Alpha Decay Alpha Decay Energy relations S ( A , Z ) = Q ( A , Z - - PowerPoint PPT Presentation

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Jan 15, 2023 267 likes •317 views

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 % ( % $ ' M

SLIDE 1

Alpha Decay

SLIDE 2

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 3

SLIDE 4

P = χ III

χ I

2 ∝exp −2

k(r)dr

∫

$ % & & ' ( ) )

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.

Alpha Decay

Alpha Decay

Energy relations

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 # $ % & ' (

http://www.nndc.bnl.gov/chart/reColor.jsp?newColor=qa +electron screening +bremsstrahlung

P = χ III

χ I

k(r)dr

∫

$ % & & ' ( ) )

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