NIC IX Summer School

Homework problemThe 15O( , )19Ne reaction rate is dominated by the contribution from the 4.03 MeV (Ecm=504 keV, J =3/2+) resonance in 19Ne. I argued that this rate does not compete with 15O beta decay (t1/2=122 s) in novae. Show that this is true by plotting the density as a function of temperature where the 15O( , )19Ne rate is equal to the beta decay rate. Use the narrow-resonance approximation for the 15O( , )19Ne reaction rate:

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σv ≈h2 2πμkT ⎛ ⎝ ⎜

⎞ ⎠ ⎟

3/2

ωγ( )re−Er / kT( )

The number of alpha particles/cm3, N , is given by:

where is t he density (g/cm3), A is Avogad ’ ro s number, and w isthe molecular wei ght of helium (4 /g mole). Take the mass fraction of 4 , He X , to be 0.25 €

Nα=ρXα Awα

Take the alpha-decay branching ratio of the 4.03 MeV resonance to be the upper limit (4x10-4) from Davids et al., PRC 67 (2003). The 15O ground state has J =1/2-. What is the orbital angular momentum of the captured alpha particle?

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k=8.62x10−5eV⋅K

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h=6.6x10−16eV⋅s

NIC IX Summer School

Homework solution

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σv ≈ h2 2π

μkT

⎛

⎝ ⎜

⎞

⎠ ⎟

3 / 2

ωγ( )re−Er / kT( )

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ωγ( )r =2J +1

2Jα +1( ) 2J15 +1( )

ΓαΓγΓ

=2

3

2+1

2 ⋅0 +1( ) 21

2+1

⎛

⎝ ⎜

⎞

⎠ ⎟4 ×10−4

( ) 50meV( ) = 4 ×10−5eV

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NαN15 σv = λN15

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ρXαA

wασv =

ln2

122s

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σv ≈ 6.6 ×10−16eV ⋅s( )2

2π( )3 / 2 19

60 × 0.931×109eV

⎛

⎝ ⎜

⎞

⎠ ⎟3 / 2

3×1010cm /s( )3

4 ×10−5eV( ) kT( )−3 / 2e−Er / kT( )

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σv ≈ 4.64 ×10−17 kT( )−3 / 2e−Er / kT( )cm3 /s (with kT in eV)

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ρ 0.25( )6 ×1023

4g4.64 ×10−17 kT( )

−3 / 2e−Er / kT( )cm3 /s = 0.0057 /s

/ see slide 32 in lecture notes

γ~ see slide 33 in lecture notes

NIC IX Summer School

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ρ =3.3×10−9 kT( )3 / 2e504000 / kT( )g /cm3 With kT in eV

Temperature (109 K)

Den

sity

(g/

cm3)

nova

e

T (109 K) ρ (g/cm3)

0.1 6.5x1022

0.2 3.7x1010

0.5 3500

1.0 2915 O(,γ)1

9 Ne

-decay