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This is from The Nuclear Weapons Archive:2.1.4.1.2 Gun AssemblyAssembling a critical mass by firing one piece of fissionable material at another is an obvious idea and was the first approach developed for designing atomic bombs. But it is probably not obvious how you take two subcritical masses and obtain the equivalent of three critical masses by bringing ...


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I suppose that what is meant is the following: We can consider the neutron and proton as 2 states of the same particle, the nucleon N (regarding the strong interaction, not electromagnetism). Since neutrons and protons are fermions, the wave function of 2 identical particles (here 2 nucleons) must be anti-symetric because of Pauli principle. If the 2 ...


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You'd need to use lead-204, which has a natural abundance of 1.4%. If you want to use one of the more common isotopes of lead, you'll have a few extra neutrons. You could let them fly off and then decay into hydrogen.


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If a nuclear bomb explodes very near the ground, the fallout is increased, as the bomb irradiates the surrounding dirt, however, the blast radius is decreased. On the other hand, if the bomb is detonated a reasonable distance above ground, then the fallout is greatly decreased, and the blast radius is increased. A bonus is the "double shockwave" AKA "Mach ...


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Electron capture: $p+e^-\rightarrow n+\nu_e$ Beta plus decay: $p\rightarrow n+e^++\bar{\nu_e}$ Let's check the masses of both sides of the processes: Electron capture, initial state: $m_p+m_e=938.78 \frac{MeV}{c^2}$ Final state: $m_n=939.56 \frac{MeV}{c^2}$ The difference: (Final minus initial) $0.78 \frac{MeV}{c^2}$ Beta plus decay: (Positron emission) ...


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if it's a singlet then its spin is 0. the SU(3) rep 3x3x3x3x3' (dim=243) decomposes into a bunch of 1,8,10,27, and 35 multiplets with many possible spins (mathematically speaking at least)


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The typical way to handle such things is the "sudden approximation". The time scale of the decay/capture process is assumed to be much smaller than the time scale of the evolution of the electron shell. The probabilities of the new states will then just be the projections of the old state to the new stationary states. (The typical analytically solvable ...


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I can't answer your question per se, but I hope I can offer something that's of some use: See this report: "The LHCb team is confident that the particles are indeed pentaquarks that comprise two up quarks, one down quark, one charm quark and one anticharm quark. "Benefitting from the large data set provided by the LHC, and the excellent precision of our ...


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One example would be 'Nuclear Magnetic Resonance of F$^{20}$ by Polarized Neutron Capture and $\beta$-Decay Anisotropy, TUng Tsang and Donald Connor, Physical Review 132(3) 1141-1146 (1963). Briefly, they irradiated fluorine compounds with polarized neutrons to make the F$^{20}$ nuclei. The beta decay under NMR conditions (magnetic field + RF) then shows ...


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The force between nucleons is rather complicated, so let's consider the simpler example of assembling a hydrogen atom from a proton and an electron. We start with the proton and electron at rest and a long way apart. The kinetic energy is zero (because they're at rest) and the potential energy is given by: $$ V = -k\frac{Q_1Q_2}{r^2} \tag{1} $$ where ...


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The answers by Alex and userTLK are correct but incomplete. It is true that whilst the strong force essentially only acts between nearest neighbours, whilst coulomb repulsion acts between all protons, it is actually the weak force that prevents the building of extremely large nuclei. For example, one must explain why you can't build nuclei with more and ...


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The reason is because the strong force isn't cumulative but the electromagnetic force is. Now, the strong force is a bit more complicated as it does change based on the number of protons and neutrons, but it doesn't build continuously as more protons or neutrons are bound to the nucleus, but the electromagnetic force does. Say you have a helium atom, 2 ...


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Although there are more neutrons than protons to counteract the electrostatic repulsion, there still is proton repulsion. This repulsion grows with larger and larger atoms. By emitting alpha radiation or helium nuclei, an atom can transition from a high energy state to a lower energy state. This is why it is favored as a decay. The more protons the more ...


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In your scheme, other processes, such as Coulomb scattering, are much more probable than fusion, so you will have net energy loss, as dmckee noted.


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Walecka's graduate level text Theoretical Nuclear and Subnuclear Physics focuses on developing the theory in detail. He follows the historical development most of the time, but skips over big chucks that don't lead in the direction he is going. That direction is actually about laying the foundation for being able to do JLAB-style transition-regime physics ...


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Have you tried Samuel Wong's book "Introductory Nuclear Physics"? I used in for both undergraduate and graduate nuclear physics and found it very useful. It seemed to have less handwaving than many other traditional texts.


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Would a photon be released due to changes in the electron orbital energy in this case or would it be transparent or absorbed elsewhere in other state reconfiguration? If something causes a vacancy in an electron inner shell, there can be a photon emitted. Two processes related to nuclear decay in which this often happens are electron capture and ...


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All parties are to be congratulated for asking such 'key' question and exploring answers regarding 'conventionally' calculated 'binding energy' of H-3 and He-3 and the stability of each! I suggest (like Sears & Zemansky illustrated in their standard, much used old 'College Physics' textbook) that H-1, the neutral Bohr Atom mass, and specifically its ...


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This would certainly result in a change in energy levels if the nucleus changes charge; in addition, you would expect a electron ejected as well to remain overall neutral (though there are some stable ions). The simplest quantum model that would give you an idea of the energy changes for an atom are described by so-called "Hydrogen-Like" atoms. The ...



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