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There are quite a few different versions of the Born potential. Most popular ones include the Bonn-A, B, C (with different strength for tensor force) and the Charge dependent Bonn (CD-Bonn) potential. CD-Bonn 2000 is the most recent one. None of the three potentials you mentioned are chiral potential. They are based on meson exchange potential.

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Assuming that by escape velocity you mean exhaust velocity, then the velocity comes from the Maxwell-Boltzmann distribution. This gives the velocity distribution of the particles in a gas as a function of temperature. For our purposes we can use the most probably speed, i.e. the peak in the distribution, as a rough estimate and this is given by: $$V_e = ... 0 Can we use heating to separate electrons from their nucleus? With the following 'conversion factor' between temperature and energy:$$ {1 \over k_{\text{B}}} \approx 11600 \, \text{K/eV} $$you'll see that 1'000'000 Kelvin corresponds an average energy of about 86 eV, much more than enough to fully ionize (separate the electron from the nucleus) e.g. ... 0 To produce electrons one simply heats up a piece of metal, and they come boiling off. If you want a beam of electrons, you just set up a positively charged plate nearby, to attract them over, and poke a small hole in it; the electrons that make it through the hole constitute the beam. Such an electron gun is the starting element in a television tube or an ... 3 By a high temperature we just mean that the particles in our gas are moving rapidly. The velocity of the particles is related to the temperature by the Maxwell-Boltzmann distribution (though note this only applies to temperatures where the velocities are non-relativistic). Anyhow, once the velocities of the atoms are high enough that the collision energy is ... 5 Heavy water is an effective moderator for the production of ^{239}Pu, which is a possible active ingredient of a fission bomb. The heavy water itself is not used in the final weapon. The allies realized that this might work - and they decided to set back any attempt by the Nazis to create a plutonium based atom bomb by depriving them of the moderator ... 2 I am not an expert on nuclear technology or weapons but the Wikipedia page on the plant and it's destruction provides some clues. Ultimately there were many potential methods that might be used to design a weapon. It was known at the start of the war that bombarding Uranium with neutrons resulted in nuclear fission which could be chained together. Heavy ... 2 Russia's BN-800 is an 880 MW IFR that can burn nuclear waste. It is operational now and ready for commercialization. Here's what's it's brochure says: BN-800 Power Unit is designed primarily for the production of heat and energy. The Power Unit as part of the grid operates with constant rated load (basic mode). However, BN-800 characteristics and ... 2 Because stars are not confined. As @MariusMatutiae says the fusion in a star is maintained at equilibrium by the thermostat of pressure versus gravity. An even more apt appliance for analogy is a nuclear power plant. In nuclear fission power, control rods or other mechanisms adjust concentration so as to prevent explosion. The fissionable material is ... 3 Yes, it's possible and it's been done, in the form of Mixed Oxides (the mix being plutonium and uranium). Until it went through prolonged shutdowns due to huge technical, safety and economic setbacks (it may have some use up to 2018, if these issues can be resolved), Thorp (THermal Oxide Reprocessing Plant) has been on of several plants that take spent ... 2 Transuranic elements are typically unstable with respect to both alpha decay and spontaneous fission. The log of the half-life for alpha decay varies approximately linearly with E^{-1/2}, where E is the energy of the alpha. If you know the binding energies of the parent and daughter nuclei, then you can calculate E from conservation of energy. ... 3 I agree to many points mentioned in the previous post but the answer is: "Yes, the weight(mass) would change." My reasons are simple. You state that an unbreakable container is being used, so that the products of the explosion will be contained inside. However unbreakable this container may be it will not be able to contain all of the radiation produced in ... 9 None of these answers seems to explain correctly why the Sun differs from a nuclear bomb. The reason is that any star, including the Sun, acts as a thermostat. If the Sun were to produce more energy than it can radiate away, the energy thus freed would make it hotter; a hot gas expands, and simultaneously cools. Both factors (lower densities and lower ... 1 You make many questions in one, all of them have their own answer. Just to clarify, nuclear decay and nuclear reaction are two totally separated and different things. Radioactivity occurs naturally, spontaneously. You have to sit and wait for the nucleus to decay. A nuclear reaction is forced, is something you obtain by, for example, shooting a particle ... 0 The simplest way to think of a form factor is to interpret the meaning of the term. It is a multiplicative factor that modifies an expected area/crossection. . One then can design an experiment to measure the area seen by this scattering. This area is the crossection of the target. In classical physics when hitting a target and measuring the recoils one ... 10 Fusion, as it occurs within stars, is in fact very unlike what happens in a bomb. An "H-bomb" is actually a mixture of fission and fusion. The fission part works on a chain reaction: when a fissile nucleus absorbs a neutron, it vibrates madly and then splits into several components, in particular two or three neutrons. These extra neutrons go on breaking ... 30 The conditions at the core of the Sun are very different from those in a thermonuclear bomb. The first thermonuclear bomb used deuterium as the secondary. The Sun has to create deuterium before getting to this stage. It's the creation of deuterium that's the bottleneck in the fusion that occurs inside the Sun. Later bombs used lithium deuteride, which is ... 67 This is an answer that I made, as suggested by John Rennie, by cutting and pasting his answer and dmckee's and adding a little more material. There are four factors involved: Velocity distribution of the nuclei Small geometrical cross-section for head-on collisions of nuclei Quantum-mechanical tunneling probability For the p-p reaction, a weak-force ... 71 The bottleneck in Solar fusion is getting two hydrogen nuclei, i.e. two protons, to fuse together. Protons collide all the time in the Sun's core, but there is no bound state of two protons because there aren't any neutrons to hold them together. Protons can only fuse if one of them undergoes beta plus decay to become a neutron at the moment of the ... 8 The interior of a star is a hot ionized gas at high pressure and temperature. High temperature means high average kinetic energy per particle, so all the nuclei of the atoms are whizzing around very fast (though for relatively short distance between collisions because the gas is so dense). The thing is that they are not all whizzing around at the same ... 25 The premise that the sun has the same conditions all throughout is incorrect. For the most part the conditions (Temperature and Pressure) necessary for nuclear fusion to occur are only found within a small region in the core. For example, when hydrogen fusion occurs and creates helium, since that helium is heavier it tend to coalesce as the core. In ... 3 According to the Section 1.1 of the ENDF-6 Formats Manual, each header of the raw ENDF/B-VII.1 Incident-Neutron Data file available in the LANL Data area has the following format: ZA, AWR, LRP, LFI, NLIB, NMOD ELIS, STA, LIS, LISO, 0, NFOR AWI, EMAX, LREL, 0, NSUB, NVER TEMP, 0.0, LDRV, 0, NWD, NXC where the TEMP field denotes ... 3 For the most part, the temperature of the medium doesn't matter. Thermal energies are typically around kT=25\,\mathrm{meV}, while nuclear reactions typically have energies of a few MeV. A factor of a billion in energy is a big difference. A skim of the explanatory text in the datafile corresponding to your plots reveals no mention of temperature. If ... 3 For those of you are perhaps a little behind on the background, these cross section libraries need to be processed before they can be applied to a realistic material. This would be appropriate for situations like a simple source term with the neutron tracks propagating throughout the material. You know the temperature, and you have the library specific to ... 2 I think I can clear up most of this. Maybe someone whose qm chops are better than mine could help with the parts I'm fuzzy on. Suppose an even-even nucleus has a prolate deformation (like an American football). This is very common, and basically occurs for any nucleus whose N and Z are both far from any magic numbers. What we really mean when we say that ... 0 Some additional information about shielding with high Z materials: According to Table 3 in http://iopscience.iop.org/0022-3700/8/12/014/pdf/jbv8i12p2015.pdf, there is a strong relationship between the atomic number of a material, and the photoelectric cross section (the probability of a PE interaction). Plotting the numbers for incident gammas of 25 keV, ... 4 Rob's explanation of how we know is bang on, but I wanted to address a part of your question that might point to a basic misunderstanding. What is special relativity inside the nucleus? Everything is always relativity. Everything. Always. All those Newtonian equations like T = \frac12 m v^2 for the kinetic energy can be properly understood as ... 5 The relationship between nuclear masses and mass differences and binding energies has been confirmed by many decades of careful nuclear spectroscopy. It's possible to measure an atom's mass by purely mechanical means: you ionize the atoms, accelerate them to a known energy, and use a magnetic field to measure their momentum. This lets you come up with an ... 1 The static magnetic moment of Li-6$$\mu_{6Li} = 0.822 \mu_N comes from its nuclear spin $I^\pi = 1^+$, with positive parity $\pi$, so in the ground state of Li-6, only even values of $l = 0, 2, ..$ would be allowed, neglecting the paired $2p$ plus $2n$ in the $s_{1/2}$-state core with net $I=0$. The nuclear spin then comes from $L$-$S$ coupling of ...

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Here's the Feynman response I read in his opening paragraphs in his Feynman lectures: The reason a proton and electron simply don't crash into each other is that if they did, we would exactly know their position-assuming one of them is stable, which one is (the proton). If we knew their position, we would be highly unaware of the momentum, meaning it could ...

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