1274_Physics ProblemsTechnical Physics

# 1274_Physics ProblemsTechnical Physics - Chapter 45 P45.58...

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Chapter 45 615 P45.58 (a) The mass of the pellet is mV == × F H G I K J L N M M O Q P P ρ π 0200 4 3 150 10 353 10 2 3 7 . . . gcm cm 2 g 3 ej . The pellet consists of equal numbers of 2 H and 3 H atoms, so the average molar mass is 2.50 and the total number of atoms is N = × F H G I K J ×= × 250 6 02 10 8 51 10 7 23 16 . . .. g gmo l atoms mol atoms . When the pellet is vaporized, the plasma will consist of 2 N particles ( N nuclei and N electrons). The total energy delivered to the plasma is 1.00% of 200 kJ or 2.00 kJ. The temperature of the plasma is found from EN k T = 2 3 2 af B as T E Nk × ×× 3 200 10 8 51 10 1 38 10 568 10 3 16 23 8 B J 3 J K K . . e j . (b) Each fusion event uses 2 nuclei, so N 2 events will occur. The energy released will be E N Q = F H G I K J = × F H G I K J × = 2 851 10 2 17 59 1 60 10 1 20 10 120 16 13 5 . . MeV J MeV J kJ . P45.59 (a) The solar-core temperature of 15 MK gives particles enough kinetic energy to overcome the Coulomb-repulsion barrier to 1 1 2 3 HH e H e e 2 4 +→+ + + ν , estimated as ke e r e a f 2 . The Coulomb barrier to Bethe’s fifth and eight reactions is like r e 7 , larger by 7 2 times, so the required temperature can be estimated as 7 2 15 10 5 10 67 ×≈ × K K . (b) For 12 1 13 CH N +→ + Q , Q 1 12 000 000 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 .

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