CHAPTER 43 - 7 - N =(7.50 10

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  4 H 1 1 He 2 4 + 2e + + 2 ν   C 6 12 + H 1 1 N 7 13   N 7 13 C 6 13 + e +   C 6 13 + H 1 1 N 7 14   N 7 14 + H 1 1 O 8 15   O 8 15 N 7 15 + e + N   = (7.50 × 10 25  MeV/h)/[(7.30 MeV)/(4 nuclei)](2 nuclei/molecule)(0.015 × 10 –2 = 1.37 × 10 29  molecules/h. The required mass of water is m   = [(1.37 × 10 29  molecules/h)/(6.02 × 10 23  molecules/mol)](18 g/mol)  = 4.1 × 10 6  g/h =      4.1 × 10 3  kg/h . 37. The number of deuterium nuclei in 1.00 kg of water is N   = [(1.00 × 10 3  g)/(18 g/mol)](6.02 × 10 23  molecules/mol)(2 nuclei/molecule)(0.015 × 10 –2 = 1.00 × 10 22  nuclei. Two deuterium nuclei release 4.03 MeV, so the total energy is E  = (1.00 × 10 22  nuclei)(4.03 MeV)(1.60 × 10 –13  J/MeV)/(2 nuclei) =      3.23 × 10 9  J . This is       65 ×       the energy from burning 1.0 kg of gasoline. 38. (
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This note was uploaded on 09/13/2010 for the course PHYSICS 7 taught by Professor ? during the Spring '08 term at University of California, Berkeley.

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CHAPTER 43 - 7 - N =(7.50 10

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