Quantum Physics

950 Chapter 29 Nuclear PhysicsEXAMPLE 29.5 Decaying RadiumGoal Calculate the energy released during an alpha decay.22688 Ra22286 RnProblem We showed that the nucleus undergoes alpha decay to (Eq. 29.10). Calculate the amount226222of energy liberated in this decay. Take the mass of 88 Ra to be 226.025 402 u, that of 86 Rn to be 222.017 571 u, and4that of 2He to be 4.002 602 u, as found in Appendix B.Strategy This is a matter of subtracting the neutral masses of the daughter particles from the original mass of theradon nucleus.SolutionCompute the sum of the mass of the daughter particle,m d , and the mass of the alpha particle, m :Compute the loss of mass, m, during the decay by subtractingthe previous result from M p , the mass of theoriginal particle:Convert the loss of mass m to its equivalent energyin MeV:m d m 222.017 571 u 4.002 602 u 226.020 173 um M p (m d m ) 226.025 402 u 226.020 173 u 0.005 229 uE (0.005 229 u)(931.494 MeV/u) 4.871 MeVRemark The potential barrier is typically higher than this value of the energy, but quantum tunneling permits theevent to occur, anyway.Exercise 29.5Calculate the energy released when84Be splits into two alpha particles. Beryllium-8 has an atomic mass of 8.005 305 u.Answer0.094 1 MeVEXAMPLE 29.6GoalBeta DecayWhen a radioactive nucleus undergoes beta decay, the daughter nucleus has the samenumber of nucleons as the parent nucleus, but the atomic number is changed by 1:AAZX : Z1 Y e[29.11]AAZX : Z1Y e [29.12]Again, note that the nucleon number and total charge are both conserved in thesedecays. However, as we will see shortly, these processes are not described completelyby these expressions. A typical beta decay event is146 C : 14 7 N e[29.13]The emission of electrons from a nucleus is surprising, because, in all our previousdiscussions, we stated that the nucleus is composed of protons and neutronsonly. This apparent discrepancy can be explained by noting that the emitted electronis created in the nucleus by a process in which a neutron is transformed intoa proton. This process can be represented by the equation10 n : 1 1 p e[29.14]Consider the energy of the system of Equation 29.13 before and after decay.As with alpha decay, energy must be conserved in beta decay. The next example14illustrates how to calculate the amount of energy released in the beta decay of 6 C.The Beta Decay of Carbon-14Calculate the energy released in a beta decay.Problem Find the energy liberated in the beta decay of to , as represented by Equation 29.13. Thatequation refers to nuclei, while Appendix B gives the masses of neutral atoms. Adding six electrons to both sides ofEquation 29.13 yields14 146 C atom : 7 N atom146 C147 N