ch43-p045 - 45 Since the mass of a helium atom is(4.00 u(1...

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N = 6.92 × 10 58 /3 = 2.31 × 10 58 . If Q is the energy released in each event and t is the conversion time, then the power output is P = NQ / t and t NQ P == ×× × × = × 2 31 10 7 27 10 160 10 53 10 507 10 16 10 58 6 19 30 15 8 .. . . . ch c h c h eV J eV W sy 45. Since the mass of a helium atom is (4.00 u)(1.661 × 10 – 27 kg/u) = 6.64
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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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