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10

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 ...


7

This creates a point of extremely focused energy at the middle point where the bubble collapses. In theory, this point focuses enough energy to trigger nuclear fusion. It is not currently accepted mainstream science to say that collapsing bubbles focus energy enough to cause nuclear fusion. Temperatures over 10,000K can be acheived, but are still well ...


5

The basic problem of modelling a star is covered in a number of textbooks and lecture notes. Try searching for "stellar structure and evolution" or something along those lines. The best readily available lecture notes, IMO, are those of Onno Pols, available here. There was also a similar post on Quora, which you can read too. In the mean time, here's the ...


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

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 ...


4

In a very general sense a lot of reaction that are written in one step can also be written in two. I.e. alpha capture on carbon-131 is often written $$\alpha + ^{13}\!\mathrm{C} \to ^{16}\!\mathrm{O} + \text{various photons and leptons} \,,$$ but may be written as one of $$\begin{align} \alpha + ^{13}\!\mathrm{C} &\to ^{16}\!\mathrm{O} \\ \alpha + ...


4

A seemingly problematic aspect of the proposed mechanism is that it allegedly requires two hot deuterons. (By contrast, U-235 fission requires just one neutron.) Why is that so problematic? If $n$ is the number of 20keV particles (i.e. hot deuterons, or K-shell holes, or some superposition of them), then we expect something like: $$dn/dt = An^2 - Bn$$ ...


4

Here they say that there is no waste per se only that some parts can become contaminated and they'll refurbish them onsite. The rest will be handed over to the authorities. https://www.iter.org/mach/hotcell The Hot Cell Facility will be necessary at ITER to provide a secure environment for the processing, repair or refurbishment, testing, and ...


3

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.


3

Hydrogen atoms fuse to form helium through the proton-proton chain which fuses four protons into one alpha particle (nucleus of ${}^{4}He$) and releases two neutrinos, two positrons and energy in the form of gamma photons. Although photons travel at the speed of light, the random motions they experienced inside the sun takes them thousand of years to leave ...


3

According to this talk by Neil 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 ...


3

There is a claim often made about cold fusion, that it is excluded theoretically. The main theoretical argument is that electronic energies are too low to overcome the Coulomb barrier, since d-d fusion only takes place at KeV energies, while chemistry is at eV energies. This is belied by inner shells, which in Palladium store 3 or 20 KeV of energy ...


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

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

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

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. ...


3

Very high temperature, on the order of 10 keV, 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 (due to plasma turbulence) causes heat losses. Making the system larger allows increasing the heating power (due to fusion reactions, plasma ...


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 ...


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

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 ...


2

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, ...


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 $R$ ...


2

To recap: a classic (not as in "classic versus quantum") picture is the one-dimensional model of $\alpha$ decay by Gamow and Gurney/Condon. $Q$ represents the energy of the particle within the well, which in this example will be the "disintegration energy" of the system, i.e. the energy the escaping particles are seen to have after a decay. $r<a$ ...


2

That's right. I'm really out of the loop regarding nuclear fusion shielding so feel free to correct me, but the only radioactive waste will be the reactor's inner walls (because of the radiation). The only other 'waste' that a fusion reactor produces is helium.


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 simplest reaction deuterium and tritium. Tritium is common in big labs (like NIF, JET, Omega) [1]. Tritium sucks - practically speaking. It is expensive, radioactive, and hard to stockpile. Omega spent millions and years on a tritium facility. It may even never be used in fusion power [2]. The next easiest reaction is deuterium with itself. This ...


2

The Wikipedia article answers most of your questions. What are the requirements for hydrogen atoms to go through fusion? Two atoms must overcome the coulomb barrier, which can be done by forcing two atoms very close together, or by leaving them moderately close for long periods of time, which allows them to tunnel through the barrier. Is it a ...


1

Muon mean lifetime is 2.2 ┬Ás. There's your problem. Muons mass 105.7 MeV/c2, about 200 times that of the electron. If you wanted to ionize a hydrogen atom, you would need 13.6 eV. If you wanted to ionize a muonic hydrogen atom, you would need about 2813 eV or about a 0.441 nm photon. Start building your laser.


1

Carbon has to be produced by the triple-alpha process because there is no stable nucleus with 8 or 5 nucleons. The probability of this is very low, because it requires three different particles to be in the same place at the same time. You'll note that the Wikipedia article says: One consequence of this is that no significant amount of carbon was ...



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