p44_049 - 10 58 . If Q is the energy released in each event...

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49. Since the mass of a helium atom is (4 . 00 u)(1 . 661 × 10 27 kg / u) = 6 . 64 × 10 27 kg, the number of helium nuclei originally in the star is (4 . 6 × 10 32 kg) / (6 . 64 × 10 27 kg) = 6 . 92 × 10 58 . Since each fusion event requires three helium nuclei, the number of fusion events that can take place is N =6 . 92 × 10 58 / 3= 2 . 31
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Unformatted text preview: 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 58 )(7 . 27 10 6 eV)(1 . 60 10 19 J / eV) 5 . 3 10 30 W = 5 . 07 10 15 s = 1 . 6 10 8 y ....
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