580 Nuclear StructureP44.34(a)A gamma ray has zero charge and it contains no protons or neutrons. So for a gamma rayZ=0 and A=0. Keeping the total values of Zand Afor the system conserved then requiresZ=28 and A=65 for X. With this atomic number it must be nickel, and the nucleus must bein an exited state, so it is 2865Ni*.(b)α=24He has Z=2andA=4so for Xwe requireZ=−=84 282for PbandA=−=215 4211, X=82211Pb.(c)A positron ee+=10has charge the same as a nucleus with Z=1. A neutrino 00νhas no charge.Neither contains any protons or neutrons. So Xmust have by conservation Z=+=26 127. Itis Co. And A=+=55 055. It is 2755Co.Similar reasoning about balancing the sums of Zand Aacross the reaction reveals:(d)−10e(e)11H (or p). Note that this process is a nuclear reaction, rather than radioactive decay. We cansolve it from the same principles, which are fundamentally conservation of charge andconservation of baryon number.P44.35NC=FHGIKJ×00210602 1023..g12.0 g molmolecules molejNC=×105 1021.carbon atomsof which 1 in 7 70 10
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This note was uploaded on 12/15/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .