Find the energy emitted in the - decay of 60 Co. Strategy and Concept As in

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Find the energy emitted in the β- decay of 60Co.


Strategy and Concept

As in the preceding example, we must first find Δm, the difference in mass between the parent nucleus and the products of the decay, using masses given in Appendix A. Then the emitted energy is calculated as before, using E = (Δm)c2. The initial mass is just that of the parent nucleus, and the final mass is that of the daughter nucleus and the electron created in the decay. The neutrino is massless, or nearly so. However, since the masses given in Appendix A are for neutral atoms, the daughter nucleus has one more electron than the parent, and so the extra electron mass that corresponds to the β- is included in the atomic mass of Ni. Thus,

m = m(60Co)=m(60Ni ). 31.24

Data from Appendix A

Atomic Masses Atomic Number, Z 1 2 3 Name neutron Hydrogen Deuterium Tritium Helium Lithium Atomic Massst 4 5 6 Beryllium Boron Carbon 7 7 9 10 11 11 12 Li 7 Be 9 Be 10B 11B 11C 12C 7.016 003 7.016 928 9.012 182

7 8 9 Nitrogen Oxygen Fluorine 13 14 13. 14 15. 15 16 18 18 13 C 14C 13 N 14N 15 N 150 160 180 18 F 13.003

10 11 Neon Sodium 19 20 22 22 23 24 19 F 20 Ne 22 Ne 22 Na 23 Na 24 Na 18.998 403 19.992 435 21.991 383

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