An isolated neutron decays into a proton, electron, and antineutrino, with Q = 782 keV energy shared among the three particles. However, a bound neutron can be stable, as in a nucleus or a neutron star. In a nucleus, there is a balance between neutrons and protons, and along a given isobar (equal mass number), there is a minimum energy nucleus. Nuclei with excess neutrons undergo beta decay by emitting electrons, those with excess protons capture an orbiting electron or emit a positron. Even mass isobars are a bit more complicated since even-even nuclei are more stable than odd-odd, and there can be local energy minima. Three minima occur for mass 124: tin, tellurium, and xenon. Even-even nuclei trapped in such a local minimum can only decay by double beta decay (or alpha decay, if allowed), with half-life > 10^19 years. The atomic electron cloud can affect nuclear stability. Fully ionized dysprosium-163 nuclei decay by electron emission to holmium-163 with a 48-day half life. But neutral Ho-163 atoms capture an electron forming Dy-163 with a 4570-year half-life. In a neutron star, neutrons, protons, and electrons are in equilibrium. The simplest model takes these particles as free and at zero temperature. Being fermions, they occupy the lowest allowed energy states from zero up to the Fermi energy. The Fermi energy εF is related to the number density n by εF = (1/2)(3π2)2/3 ℏ2 n2/3 / m (non-relativistic), εF = (3π2)1/3 ℏ c n1/3 (ultra-relativistic). The equilibrium is controlled by the neutron and electron Fermi energies. If we remove a neutron from its top occupied state, we remove energy mnc2 + εFn If we insert a proton and electron at their top occupied states, we must provide energy mpc2 + mec2 + εFp + εFe These must be equal at equilibrium. (mn - mp - me)c2 = Q = εFe + εFp - εFn Suppose the neutron density is nn = 1045 m-3. Then from the non-relativistic formula, the neutron Fermi energy is εFn = 3.2×10-11 J = 200 MeV. This is reasonably non-relativistic since the neutron mass is 930 MeV. This swamps out the proton Fermi energy, the electron mass, and Q, leaving εFe = εFn and from the ultra-relativistic formula, the electron density is ne = 0.04×1045 m-3 So the electron density is 4% of the neutron density, and by electrical neutrality the same holds for the proton density. Note that the electron density scales as the square of the neutron density, as long as the neutrons remain non-relativistic. So if we assume instead a neutron density of 1044 m-3, the electron density will be 0.4% of the neutron density. Also note that the Fermi energy of 200 MeV corresponds to a temperature of 2 trillion K, so the star's thermodynamic temperature of a billion K can be ignored. -- Gene
________________________________ From: meekerdb <meekerdb@verizon.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Wednesday, June 25, 2014 1:49 PM Subject: Re: [math-fun] Neutron stars as atoms
I would suppose that a neutron star would continually have neutrons undergo beta decay, producing a proton and electron which are captured by gravity and a neutrino which isn't.
Brent Meeker