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Unformatted text preview: 10. First, we ﬁgure out the mass of U235 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 435; 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 U235 nuclei ﬁssion, the energy release (using the result of Eq. 446) 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 100W
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 441, 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.
 Fall '08
 SPRUNGER
 Physics, Mass

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