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

Alpha Decay

Alpha Decay Energy relations S α ( A , Z ) = − Q α ( A , Z ) = B ( A , Z ) − B ( A − 4, Z − 2) − 28.3MeV Q α = T α + T d = experimental binding # & # & energy of 4 He M D + M α A % ( T α ( ≈ T α % ( % $ ' M D A − 4 $ ' +electron screening recoil term effect +bremsstrahlung http://www.nndc.bnl.gov/chart/reColor.jsp?newColor=qa

2 $ ' r P = χ III 2 T ∝ 1 ∫ 2 ∝ exp − 2 k ( r ) dr & ) & ) χ I P % ( r 1 In the case of the Coulomb barrier, the above integral can be evaluated exactly. b log T = a + Q α 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=Q α , and that the outer turning point is much greater than the potential radius.