sv>, we note that:
sv 


0
sv r P v r dv r
Where P(vr) is the Maxwell-Boltzmann distribution given
by:
P v r dv r  T
3
2


exp  E kT dE
Fusion Reaction Rates
– Including the function for s found earlier,
the total reaction rate is then:
1
R AB
1

 8   1  
E
 EG  2 
 n An B 
   dE
   0 S (E ) exp 
 kT  E  
 m r   kT 
2
3
2
Concentrating on the integral, we note that for a given
impact energy, E, there is a competition between the
Boltzmann term and the Gamow energy term
Fusion Reaction Rates
– Proton-proton reactions

T = 2x107 K, EG = 290kT
exp(-E/kT-(EG/E)1/2)
exp(-E/kT)
1
2
exp(-(EG/E)1/2)
3
4
5
6
7
8
Impact Energy (units of kT)
9
10
Fusion Reaction Rates
– Note there is a range of energies in which
fusion rates peak
– impact energy for peak rate, E0,
 E G (kT )2 
E0  

4


1
3
Width given by:(Via a Taylor expansion)
1
5
4
DE  12 13 E G 6 (kT ) 6
3 2
Fusion Reaction Rates
– For the proton-proton reaction shown
earlier,
E0 = 4.2kT = 7.2 keV
DE = 4.8kT = 8.2 keV
Fusion Reaction Rates
– The total reaction rate is found from the
integral shown earlier.
1
R AB
1

 8   1  
E
 EG  2 
 n An B 
   dE
   0 S (E ) exp 
 kT  E  
 m r   kT 
2
3
2
This gives:
R AB
  E  13 
 n An B S (E ) exp  3 G  
  4kT  
Fusion Reaction Rates
– Fusion reaction rates are strongly temperature
dependent

e.g, p-d reaction, EG = 0.657 MeV, around 2x107 K
dR pd  E G 


 4kT 
dT
1
3
R pd
R pd

 4.6
T
T
This implies the rate varies as T 4.6
Fusion in Stars I

Hydrogen fusion mechanisms in main
sequence stars
– Proton-Proton Chain
– CNO Cycle
– Relative rates and temperature
dependencies
Hydrogen Burning

In a main sequence star, the principal
source of power is fusion of protons
into helium nuclei
– 4p  4He + 2e+ + 2ne
– Relies on weak nuclear force to mediate
reaction:
p  n + e+ + ne
– Total energy release (including annihlation
of positrons) 26.73 MeV
Hydrogen Burning

Four particle reaction unlikely, hence might
expect a three-step process:
– 2p  2He +g
– 2He  d + e+ + ne
– 2d  4He +g

Problem:
– No bound state of 2He
– Hence, it looks like hydrogen burning is slow
Proton-Proton Chain

A possibility:
– Fuse protons via the weak nuclear force to
give deuterium
– p  n + e+ + ne

Requires 1.8 MeV
–p+nd

Releases 2.2 MeV
– Net Result:
p + p  d + e+ + ne
Proton-Proton Chain
– Recall the rate of a reaction is given by:
1
R AB
1

 8   1  
E
 EG  2 
 n An B 
   dE
   0 S (E ) exp 
 kT  E  
 m r   kT 
2
3
2
This integral gives (see Phillips secn 4.1)
R AB  6.48  10
24
n An B
 EG 
S (E 0 )

 4kT 
AZ A Z B
2
3
  E  13 
exp  3 G   m-3 s 1
  4kT  
The symbols have their previous meaning, A = reduced
mass in au
S(E) is in keV barns
Proton-Proton Chain
– The reaction
p + p  d + e+ + ne
is mediated by the weak force and is hence
slow
– S ~ 4x10-22 keV barns (calculated - too
small to measure!)
– What is the rate of proton-proton fusion in
the core of the sun?
Proton-Proton Chain

Assume a typical model of the sun’s core:
– T = 15x106 K
– r = 105 kgm-3
– Proton fraction = 50%
– hence proton density np =3x1031 m-3
– Also S ~ 4x10-22 keV barns and EG = 494 keV

These numbers give a rate of:
Rpp = 5x1013 m-3s-1
Proton-Proton Chain

Mean lifetime of a proton in the sun’s
core:
– Rate of any two protons fusing
f = Rpp / (1/2 np)~3.3x10-18 s-1
– Hence, mean time for a proton pair to fuse
is:
t = 1/f = 3x1017 s ~ 9x109 years
– Slow proton-proton rate sets timescale for
stellar lifetimes
Proton-Proton Chain
Following p-p fusion, further reactions
to produce 4He are rapid
 The proton-proton chain can follow
three branches (pathways):

Proton-Proton Chain
p + p  d + e+ + ne
p + d  3He +g
3He
+ 4He  7Be + g
p + 7Be  8B + g
e- + 7Be  7Li + ne
8B
2 3He  4He + 2p
p + 7Li  4He + 4He
8Be
Qeff = 26.2 MeV
Qeff = 25.7 MeV
85%
15%
 8Be + e+ + ne
 4He + 4He
Qeff = 19.1 MeV
0.02%
Proton-Proton Chain

Average energy release per p-p fusion:
– Take into account:
– two p-p fusions per branch
– weightings of each branch
– 15 MeV per p-p fusion
– Given number of fusions per m-3 calculated
earlier, energy production rate ~ 120 Wm-3
The CNO Cycle
The proton-proton chain has a
temperature dependence of ~ T 4
 Internal temperatures of more massive
stars are only moderatly higher
 Luminosities much greater than can be
explained by the T 4 dependence

The CNO Cycle

Implications:
– Another mechanism must be at work
– This mechanism must have a higher order
temperature dependence
– Implies a higher Coulomb barrier

Recall power of dependence  EG1/3
and EG  (ZAZb)2
– Such a mechanism is the CNO cycle
The CNO Cycle
– Reaction Pathway:
S = 1.5 keV barns p + 12C  13N + g
EG = 32.8 MeV
S = 5.5 keV barns p + 13C  14N + g
EG = 33 MeV
S = 3.3 keV barns p + 14N  15O + g
EG = 45.2 MeV
S = 78 keV barns p + 15N  12C + 4He
EG = 45.4 MeV
Qeff = 23.8 MeV
13N

13C
+ e+ + ne
15O

15N
+ e+ + ne
The CNO Cycle

Notice that:
– 12C acts as a catalyst
– Rate governed by slowest step


in p-p, the first p-p fusion step
In CNO, if one considers all parameters, the
p-14N step is slowest
– Abundance of
14N~0.6%
The CNO Cycle
– Rate of p-14N fusion in the sun
– Abundance of 14N~0.6%

gives 2.6x1028
14N
m-3
S = 3.3 keV barns, EG = 45.2 MeV
Other parameters same as for p-p fusion:
RpN = 1.6x1012 s-1m-3
CNO cycle contributes at most a few % to the
power of a sun-like star
– Mean lifetime of a 14N nucleus in the sun
~ 5x108 yrs
–
–
–
–
The CNO Cycle

The CNO cycle is strongly temperature
dependent
– Using ideas from previous lecture, at the
temperature of a sun-like star, and
considering the p-14N step,
RpN  T
20
– We can compare the rate of p-p vs. CNO as
a function of temperature
CNO vs. p-p
1.E+26
-1
-3
Fusion rate (s m )
1.E+24
p-p
CNO
1.E+22
1.E+20
T 15.63
1.E+18
1.E+16
1.E+14
T 2.96
1.E+12
1.E+10
1.E+08
1.E+07
sun
Temperature (K)
1.E+08
CNO vs. p-p

We can see that:
– CNO contributes a few % of the sun’s
output
– In a moderately hotter stellar core
(~1.8x107 K) CNO ~ p-p
– In hot (>2x107) cores, CNO > p-p
A requirement for CNO

The CNO cycle requires the heavy
elements C, N and O.
– These elements have negligible abundance
from the Big Bang
– They are not formed in the p-p chain
– What is the origin of heavy elements?
– See Next Lecture - Fusion in Stars II
Next Week
Heavy element production
 Star formation
