Tag Info

Hot answers tagged

27

Jupiter's mass is too small to produce nuclear fusion. Jupiter would need to be about 75 times as massive to fuse hydrogen and become a star http://en.wikipedia.org/wiki/Jupiter This wikipedia page explains the detailed requirements of nuclear fusion: http://en.wikipedia.org/wiki/Nuclear_fusion


15

There is a minor change to yyahn's answer. The isotopes Deuterium, and Lithium7 which are present in small amounts -left over from the big bang, can fuse at lower mass than pure hydrogen burning. The estimate is at around 65 mass Jupiter Lithium fuses with hydrogen to form two helium nuclei. Brown dwarfs as light as 13 mass Jupiter can ignite Deuterium ...


14

To answer the question simply, $E=mc^2$. Energy is a manifestation of mass, and mass is a manifestation of energy. In a fusion or fission process, the total "energy" of the system remains constant, it just changes shape. By "energy" I mean the totality of the already present energy, and the bound energy of the mass that takes part in the reaction.


13

The analogy is facile. Helium fuses at a temperature ($10^8\ \text{K}$) roughly ten times higher than hydrogen ($10^7\ \text{K}$), so a better analogy would be alcohol and thermite. That higher temperature is achieved only by massive gravitational contraction after hydrogen fusion [EDIT: in the core] is exhausted. EDIT: To expand, different mass stars ...


12

This'll be a very rough order of magnitude estimate, but as you'll see it's good enough. Suppose that two hydrogen atoms bump into each other. In order to fuse, the nuclei have to tunnel to within about a nuclear distance of $10^{-15}$ m of each other. The tunneling probability is something like $e^{-(2mE)^{1/2}L/\hbar}$, where $E$ is the energy gap, $m$ ...


12

There are quite a few novel energy technologies coming through. I guess that without quantification, "breakthrough" is a subjective term. Below, I've tried to list all the energy technologies that I know of, that are not yet at commercialisation, but could be within 50 years, and that could offer at least tens of gigawatts of power. They are, in descending ...


11

You actually make reference to something which is of crucial importance to the answer to this question: "With a tokamak, I imagine that if you double the linear dimensions, the plasma volume (and hence the power production) will increase eightfold, whereas area that you have to protect against fast neutrons will only quadruple. So once you master the ...


10

The WKB approximation states that in one dimension, the tunneling probability $P$ can be approximated as $\ln P=-\frac{2\sqrt{2m}}{\hbar}\int_a^b \sqrt{V-E} dx$ , where the limits of integration $a$ and $b$ are the classical turning points, $m$ is the reduced mass, the electrical potential $V$ is a function of $x$, and $E$ is the total energy. Setting ...


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 ...


9

This is a very difficult question to answer. There are (at least) two reasons. First, we have detailed, numerically exact wave functions for stable, light nuclei only up to, just recently, $A=12$ (like $^{12}C$). The Argonne-Los Alamos-Urbana collaboration uses quantum Monte Carlo (QMC) techniques to evaluate the ground and excited states of bound nucleons ...


9

The reaction rate doesn't increase that quickly with temperature, but pressure does. If you perturb a solar model, making a zone near the core marginally hotter, the increased pressure will rapidly (at roughly the soundspeed divided by a characteristic length) cause it to expand. That lowers the pressure and temperature enough to substantially quench the ...


9

As far as a catastrophic plasma disruption is concerned, it might be a problem for the first wall, and hence the ability to use the reactor. But despite the high energy per nucleon, the desity of the plasma is extremely low, the total energy is dominated by the energy in the magnetic fields, and thats not tremendous. Of course you do have the radioactivity ...


9

These are valid questions. I am a graduate student in the program so it won't surprise you that I will advocate for it, nonetheless, if you'd like to double check my claims, you're welcome to. Let's go in order: We are hoping that the next major breakthrough does come from ITER: the first burning plasma. But it is BECAUSE of the fact that a machine like ...


9

The next serious advance that is not an speculative/fringe idea is most likely to be fusion power. Harnessing the power of nuclear fusion has long been a goal for energy production since the first hydrogen bomb was created in the 1950s. Creating controlled fusion, rather than the chaotic variety has proven a rather challenging task here on Earth however. ...


8

No, Jupiter is simply not massive enough to sustain nuclear fusion (and effectively become a star). Off the top of my head, I remember the statistic that it needs to be roughly 10 times bigger in order to generate the necessary gravitational forces at the core to initiate fusion and become a star.


8

This question is a near duplicate of Does the "Energy Catalyzer" generate energy by converting Nickel to Copper? , but perhaps it is ok, because some time has passed, and there is more confidence in the assessment. It is not reasonable to reject Rossi out of hand, because there are observed unexplained nuclear reactions in Palladium/Deuterium ...


8

I will give you an analogy. Molotov bombs are made by filling a plastic bottle with gasoline, attaching a wick ingeniously, lighting it and throwing it on a target, usually a car or a policeman controlling a demonstration. Now there are a number of BTUs of energy in this bottle of gasoline. I can ask paraphrasing you Could we put these Molotov s to ...


8

For some recent information on the running battle between cold fusion researchers and myself over my proposed conventional (non-nuclear) explanation of the Fleishmann-Pons(-Hawkins) effect, you might want to look here: https://docs.google.com/open?id=0B3d7yWtb1doPc3otVGFUNDZKUDQ (referenced in this: ...


7

The big problem with controlled fusion is that the equations governing the plasma are highly non-linear. So each time the physicist increase the size of the Tokamak, new effects are discovered. So I guess that the answer is no-one really knows the correct scaling laws ! This contrasts a lot with fission reactors, where the relevant equations are ...


7

The binding energy curve for nucleons in nuclei shows which atoms can take part in fusion, releasing energy in the process. Fusion happens as one goes from left to right, until reaching Fe, iron. From there to the right it is fission that will release extra energy This is an example of a fusion reaction, the one that is actually being materialized in ...


7

I find the idea of a breakthrough technology presented by the other answers to be extremely modest. I'm almost positive that I'll get downvoted and flamed for my answer, but I have a strong desire to impress the potential for, and the implications of, a truly breakthrough energy technology. I want to begin calling attention to Jevons Paradox, as well as ...


7

The Sun's energy, of course, comes from fusion. I think there's a small and totally insignificant amount of fission going on as well. The majority of the Sun's mass is hydrogen, and the vast majority of what isn't hydrogen is helium (with the ratio changing over billions of years as hydrogen is fused into helium). But since the Sun formed from the same ...


7

From what I have read in "American Prometheus: The Triumph and Tragedy of J. Robert Oppenheimer" Teller was the first one to express this concern before the Trinity test. Also quoting from: http://www.sciencemusings.com/2005/10/what-didnt-happen.html Physicist Edward Teller considered another possibility. The huge temperature of a fission explosion -- ...


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 ...


7

[Rewrote the answer because I found out that my initial approximation was too crude.] In the WKB approximation, the tunneling probability is $\exp[-\int_a^b dx \sqrt{(2m/\hbar^2)(V-E)}]$, where the integral is over the classically forbidden region from $a\sim10^{-15}$ m to $b\sim 10^{-10}$ m. The first obvious thing to try is approximating the integrand as ...


6

The problem with proton-proton fusion is that there is no bound state of two protons. For the fusion to occur one of the protons has to turn into a neutron by beta plus decay. This is mediated by the weak force so it's a slow process and the probability of it happening while the protons are close enough to form a deuteron is very low. By contrast a deuteron ...


6

Well, if you search the internet it seems there are kids out there that make the claim of having built a fusion reactor . I watched this link. Note that in .56 minute he gives a small description, and does not claim breaking even, but that he demonstrated fusion. It is a plasma that he obviously creates and manages to fusion some deuterium that is not ...


6

There are other interactions to consider besides the Coulomb interaction. A very nice model of the nucleus is the liquid drop model, in which one models it as a constant-density liquid with various interparticle interactions. The result is known as the semi-empirical mass formula, which I summarize here. Let $Z$ be the number of protons, $N$ the number of ...



Only top voted, non community-wiki answers of a minimum length are eligible