CHAPTER 42 - 4 - H 1 3 He 2 3 + e 1 + n 1 p 1 1 + e 1 + C 6...

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Unformatted text preview: H 1 3 He 2 3 + e 1 + n 1 p 1 1 + e 1 + C 6 11 B 5 10 + p 1 1 Na 11 22 Mg 12 22 Na 11 22 Ne 10 22 + + + 20. The decay is . When we add an electron to both sides to use atomic masses, we see that the mass of the emitted particle is included in the atomic mass of 3 He. Thus the energy released is Q = [ M ( 3 H) M ( 3 He)] c 2 = [(3.016049 u) (3.016029 u)] c 2 (931.5 MeV/u c 2 ) = 0.0186 MeV = 18.6 keV . 21. The decay is . We take the electron mass to use the atomic mass of 1 H. The kinetic energy of the electron will be maximum if no neutrino is emitted. If we ignore the recoil of the proton, the maximum kinetic energy is K = [ m ( 1 n) M ( 1 H)] c 2 = [(1.008665 u) (1.007825 u)] c 2 (931.5 MeV/u c 2 ) = 0.782 MeV ....
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CHAPTER 42 - 4 - H 1 3 He 2 3 + e 1 + n 1 p 1 1 + e 1 + C 6...

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