Unbound states near the proton emission threshold. The case of 16 F. I. Ştefan (1), F. de Oliveira...

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Unbound states near the proton emission threshold. The case of 16F.

I. Ştefan(1) , F. de Oliveira Santos(2)

(1) IPN Orsay(2) GANIL

Nova explosions

X-ray bursts15O

14N(p,γ)

15O(α,γ)19Ne

15O(pp,γ)17Ne

15O + p <=> 16F (unbound)15O – decay β+ ( T1/2=122 s )

The case of 16F

β+ (Qβ=15 MeV)

15O+p

16F

E0= 0.536 MeV

p

1/2-

0- Γp=25 keV

16O

β+

Qβ=15 MeV

β+ (Qβ=15 MeV)

15O+p

16F

E0= 0.536 MeV

p

1/2-

0- Γp=25 keV

16O

β+

Qβ=15 MeV

Mirror nucleus - 16N

Γβ=6.6 10-22 MeV

β+ (Qβ=15 MeV)

15O+p

16F

E0= 0.536 MeV

p

1/2-

0- Γp=25 keV

16O

β+

Qβ=15 MeV

(pp,γ) – Gorres et al 1995

(α,γ) – Taux estimé

Vc (

MeV

)

r (fm)

E0

16Fgs

Coulomb barrier

ECM (MeV)

)()( EPE lp

MeV2210*6.6

Wid

th (

MeV

)

15O+p 16F

15O+p

16O+β++ν

r (fm)

p

β+

E0

Vc (

MeV

)

ECM (MeV)

High ECM for the 15O – p system:

•Γp >> Γβ

•High probability to form 16F compound nucleus

•Beta decay improbable

)()( EPE lp

MeV2210*6.6

Wid

th (

MeV

)

15O+p 16F

15O+p

16O+β++ν

ECM (MeV)

Low ECM for 15O – p system:

•Γp << Γβ

•Decreased probability to form the 16F compound nucleus

•Beta decay branching ratio increased

)()( EPE lp

MeV2210*6.6

r (fm)

p

β+

E0

p

β+

Vc (

MeV

)

Wid

th (

MeV

)

γ

15O+p 16F

15O+p

16O+β++ν

r (fm)

Vc (

MeV

)

E0

2201 )

2

)(()(

EEEE

ctresonancetheofShape

tot

Hypothesis:

β+ (Qβ=15 MeV)

15O+p16F

E0= 0.536 MeVγ

dEEEPEPEE pppp

),()(),(),())(,(

γ

1/2-

0- Γp=25 keV

16O

pE1

p

Γp=70 keV

E0= 0.729 MeV

1-

β+

β+

Qβ=15 MeVβ+

Vc (

MeV

)

E0

γ

dEEEPEPEE pppp

),()(),(),())(,(

β+

β+

pE1

Facteur 109 !!

Vc (

MeV

)

E0

γ

dEEEPEPEE pppp

),()(),(),())(,(

β+

β+

pE1

Factor 109 !!

r (fm)

2201 )

2

)(()(

EEEE

ctresonancetheofShape

tot

Hypothesis:

Vc (

MeV

)

E0

γ

dEEEPEPEE pppp

),()(),(),())(,(

β+

β+

pE1

Factor 109 !!

r (fm)

2201 )

2

)(()(

EEEE

ctresonancetheofShape

tot

Hypothesis:

Let’s check it!

Let’s check it!!

• EURISOL 15O, 14O beam ?

• Different consequences?

p + p ↔ 2He → d + e+ +

p + 4He ↔ 5Li → γ + e+ + ν +4He + n

15O intensity = 1010 pps

E = 0.8 A MeV

N reactions = 8 / day.Continuum Shell Model

• Astro?

• Unbound state isomer !

16F radioactive??

•Non exponential decay!

•More surprises to come …

Future

Application for other unbound nuclei

- 15F (very interesting case)

- 5Li

- 4Li

- 19Na

…….

Outlook

Proposition of a new reaction pathway - 15O(p,γ)(β+)16O

- Consequences in different astrophysical scenarios

- Sequential decay ?

- Shape of the resonance ?

- γ decay between two unbound states ? σ(p,γ)~ 1 μb

- Experimental Confirmation

Questions related with the model

End

β+ (Qβ=15 MeV)

15O+2p 16F+p

0.536 MeV

p

1/2-

0-

0.732 MeV

16O+p

17Ne

2-

5/2-

1/2+ E1= 0.892 MeV

0.324 MeV

0 MeV

-0.958 MeV

-15 MeV

γ γ

γ

β+

15O(p,γ)(p,γ)17Ne

1/2-

15O(p,γ)(β+)16O

15O(β+)15N (2 min)15O(α,γ)19Ne 15O(p,γ)(β+)16O15O(p,γ)(p,γ)17Ne

Conclusions

Futur

• Astro à revoir ?

• Expérience SPIRAL 2 ?

• D’autres conséquences en astro ?p + p ↔ 2He → d + e+ +

15O intensité = 1010 pps

E = 0.8 A MeV

N réactions = 8 / jour.Continuum shell model

• Intrastate decay !

• Isomère de noyaux non liés ! 16F radioactif ??

• Décroissance non exponentielle !

• Test Heisenberg ?????

• D’autres surprises …

Futur

En physique nucléaire 1%

Ex(keV) Γp (keV) Jπ

0 25 ± 10 0-

193 (10) 70 ± 5 1-

424 (2) 6 ± 3 2-

721 (4) 15± 5 3-

1H(15O,p)15O

~1.4 107 pps 15O1+ SPIRAL beam

E = 1.2 MeV/A

Thick polypropylene target (CH3)

e-print at the address:

http://arxiv.ccsd.cnrs.fr/abs/nucl-ex/0603020

General context

Astrophysical motivation

Experimental measurements

Implications Conclusions

β+ (Qβ=15 MeV)

15O+2p 16F+p

0.536 MeV

p

1/2-

0-

0.732 MeV

16O+p

17Ne

2-

5/2-

1/2+ E1= 0.892 MeV

0.324 MeV

0 MeV

-0.958 MeV

-15 MeV

γ γ

γ

β+

15O(p,γ)(p,γ)17Ne

1/2-

15O(p,γ)(β+)16O

15O(p,β+)16O

Contexte général

Motivation astrophysique

Mesures expérimentales

Résultats obtenus

Conséquences Conclusions

β+ (Qβ=15 MeV)

15O+p

16F

E0= 0.536 MeV

p

1/2-

0- Γp=25 keV

16O

β+

Qβ=15 MeV

E (MeV)ECM (MeV)

σ(p,β)=(Facteur Géométrique) Γp(E)(Forme de la résonance)Γβ

22

2)(

1

rEE

résonanceladeForme )()( EPE lp

MeV2210*6.6

ECM (MeV)

15O(p,γ)(β+)16O15O(p,β+)16O

Contexte général

Motivation astrophysique

Mesures expérimentales

Résultats obtenus

Conséquences Conclusions

p

r (fm)

p

β+

E0

p

β+

γE1

Vc (

MeV

)

Lar

geur

(M

eV)

Contexte général

Motivation astrophysique

Mesures expérimentales

Résultats obtenus

Conséquences Conclusions

Théorie matrice R

Canal d’entré

15N+p Canal sortie 15N+p

Canal sortie

12 C+αRégion intérieure

16O

Paramètre de la matrice R (ra)

• r > ra les seules forces sont les forces coulombiennes;

• r < ra les forces nucléaires et coulombiennes;