SLIDE 1
Alpha decay Alpha Decay Alpha Decay Energy relations S ( A , Z ) = - - PowerPoint PPT Presentation
Alpha decay Alpha Decay Alpha Decay Energy relations S ( A , Z ) = - - PowerPoint PPT Presentation
Alpha decay Alpha Decay Alpha Decay Energy relations S ( A , Z ) = Q ( A , Z ) = B ( A , Z ) B ( A 4, Z 2) 28.3MeV Q = T + T d = experimental binding # & # & energy of 4 He M D + M A % ( T ( T
SLIDE 2
SLIDE 3
Alpha Decay
Energy relations
experimental binding energy of 4He
Sα(A, Z) = −Qα(A, Z) = B(A, Z)− B(A− 4, Z −2)−28.3MeV
Qα = Tα + Td = Tα MD + Mα MD # $ % % & ' ( ( ≈ Tα A A − 4 # $ % & ' (
recoil term effect
http://www.nndc.bnl.gov/chart/reColor.jsp?newColor=qa +electron screening +bremsstrahlung
SLIDE 4
Theory of Alpha decay: Gamow 1928
At t=0, alpha particle is localized inside the nucleus. It can be represented by a wave packet. At large times, the wave function is an outgoing wave. Coulomb potential Attractive nuclear potential
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Two potential approach to tunneling
(decay width and shift of an isolated quasistationary state)
- Phys. Rev. A 38, 1747 (1988); Phys. Rev. A69, 042705 (2004)
V r
( ) = U r ( ) + W r ( )
- pen
closed scattering
˜ W = W + V0
Fermi’s golden rule!
SLIDE 6
SLIDE 7
P = χ III
2
χ I
2 ∝exp −2
k(r)dr
r
1
r
2
∫
$ % & & ' ( ) )
In the case of the Coulomb barrier, the above integral can be evaluated exactly.
logT = a + b Qα
Geiger-Nuttall law of alpha decay 1911
For the Coulomb barrier above, derive the Geiger-Nuttal law. Assume that the energy of an alpha particle is E=Qα, and that the outer turning point is much greater than the potential radius.
T ∝ 1 P
SLIDE 8
10-6 10-3 100 103 106 109
0,34 0,36 0,38 0,40 0,42 0,44 0,46
0,1 1 10 Hg Pt Os W Hf Yb Ra
Po
Pb Rn
T
1/2 [sec]
186Po 190Po 186Po 188Po
g.s.->g.s. decays
186-208Po
T1/2(exp)/T1/2 (GN)
(a) (b) Qa
- 1/2
- Phys. Lett. B 734 203 (2014)