1

9. Direct reactions - for example direct capture:

Direct transition from initial state |a+A> to final state <f| (some state in B)

a + A -> B +

)(22 EPAaHf la

geometrical factor(deBroglie wave length of projectile - “size” of projectile)

mE

h

p

h

2

Interaction matrixelement

Penetrability: probabilityfor projectile to reach the target nucleus forinteraction. Depends on projectileAngular momentum land Energy E

)(1 2

EPAaHfE l III.25

2

Penetrability: 2 effects that can strongly reduce penetrability:

1. Coulomb barrier

V

rR

Coulomb Barrier Vc

R

eZZVc

221

Pote

ntia

lfor a projectile with Z2 anda nucleus with Z1

)(2.1

fm][44.1MeV][

3/12

3/11

2121

AA

ZZ

R

ZZVc

or

Example: 12C(p,) 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

3

2. Angular momentum barrier

Incident particles can have orbital angular momentum L

pd

Classical: Momentum pImpact parameter d

pdL

In quantum mechanics the angular momentum of an incident particle can havediscrete values:

)1( llL With l = 0l = 1

l = 2

s-wavep-wave

d-wave…

For radial motion (with respect to the center of the nucleus), angular momentumconservation (central potential !) leads to an energy barrier for non zero angularmomentum.

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.

And parity of thewave function: (-1)l

4

2

2

2

)1(

r

llVl

reduced mass of projectile-target system

Peaks again at nuclear radius (like Coulomb barrier)

Energy E of a particle with angular momentum L (still classical)

2

2

2mr

LE

Similar here in quantum mechanics:

Or in MeV using the nuclear radius and mass numbers of projectile A1 andtarget A2:

)(

)1(12]MeV[

3/12

3/11

21

21 AAAA

AA

llVl

5

9.1. Direct reactions – the simplest case: s-wave neutron capture

No Coulomb or angular momentum barriers: Vl=0

VC=0

But, change in potential still causes reflection – even without a barrierRecall basic quantum mechanics:

Incoming wave transmitted wave

Reflected wave

Potential

Transmission proportional to E

s-wave capture therefore always dominates at low energies

6

Example: 7Li(n,)

Penetrability EEPl )(

Therefore, for direct s-wave neutron capture:

Cross section (use Eq. III.19): E

1 Or

v

1

~1/v

Deviationfrom 1/vdue to resonantcontribution

thermalcross section<>=45.4 mb(see Pg. 27)

7

Why s-wave dominated ? Level scheme:

0.981

0

8Li

7Li + n

2.063

2+

1+

3/2- + 1/2+

Entrance channel 7Li + n : 3/2 + 1/2 + l(-1) = 1, 2

Exit channel 8Li +

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

Angular momentum and parity conservation:

( l=0 for s-wave )

2+ + ? (photon spin/parity)

l

8

Also recall:

E.M. Transition strength increases:• for lower L• for E over M• for higher energy 12 LE

Entrance channel 7Li + n : 3/2 + 1/2 + l(-1) = 1, 2

Exit channel 8Li +

( l=0 for s-wave )

2 + 1 = 1, 2, 3

E1 photonlowest EL that allows to fulfillconservation laws

match possible

Same for 1+ state

“At low energies 7Li(n,) is dominated by (direct) s-wave E1 capture”.

9

Stellar reaction rate for s-wave neutron capture:

Because v

1 vconst v

Thermal neutron cross section:

Many neutron capture cross sections have been measured at reactors usinga “thermal” (room temperature) neutron energy distribution atT= 293.6 K (20 0C), kT=25.3 meV

The measured cross section is an average over the neutron flux spectrum (E) used:

dEE

dEEE

)(

)()( (all Lab energies)

For a thermal spectrum kT

E

EE

e)( so

dEE

dEEE

kT

E

kT

E

e

e)(th

10

Why is a flux of thermalized particles distributed as kT

E

EE

e)( ?

The number density n of particles in the beam is Maxwell Boltzmann distributed

kTEEdE

dn /e

BUT the flux is the number of neutrons hitting the target per second and area. Thisis a current density j = n * v

thereforekTEE

dE

dnv

dE

dj /e

The cross section is averaged over the neutron flux (the number of neutronshitting 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 particlesis M.B. distributed, but the number of particles passing through an area per secondis kTEE /e distributed, and so is the stellar reaction rate !

11

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

thTthT

2

vvv

with kT

vT

2 (most frequent velocity, corresponding to ECM=kT)

for reactor neutrons (thermal neutrons) vT=2.20e5 cm/s

)(th kT

Tthth

2

v

v

and that’s usually tabulated as“thermal cross section”

which is for example useful to read cross section graphs

With these definitions one can show that the measured averaged cross sectionand the stellar reaction rate are related simply by

12

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

ll EEP 2/1)(

and therefore the cross section (Eq III.19)

2/1 lEfor 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 atthermal 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 wouldthen require higher gamma-ray multipolarities)

13

Example: p-wave capture in 14C(n,)15C

E

(from Wiescher et al. ApJ 363 (1990) 340)

14

14C+n

15C

0

0.74

1/2+

5/2+

Entrance channel:

s-wave 0+ 0+ 1/2+ 1/2+

p-wave 1- 0+ 1/2+ 1/2- 3/2-

Exit channel (15C + )

E1

M1

E2

1-

1+

2+

1/2- 3/2-

1/2+ 3/2+

3/2+ 5/2+

total to 1/2+ total to 5/2+

3/2- 5/2- 7/2-

3/2+ 5/2+ 7/2+

1/2+ 3/2+ 5/2+ 7/2+ 9/2+

Why p-wave ?

l 14C n total

strongest !

strongest possible Exit multipole

into 1/2+ into 5/2+

M1 E2

E1 E1

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 herefor example, the thermal cross section is exceptionally low (<1b limit known)

15

This depends on cross section shape and temperature:

s-wave n-capture:

9.2.1. What is the energy range the cross section needs to be known to determine the stellar reaction rate for n-capture ?

0 20 40 60 80energy (keV)

0

1

2

3

4

arbi

trar

y un

its

Example: kT=10 keV

M.B. distribution (E)

EdEEEvdvvvv )()()()(

(E)

(E) (E) E = E1/2 exp(-E/kT) relevant for stellar reaction rate

of the order of KT (somewhat lower than MB distribution)

16

0 20 40 60 80energy (keV)

0

1

2

3

4

arb

itra

ry u

nits

p-wave n-capture:

Example: kT=10 keVM.B. distribution (E)

(E)

(E) (E) E = E3/2exp(-E/kT)relevant for stellar reaction rate

of the order of KT (close to MB distribution)

17

9.2.2 The concept of the astrophysical S-factor (for n-capture)

recall:2

)(1

AaHfEPE l III.25

“trivial” strongenergydependence

“real” nuclear physicsweak 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 !!!

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 !!!

18

For neutron capture with strong s-wave dominance with corrections. Then define S-factor S(E)

)(1

ESv

and expand S(E) around E=0 as powers of E

ES

SE

S

S

v

S

ESESSv

)0(

)0(

2

1

)0(

)0(1

)0(

)0(2

1)0()0(

1

2/1

2/1

with denotingE

in practice, these are tabulated fitted parameters

typical S(E) units with this definition: barn MeV1/2

19

kTS

SkT

S

SSv

4

3

)0(

)0()(

2

)0(

)0(1)0( 2/1

and for the astrophysical reaction rate (after integrating over the M.B. distribution)

of course for pure s-wave capture

)0(Sv

for pure s-wave capture the S-factor is entirely determined by the thermal cross section measured with room temperature reactor neutrons:

sS /cmbarn][1020.2)0( 3th

19

)0(Tth Svv using one finds (see Pg. 35)

20

For neutron capture that is dominated by p-wave, such as 14C(p,) one can definea p-wave S-factor:

)(ESE

which leads to a relatively constantS-factor because of

E

or E

ES

)(

(typical unit for S(E) is then barn/MeV1/2)

S-factor

21

9.3. Charged particle induced direct reactions

2)(

1AaHfEP

E l again

but now incoming particle has to overcome Coulomb barrier. Therefore

(for example proton capture - such as 12C(p,) in CN cycle)

incoming projectile Z1 A1 (for example proton or particle)target nucleus Z2 A2

2e)( EPl with

221 e

2

ZZ

E

(from basic quantum mechanical barrier transmission coefficient)

9.3.1Cross section and S-factor definition

22

so the main energy dependence of the cross section (for direct reactions !)is given by

E

b

e1

E

keV31.28 2/121 AZZb

U21

21

mAA

AAA

therefore the S-factor for charged particle reactions is defined via

)(e1 Eb/- ESE

So far this all assumed s-wave capture. However, the additional angular momentumbarrier leads only to a roughly constant addition to this S-factor that strongly decreaseswith l

Therefore, the S-factor for charged particle reactions is defined independentlyof the orbital angular momentum

typical unit for S(E): keV barn

23

Given here is the partial proton width

2)(2

llp EPR

v

R=nuclear radiusreduced width (matrix element)

(from Rolfs & Rodney)

24

Example:12C(p,) cross section

need cross sectionhere !

25

Need rateabout here

S-Factor:

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

26

9.3.2. Relevant cross section - Gamov Window

for charged particle reactions

dEESkTdEEEkTv kT

E

E

b

-

2/3kT

E2/3 e)()(

8e)()(

8

Gamov Peak

Note: relevant cross sectionin tail of M.B. distribution, much larger thankT (very differentfrom n-capture !)

27

The Gamov peak can be approximated with a Gaussian

2

2E/

EE 0

ee

kT

E

E

b

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:

MeV12204.02

3/29

3/122

21

2/3

0 TAZZbkT

E

MeV23682.03

4 6/59

6/122

210 TAZZkTEE

with A “reduced mass number” and T9 the temperature in GK

28

Example:

Note:kT=2.5 keV !

29

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:

3/1

9

22

212487.4

0

3/1

29

219 ebarn] MeV)[(1083.7

T

AZZ

A ESAT

ZZvN

Equation III.53

(A reduced mass number !)

30

better (and this is often done for experimental data) one expands S(E) around E=0as powers of E to second order:

)0(''2

1)0(')0()( ESESSES

If one integrates this over the Gamov window, one finds that one can useEquation III.46 when replacing S(E0) with the effective S-factor Seff

kTEE

S

SkTE

S

SSSeff 0

200 36

89

)0(

)0(''

2

1

36

35

)0(

)0('

12

51)0(

kT

E03with and E0 as location of the Gamov Window (see Pg. 51)