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A brief history of what science thought about the sun can be found here . It is reasonable that once thermodynamics advanced to the point of measuring and calculating energies the discrepancy between heat output of the sun and the age of the earth had to be explained. They tried with gravitation, but until the discovery of nuclear energy and E=m*c^2 it ...


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What you're missing is the difficulty of actually getting the nuclei that you are working with to actually hit each other. Nuclei are tiny, so if you try to aim them at each other, you will probably miss. This page suggests that at the energies in the core of the sun, only 1 in every $10^{26}$ collision events actually fuses. Now this isn't pure D-D, and ...


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The amount of energy liberated per gram of material per second in the fusion reactions depends on the density, the mass fraction (hydrogen, $X$, helium, $Y$, and all others $Z$) and temperature: $$ \epsilon = \epsilon(\rho,X, Y, Z, T) $$ Typically we express the energy generation rate as a power law, $$ \epsilon\propto\rho^\alpha T^\delta. $$ though the ...


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The main problem with boron (relative to 3He) is that the atomic number is high. This means that the plasma must run at a considerably higher temperature, about a factor of 10, in order to overcome the Coulomb barrier. Higher temperature means faster electrons in the plasma. Faster electrons means more radiation when the electrons "hit" the walls. This ...


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Splitting a helium atom requires energy, whereas fusing two deuterium atoms into helium liberates energy. As it can be seen from this graph: the energies you were talking about will be the same (since they involve the same number of nucleons), but the sign will be different. Note that for small nuclei, energy is released by fusing them, while for large ...


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The energy generated during fusion or fission can be seen with this graph: When a light atom is made into a heavier one by adding nucleons, it will lead to a greater output in energy; but when you reach Iron you can no longer gain energy through fusion. For heavier elements, you begin to lose energy when you fuse them and the way to gain energy is to split ...


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Note that this is not a nuclear reactor, but a Fusor. Fusors work by using a voltage drop to accelerate ions to high velocities and a few may fuse. Due to inefficiencies, they cannot be used for power generation. They are relatively simple to build (they are basically beefed-up plasma globes).


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$^{56}Ni$ is produced in silicon-fusion stars. The fusion process doesn't "stop" at $Fe$. Several A=56 nuclides show up. See the Wiki-pedia article on :Silicon burning. Also, Introductory Nuclear Physics by Krane, Chapter 19, Section 4.


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From Wikipedia: Because the nuclear force is stronger than the Coulomb force for atomic nuclei smaller than iron and nickel, building up these nuclei from lighter nuclei by fusion releases the extra energy from the net attraction of these particles. For larger nuclei, however, no energy is released, since the nuclear force is short-range and ...


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I think you already know the answer... Pop III stars, by definition, are born from primordial gas that is basically Hydrogen, Helium with trace amounts of deuterium, tritium, lithium and beryllium; they initially contain almost no C, N, or O. Therefore the primary fusion in massive Pop III stars has to be (well, initially the deuterium is burned but this is ...



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