ch43-p005 - 10 13 J. (c) If P is the power requirement of...

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N = m / m 0 = (1.0 kg)/(3.90 × 10 – 25 kg) = 2.56 × 10 24 2.6 × 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. 42-21. (b) The energy released by N fission events is given by E = NQ , where Q is the energy released in each event. For 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 8.2
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Unformatted text preview: 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. 5. (a) The mass of a single atom of 235 U is m = (235 u)(1.661 10 27 kg/u) = 3.90 10 25 kg, so the number of atoms in m = 1.0 kg is...
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This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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