The law of the storage of zero energy for pseudo e-neutrinos around charged matter as electrons protons or neutrons. How to show this?

Take for example the H atom. This atom has an ionisation energy of 13.6 eV. As a guess the labour displacement is {13.6 / (2 x 1.34) = 5.075 N_{e} }. Twice of N_{e} at ½c effective gives N_{e} interactions. Paired pseudos can only have ½c effective which explains the factor 2. Since the electron mass is involved giving N_{e} =3.813364 10^{5} interaction at pairing of 1.34 eV.

It shows the production of 5.0725 N_{e} of pseudo interactions for the ionisation labour. The production of that many interactions is absurd. Reverse the reasoning at zero energy of the atom there are stored: (1 /5.0725)N_{e} = 7.515 10^{4} needed for ionisation which is reasonable. So what does it mean energy storage?

Take the electron as a resonance cavity surrounded by a sea of pseudo vector neutrinos then the ratio 144 /137.036 = 1.050818 can be seen as a c-cavity for the pseudo sea based on the fine structure constant of 1 /137.036 which is a ratio.

The relativistic energy of m/m_{o} = 1.050818 giving a β = 0.30722c and less than ½c eff. The number of interacting pseudos is 5.0818 10^{-2} N_{e} = 1.938 10^{4} . Then 1.34 /1.050518 is the binding energy for the pseudo cluster interacting overall to the electron. Now one discovers the zero state which is that the number of active interactions adding up to N_{e} with bound pseudos of 1.2752 eV. So the fraction of 1.938 10^{4} interactions with 0.30722 c is zero for a sea of the paired state at 1.34 eV and exchanging at ½ c effective. Secondly all of the 1.938 10^{4} should one way or the other contribute to electric charge due to the long wave interference of virtual of hν-photons of 1.275 eV in respect to the zero photon state of 1.34 eV.

Consider the proton 1836 /1728 = 1.062500 as c- cavity the 17^{th} τ-pseudo of 108 m_{e} in end cap geometry interacts to the e-pseudo sea with 1728m_{e }as the conserved internal quark cell. The actual rest mass of the proton is close to 1836.153m_{e} in which 0.153m_{e} is the fraction of pseudo labour to generate triplet states, ortho states of the e-pseudos around the c-cavity derived in a following blog. Again the factor m/m_{o} = 1.062500 or 1.062588 including the triplet formation. Then 6.2500 10^{-2} N_{e} is the fraction of active interactions for the bound state of (1.34 /1.062500) eV. The electric charge induction is again due to the fraction calculated from N_{e} equal to the electric charge of the electron.

For the neutron of 1838.684 m_{e }giving the ratio of 1.064053 to the conserved quark cell and the same calculation is valid from which in another blog the decay time for the e-neutrinos can be calculated.

For the more complex atoms the overall fraction of active interactions can be assessed by reducing the atom involving the number of neutrons and protons to a one proton state including the nucleon binding energy of the overall atomic rest mass divided by the number of nucleons e.a. 1824.234 /1728 the value for Helium with He of two protons and two neutrons and 4 x 1824.234 the atom rest mass, etc.

*Energy storage pseudo ‘vacuum’ in A4 format: *