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White dwarfs and neutron stars
White dwarfs and neutron stars

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5.2 Electron capture

The second mechanism by which energy is absorbed in the iron core of a star is that of electron capture, allowing energy to be carried away by neutrinos. The conversion of protons (in nuclei) to neutrons by electron capture is possible if the gas is sufficiently dense for degenerate electrons to have an energy greater than the 1.29 MeV mass-energy excess of neutrons relative to protons ( m sub n times c squared equals 939.56 MeV , while m sub p times c squared equals 938.27 MeV ). Actually, the energy excess required for electron capture by bound nuclear protons, rather than free protons, is usually somewhat higher and depends on the nucleus, but is nevertheless usually just a few MeV.

The general reaction may be written as

e super minus postfix plus p long right arrow n prefix plus of nu sub e

and is sometimes referred to as neutronisation.

At densities above about 1012 kg m-3, iron-56 nuclei will capture electrons in the following reaction which produces a neutron-rich manganese nucleus:

e super minus plus super 56 Fe long right arrow super 56 Mn prefix plus of nu sub e full stop

where nu sub e is a neutrino. Following this, further electron-capture reactions occur in successive nuclei, converting more and more protons into neutrons.

Because electrons are rapidly used up by these reactions, the pressure support provided by degenerate electrons quickly disappears, and the core collapses rapidly, as noted earlier. The neutrinos produced by the electron-capture reactions carry away most of the energy. Although the passage of the neutrinos is hindered by the high densities in the core, which raises the likelihood of interactions, all of the neutrinos are still able to escape within a few seconds.

Activity 9

If each neutrino produced in the reactions described above carries away 10 MeV of energy, how much energy (in joules) is removed by neutrinos if the whole core (comprising 1.4 M of iron-56) undergoes neutronisation? (The mass of an iron-56 nucleus is 56u, where u equals 1.66 multiplication 10 super negative 27 kg is the atomic mass unit.)

Comment

The number of nucleons contained in the stellar core is 1.4 times cap m sub circled dot operator solidus u where u is the atomic mass unit. Because an iron-56 nucleus contains 26 protons (and 30 neutrons), the number of protons in the core is left parenthesis 26 solidus 56 right parenthesis multiplication 1.4 times cap m sub circled dot operator solidus u . This is evaluated as

left parenthesis 26 solidus 56 right parenthesis multiplication 1.4 multiplication 1.99 multiplication 10 super 30 kg divided by 1.66 multiplication 10 super negative 27 kg almost equals 7.8 multiplication 10 super 56

Assuming charge neutrality, there will also be almost equals 7.8 multiplication 10 super 56 electrons. If every proton and electron undergoes neutronisation, this will produce almost equals 7.8 multiplication 10 super 56 neutrinos.

If each neutrino carries away 10 MeV of energy, then the amount of energy removed by each neutrino in joules is 10 multiplication 10 super six eV prefix multiplication of 1.60 multiplication 10 super negative 19 cap j eV super negative one equals 1.6 multiplication 10 super negative 12 cap j . So the total amount of energy removed is 1.6 multiplication 10 super negative 12 cap j prefix multiplication of 7.8 multiplication 10 super 56 equals 1.2 multiplication 10 super 45 cap j .

As the result of the previous activity shows, neutronisation, like nuclear photodisintegration, can also remove ~ 1045 J of energy from the collapsing stellar core within a few seconds!