p44_001 - each event. ±or 1 . 0 kg of 235 U, E = (2 . 56...

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1. (a) The mass of a single atom of 235 U is (235 u)(1 . 661 × 10 27 kg / u) = 3 . 90 × 10 25 kg, so the number of atoms in 1 . 0kgis(1 . 0kg) / (3 . 90 × 10 25 kg) = 2 . 56 × 10 24 . An alternate approach (but essentially the same once the connection between the “u” unit and N A is made) would be to adapt Eq. 43-20. (b) The energy released by N Fssion events is given by E = NQ ,whe re Q
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Unformatted text preview: each event. ±or 1 . 0 kg of 235 U, E = (2 . 56 × 10 24 )(200 × 10 6 eV)(1 . 60 × 10 − 19 J / eV) = 8 . 19 × 10 13 J. (c) If P is the power requirement of the lamp, then t = E/P = (8 . 19 × 10 13 J) / (100 W) = 8 . 19 × 10 11 s = 2 . 6 × 10 4 y. The conversion factor 3 . 156 × 10 7 s / y is used to obtain the last result....
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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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