Quantum Physics

29.6 Nuclear Reactions 957EXAMPLE 29.9 Synthetic ElementsGoal Construct equations for a series of radioactive decays.Problem (a) A beam of neutrons is directed at a target of . The reaction products are a gamma ray and anotherisotope. What is the isotope? (b) The isotope 92 U is radioactive and undergoes beta decay. Write the equation239symbolizing this decay and identify the resulting isotope.23892 UStrategyBalance the mass numbers and atomic numbers on both sides of the equations.Solution(a) Identify the isotope produced by the reaction of a238neutron with a target of 92 U, with production of agamma ray.Write an equation for the reaction in terms of theunknown isotope:Write and solve equations for the atomic mass andatomic number:Identify the isotope:10 n 23892 U : AZ X A 1 238 239; Z 0 92 92AZ X 23992 U, iden-(b) Write the equation for the beta decay oftifying the resulting isotope.23992 U23992 Uby beta emis-Write an equation for the decay ofsion in terms of the unknown isotope:23992U :AZ Y e Write and solve equations for the atomic mass andcharge conservation (the electron counts as 1 on theright):A 239; 92 Z 1 : Z 93Identify the isotope: Z Y 239 93 Np (neptunium)Remarks The interesting feature of these reactions is the fact that uranium is the element with the greatest numberof protons (92) which exists in nature in any appreciable amount. The reactions in parts (a) and (b) do occur occasionallyin nature; hence, minute traces of neptunium and plutonium are present. In 1940, however, researchersbombarded uranium with neutrons to produce plutonium and neptunium. These two elements were the first elementsmade in the laboratory. Since then, the list of synthetic elements has been extended to include those up toatomic number 112. Recently, elements 113 and 115 have been observed, but as of this writing, their existence hasnot yet been confirmed.Exercise 29.9The isotope23893 UAnswer23994 Puis also radioactive and decays by beta emission. What is the end product?AQ ValuesWe have just examined some nuclear reactions for which mass numbers andatomic numbers must be balanced in the equations. We will now consider the energyinvolved in these reactions, because energy is another important quantity thatmust be conserved.We illustrate this procedure by analyzing the following nuclear reaction:21 H 14 7 N : 12 6 C 4 2He[29.22]