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11

If they didn't release energy, they wouldn't happen. The alternative, nuclear reactions that require energy, clearly need said amount of energy, which has to come from somewhere, e.g. kinetic energy involved in the collision of two nuclei (even ones that release energy usually have a "barrier" and some amount of initial kinetic energy is needed to overcome ...


8

I think the colloquial term for that type of plot is "spaghetti diagram" because you have a bunch of lines running across it. It's really the mass fraction as a function of interior mass. From our stellar structure equations, we have that $$ \frac{dm}{dr}=4\pi r^2\rho, $$ which is derived from the mass-continuity equation, so you can relate the radius, $r$ ...


7

Elements heavier than iron are produced mainly by neutron-capture inside stars, although there are other more minor contributors (cosmic ray spallation, radioactive decay or even the collision of neutron stars). Neutron capture can occur rapidly (the r-process) and occurs mostly inside supernova explosions. The free neutrons are created by electron capture ...


5

When you mention solar power, it makes me think you are thinking about photo-voltaic power or power extracted from solar panels. The power put out by the sun is about $3.95*10^{26}W$ per second. But solar panels can only capture a fraction of that energy. Even so, in 2008 humans used about $4*10^{13}W$ per second which is many orders of magnitude less ...


4

Very interesting question! In chemistry you spend lots of time discussing exothermic and endothermic reactions: when you put your reagents together, sometimes the reaction heats things up, and sometimes the reaction cools things down. Nuclear reactions are very different, in that essentially all spontaneous reactions studied in laboratories are exothermic. ...


4

The article The path to metallicity: synthesis of CNO elements in standard BBN attempts to quantify the amount of carbon produced during big bang nucleosynthesis. It concludes that the ratio of carbon-12 formed to hydrogen was $~4 \times 10^{-16}$ with lesser amounts of carbon-13 and carbon-14.


4

Very high temperature, on the order of 10 keV (100 million degrees Kelvin), is needed for fusion reactions to start to happen at appreciable rates. However, in magnetic fusion devices (tokamak, stellarator) the transport of heat across the plasma (mainly due to plasma turbulence) causes heat losses. Making the system larger allows increasing the heating ...


4

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 ...


3

$^{56}Ni$ 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.


3

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 ...


3

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 ...


3

Some rough estimates (you can dig up more accurate numbers): The oceans contain about 321 million cubic miles of water (source: http://oceanservice.noaa.gov/facts/oceanwater.html), or 3.5e20 U.S. gal. 1 gal seawater contains roughly enough deuterium to provide the same energy as 300 gal of gasoline (maybe slightly less - that's the part for your homework!), ...


3

I wonder if your number 48% comes from the typical Carnot efficiency of a heat engine - see for example a detailed description at http://www.visionofearth.org/industry/fusion/how-do-we-turn-nuclear-fusion-energy-into-electricity/ When you want to use heat to create electricity, you typically convert the heat into motion (for example by rotating a turbine, ...


3

I personally doubt that the Compact Fusion Reactor as presented by Lockheed Martin last week can work, but I haven't seen enough information to be certain. And to some extent, you never know until you try. (As I understand it, they only have a very early prototype, I mean try as in a full scale prototype.) What I think I can say with certainty, is that it ...


3

Well, a large number of countries, after the break even ( actually 60% of output over input energy) in energy of the prototype tokamak in JET joined into creating ITER, a prototype Tokamak design designed to have output energy in megawats. If interested you should go to the FAQ of the link given for ITER . There exist alternate projects: Of the ...


3

The sun in powered by nuclear fusion of hydrogen to helium. There is plenty of hydrogen remaining to keep the sun going for billions of years. So, solar energy is not infinite, but it will last for billions of years.


3

Since your plasma is in a vacuum environment, the only way for it to loose energy is by radiation (conduction transfer through the magnets are neglected). You have thus to consider which bodies are surrounding your plasma and which radiative model is the best ton consider for them. I guess you can consider a black body with the simple Stefan equation. The ...


3

The sun gets its energy from the pp-chain. The first step is the two protons forming the diproton (Helium-2): $$ \,^1_1H+\,^1_1H\to\,^2_2He+\gamma $$ where the $\gamma$ is the photon (of energy about half an MeV). This quickly $\beta^+$-decays into a deuterium by converting a proton into a neutron: $$ \,^2_2He\to\,^2_1D+e^++\nu_e $$ where $e^+$ is the ...


3

There isn't exactly a mathematical relationship, but there is a physical one. It is the gravitational compression that causes the increase in temperature in the core of the gas cloud that becomes a star. When the temperature reaches a critical value (in the millions of Kelvin range), hydrogen fusion can occur. This is because the temperature is great enough ...


3

Oh, but we do! I'm assuming you mean using the fields to simply collide particles with each other, right? Then that's already being done. For example, take this neat little machine: http://en.wikipedia.org/wiki/Fusor This one runs on the exact same principle you described (though I'm not quite familiar with the inner workings of the LHC). For energy ...


3

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 ...


3

The simple answer is No. Fusion happens at nuclear energies between particles to be fused, i.e. MeVs, because it is at the framework of nuclear bound states. LHC particles start with energies of TeV, so particle particle interactions are way over any nuclear bound state levels. Even if one accelerates deuterium nuclei the phase space is way over the ...


2

No, because the LHC puts too much energy into its particles for them to fuse. While we need enough energy to fuse particles, too much will stop it from happening.


2

This might not be on-topic for this site. Specific degree programs vary from school to school, so there's no way to know offhand. Certainly there may be overlap between "physics" at one school and "engineering" at another. At a larger school with both programs, the overlap may extend to shared faculty, or else the "engineering" program may be geared to ...


2

To get a lot of energy from a fusion reactor you need lots of D-T fusion events per second, and this means a combination of reasonably high density and very high temperature. This is extraordinarily difficult to achieve. In particular as you try and increase the plasma density it gets increasingly difficult to maintain the plasma in a stable state. There ...


2

The motivation for pursuing fusion is clear, but there are currently several main physics and engineering challenges: Confinement time: An operational reactor requires a long energy confinement time, $\tau_E$. An empirical scaling law for confinement time has been found to depend on the size of the tokamak as $\tau_E \propto R^{2.04} a^{1.04}$, where $R$ ...


1

The comments basically answer the question. Another answer is found on HyperPhysics by Rod Nave: While magnetic confinement seeks to extend the time that ions spend close to each other in order to facilitate fusion, the inertial confinement strategy seeks to fuse nuclei so fast that they don't have time to move apart. Directed onto a tiny ...


1

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 ...


1

Conservation of charge would not allow the reaction $$ p+p\to p+p+p $$ because you have +2q charge on the left and +3q charge on the right.


1

I will answer 3) as seen in this encyclopedic entry: Practical efforts to harness fusion energy involve two basic approaches to containing a high-temperature plasma of elements that undergo nuclear fusion reactions: magnetic confinement and inertial confinement. A much less likely but nevertheless interesting approach is based on fusion catalyzed by ...



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