http://en.wikipedia.org/wiki/Internal_conversion
*Internal conversion* is a radioactive decay<http://en.wikipedia.org/wiki/Radioactive_decay>process where an excited nucleus <http://en.wikipedia.org/wiki/Atomic_nucleus> interacts electromagnetically <http://en.wikipedia.org/wiki/Electromagnetism> with an electron <http://en.wikipedia.org/wiki/Electron> in one of the lower atomic orbitals <http://en.wikipedia.org/wiki/Atomic_orbital>, causing the electron to be emitted (ejected) from the atom.[1]<http://en.wikipedia.org/wiki/Internal_conversion#cite_note-Loveland-1>Thus, in an internal conversion process, a high-energy electron is emitted from the radioactive atom, but not from a nucleon in the nucleus. Instead, the electron is ejected as a result of an interaction between the entire nucleus and an outside electron that interacts with it. For this reason, the high-speed electrons from internal conversion are not beta particles<http://en.wikipedia.org/wiki/Beta_particle>, since the latter come from beta decay<http://en.wikipedia.org/wiki/Beta_decay>, where they are newly created in the process. Since no beta decay takes place during internal conversion, the element atomic number does not change, and thus (as is the case with gamma decay<http://en.wikipedia.org/wiki/Gamma_decay>) no transmutation of one element to another is seen. However, since an electron is lost, an otherwise neutral atom becomes ionized<http://en.wikipedia.org/wiki/Ionization>. Also, no neutrino is emitted during internal conversion. Internally converted electrons do not have the characteristic energetically spread spectrum of beta particles, which results from varying amounts of decay energy being carried off by the neutrino (or antineutrino) in beta decay. Internally converted electrons, which carry a fixed fraction of the characteristic decay energy, have a well-specified discrete energy. The energy spectrum of a beta particle is thus a broad hump, extending to a maximum decay energy value, while the spectrum of internally converted electrons has a sharp peak. If these internal conversion electrons are dipole electrons and have been absorbed inside the solation, the nuclear decay energy would be transferred directly into the solation and be thermalized by resonance. On Thu, Feb 13, 2014 at 10:44 PM, Axil Axil <janap...@gmail.com> wrote: > One big limitation of gamma decay is for nuclear states of zero spin. This > is the usual case in LENR. A state of zero spin cannot transition to > another state of zero spin by emitting a photon. As discussed in chapter > this violates conservation of angular momentum. > > But there are other ways that a nucleus can adjust energy besides emitting > an electromagnetic photon. One way is by kicking an atomic electron out of > the surrounding atom. This process is called "internal conversion" > because the electron is outside the nucleus. It allows transitions between > states of zero spin. > > For atoms, two-photon emission is a common way to achieve decays between > states of zero angular momentum. However, for nuclei this process is less > important because internal conversion usually works so well. > Internal conversion is also important for other transitions. Gamma decay > is slow between states that have little difference in energy and/or a big > difference in spin. For such decays, internal conversion can provide a > faster alternative. > > Internal conversion may be where the excess elections come from in LENR > systems. Electrons could be carrying away spin in a zero spin nuclear > reaction. > > I am sure that in a complex cluster fusion/fission reaction nature will > balance the spin books correctly. > > > > On Thu, Feb 13, 2014 at 10:28 PM, Bob Cook <frobertc...@hotmail.com>wrote: > >> Ed --Bob Here- >> >> I have assumed spin--angular momentum--is conserved. Are you saying >> forget about that conventional thinking--that angular momentum is not >> conserved in the lenr new nuclear process? >> >> Bob >> >> >> -----Original Message----- From: Bob Cook >> Sent: Thursday, February 13, 2014 6:52 AM >> To: vortex-l@eskimo.com >> Subject: [Vo]:Re: a note from Dr. Stoyan Sargoytchev >> >> >> Ed--Bob here-- >> >> The protons are fermi particles with a spin of 1/2, so 2 protons would >> create a new particle spin of 1 plus the 1/2 from the electron for a total >> of +1-1/2. >> I think deuteron's are Bose particles with a spin of 0. Correct me if I >> am >> wrong. >> >> What happens to the excess spin? >> >> Bob >> -----Original Message----- >> From: Edmund Storms >> Sent: Thursday, February 13, 2014 6:30 AM >> To: vortex-l@eskimo.com >> Subject: Re: [Vo]:Re: a note from Dr. Stoyan Sargoytchev >> >> Bob, these three particles create a deuteron after all of the excess mass >> energy has been emitted as photons. The neutrino has very little energy >> because very little remains when the d forms. The creation process is >> unique >> to lenr and applies to all the isotopes of hydrogen, at least that is my >> model. if lenr is to be explained, you need to stop thinking in >> conventional >> terms. This is a new kind of nuclear process. >> >> Ed Storms >> >> Sent from my iPad >> >> On Feb 12, 2014, at 3:00 PM, "Bob Cook" <frobertc...@hotmail.com> wrote: >>> >>> Jones--Bob Cook Here-- >>> >>> Can you show how the p-e-p reaction as you understand it conserves spin? >>> >>> I would think that the newly fused particle, whatever it is, would have >>> 1/2 or 3/2 spin--I do not know. >>> >>> If a positron is emitted, its spin would be -1/2 I think. That would >>> make the new particle have 0 or 1 spin. >>> >>> The reaction of the positron and electron give photons with 0 spin. >>> >>> Bob >>> >>> >>> . >>> >>> -----Original Message----- From: Jones Beene tt >>> Sent: Wednesday, February 12, 2014 1:10 PM >>> To: vortex-l@eskimo.com >>> Subject: RE: [Vo]:a note from Dr. Stoyan Sargoytchev >>> >>> >>> >>> -----Original Message----- >>> From: mix...@bigpond.com >>> >>> The most elegant answer begins with the obvious assertion that there >>>> are no >>>> >>> gammas ab initio, which means that no reaction of the kind which your >>> theory >>> proposes can be valid because gammas are expected. >>> >>> Actually not only would I not expect to detect any gammas from a p-e-p >>> reaction, I wouldn't expect to detect any energy at all. That's because >>> the >>> energy of the p-e-p reaction is normally carried away by the neutrino, >>> which >>> is almost undetectable. >>> >>> Hi, >>> >>> Not so - the reaction produces a positron, which annihilates with an >>> electron producing 2 gammas. They net energy is over 1 MeV and easily >>> detectable. >>> >>> Jones >>> >>> >> >
