# Tag Info

4

THe U238 nucleus is unstable and does undergo "ordinary" radioactive decay. For a U238 nucleus to undergo fission you need to provide the nucleus with something akin to an activation energy of about 7 MeV. One model is that the nucleus is like a liquid drop and adding a neutron can make the drop oscillate so much that it breaks up into two smaller pieces ...

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Nuclei belong to the quantum mechanical framework, the underlying network of all natural forces. They are composed out of protons and neutrons . Protons and neutrons are composed out of 3 quarks each , between quarks , the strong force described by quantum chromodynamics generates the "bag" where the quarks are tightly bound and exist in a sea of gluons ( ...

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My intuitive thinking is that the four nucleons form could be modeled as a tetrahedron--with each nucleon at a vertex and pressed tightly against each neighbor. If you place each of the four particles at the vertices, then each has one like-particle neighbor, and two unlike-particle neighbors. Such a composite particle is known as an alpha particle--and is ...

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This was actually considered under the Plowshare project The idea being to use an underground nuke to heat the surrounding rock and then run it like a geothermal resource. However, radiation problems.

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As you say, the problem is confinement (aka containment) - fusion is relatively easy; controlled, useful fusion turns out to be incredibly difficult and expensive. So the easiest way to harness it, and the only technically and economically viable way (for decades from now, and possibly a century or more into the future), is to have a large fusion reactor in ...

0

In principle, the answer should be yes. At any given temperature, the particles will have a distribution of speeds. Those in the tail of the distribution might have enough energy to fuse. However, the probability of this event would be extremely low because the number of particles with the required (HIGH!) energy is very low.

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a giant air tank. if the tank is large enough, we could set off a bomb inside without exceeding the containment strength of the perimeter, then harvest power from the increased pressure in the tank. i imagine making a large air-tight tank might be cost-prohibitive, although a 19 million cubic foot tank is already in use. perhaps we could build a ...

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No... In a nuclear bomb, energy does not last long enough... For continuous power, simply not enough...... But it might be possible to power a satellite launch vehicle with a nuclear bomb... But hydrogen fuel cells have an excellent efficiency which (probably) makes it a better option... All in all, I would say no... Regards, Pradyoth Shandilya

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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 ...

1

You can only have an inelastic collision between two bodies if one or both of the bodies have some internal degrees of freedom that can absorb energy. For example if you have a rigid sphere then the only type of energy it can possess is kinetic energy. If we collide two rigid spheres then conservation of energy means the sum of the kinetic energies before ...

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It's rarer than that, more like 1 in 6000. Because each molecule of water has two atoms of hydrogen, then about every 2 in 6000 (1 in 3000) has a single atom of deuterium (DHO). And would be closer to 1 in every 6000$^2$ for a molecule of D2O. The linked question Deuterium density in seawater gives sources that show deuterium is well-mixed in the ocean. ...

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A helium nucleus is not two protons, it is two protons and two neutrons. Instead you can compare two deuterium nuclei (one proton and one neutron) with one helium nucleus, so you have the same number of nucleons of the same types. And the answer is (as far as I am aware): no, there is no experimental evidence for the magnitude gravitational forces produced ...

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Beta decay occurs, approximately, in nuclei where the Fermi energy of one species of nucleon is higher than the first unoccupied orbital for the other species. In these nuclei energy can be liberated by turning one type of nucleon into another --- the new nucleon moves into the available, lower-energy orbital. By definition, a nucleus with neutron excess ...

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The spin orbit coupling can be derived from the nonrelativistic limit of the dirac equation and is given by $$H_{\text{s-p}} = \frac{\varepsilon_0}{2m_e^2c^2}\mathbf{\hat{s}}\cdot\left(\mathbf{E}\times\mathbf{\hat{p}} \right)$$ $\mathbf{E}$ is the total electric field acting on an electron, which consist of a microscopic electric field ...

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$^1_1H$ is a special case as the proton does not have any other nucleons to bind onto. I suppose that you could call the binding energy per nucleon zero which means that you require no energy to split up the nucleus of $^1_1H$ into its constituent parts? Note also that the binding energy per nucleon is not necessarily the full measure of whether a nucleus ...

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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 ...

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Well, we are all aware of the fact that every system tries to minimize its potential energy.Now, potential energy of of heavy nuclei is greater than that of light nuclei. If it were to encounter FUSION, that would result in decrement of binding energy AS PER THE BINDING ENERGY CURVE Therefore fission would result in higher binding energy, thus is ...

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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, ...

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One has to think about what happens in the explosion. In a conventional explosion, a chemical reaction creates a whole lot of hot gas - that volume is initially contained inertially, and it expands as a shock wave travels outward. Any object on the boundary will experience a larger thermal and pressure gradient; the pressure and flow of matter behind the ...

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short answer no longer answer it depends on what you mean by survive. The exposure to the intense heat is going to melt anything that is small enough to "tape to a Tsar-Bomba-yield nuclear warhead" the temperatures of fission reactions are used to induce fusion temperatures they are going to far exceed any threshold for a small piece of material and the ...

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Standard way to evaluate the risk associated with eating the contaminated food is to start with the ingested activity (Bq) and then estimate the effective dose using the conversion coefficients. Many parameters must be considered in this calculation- for instance the age of the guy who eat such food, chemical form of the radionuclide... Talking about the ...

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I would note that Mn54 has a 300 day half life, and also emits at 834keV - you are likely seeing activation of material from the past neutron activity (Mn is a fairly standard alloying addition in steels and aluminum alloys).

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Weapons grade $U$ has to be highly enriched to boost its levels of $U^{235}$ from the natural level (about 0.7 w%) to > 90 w%. That's because $U^{235}$ is the fissionable isotope and its concentration in the bomb core has to be sufficiently high for an explosive nuclear chain reaction to be able to occur. Weapons grade $Pu$ is not obtained by enrichment. It ...

1

If a quantum system has angular momentum of zero, then it is necessarily spherical. One can see this as rob mentioned by looking at the Wigner-Eckart theorem and seeing that only scalar moments are allowed. Another way of seeing this is remembering that the total angular momentum operator $J$ is the generator of rotation transformations. For some rotation ...

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You already do this to some extent when your power supply is an electrolyte-based battery (a, b) --- there you have motion of both negative and positive ions in the electrolyte. Electrons are free to move in conductors when the Fermi energy falls within a band of energy levels rather than in a gap when there are no allowed energies. However, the band/gap ...

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The Nuclear Wallet Cards make this process a little easier by listing mass excesses rather than binding energies. For your reaction we have $$\begin{array}{cr} \text{nuclide} & \Delta\,\rm (MeV) \\\hline \rm n & +8.1 \\ \rm^{140}Cs & -77.1 \\ \rm^{93}Rb & -72.6 \\ \rm^{235}U & +40.9 \end{array}$$ So your initial state has mass excess ...

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In general an atom undergoing fission breaks up into other, smaller nuclei and stable particles: photons (x rays and gamma rays), electrons(beta decays) , alpha particles an other lower mass nuclei. Gluons are never free, because of QCD, and always inside a proton or a neutron. Their spill over attractive force is due to virtual gluons which will never ...

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The energy released in (mainly) neutrinos, along with light and the kinetic energy of the exploding envelope is around $10^{46}$ J. This is equivalent to 0.05 solar masses being converted into energy. This is less than a percent of the progenitor mass. The energy source is gravitational potential energy. At the heart of a supernova, there was an Earth-sized ...

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Your result for Q in terms of the Bs is okay, but you're not going to find B's tabulated directly. You will find B/A for each nuclide. I find it easier to use either atomic masses or mass defects, since both of those are tabulated somewhere. The SEMF is a gross approximation and doesn't fit individual nuclides well enough for better than a 10-20% ...

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