ph135cS9 - Ph135c. Solution set #9, 6/3/10 1. 12.2 In the...

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/04/2011 for the course PHY 7070 taught by Professor Smith during the Spring '11 term at University of Wisconsin.

Page1 / 4

ph135cS9 - Ph135c. Solution set #9, 6/3/10 1. 12.2 In the...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online