1273_Physics ProblemsTechnical Physics

1273_Physics ProblemsTechnical Physics - 614 Applications...

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614 Applications of Nuclear Physics (d) E n the number of C atoms in 1.00 kg 4.20 eV E n = × F H G I K J ×× = −− 602 10 12 4 20 10 4 44 10 9 36 26 62 0 . .. . MeV kWh ej e j (e) Coal is cheap at this moment in human history. We hope that safety and waste disposal problems can be solved so that nuclear energy can be affordable before scarcity drives up the price of fossil fuels. P45.56 Add two electrons to both sides of the given reaction. Then 4 1 1 H atom He atom 2 4 →+ Q where Qm c == = a f bg b g 2 4 1 007 825 4 002603 26 7 . u 931.5 MeV u MeV or Q = × 26 7 1 60 10 4 28 10 13 12 . MeV J MeV J af . The proton fusion rate is then rate power output energy per proton Js J 4 protons protons s × × 377 10 428 10 353 10 26 12 38 . . . . P45.57 (a) QMMMM c IA B C E =+ 2 , and QM M M M c II C D F G 2 QQ Q M M M M M M M M c net I II A B C E C D F G =
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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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