1. PHL424: Semi-classical reaction theory

2. Semi-classical reaction theory
elastic scattering b

3. Semi-classical reaction theory
elastic scattering b

4. Semi-classical reaction theory
elastic scattering b

5. Semi-classical reaction theory
elastic scattering b
nucleon transfer

6. Semi-classical reaction theory
elastic scattering
b≈0
nucleon transfer

7. Semi-classical reaction theory
elastic scattering
b≈0
nucleon transfer

8. Semi-classical reaction theory
elastic scattering
b≈0
compound nucleus
formation
nucleon transfer

9. Nuclear reaction cross sectiond
Consider a beam of projectiles of intensity Φa particles/sec which hits a thin foil of target nuclei with the result that the beam is attenuated by reactions in the foil such that the transmitted intensity is Φ particles/sec.
The fraction of the incident particles disappear from the beam, i.e. react, in passing through the foil is given by
The number of reactions that are occurring is the difference between the initial and transmitted flux A va
𝑑Φ=−Φ∙ 𝑛 𝑏 ∙𝜎∙𝑑𝑥
Φ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − Φ 𝑡𝑟𝑎𝑛𝑠 = Φ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 1−𝑒𝑥𝑝 − 𝑛 𝑏 ∙𝑑∙𝜎
≈ Φ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 ∙ 𝑁 𝑏 ∙𝜎
(for thin target)
Example:
Φa = na·va
Nb = nb·A·d
A particle current of 1 pnA consists of 6·109 projectiles/s.
A 132Sn target (1 mg/cm2) consists of 5·1018 nuclei/cm2
6∙ ∙ 10 −3 𝑔 𝑐𝑚 𝑔 =4.5∙ 𝑡𝑎𝑟𝑔𝑒𝑡 𝑛𝑢𝑐𝑙𝑒𝑖 𝑐𝑚 2
Luminosity = projectiles [s-1] · target nuclei [cm-2]
Luminosity (projectile → 132Sn) = 3·1028 [s-1cm-2]
Reaction rate [s-1] = luminosity · cross section [cm2]
= projectiles [s-1] · target nuclei [cm-2] · cross section [cm2]

10. Scattering parametersD
impact parameter:
𝑏=𝑎∙𝑐𝑜𝑡 𝜃 𝑐𝑚 2
θcm
half distance of closest approach
in a head-on collision (θcm=1800):
distance of closest approach:
𝑎= 0.72∙ 𝑍 1 𝑍 2 𝑇 𝑙𝑎𝑏 ∙ 𝐴 1 + 𝐴 2 𝐴 𝑓𝑚
𝐷=𝑎∙ 𝑠𝑖𝑛 −1 𝜃 𝑐𝑚 2 +1 b D
asymptotic wave number:
𝑘 ∞ =0.219∙ 𝐴 2 𝐴 1 + 𝐴 2 ∙ 𝐴 1 ∙ 𝑇 𝑙𝑎𝑏 𝑓𝑚 −1
orbital angular momentum:
ℓ= 𝑘 ∞ ∙𝑏=𝜂∙𝑐𝑜𝑡 𝜃 𝑐𝑚 2
Sommerfeld parameter:
𝜂= 𝑘 ∞ ∙𝑎=0.157∙ 𝑍 1 𝑍 2 ∙ 𝐴 1 𝑇 𝑙𝑎𝑏
center of mass system

11. Scattering theory 2πb db 2π sinθ dθ 𝑑𝜎 𝑑Ω = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2
ring area:
2πb db
solid angle:
2π sinθ dθ
𝑑𝜎 𝑑Ω = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2

12. Scattering theory 𝑑𝜎 𝑑ℓ = 2𝜋 𝑘 ∞ 2 ∙ℓ
angular momentum and scattering angle:
𝑘 ∞ = 2∙𝑚∙𝑇 ℏ
ℓ=𝑏∙𝑝=𝑏∙ 2∙𝑚∙𝑇
ℓ=𝜂∙𝑐𝑜𝑡 𝜃 2
𝜂= 𝑘 ∞ ∙𝑎
𝑑𝜎 𝑑ℓ = 𝑑𝜎 𝑑Ω ∙ 𝑑Ω 𝑑ℓ = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2 ∙ 𝑑Ω 𝑑ℓ
𝑑Ω 𝑑ℓ =2𝜋∙𝑠𝑖𝑛𝜃∙ 𝑑𝜃 𝑑ℓ
𝑑ℓ 𝑑𝜃 = 𝜂 2 ∙ 𝑠𝑖𝑛 −2 𝜃 2
𝑑Ω 𝑑ℓ =2𝜋∙2∙𝑐𝑜𝑠 𝜃 2 ∙𝑠𝑖𝑛 𝜃 2 ∙ 2∙ 𝑠𝑖𝑛 2 𝜃 2 𝜂
𝑐𝑜𝑠 𝜃 2 = ℓ 𝜂 ∙𝑠𝑖𝑛 𝜃 2
𝑑𝜎 𝑑ℓ = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2 ∙ 8𝜋∙ℓ 𝜂 2 ∙ 𝑠𝑖𝑛 4 𝜃 2
𝑑𝜎 𝑑ℓ = 2𝜋 𝑘 ∞ 2 ∙ℓ

13. Scattering theory 𝑑𝜎 𝑑𝐷 = 𝑑𝜎 𝑑Ω ∙ 𝑑Ω 𝑑𝐷 = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2 ∙ 𝑑Ω 𝑑𝐷
distance of closest approach and scattering angle:
𝐷=𝑎∙ 𝑠𝑖𝑛 −1 𝜃 2 +1
𝑠𝑖𝑛 𝜃 2 = 𝑎 𝐷−𝑎
𝑑𝜎 𝑑𝐷 = 𝑑𝜎 𝑑Ω ∙ 𝑑Ω 𝑑𝐷 = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2 ∙ 𝑑Ω 𝑑𝐷
𝑑𝐷 𝑑𝜃 = 𝑎 2 ∙ −𝑐𝑜𝑠 𝜃 2 𝑠𝑖𝑛 2 𝜃 2
𝑑Ω 𝑑𝐷 =2𝜋∙𝑠𝑖𝑛𝜃∙ 𝑑𝜃 𝑑𝐷
𝑑Ω 𝑑𝐷 =2𝜋∙2∙𝑐𝑜𝑠 𝜃 2 ∙𝑠𝑖𝑛 𝜃 2 ∙ 2 𝑎 ∙ 𝑠𝑖𝑛 2 𝜃 2 𝑐𝑜𝑠 𝜃 2
𝑑Ω 𝑑𝐷 = 8𝜋 𝑎 ∙ 𝑠𝑖𝑛 3 𝜃 2
𝑑𝜎 𝑑𝐷 = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2 ∙ 8𝜋 𝑎 ∙ 𝑠𝑖𝑛 3 𝜃 2 = 2𝜋∙𝑎 𝑠𝑖𝑛 𝜃 2
𝑑𝜎 𝑑𝐷 =2𝜋∙ 𝐷−𝑎

14. Summary 𝑑𝜎 𝑑Ω = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2 𝑑𝜎 𝑑ℓ = 2𝜋 𝑘 ∞ 2 ∙ℓ 𝑑𝜎 𝑑𝐷 =2𝜋∙ 𝐷−𝑎
impact parameter and scattering angle:
𝑏=𝑎∙𝑐𝑜𝑡 𝜃 2
𝑎= 𝑍 𝑝 ∙ 𝑍 𝑡 ∙ 𝑒 2 2∙ 𝐸 𝑐𝑚
𝑑𝜎 𝑑Ω = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2
angular momentum and scattering angle:
𝑘 ∞ = 2∙𝑚∙ 𝐸 𝑐𝑚 ℏ
ℓ=𝜂∙𝑐𝑜𝑡 𝜃 2
𝜂= 𝑘 ∞ ∙𝑎
𝑑𝜎 𝑑ℓ = 2𝜋 𝑘 ∞ 2 ∙ℓ
distance of closest approach and scattering angle:
𝐷=𝑎∙ 𝑠𝑖𝑛 −1 𝜃 2 +1
𝑑𝜎 𝑑𝐷 =2𝜋∙ 𝐷−𝑎

15. Summary 𝑑𝜎 𝑑Ω = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2 𝑑𝜎 𝑑ℓ = 2𝜋 𝑘 ∞ 2 ∙ℓ 𝑑𝜎 𝑑𝐷 =2𝜋∙ 𝐷−𝑎
impact parameter and scattering angle:
𝑏=𝑎∙𝑐𝑜𝑡 𝜃 2
𝑑𝜎 𝑑Ω = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2
angular momentum and scattering angle:
ℓ=𝜂∙𝑐𝑜𝑡 𝜃 2
𝑑𝜎 𝑑ℓ = 2𝜋 𝑘 ∞ 2 ∙ℓ
distance of closest approach and scattering angle:
𝐷=𝑎∙ 𝑠𝑖𝑛 −1 𝜃 2 +1
𝑑𝜎 𝑑𝐷 =2𝜋∙ 𝐷−𝑎

16. Nuclear radius ρ/ρ0 ρ/ρ0 C R 𝑅 𝑖 =1.28∙ 𝐴 𝑖 1/3 −0.76+0.8∙ 𝐴 𝑖 −1/3 𝑓𝑚
1.0
0.9
0.5
0.1 C R
nuclear radius of a homogenous charge distribution:
𝑅 𝑖 =1.28∙ 𝐴 𝑖 1/3 − ∙ 𝐴 𝑖 −1/ 𝑓𝑚
nuclear radius of a Fermi charge distribution: r
𝐶 𝑖 = 𝑅 𝑖 ∙ 1− 𝑅 𝑖 − 𝑓𝑚

17. Elastic scattering and the nuclear radius
Nuclear density distributions at the nuclear interaction radius
𝑅 𝑖𝑛𝑡 = 𝐶 𝑝 + 𝐶 𝑡 +4.49− 𝐶 𝑝 + 𝐶 𝑡 𝑓𝑚
𝐶 𝑖 = 𝑅 𝑖 ∙ 1− 𝑅 𝑖 − 𝑓𝑚
𝑅 𝑖 =1.28∙ 𝐴 𝑖 1/3 − ∙ 𝐴 𝑖 −1/ 𝑓𝑚
Nuclear interaction radius: (distance of closest approach)
θ1/4 = 600 → Rint = 13.4 [fm]
𝑅 𝑖𝑛𝑡 =𝐷=𝑎∙ 𝑠𝑖𝑛 −1 𝜃 1/
→ ℓgr = 152 [ħ]

18. Elastic scattering and nuclear reactions
impact parameter and scattering angle:
𝑏=𝑎∙𝑐𝑜𝑡 𝜃 2
𝑑𝜎 𝑑Ω = 𝑎 2 4 ∙ 𝑠𝑖𝑛 −4 𝜃 2
θ1/4 = 1320
angular momentum and scattering angle:
ℓ=𝜂∙𝑐𝑜𝑡 𝜃 2
𝑑𝜎 𝑑ℓ = 2𝜋 𝑘 ∞ 2 ∙ℓ
ℓgr = 206 ħ
distance of closest approach and scattering angle:
𝐷=𝑎∙ 𝑠𝑖𝑛 −1 𝜃 2 +1
𝑑𝜎 𝑑𝐷 =2𝜋∙ 𝐷−𝑎
Rint = 16.2 fm

19. Elastic scattering and nuclear reactions
impact parameter and scattering angle:
𝑏=𝑎∙𝑐𝑜𝑡 𝜃 2
𝜎 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = 2𝜋𝑎 2 ∙ 1−𝑐𝑜𝑠 𝜃 1/4 𝑐𝑚 −1 −0.5
θ1/4 = 1320
a = 7.73 fm
angular momentum and scattering angle:
ℓ=𝜂∙𝑐𝑜𝑡 𝜃 2
𝜎 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = 𝜋 𝑘 ∞ 2 ∙ ℓ 𝑔𝑟 ℓ 𝑔𝑟 +1
ℓgr = 206 ħ
k∞ = 59.9 fm-1
distance of closest approach and scattering angle:
𝐷=𝑎∙ 𝑠𝑖𝑛 −1 𝜃 2 +1
𝜎 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 =𝜋∙ 𝑅 𝑖𝑛𝑡 2 ∙ 1− 𝑉 𝐶 𝑅 𝑖𝑛𝑡 𝐸 𝑐𝑚
Rint = 16.2 fm
VC(Rint) = 656 MeV

20. Classification of heavy ion collisions
partial cross section vs. angular momentum