p44_010

# p44_010 - 10. First, we ﬁgure out the mass of U-235 in...

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Unformatted text preview: 10. First, we ﬁgure out the mass of U-235 in the sample (assuming “3.0%” refers to the proportion by weight as opposed to proportion by number of atoms): MU−235 = = (97%)m238 + (3.0%)m235 (97%)m238 + (3.0%)m235 + 2m16 0.97(238) + 0.030(235) (0.030)(1000 g) 0.97(238) + 0.030(235) + 2(16.0) (3.0%)Msam = 26.4 g . Next, this uses some of the ideas illustrated in Sample Problem 43-5; our notation is similar to that used in that example. The number of 235 U nuclei is N235 = (26.4 g)(6.02 × 1023 /mol) = 6.77 × 1022 . 235 g/mol If all the U-235 nuclei ﬁssion, the energy release (using the result of Eq. 44-6) is N235 Qﬁssion = 6.77 × 1022 (200 MeV) = 1.35 × 1025 MeV = 2.17 × 1012 J . Keeping in mind that a Watt is a Joule per second, the time that this much energy can keep a 100-W lamp burning is found to be t= 2.17 × 1012 J = 2.17 × 1010 s ≈ 690 y . 100 W If we had instead used the Q = 208 MeV value from Sample Problem 44-1, then our result would have been 715 y, which perhaps suggests that our result is meaningful to just one signiﬁcant ﬁgure (“roughly 700 years”). ...
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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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