Chapter 45613For a more accurate calculation of the kinetic energy, we should use relativistic expressions.Conservation of momentum givesγγαααnnnmmvv+=010087140026122..vvcvnn−=−αyieldingvcvcvnn22215 74614 746=−.Thenααnnmc−+−=11176bg. MeVandn=0171., implying that γ402eV .P45.54From Table A.3, the half-life of 32P is 14.26 d. Thus, the decay constant isλ====×−−−lnln.14 260 048 65 63 10127Tdd s.NR006712522 10563 10928 10××=×...decay ssnuclei1At t=10.0 days, the number remaining isNNeet×=×−−−01200486100129 28 105 71 10nucleinucleid1ejso the number of decays has been 012357 10−= ×.and the energy released isE×=−700160 1004001216...afkeVJ keVJ.If this energy is absorbed by 100 g of tissue, the absorbed dose isDoseJ.100 kgradJkgrad=FHGIKJFHGIKJ=−01104002..P45.55(a)The number of Pu nuclei in 602 10239 05100023...kgnuclei mol
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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 .