ph135cS9 - Ph135c Solution set#9 1 12.2 In the iron core...

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Ph135c. Solution set #9, 6/3/10 1. 12.2 In the iron core, the number of protons is: N p = Z Z + N 1 . 4 M sun m p 7 . 7 × 10 56 Each of these is converted into a neutron and neutrino by a process which consumes an energy of Δ E m n - m p - m e 0 . 78 MeV , for a total energy of neutronization of: E n = 6 . 1 × 10 56 MeV Since each such process creates a single electron neutrino, the number of such neutrinos produced is also 7 . 7 × 10 56 . Next consider the energy released in collapse to a smaller neutron core. The change in gravitational potential energy is: Δ < V > = - 3 5 GM 2 ± 1 R i - 1 R f ² ≈ - 1 . 7 × 10 58 MeV Assuming the virial theorem holds, this corresponds to a change in total energy of: Δ < E > = Δ( < T > + < V > ) = 1 2 Δ < V > or half this amount (where we’ve used 2 < T > + < V > = 0). So we see that only about 1% of this is necessary for neutronization. Most of the rest of the energy is released as neutrinos of all types, with a typical energy of 12 MeV , so that a total of about 3 × 10 57 of them are produced. 12.3
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ph135cS9 - Ph135c Solution set#9 1 12.2 In the iron core...

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