Chapter 45615P45.58(a)The mass of the pellet is mV==×FHGIKJLNMMOQPP=×−−ρπ020043150 10353 10237...gcmcm2g3ej.The pellet consists of equal numbers of 2H and 3H atoms, so the average molar mass is 2.50and the total number of atoms isN=×FHGIKJ×=×−2506 02 108 51 1072316....ggmolatoms molatoms.When the pellet is vaporized, the plasma will consist of 2Nparticles (Nnuclei and Nelectrons). The total energy delivered to the plasma is 1.00% of 200 kJ or 2.00 kJ. Thetemperature of the plasma is found from ENkT=232afBasTENk×××−3200 108 51 101 38 10568 10316238BJ3 JKK..ej.(b)Each fusion event uses 2 nuclei, so N2events will occur. The energy released will beENQ=FHGIKJ=×FHGIKJ×=−2851 10217 591 60 101 20 1012016135..MeVJ MeVJkJ.P45.59(a)The solar-core temperature of 15 MK gives particles enough kinetic energy to overcome theCoulomb-repulsion barrier to 1123HHeHee24+→+++ν, estimated as ke erea f2. TheCoulomb barrier to Bethe’s fifth and eight reactions is like re7, larger by 72times, sothe required temperature can be estimated as 7215 105 1067×≈×K K.(b)For 12113CHN+→+Q,Q112 000 000 1 007 825 13 005 739 931 5
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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 .