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61

Heavy elements couldn't form right after the Big Bang because there aren't any stable nuclei with 5 or 8 nucleons. Source: Wikipedia (user Pamputt) In the Big Bang nucleosynthesis, the main product was $^4He$, because it is the most stable light isotope: 20 minutes after the Big Bang, helium-4 represented about 25% of the mass of the Universe, and the ...


31

In the case of a supernova explosion it is possible to create heavy elements through fusion. Supernovae have a tremendous amount of energy in a very small volume but not as much energy per volume as there was in our early universe. So, what is the major difference? Why didn't the Big Bang create heavy elements? I just want to point out, too much ...


28

Has Musk done his homework? With regard to the basic idea of using nuclear weapons to release CO2 and thereby warm Mars, no, he hasn't. I suspect this was either Bored Elon Musk speaking, or perhaps the Elon Musk who didn't quite deny being a super villain ( 1-900-MHA-HAHA Elon Musk?) in that interview with Colbert. CO2's enthalpy of sublimation is about ...


19

This question is answered in detail by the so-called "Big Bang Nucleosynthesis", the theory about the creation of the nuclei in the early Universe. Almost out of nothing, it allows one to determine that 75% of the nuclear mass was coming in hydrogen, 25% in helium, and some small traces of lithium appeared, too. Even though Gamow used to think that all ...


17

Lithium and other light elements (e.g. beryllium) can be formed indirectly from supernovae via cosmic ray spallation, a process where protons and neutrons are ejected when a cosmic ray collides with another atom. The nuclei can then become new elements. Nakamura & Shigeyama (2004) were able to calculate yields for 6Li, 7Li, and isotopes of Beryllium and ...


13

Elements up to and including iron can be produced exothermically by fusion reactions in stars. Producing heavier elements is then endothermic. The reason for this is that the binding energy per nucleon is maximised in nuclei around the "iron peak". This means that if you tried to add something to an iron nucleus, the resulting nucleus would have a smaller ...


13

Heavier nuclei can also undergo fusion, but that's not very useful for energy production. One of the reasons is, as you've mentioned, the binding energy per nucleon. Let's have a look at the binding energy curve (image taken from Wikipedia): Iron-56 has the highest binding energy per nucleon, which means it is the most stable nucleus. Roughly speaking, ...


12

As dmckee says in his comment - Population III stars have no metals (a tiny bit of lithium and beryllium), but they are not "pure hydrogen stars", they still have the big bang fraction of Helium. Taking the second part of your question first. These "stars" will last for ever. Their final fate is to become a completely degenerate ball of helium, supported by ...


7

By "consume" we mean "convert into helium." That $6\times10^{11}\ \mathrm{kg}$ of hydrogen is part of the Sun (specifically it is found in the core of the Sun), and it is converted into $6\times10^{11}\ \mathrm{kg}$ of helium. The Sun doesn't need to suck up material from space. Note that this amount of material is miniscule compared to the $2\times10^{30}\ ...


7

Suppose you start with a linear solenoid. Due to the Lorentz force charge particles travel in circles (or helices) inside the solenoid so they can't reach the walls of the solenoid. But obviously the trouble is that they will leak out of the ends. Now we curve the solenoid round and join its ends together to make a torus so now the particles can't leak out ...


6

An analogy: nucleons stick together because of some powerful glue (because otherwise they would fly apart, especially protons because they have positive charge and thus repel each other). Bigger nuclei need more glue, but, until you reach 26 protons or so (i.e. iron), the amount of glue for each extra nucleon is decreasing. Past that point, it starts ...


5

OK, lets run some numbers. Assume you had a setup where every single muon actually catalyzed a D-T fusion event (note - that won't happen by a large factor). The energy released in D-T fusion is about 18MeV, of 2.88fJ (yes, femtoJoules). Across a 1km by 1km array, that would yield 28.8 mJ per minute, or 0.5 mW of power. In to a one square kilometer array. ...


4

The deuterium-tritium fusion reaction cross-section is highly temperature dependent and peaks at temperature of about $8\times 10^{8}$ K, so I suppose these are the temperatures to aim for in a controlled nuclear fusion experiment. In fact according to this, the operating temperatures are at least $10^{8}$ K. The density of the fusion plasma is a factor - ...


4

This is really just a footnote to Rob's answer. The Sun is an absolutely terrible fusion reactor. It uses a reaction $p + p \rightarrow d$ that is hopelessly inefficient. The $d + t \rightarrow He + n$ reaction that we use in fusion reactors is (up to) 26 orders of magnitude faster. As Rob says in his answer, the power produced per cubic metre in the Sun is ...


4

Converting my comment to an answer: Well, I'd start up a new plasma system using an inert gas - I'd rather find out any problems with a non-explosive gas mixture. Once I know that everything works (vacuum, valves, no leaks, plasma actually ignites, ...) I would move on to the more exciting mixtures. Remember - for any non-trivial system (and even many ...


4

You may have things a bit mixed up. Plasma is not something that plays a role in fusion as if it were a tool or an instrument for its achievement. It is instead the only possible medium where nuclear fusion can occur: very basically, high enough temperature for protons to overcome the Coulomb repulsion, and high enough density for increased chances of ...


3

His name was Robert Bussard. This is what you probably saw: https://www.youtube.com/watch?v=rk6z1vP4Eo8 He passed away a few years ago. EMC2 was carrying on the work but it looks like they stalled in 2014. Lookup www.emc2fusion.org Fun Fact: Those red tips on the end of warp nacelles in Star Trek are called "Bussard Collectors" from a paper Robert Bussard ...


3

Think of 2 hydrogen atoms or, protons more accurately since at those temperature the atoms don't have electrons, it's more of a soup. So, 2 protons, both positively charged so they repel each other, crash into each other pretty rarely, cause it's still a lot of empty space, but they do make contact every so often. The energy required to get 2 protons or ...


3

Is it carried away as momentum imparted on the [product] atom? Is it carried away in neutrinos? Is it carried away as gamma rays? All of these can happen, and in general nuclear reactions will output their energy via a combination of these. The specific combination, of course, depends on the specific reaction. Also, if neutrinos are massless, can they ...


3

Electron degeneracy does not lead to an infinitely hard equation of state. The Pauli exclusion principle does not say that two fermions cannot occupy the same space; it says they cannot occupy the same quantum state. What this means is that as you squish the electrons together they have to occupy higher and higher momentum states. It is this non-zero ...


3

First, note that Population III stars are expected to be massive, not tiny, with masses upwards of $10^6\,M_\odot$. The reason for this is due to the Jeans criteria, where the mass follows $$ M_J\propto T^{3/2} $$ In the early universe (~ 1 Myr), the temperature was around 10,000 K; so in a pure-hydrogen environment, no cloud with a mass less than about a ...


3

Short answer - almost all of it. Your body is a mixture of chemical elements. By number, most of the nuclei are hydrogen, by mass its mostly oxygen. In cremation, most of this oxygen, along with most of the other combustible things like carbon, hydrogen ends up going up the chimney. Hence the low residual mass. Energy (including rest mass energy) is ...


3

Theoretically, yes. The problem is that there's no way to build a "small" thermonuclear warhead. Fusion isn't as simple as fission, the latter being as easy as smacking the right amount of Plutonium together. It is technically energetically favorable to fuse heavy isotopes of hydrogen into helium, but the conditions required to do so include giving the ...


3

The reaction has been studied in accelerators A bubble chamber study of proton-proton interactions at 4 GeV/c Part I—Elastic scattering, single-pion and deuteron production .Summary Elastic scattering, single-pion and deuteron production have been investigated. The cross-section for elastic scattering is σelastic = (13.5±0.3) mb. The angular ...


2

other people far smarter than I have since published more detailed reasons why the concept presented by Lockheed Martin can't work. One example is from two professors of plasma physics on the website of the Max Planck Institute for Plasma Physics. To summarise coils inside the plasma need connections for a power supply and coolant. These connections will ...


2

Iron fusion can take place in stars - what you need is lots of iron and very high temperatures. These conditions exist in the cores of massive stars near the ends of their lives. For example alpha particles can fuse with an iron-56 nucleus to produce nickel-60 and then zinc-64; these reactions are barely endothermic. The problem is that there are competing ...


2

To small numerical factors, the fusion reaction rate is $r_{AB} \propto n_A n_B <\sigma_{AB}\, v>$, where $<\sigma_{AB}\, v>$ is the (temperature-dependent) "reactivity" for the reaction, formed from the averaging the cross-section over an appropriate Maxwellian velocity distribution, and $n$ are the number densities of the reactants. The proton ...


2

Look up Lithium deuteride: a form of lithium hydride where the hydrogen is all deuterium: this will give you most of your answers. The main fusion[1] reaction that lets slip most of the energy and thus the horrendous blast in a fusion bomb is: $$_1^2 D + _1^3 T \rightarrow _2^4 He (3.5{\rm MeV}) + _0^1 n (14.1{\rm MeV})\tag{1}$$ Here I've written $D$ for ...


2

Generally, you are correct, and trying to store gaseous hydrogen for long periods of time without significant losses doesn't work. The usual way around this is use lithium deuteride, a solid compound, as the main fuel in the fusion portion of the device (the secondary). Deuterium/tritium is also used in the fission primary. According to http://...


2

The amount of energy per cubic metre in a degenerate gas depends on the density and composition of that material. I can only give some examples - any specific mixtures or densities would require individual calculation. The energy density of an ideal, completely degenerate gas of fermions is given by $$u = \frac{\pi m^4 c^5}{h^3} \left[ x(1+2x^2)(1+x^2)^{1/...



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