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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 ...


9

The Sun obviously produces far more energy per second than is required to fuse an iron nucleus with some other nucleus. The problem is concentrating all that energy on the iron nucleus. It's not enough to known that it takes the energy from $n$ hydrogen fusions to fuse one iron nucleus, it's getting the energetic products from those $n$ hydrogen fusion ...


7

The problem with attempting to fuse two protons is that there is no bound state $^2$He, for the rather obvious reason that there are no neutrons present to hold the two protons together. The fusion of two protons requires one of them to undergo beta plus decay while the two protons are close, and the probability of this is vanishingly small. It happens in ...


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

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 ...


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 ...


4

As you correctly stated in normal situation the star cannot sustain the process. This doesn't mean that there are no such reactions going on in the core. The difference is that during the pre-supernova phase of the star the production of iron is negligible compared to the star. When it goes supernova, it produces a comparable amount of iron.


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 ...


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: $$dn/dt = An^2 - Bn$$ ...


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

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

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

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

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 ...


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

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 ...


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

No, it is not true. This is not how fusion in stars like the sun works. The sun is fusing hydrogen to make helium. At this point any other reactions are very rare, not sustaining, and irrelevant. It is not true that a whole chain of higher elements are being created at this point. It is also not true that iron is too heavy to allow nuclear reations. ...


2

First, a minor correction: iron is not too heavy to allow the reaction to continue, it is incredibly stable and therefore cannot produce anything else (it would take $\sim10^{22}$ years for it to decay into chromium and its binding energy is the highest per nucleon, so it would "cost" more to produce something heavier than iron has in it). Research shows ...


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

You're totally misinformed... atomic bombs raise from an exponentially expanding interaction, where in Uranium, for examples, a neutron is fired initially, and each neutron hits another Uranium atom and releases another 3 neutrons, and the energy released grows exponentially to create a horrible explosion. And about the speed you can reach, this depends ...


2

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 ...


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 ...


1

I do not trust Dr Bussards scaling. He did not have enough data to make those scaling claims. The University of Sydney (Gummersall, 2013) scaled some factors in their simulations: A. Current in the rings (Amps or AmpTurns) B. Size of the rings (Meters) C. Energy of electrons (KeV) But, they were looking at how many electrons were trapped - not fusion ...


1

The answer lies in this diagram It shows the binding energy per nucleon over the number of nucleons n in a nucleus. As you can see, it has a maximum around Iron (n=56), which means that left from iron energy can be released by fusion. For elements heavier than iron you need to put in energy to fuse them together, or equivalantly that fission is ...


1

The term "Tokamak" refers to a design, not a size. The planed ITER reactor has the goal of 500 megawatts output. So it would take approximately 300,000,000 such reactors to produce the same power as the solar energy reaching the Earth. https://www.iter.org/factsfigures


1

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. $r<a$ ...



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