p44_044 - 235 U nuclei is N 235 = 1000 g 235 g / mol ( 6 ....

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44. (a) We are given the energy release per fusion (calculated in § 44-7: Q =26 . 7MeV = 4 . 28 × 10 12 J) and that four protons are consumed in each fusion event. To Fnd how many sets of four protons are in the sample, we adapt Eq. 43-20: N 4 p = M sam 4 M H N A = µ 1000 g 4(1 . 0g / mol) ( 6 . 02 × 10 23 / mol ) =1 . 5 × 10 26 . Multiplying this by Q gives the total energy released: 6 . 4 × 10 14 J. It is not required that the answer be in SI units; we could have used MeV throughout (in which case the answer is 4 . 0 × 10 27 MeV). (b) The number of
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Unformatted text preview: 235 U nuclei is N 235 = 1000 g 235 g / mol ( 6 . 02 10 23 / mol ) = 2 . 56 10 24 . If all the U-235 nuclei Fssion, the energy release (using the result of Eq. 44-6) is N 235 Q fssion = ( 2 . 56 10 22 ) (200 MeV) = 5 . 1 10 26 MeV = 8 . 2 10 13 J . We see that the fusion process (with regard to a unit mass of fuel) produces a larger amount of energy (despite the fact that the Q value per event is smaller)....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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