Transcript Document

III. Nuclear Reaction Rates
Nuclear reactions
• generate energy
• create new isotopes and elements
Notation for stellar rates:
p
12C
13N
g
12C(p,g)13N
The heavier
“target”
nucleus
(Lab: target)
the lighter
“incoming
projectile”
(Lab: beam)
the lighter
“outgoing
particle”
(Lab: residual
of beam)
the heavier
“residual”
nucleus
(Lab: residual of target)
(adapted from traditional laboratory experiments with a target and a beam)
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Example for a sequence of reactions in stars: the CN cycle
Ne(10)
F(9)
O(8)
N(7)
C(6)
3
4
5
6
7
8
9
neutron number
12C(p,g)13N (b+)13C(p,g)14N
(p,g)15O (b+)15N(p,a)12C
Net effect: 4p -> a + 2e+ + 2ve
But requires C or N as catalysts (second and later generation star)
Need reaction rates to:
• establish existence of cycle
• determine total rate of cycle
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Sun
pp reaction chain:
3
The 4 CNO cycles
4
Competition between the p-p
chain and the CNO Cycle
To understand this (and else):
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1. cross section s
bombard target nuclei with projectiles:
relative velocity v
Definition of cross section:
# of reactions
per second and target nucleus
or in symbols:
l=sj
=
s
.
# of incoming projectiles
per second and cm2
with j as particle number current density.
Of course j = n v with particle number density n)
Units for cross section:
1 barn = 10-24 cm2 ( = 100 fm2 or about half the size (cross sectional area) of a
uranium nucleus)
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2. Reaction rate in the laboratory
beam of particles
hits target at rest
area A
j,v
thickness d
assume thin target (unattenuated beam intensity throughout target)
Reaction rate (per target nucleus):
Total reaction rate (reactions per second)
l s j
R  l AdnT  s IdnT
with nT
: number density of target nuclei
I =jA : beam number current (number of particles per second hitting the target)
note: dnT is number of target nuclei per cm2. Often the target thickness is
specified in these terms.
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3. Reaction rate in stellar environment
Mix of (fully ionized) projectiles and target nuclei at a temperature T
3.1. For a given relative velocity v
in volume V with projectile number density np
l  s n pv
R  s n p vnT V
so for reaction rate per second and cm3:
r  np nTs v
This is proportional to the number of p-T pairs in the volume.
If projectile and target are identical, one has to divide by 2 to avoid double counting
r
1
1   pT
n p nT s v
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3.2. at temperature T
for most practical applications (for example in stars) projectile and target nuclei
are always in thermal equilibrium and follow a Maxwell-Bolzmann velocity
distribution:
then the probability F(v) to find a particle with a velocity between v and v+dv is
 m 

F (v)  4 
 2 kT 
3/ 2
v2 e
m v2

2 kT
with
 F (v)dv  1
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example: in terms
of energy
E=1/2 m v2
arbitrary units
3
max at
E=kT
2
1
0
0
20
40
energy (keV)
60
80
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Temperature in Stars
The horizontal axis is the velocity in cm/sec and the vertical axis
is proportional to the probability that a particle in the gas has
that velocity.
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Maxwell-Boltzmann distribution
Boltzmann distribution = probability that any one molecule will be
found with energy E:
 E / kt
F( E)  Ae
If this distribution is applied to one direction of velocity for a molecule
in an ideal gas, it becomes
m mv z2 / 2 kT
F(vz ) 
e
2kT
Converting this relationship to probability in terms of speed in three
dimensions:
 m 

F (v)  4 
 2 kT 
3/ 2
2
v e
m v2

2 kT
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Details of calculation:
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one can show (Clayton Pg 294-295) that the relative velocities between two particles
are distributed the same way:
  

F (v)  4 
 2 kT 
3/ 2
2
v e

 v2
2 kT
with the mass m replaced by the reduced mass  of the 2 particle system

m1m2
m1  m 2
the stellar reaction rates has to be averaged over the distribution F(v)
r
or short hand:
1
1   pT
r
n p nT  s (v)F(v)vdv
1
1   pT
typical strong
velocity dependence !
n p nT  sv 
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expressed in terms abundances
r
l
1
1   pT
1
1   pT
YT Yp  2 N 2A  sv 
Yp  N A  sv 
reactions per s and cm3
reactions per s and target
nucleus
this is usually referred to
as the stellar reaction rate
of a specific reaction
units of stellar reaction rate NA<sv>: usually cm3/s/mole, though in fact
cm3/s/g would be better (and is needed to verify dimensions of equations)
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4. Abundance changes, lifetimes
Lets assume the only reaction that involves nuclei A and B is destruction
(production) of A (B) by A capturing the projectile a:
A + a -> B
Again the reaction is a random process with const probability (as long as the
conditions are unchanged) and therefore governed by the same laws as
radioactive decay:
dnA
  n Al   n AYa  N A  s v 
dt
dnB
  n Al
dt
consequently:
n A (t )  n0 A e  l t
nB (t )  n0 A (1  e l t )
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and of course
YA (t )  Y0 A e  l t
after some time, nucleus A
is entirely converted to nucleus B
YB (t )  Y0 A (1  e  l t )
Example:
0.007
abundance
0.006
0.005
Y0A
A
B
0.004
0.003
same
abundance
level Y0A
0.002
Y0A/e
0.001
0
-2
10
time
-1
10
0
10
1
10

2
Lifetime of A (against destruction via the reaction A+a) :
(of course half-life of A T1/2=ln2/l)
3
10
10

1
l

4
10
5
10
1
Ya  N A  s v 
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5. Energy generation through a specific reaction:
Again, consider the reaction A+a->B
Reaction Q-value: Energy generated (if >0) by a single reaction
in general, for any reaction (sequence) with nuclear masses m:



Q  c   mi   m j 
final nuclei j 
 initialnucleii
2
Energy generation:
Energy generated per g and second by a reaction:
rQ
1

Q
YAYa N 2A  s v 

1   aA
6. Reaction flow:
abundance of nuclei converted in time T from species A to B via a specific reaction
 dYA 
F  
dt   l (t )YA (t )dt

dt  via specific reaction
0
0
T
T
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7. Multiple reactions destroying a nuclide
example: in the CNO cycle, 13N
can either capture a proton or b decay.
14O
(p,g)
13N
each destructive reaction i has a rate li
(b+)
13C
7.1. Total lifetime
the total destruction rate for the nucleus is then
its total lifetime

1
l

l   li
i
1
l
i
i
7.2. Branching
the reaction flow branching into reaction i, bi is the fraction of destructive flow
through reaction i. (or the fraction of nuclei destroyed via reaction i)
li
bi 
lj
j
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8. Determining nuclear reaction rates - Introduction
Needed is the cross section as a function of energy (velocity)
The stellar reaction rate can then be calculated by integrating over the
Maxwell Boltzmann distribution.
The cross section depends sensitively on the reaction mechanism and
the properties of the nuclei involved. It can vary by many (tens) orders of magnitude
It can either be measured experimentally or calculated. Both are difficult.
Experiments are complicated by extremely small cross sections that
prevent direct measurements of the cross sections at the relevant
astrophysical energies (with a some exceptions)
Typical energies for astrophysical reactions are of the order of kT
Sun
T ~ 10 Mio K
Si burning in a massive star: T ~ 1 Bio K
There is no nuclear theory that can predict the relevant properties of
nuclei accurately enough.
In practice, a combination of experiments and theory is needed.
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Nuclear properties that are relevant for reaction rates:
Nucleons in the nucleus can only have discrete energies. Therefore, the nucleus as
a whole can be excited into discrete energy levels (excited states)
Excitation energy (MeV)
Spin
Excitation
energy
0
Parity (+ or - )
5.03
3/2 3rd excited state
4.45
5/2
2.13
1/2 1st excited state
0
3/2
2nd excited state
ground
state
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Each state is characterized by:
• energy (mass)
• spin
• parity
• lifetimes against g,p,n, and a emission
The lifetime is usually given as a width as it corresponds to a width in the
excitation energy of the state according to Heisenberg:
E  t  
therefore, a lifetime  corresponds to a width G:
G


the lifetime against the individual “channels” for g,p,n, and a emission are
usually given as partial widths
Gg, Gp, Gn, and Ga
with
G  Gi
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A Real Example
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Basic reaction mechanisms involving strong or electromagnetic interaction:
Example: neutron capture A + n -> B + g
I. Direct reactions (for example, direct capture)
En
Sn
direct transition into bound states
g
A+n
B
II. Resonant reactions (for example, resonant capture)
Step 1: Coumpound nucleus formation
(in an unbound state)
G
En
Sn
A+n
B
Step 2: Coumpound nucleus decay
G
g
B
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or a resonant A(n,a)B reaction:
Step 1: Compound nucleus formation
(in an unbound state)
Sn
Step 2: Compound nucleus decay
a
G
En
A+n
C
C
B+a
Sa
B
For resonant reactions, En has to “match” an excited state (but all excited states
have a width and there is always some cross section through tails)
But enhanced cross section for En ~ Ex- Sn
more later …
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