Question

Consider the alpha particle decay 230 90 Th →226 88 Ra+α anduse the following expression to calculate the values of the binding energy B for the two heavy nuclei involved in this process
B(Z,A) = aVA−aSA2/3 −aCZ2A−1/3 −aAZ − A22A−1+ (−1)Z +(−1)N 2
aV = 15.85MeV/c2,aS = 18.34MeV/c2,aC = 0.71MeV/c2
aA = 92.8MeV/c2,aP = 11.46 MeV/c2. apA−1/2
Given that the total binding energy of the alpha particle is 28.3MeV, find the energy Q released in the decay.
(b) This energy appears as the kinetic energy of the products of the decay. If the original thorium nucleus was at rest, use conservation of momentum and conservation of energy to nd the kinetic energy of the daughter nucleus 226Ra.

Answer 1