complete unified theory, Nirmalendu Das

Why Curie Particle ? Why it is Mother Particle ? Nirmalendu Das Written in the book “Introduction to Nuclear Physics, (page-231)”, (Harald enge, Massachusetts Institute of Technology, Addition – Wesey publishing Company, Tenth printing, 1981) that ----- Knowing the disintegration scheme for a given radioactive, we can, however, calculate the amount of gamma ray dose4ge in roentgens received over a given time internal at a given distance from a 1- Curie source. Consider as an example a hypothetical nucleus that emits a beta ray and a 1 Mev gamma quantum. Assume also that there is no appreciable competition from internal conversion, in other words, for each ß – disintegration, one gamma ray is emitted. At a distance of 1-meter from the source, this gives a flux of photons equal to ф = 3.7x10^10 / 4πx10^4 = 2.94x10^5 photons-cm-2-sec-1 “ According to above reference, Point -1. If we consider ф = 2.94x10^5 (actual calculated value is 2.944366447x10^5 photons, when, π = 3.141592654.) I calculated the mass of a photon, σ = 1.659619614x10^-54 gram or energy = 9.309779229x10^-22 ev. So, mass of these photons is 4.996528306x10^-49 gm or 2.741140159x10^-16 ev. Point -2. The energy of a photon is = 9.309779229x10^-22 ev. and Planck constant, h/[e] = 4.1356692x10^-15 ev-sec. Therefore, one Planck contains 4.442284933x10^6 photons. With respect to the mass of one photon one atom contains 1.00055469x10^30 photons, so, one atom is 2.263580379x10^23 times larger than Planck photons, if 4.44x10^6 photons is able to form a packet, thus, E = h (when ν = 1 Hz ) then, 2.26x10^23 packets will responsible to produce same Hz frequency. If one photon realized from a packet, then, 2.26x10^23 photons in terms of energy will 5.095319433x10^16 Hz or 210.73433 ev. Again, ratio of number of photons in atom and Avogadro numbers (NA) of photons is 1.661460905x10^6 or packets or photons. 2.944862779x10^6 photons which is 10 times of Curie photons or 10 ф. Now, 3/2 10 ф in terms of energy is 4.121767436x10^-15 ev, where 3/2 is angular quantum number. j = l ± ½, l = 1 But Planck constant is 4.1356692x10^-15 ev-sec, both the results is same as we can say. Therefore, we can consider Avogadro number of ф or NAф photon, in terms of energy is 1.773137724x10^29 photons x energy of a photons = NAф = 1.650752075x10^8 ev or 165 Mev. Now we can consider this energy for Curie as the mass of “Curie Particle or Mother Particle”. Because, Birth of sub-atomic particles from the Curie mass. 1) (√3 / √2) x 165 Mev = 202.1750138 Mev. 165 Mev is the K.E. of fission fragments (Ref: 165 ± 5 Mev) 202 Mev is the total energy of fission product ( Ref: 200 ± 6 Mev) 2) NAф / (√3 /√2) x mass of a photon = 134.78334 Mev. ( Ref: mass of neutral pion ( π^0 ) is 134.976 Mev ). 3) { [ Σm – √2 x NAф / (√3 / √2) ] x mass of a photon } √2 = 493.5628 Mev. (Ref: K = 493.68 Mev, Kaon K^± ) 4) { [ Σm – NAф / (√3 / √2) ] x mass of a photon } √3/2 = 547.617 Mev. ( Ref: η = 547.5 Mev , Eta ) 5) √3/√2 ( Σm – NAф ) mass of a photon ] = 938.6678 Mev (Ref: p = 938.2723 Mev, Proton ) 6) [ ( Σm – π √2 NAф ) ] mass of a photon / √2 = 139.974 Mev ,( Ref: 139.570 Mev, Pi, π± ) 7) But, ( Σm – NAф ) mass of a photon = 766.419125 Mev, need to mass of a new particle.’ From the above calculated results, we can say, Curie particle is there in an atom which is acting as the MOTHER PARTICLE to form other particles, when we will we will tack action to find the mass of other particles.