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Nucleons get close to each other to form nuclei so this blanket statement of repulsion needs qualification and I am puzzled why the Pauli exclusion principle is involved. Protons will repulse protons at one fermi due to the same charge repulsion. That is the reason that in order to bind into nuclei of more than one nucleon neutrons are necessary so that ...

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The number of neutrons is even, so it indeed means that they contribute spin zero and positive parity. The spin and parity comes from the "last proton" because the number of protons is odd. The dependence of the energy on the angular momentum is such that the pairs at a high value of $J$ are preferred (lower in energy) due to the special, spin-dependent ...

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According to this talk by Niel Mitchell (slide 4) there are two different strategies. EU, US, and Japan are planning a "pure fusion" reactor in which the $14~MeV$ neutrons resulting from the fusion reaction are directly converted to heat. China and Korea are studying for a hybrid reactor in which those neutrons will be used to catalyse a fission reaction ...

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Usually, Cronin effect is given in terms of the central-to-peripheral nuclear modification factor for $dAu$ collisions at midrapidity $$R^h_{CP}(p_t)= > \frac{(1/N^C_{coll})dN^h/p_tdp_t(C)}{(1/N^P_{coll})dN^h/p_tdp_t(P)}$$ where $C$ central, $P$ reipheral, $N_{coll}$ the average number of inelastic $NN$ collisions.If hadronization is ...

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What happens after the blast does not depend on the speed of the object - it is the principle of relativity. Assuming that the object moves at 9/10 of the speed of light and that the blast accelerates the particles at 1/10 of the speed of light in all directions, the 'at rest' observer would see the particles in front of the object moving at a speed of: $$... 2 The most general relationship is$$c(b) = \frac{\int_0^b \frac{\mathrm{d}\sigma}{\mathrm{d}b}\mathrm{d}b}{\int_0^\infty \frac{\mathrm{d}\sigma}{\mathrm{d}b}\mathrm{d}b} = \frac{1}{\sigma_\text{inel}}\int_0^b \frac{\mathrm{d}\sigma}{\mathrm{d}b}\mathrm{d}b\tag{1}$$(source, one of many). In practice, we usually use the Glauber model to describe heavy ion ... 0 It looks like the following article is relevant: http://arxiv.org/abs/hep-ph/0402256 (published in Nucl. Phys. A). (The phrase you quote is probably from lectures http://www.physik.uni-bielefeld.de/~borghini/Teaching/HIC-Seminar/SoSe2013/Francois_SPhT2006-1.pdf by one of the authors of the article): "The Cronin effect was discovered in proton-nucleus ... 2 Have a look at the binding energy per nucleon curve: There are many stable configurations below iron, so the binding energy is not the only criterion for stability. Graph of nuclides (isotopes) by type of decay. Orange and blue nuclides are unstable, with the black squares between these regions representing stable nuclides. The unbroken line passing ... 0 Here's my understanding based on my QM class and a lab I did a while back. I've tried to set it up in an intuitive way. This is my first answer so I'm sure there will be room for improvement but hopefully it gives you a general idea how it works: Say we apply a constant magnetic field to a sample of hydrogen nuclei along the z direction, the Hamiltonian is ... 5 No, quantum chromodynamics is very non-linear and quantum effects are very strong. What you have is an approximation. A better approximation is given by the Semi-emprical mass formula. 3 As Mitchell says in his comment, this is related to the uncertainty principle. The uncertainty principle states that if you have some system with a position x and a momentum p then there is an uncertainty in the position, \Delta x, and an uncertainty in the momentum, \Delta p, related by Heisenberg's uncertainty principle:$$ \Delta x \Delta p ...

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Here you have a list of non-oxygen containing 4-element pnictides. It might not be a thorough list, but it might help.

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OpenMC (github) The OpenMC project aims to provide a fully-featured Monte Carlo particle transport code based on modern methods. It is a constructive solid geometry, continuous-energy transport code that uses ACE format cross sections. The project started under the Computational Reactor Physics Group at MIT. You can find more background information in ...

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MHV amplitudes are not really any more important than next-to maximal helicity violating amplitudes ($NMHV$) or $N^kMHV$ amplitudes. You need all of them to compute a general scattering amplitude. Basically, scattering amplitudes for non-Abelian Yang Mills theories are very complicated to compute for more than 4 particles, so people work on formulating ...

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Just to expand on the stars->gold thing. The lightest elements (hydrogen, helium, deuterium, lithium) were produced in the Big Bang nucleosynthesis. The temperatures in the early universe were so high that fusion reactions could take place. Nuclear fusion in stars converts hydrogen into helium in all stars. More massive stars have further reactions that ...

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Gravity is not needed in any way (it only helps to increase the pressure inside the stars but the pressure may be "mimicked" in other ways) and the energy needed for these transmutations isn't extremely high. It's just the nuclear energy conditions. See https://en.wikipedia.org/wiki/Synthesis_of_precious_metals Consequently, one may produce gold in ...

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Here is a link to a paper from October 13th 2013 for the ICCF-18 conference, co-authored by : Prof. Yeong E. Kim, Department of Physics, Purdue University, West Lafayette, Indiana 47907, USA, John Hadjichristos, Defkalion Green Technologies Corporation, 1140 Homer Street, Suite 250, Vancouver BC V682X6, Canada http://www.physics.purdue.edu/... Abstract: ...

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