A ²²⁸ Th nucleus emits an alpha particle, reducing to ²²⁴ Ra . Calculate the kinetic energy of the alpha particle emitted during the following decay: ²²⁸ Th ²²⁴ Ra ^ * + ²²⁴ Ra ^ * ²²⁴ Ra + ;(217 keV ) . The atomic mass of ²²⁸ Th is 228.028726 u , that of ²²⁴ Ra is 224.020196 u , and that of ⁴₂ He is 4.00260 u .

Step-by-step solution

The mass defect for the decay to the ground state is given by: m = m(²²⁸ Th ) - m(²²⁴ Ra ) - m(⁴₂ He ) m = 228.028726 - 224.020196 - 4.00260 = 0.005930 u The total energy released (Q-value) in the decay to the ground state is: Q = m 931 MeV/u Q = 0.005930 931 5.5208 MeV Since the decay leaves the ²²⁴ Ra nucleus in an excited state which subsequently emits a 217 keV = 0.217 MeV gamma ray, the energy available for the kinetic energy of the alpha particle and the recoiling nucleus is: Q' = Q - E_ Q' = 5.5208 MeV - 0.2