Transcript Document

9. Direct reactions - for example direct capture:
a + A -> B + g
Direct transition from initial state |a+A> to final state <f| (some state in B)
2
    f H a  A  Pl ( E )
2
a
geometrical factor
(deBroglie wave length
of projectile - “size” of
projectile)

Penetrability: probability
for projectile to reach
the target nucleus for
interaction.
Depends on projectile
Angular momentum l
and Energy E
Interaction matrix
element
h
h

p
2m E
1
   f H a A
E
2
 Pl ( E )
III.25
25
Penetrability: 2 effects that can strongly reduce penetrability:
1. Coulomb barrier
V
Coulomb Barrier Vc
R
Z1 Z 2 e 2
Vc 
R
Example: 12C(p,g)
or
for a projectile with Z2 and
a nucleus with Z1
r
Vc [MeV]  1.44
Z1Z 2
ZZ
 1.2 1/ 3 1 2 1/ 3
R [fm]
( A1  A2 )
VC= 3 MeV
Typical particle energies in astrophysics are kT=1-100 keV !
Therefore, all charged particle reaction rates in nuclear astrophysics
occur way below the Coulomb barrier – fusion is only possible
through tunneling
26
2. Angular momentum barrier
Incident particles can have orbital angular momentum L
Classical:
p
d
Momentum p
Impact parameter d
L  pd
In quantum mechanics the angular momentum of an incident particle can have
discrete values:
L  l (l  1) 
With
l=0
l=1
l=2
…
s-wave
p-wave
d-wave
And parity of the
wave function: (-1)l
For radial motion (with respect to the center of the nucleus), angular momentum
conservation (central potential !) leads to an energy barrier for non zero angular
momentum.
Classically, if considering the radial component of the motion, d is decreasing,
which requires an increase in momentum p and therefore in energy to conserve L.
27
Energy E of a particle with angular momentum L (still classical)
L2
E
2m r2
Similar here in quantum mechanics:
l (l  1) 2
Vl 
2r 2
 : reduced mass of projectile-target system
Peaks again at nuclear radius (like Coulomb barrier)
Or in MeV using the nuclear radius and mass numbers of projectile A1 and
target A2:
Vl [MeV]  12
l (l  1)
 A1 A2  1/ 3

( A1  A21/ 3 )
 A1  A2 
28
9.1. Direct reactions – the simplest case: s-wave neutron capture
No Coulomb or angular momentum barriers:
Vl=0
VC=0
s-wave capture therefore always dominates at low energies
But, change in potential still causes reflection – even without a barrier
Recall basic quantum mechanics:
Incoming wave
transmitted wave
Reflected wave
Potential
Transmission proportional to
E
29
Therefore, for direct s-wave neutron capture:
Pl ( E)  E
Penetrability
Cross section (use Eq. III.19):
1

E
Or
1

v
Example: 7Li(n,g)
thermal
cross section
<>=45.4 mb
(see Pg. 27)
~1/v
Deviation
from 1/v
due to
resonant
contribution
30
Why s-wave dominated ? Level scheme:
2.063
3/2- + 1/2+
7Li
E1 g
+n
E1 g
0.981
1+
0
2+
8Li
Angular momentum and parity conservation:
Entrance channel 7Li + n :
Exit channel
8Li
l
3/2- + 1/2 + l(-1) = 1-, 2-
( l=0 for s-wave )
+ g : 2+ + ? (photon spin/parity)
Recall: Photon angular momentum/parity depend on multiploarity:
For angular momentum L (=multipolarity) electric transition EL parity (-1)L
magnetic transition ML parity (-1)L+1
31
Also recall:
E.M. Transition strength increases:
• for lower L
• for E over M
• for higher energy  E 2 L1
g
Entrance channel 7Li + n :
Exit channel
8Li
3/2- + 1/2 + l(-1) = 1-, 2-
+ g : 2 + 1E1 photon
lowest EL that
allows to fulfill
conservation laws
( l=0 for s-wave )
= 1-, 2-, 3match possible
Same for 1+ state
“At low energies 7Li(n,g) is dominated by (direct) s-wave E1 capture”.
32
Stellar reaction rate for s-wave neutron capture:
Because
1

v
v  const  v 
Thermal neutron cross section:
Many neutron capture cross sections have been measured at reactors using
a “thermal” (room temperature) neutron energy distribution at
T= 293.6 K (20 0C), kT=25.3 meV
The measured cross section is an average over the neutron flux spectrum F(E) used:
 ( E )F( E )dE

  
 F(E)dE
For a thermal spectrum
F( E )  E e
   th
-
E
kT so
 (E)E e


Ee
-
(all Lab energies)
E
kT
-
E
kT
dE
dE
33
Why is a flux of thermalized particles distributed as
F( E )  E e
-
E
kT
?
The number density n of particles in the beam is Maxwell Boltzmann distributed
dn
 E e - E / kT
dE
BUT the flux is the number of neutrons hitting the target per second and area. This
is a current density j = n * v
therefore
dj
dn
v
 E e - E / kT
dE
dE
The cross section is averaged over the neutron flux (the number of neutrons
hitting the target for each energy bin) because that is what determines the event
rate.
This is the same situation in the center of a star. The number density of particles
is M.B. distributed, but the number of particles passing through an area per second
is  E e - E / kT
distributed, and so is the stellar reaction rate !
34
With these definitions one can show that the measured averaged cross section
and the stellar reaction rate are related simply by
 v 
with
and
vT 
2kT

2

vT    th  vT th
(most frequent velocity, corresponding to ECM=kT)
for reactor neutrons (thermal neutrons) vT=2.20e5 cm/s
 v 
 th 
   th 
vT

2
that’s usually tabulated as
“thermal cross section”
For s-wave neutron capture (which is generally the only capture mechanism for
room temperature neutrons) one can relate the thermal cross section to the
actual cross section value at the energy kT
 th   (kT )
which is for example useful to read cross section graphs
35
9.2. Direct reactions – neutron captures with higher orbital angular momentum
For neutron capture, the only barrier is the angular momentum barrier
The penetrability scales with
Pl ( E)  E1/ 2l
and therefore the cross section (Eq III.19)
  E l -1/ 2
for l>0 cross section decreases with decreasing energy (as there is a barrier present)
Therefore, s-wave capture in general dominates at low energies, in particular at
thermal energies. Higher l-capture usually plays only a role at higher energies.
What “higher” energies means depends on case to case - sometimes s-wave is
strongly suppressed because of angular momentum selection rules (as it would
then require higher gamma-ray multipolarities)
36
Example: p-wave capture in 14C(n,g)15C
 E
(from Wiescher et al. ApJ 363 (1990) 340)
37
Why p-wave ?
14C+n
0.74
5/2+
0
1/2+
15C
Exit channel (15C + g)
g
total to 1/2+
total to 5/2+
1/2- 3/2-
3/2- 5/2- 7/2-
M1
11+
1/2+ 3/2+
3/2+ 5/2+ 7/2+
E2
2+
3/2+ 5/2+
1/2+ 3/2+ 5/2+ 7/2+ 9/2+
E1
strongest !
Entrance channel:
strongest possible Exit multipole
l
14C
n
total
s-wave
0+
0+
1/2+
1/2+
M1
E2
p-wave
1-
0+
1/2+
1/2- 3/2-
E1
E1
into 1/2+
into 5/2+
despite of higher barrier, for relevant energies (1-100 keV) p-wave E1 dominates.
At low energies, for example thermal neutrons, s-wave still dominates. But here
for example, the thermal cross section is exceptionally low (<1b limit known)
38
9.2.1. What is the energy range the cross section needs to be known to determine
the stellar reaction rate for n-capture ?
This depends on cross section shape and temperature:
 v    (v)F(v)vdv    ( E )Y ( E ) EdE
s-wave n-capture:
of the order of KT (somewhat lower than MB distribution)
4
Example: kT=10 keV
arbitrary units
3
M.B. distribution Y(E)
(E) Y(E) E = E1/2 exp(-E/kT)
relevant for stellar reaction rate
2
1
0
0
(E)
20
40
energy (keV)
60
80
39
p-wave n-capture:
of the order of KT (close to MB distribution)
4
M.B. distribution Y(E)
arbitrary units
3
Example: kT=10 keV
(E) Y(E) E = E3/2exp(-E/kT)
relevant for stellar reaction rate
2
1
0
0
(E)
20
40
energy (keV)
60
80
40
9.2.2 The concept of the astrophysical S-factor (for n-capture)
recall:
1
   Pl ( E )  f H a  A
E
“trivial” strong
energy
dependence
2
III.25
“real” nuclear physics
weak energy dependence
(for direct reactions !)
S-factor concept: write cross section as
strong “trivial” energy dependence
X
weakly energy dependent S-factor
The S-factor can be
• easier graphed
• easier fitted and tabulated
• easier extrapolated
• and contains all the essential nuclear physics
Note: There is no “universally defined S-factor - the S-factor definition depends on
Pl(E) and therefore on the type of reaction and the dominant l-value !!!
41
For neutron capture with strong s-wave dominance with
corrections. Then define S-factor S(E)
1
  S (E)
v
E
and expand S(E) around E=0 as powers of

1
1 

1/ 2
   S ( 0)  S ( 0) E  S ( 0) E 
v
2





S (0)  S (0) 1/ 2 1 S (0) 

1
E 
E

v  S ( 0)
2 S (0) 


with
denoting

 E
in practice, these are tabulated fitted parameters
typical S(E) units with this definition: barn MeV1/2
42
and for the astrophysical reaction rate (after integrating over the M.B. distribution)




S
(
0
)
2
S
(
0
)
3


 v  S (0)1 
(kT )1/ 2 
kT 
S (0) 4 
 S (0) 


of course for pure s-wave capture
 v  S (0)
for pure s-wave capture the S-factor is entirely determined by the thermal
cross section measured with room temperature reactor neutrons:
using
  v   thvT  S (0)
one finds (see Pg. 35)
S (0)  2.2010-19  th[barn] cm3 / s
43
For neutron capture that is dominated by p-wave, such as 14C(p,g) one can define
a p-wave S-factor:
  E S (E)
or
S (E) 

E
which leads to a relatively constant
S-factor because of
 E
(typical unit for S(E) is then barn/MeV1/2)
S-factor
44
9.3. Charged particle induced direct reactions
9.3.1Cross section and S-factor definition
(for example proton capture - such as 12C(p,g) in CN cycle)
incoming projectile Z1 A1 (for example proton or a particle)
target nucleus
Z2 A2
again
1
   Pl ( E )  f H a  A
E
2
but now incoming particle has to overcome Coulomb barrier. Therefore
Pl (E)  e-2
with

 Z1Z 2e 2
2E

(from basic quantum mechanical barrier transmission coefficient)
45
so the main energy dependence of the cross section (for direct reactions !)
is given by
1
 e
E
-
b
E
b  31.28 Z1Z2 A1/ 2 keV
A1 A2

A

A1  A2 mU
therefore the S-factor for charged particle reactions is defined via
1 -b/
 e
E
E
S (E)
typical unit for S(E): keV barn
So far this all assumed s-wave capture. However, the additional angular momentum
barrier leads only to a roughly constant addition to this S-factor that strongly decreases
with l
Therefore, the S-factor for charged particle reactions is defined independently
of the orbital angular momentum
46
Given here is the
partial proton width
p 
2v
Pl ( E ) l2
R
R=nuclear radius
  reduced width
(matrix element)
(from Rolfs & Rodney)
47
Example:
12C(p,g) cross section
need cross section
here !
48
S-Factor:
Need rate
about here
From the NACRE compilation of charged particle induced reaction rates on
stable nuclei from H to Si (Angulo et al. Nucl. Phys. A 656 (1999) 3
49
9.3.2. Relevant cross section - Gamov Window
for charged particle reactions
 v 
8

(kT )
-3 / 2
  (E)E e
-
E
kT
dE 
8

(kT ) -3 / 2  S ( E )e
E 
 b
-


 E kT 
dE
Gamov Peak
Note: relevant
cross section
in tail of M.B.
distribution,
much larger than
kT (very different
from n-capture !)
50
The Gamov peak can be approximated with a Gaussian
e
E 
 b
-
 
 E kT 
e
 E -E0  2
-

  E/ 2 
centered at energy E0 and with 1/e width E
Then, the Gamov window or the range of relevant cross section can be
easily calculated using:
3/ 2


1/ 3 2 / 3
 bkT 
2 2
E0  
  0.12204Z1 Z 2 A T9 MeV
 2 
1/ 6 5 / 6
4
2 2
E 
E0 kT  0.23682 Z1 Z 2 A T9 MeV
3


with A “reduced mass number” and T9 the temperature in GK
51
Example:
Note:
kT=2.5 keV !
52
9.3.3. Reaction rate from S-factor
Often (for example with theoretical reaction rates) one approximates the rate
calculation by assuming the S-factor is constant over the Gamov WIndow
S(E)=S(E0)
then one finds the useful equation:
1/ 3
 Z1Z 2 
 S ( E0 )[MeV barn] e
N A  v  7.8310 
2 
 AT9 
9
1/ 3
 Z12 Z 22 A 

- 4.2487 

T
9


Equation III.53
(A reduced mass number !)
53
better (and this is often done for experimental data) one expands S(E) around E=0
as powers of E to second order:
1
S ( E )  S (0)  ES ' (0)  ES ' ' (0)
2
If one integrates this over the Gamov window, one finds that one can use
Equation III.46 when replacing S(E0) with the effective S-factor Seff

5 S ' (0) 
35  1 S ' ' (0)  2 89

Seff  S (0)1 

E

kT

E

E
kT
 0

 0

0
36  2 S (0) 
36

 12 S (0) 
with  
3 E0
kT
and E0 as location of the Gamov Window (see Pg. 51)
54