Pu (~AMU=239 or 240, or 241) is heavy, in fact it is almost 60 times heavier than the alpha particle (~AMU=4), the alpha provides a thrust kinetic energy of 5 MeV ( the typical alpha energy, and assumes all the kinetic energy can be translated into thrust energy, it cannot), your "fuel" then has a kinetic energy thrust to weight ratio (cheating to make the math simple => 1 AMU = 931 MeV using E=mc^2) of 0.005. A chemical rocket engine like the Russia RD-180 has a thrust to weight ratio of about 75! Basically this is saying that you are taking a lot of useless mass with you that you have to accelerate, and you cannot get off the ground on Earth. Now, the problem is complex, but we will simplify and make many assumptions to get an simple answer. First instead of solid Pu, make it a gas (plasma really), then contain it in a (leaky) magnetic mirror ( ), the alpha particles would shoot out the leaky side.
ALPHA PARTICLE FULL
Within the bonding distance of the Coulomb Barrier, the full time negative charge of the pions will be attracted to the positive proton charges of the nucleus while the resonating proton states ofthe alpha nucleus are only present half the time, so the alpha particle is technically bound to the positive charge of the nucleus by its pion negative charges. Since this is greater for a large surface, the alpha particle stays more closely bound for large nuclei.Īctually, it is totally doable (sort of). In helium, this 2.224 MeV 1-D bond energy is geometrically compounded to BE = 3 1/d (p x 2.224 ) n = 28.3 MeV in terms of the alpha particle's d = 3 dimensions, p = 2 protons, and n = 2 neutrons, such that the pion exchanges distribute equally between the nucleons, as do their proton and neutron states. This negative charge exchange from the neutron to the proton is a light speed pion exchange resonance, with an energy of 2E n + 1/3 E n (m e + E n)/m e = 2.224 MeV, where E n = m n - m p - m e = 0.782 MeV, is the mass difference between a neutron and its proton-electron elements. More specifically, page 10 of A Nuclear-Gravitational Electrodynamic Framework (at ) shows a pion exchange Feynmann Diagram between a deuterium nucleus proton and neutron, exchanging the particles' UpUpDown and DownDownUp quark states pion exchange.
ALPHA PARTICLE PLUS
However, in large nuclei, with a large surface area, S equals the radial degree of freedom into the nucleus, plus the four degrees of freedom of the surface, and the radial degree of freedom away from the nucleus, so there are five degrees of freedom available to keep the alpha particle within the Coulomb Barrier and only one degree of freedom outside the barrier, hence a greater probability to prevent the alpha particles from escaping. In terms of Boltzmann's P = e^S/k probability principle, for small nuclei, where k is the alpha particle and S is the available into or out of degrees of freedom, it has a higher probability of moving out of the nucleus. In nuclei with larger radii, within the Coulomb Barrier, the charge attraction on the alpha particle's protons is much greater because of the influence of the larger surface area, whereas in nuclei with smaller radii, the charge attraction is only in the direction of its center. Notice the Coulomb Barrier reverses polarity from repulsion outside the pion 1/2 wave nuclear bonding distance to attraction within it. is an excellent question Professor Elwarraki.
Sternheim, Physics( John Wilwy & Sons, Inc., Singapore, 1988 ), 3 rd ed., Sec.28.7 & 30.9 El-Sayed Abdel-Gawad El-Warraki Physics Dept., Faculty of Science, DamiettaUniv., New-Damietta, EGYPT. Yudin, Nuclear Physics ( Mir Publishers, Englishtranslaton, Moscow, 1982 ), Vol. Albright, Introduction toAtomic and Nuclear Physics(CHAPMAN AND HALL Ltd., London,1973 ), 5 th ed., Sec.14-10 &14-11. Clarify why or Explain how the Coulomb potentialbarrier exist outside the nucleus inorder to prevent escaping of alpha-particles from heavy nuclei? I think the Coulomb force in this case helpα-particles to escape, not to prevent ! 1. Their calculations of disintegrationconstant λ gives qualitative agreement with the experiment, and also theexpression of λ is considered as a one form of the empirical Geiger-Nuttal law( 1911 ). Coulomb potential barrier for alpha-particle decay Question: ءً ٠٧:١٣ ، ٢٠١٤ ، ا ٢٠ أ Theradioactive disintegration of a nucleus by alpha-particle emission was firstsuccessfully explained on the basis of wave mechanics ( tunnel effect ), byG.Gamow, Edward Condon, and R.
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