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1274_Physics ProblemsTechnical Physics

# 1274_Physics ProblemsTechnical Physics - Chapter 45...

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Chapter 45 615 P45.58 (a) The mass of the pellet is m V = = × F H G I K J L N M M O Q P P = × ρ π 0 200 4 3 1 50 10 3 53 10 2 3 7 . . . g cm cm 2 g 3 e j . 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 × = × 3 53 10 2 50 6 02 10 8 51 10 7 23 16 . . . . g g mol atoms mol atoms e j . 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 E N k T = 2 3 2 a f e j B as T E Nk = = × × × = × 3 2 00 10 8 51 10 1 38 10 5 68 10 3 16 23 8 B J 3 J K K . . . . e je 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 8 51 10 2 17 59 1 60 10 1 20 10 120 16 13 5 . . . . MeV J MeV J kJ a f e j . 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 H He He e 2 4 + + + + ν , estimated as k e e r e a fa f 2 . The Coulomb barrier to Bethe’s fifth and eight reactions is like k e e r e a fa f 7 , larger by 7 2 times, so the required temperature can be estimated as 7 2 15 10 5 10 6 7 × × K K e j . (b) For 12 1 13 C H N + + Q , Q 1 12 000 000 1 007 825 13 005 739
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