29.4 The Decay Processes 949Alpha DecayIf a nucleus emits an alpha particle ( 4 2He), it loses two protons and two neutrons.Therefore, the neutron number N of a single nucleus decreases by 2, Z decreasesby 2, and A decreases by 4. The decay can be written symbolically asAZ X : Z2 A4 Y 4 2 He[29.8]where X is called the parent nucleus and Y is known as the daughter nucleus. Asexamples, 238 U and 226 Ra are both alpha emitters and decay according to theschemes23892 U : 23490 Th 4 2 He[29.9]and22688 Ra : 22286 Rn 4 2 He[29.10]The half-life for 238 U decay is 4.47 10 9 years, and the half-life for 226 Ra decay is1.60 10 3 years. In both cases, note that the A of the daughter nucleus is four lessthan that of the parent nucleus, while Z is reduced by two. The differences are accountedfor in the emitted alpha particle (the 4 He nucleus).The decay of 226 Ra is shown in Active Figure 29.7. When one element changesinto another, as happens in alpha decay, the process is called spontaneous decay ortransmutation. As a general rule, (1) the sum of the mass numbers A must be thesame on both sides of the equation, and (2) the sum of the atomic numbers Zmust be the same on both sides of the equation.In order for alpha emission to occur, the mass of the parent must be greaterthan the combined mass of the daughter and the alpha particle. In thedecay process, this excess mass is converted into energy of other forms andappears in the form of kinetic energy in the daughter nucleus and the alphaparticle. Most of the kinetic energy is carried away by the alpha particle becauseit is much less massive than the daughter nucleus. This can be understoodby first noting that a particle’s kinetic energy and momentum p are related asfollows:pRnBefore decayKE Rn226 Ra88222 Rn86After decayKE Ra = 0p Ra = 0KEααp αACTIVE FIGURE 29.7The alpha decay of radium-226. Theradium nucleus is initially at rest. Afterthe decay, the radon nucleus haskinetic energy KE Rn and momentump: Rn , and the alpha particle has kineticenergy KE and momentum .p: Log into <strong>Physics</strong>Now at www.cp7e.comand go to Active Figure 29.7, whereyou can observe the decay ofradium-226.KE p22mBecause momentum is conserved, the two particles emitted in the decay of anucleus at rest must have equal, but oppositely directed, momenta. As a result, thelighter particle, with the smaller mass in the denominator, has more kinetic energythan the more massive particle.Quick Quiz 29.3If a nucleus such as 226 Ra that is initially at rest undergoes alpha decay, which ofthe following statements is true? (a) The alpha particle has more kinetic energythan the daughter nucleus. (b) The daughter nucleus has more kinetic energythan the alpha particle. (c) The daughter nucleus and the alpha particle have thesame kinetic energy.Applying <strong>Physics</strong> 29.2Energy and Half-lifeIn comparing alpha decay energies from a number ofradioactive nuclides, why is it found that the half-life ofthe decay goes down as the energy of the decay goes up?Explanation It should seem reasonable that thehigher the energy of the alpha particle, the morelikely it is to escape the confines of the nucleus.The higher probability of escape translates to afaster rate of decay, which appears as a shorterhalf-life.
950 Chapter 29 Nuclear <strong>Physics</strong>EXAMPLE 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
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Problems 897The probability per uni
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Problems 1005particles fuse to prod
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APPENDIX BAn Abbreviated Table of I
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An Abbreviated Table of Isotopes A.
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Some Useful Tables A.15TABLE C.3The
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IndexPage numbers followed by “f
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Current, 568-573, 586direction of,
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Index I.5Fissionnuclear, 973-976, 9
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Polarizer, 805-806, 805f, 806-807Po
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South poleEarth’s geographic, 626
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CreditsPhotographsThis page constit
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PHYSICAL CONSTANTSQuantity Symbol V