# Tag Info

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A seemingly problematic aspect of the proposed mechanism is that it allegedly requires two energetic 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. deuterons, K-shell holes, or some superposition of them), then we expect something like: $$dn/dt = An^2 - Bn$$ ...

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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 + ... 0 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 That could be a positron annihilation peak. If the halo of your beam is scraping somewhere there will be a variety of interactions between the particles and matter. If it is an electron beam then many of the interactions will be electromagnetic showers which will produce (energy allowing) some e^+--e^- pairs. When the positrons annihilate they will ... 1 The condition for double beta decay to be the preferred mode are that the energy (mass) of the three states The initial nucleus (m) The would-be single-beta-decay daughter plus the electron and the anti-neutrino (s) The double-beta-decay daughter plus the two electrons and the two anti-neutrinos (d) have a relationship like E_d < E_m < E_s ...

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Fission products are generated randomly, i.e. you can't tell what nuclei will be formed as a result of particular fission process. However, "the distribution of the fragment masses formed in fission is dependent on the mass of the fissioning nucleus and the excitation energy at which the fission occurs. At low excitation energy, the fission of such ...

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I have a formula able to calculate the binding energy of the deuteron: it needs only to apply the electric and magnetic Coulomb's laws to the deuteron. The formula is the same as above, but with the potential instead of the force, needing the empirical radius: A better theory, e.g. electromagnetic, gives the correct result by using only the fundamental ...

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Protons and neutrons are different Protons contain one elementary charge and neutrons electric charges with no net charge. The consequence is that there exists an electric attraction between a proton and a neutron equilibrated by the magnetic repulsion between nucleons. The electromagnetic interactions between nucleons are, except for the Coulomb repulsion ...

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You have to realized that the combined forces that bind the protons and neutrons together are a complex interplay between two forces: a)The electromagnetic one, where the charge of a proton repels the charge of another proton and no binding could occur b) the strong force , the force that binds the quarks into the protons and neutrons, and spills over ...

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Earth's gravitational binding energy is -1.711×10^32 J joules or 4.09×10^13 gigatons. The Tsar Bomba massed 27 tonnes to deliver 0.057 gigatons. Do the math for Earth disassembly by bomb. (Substituting depleted uranium for the used lead tamper will double the yield.) Earth orbital speed averages 30 km/s and it masses 5.97×10^24 kg. (mv^2)/2 is 2.7×10^33 ...

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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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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

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I'm guessing you're referring to the original model of proton decay that arose from the first grand unified theories: (picture from here). The proton decays to a positron and a neutral pion, and the pion then decays to two gamma particles with a half life of $8.4 \times 10^{−17}$ seconds. So to make protons you would have to get two gamma ray photons to ...

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Proton Decay is currently a completely hypothetical process, as far as recall there isn't actually any experimental evidence to back up the prediction. If however we assume that proton decay is possible into photons and a positron. Then yes it would be possible to do the reverse. The problem with this however is how to hit a positron with enough photons ...

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Start with a table of the isotopes. Look up Cs-137--the only decay channel is by $\beta^-$ (that is the conversion of a neutron into a proton with the release of an electron and a electron anti-neutrino) which makes the daughter Ba-137 in an excited state called Ba-137m. The decay of that excited state is accompanied by the production of gamma rays in the ...

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There a few more subtleties to this problem. While only a "small" fraction of neutron absorptions go to something that will be a dangerous activation product, this doesn't assure us of safety. That's because the flux coming from the fuel rods is many orders of magnitude higher than our safety limits to begin with. In fact, consider the case of boiling ...

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There aren't many radioactive isotopes of Oxygen and Hydrogen and the ones that there are aren't very radioactive. As dmckee notes, there is Deuterium formed from Hydrogen capturing a neutron, this produces D$_2$O, or heavy water. But Deuterium is stable and so doesn't cause radioactivity in itself. Heavy water is chemically a little toxic but not a ...

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Now when we operate parity operator, does that mean we are taking any physical entity at x to −x. Or we are just reverting axes of the co-ordinate system? Well, either operation should adhere to the same rules, and you mention the correct term: it depends on whether we see the operation as active or passive. Either view has the same end result: we move ...

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My previous answer was less comprehensive that what Charles Beaudette have well detailed. this books is available (courtesy of the author) on the site of University of Tsinghua http://iccf9.global.tsinghua.edu.cn/lenr%20home%20page/acrobat/BeaudetteCexcessheat.pdf Unlike the books of Huizenga and Taubes, it contains references to works done after 1989, it ...

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It is true that all even-even nuclei (hundreds of such isotopes have been measured) have spin-0 in the ground state. This is due to what is often called the pairing effect. Protons and neutrons are spin-½ particles, and they have a tendency to respectively pair up in proton-proton and neutron-neutron pairs so that their spin (and orbital) angular momentum ...

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It seems to me that you have a misconception of binding energy. If it is higher, it does not mean that there is more energy into the system, but that you need to put more energy into the system to separate the components. When the system (nuclear, chemical or even a ball on the edge of a bowl) releases some energy it drops into a more bound state, that is ...

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There's a closely related question: binding energy of a nucleus is positive? Imagine taking the nucleus and pulling it apart into individual protons and neutrons. To do this you have to put in energy, and the amount of energy you put in is equal to the binding energy - call this $E_0$. Now let the individual protons and neutrons come together again to form ...

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Let's take the simple system of a deuterium nucleus, that is the bound state of a proton and neutron. It's tempting to think of the binding energy as something that has to be added to a proton and neutron to glue them together into a deuteron, and therefore that the deuteron must weigh more than the proton and neutron because it has had something extra ...

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The binding energy doesn't compensate for the difference between the mass of the nucleons and that of the nucleus, it is that difference. $$m_\text{protons and neutrons}c^2 + \text{binding energy} = m_\text{nucleus}c^2$$ Since the mass of the nucleus is less than the mass of the protons and neutrons, the binding energy must be negative.

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Point one : The discrete energy levels for an electron in an atom is due to boundary conditions on Schrodinger equation.Consider a particle in a one dimensional box with potential at boundaries set at infinity, and solve it! Because the potential is infinite outside the box, the wave function must be zero outside the box.The wave function must be ...

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The hydrogen nucleus has exactly zero nuclear binding energy, for the reason you gave in your question. The nuclear binding energy is the energy it takes to separate all the nucleons in a nucleus from each other. Since there is only the one nucleon, it's already separated from any other nucleons. For the same reason, a bare neutron has zero nuclear binding ...

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The electron is bound to the proton by about -13.6058 eV. A naked proton is not a hydrogen atom.

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With regard to mass and energy, nuclear reactions are no different than anything else. Also, mass doesn't turn into energy. Mass is energy. $E = mc^2$ is the shortened version of the equation $E = \sqrt{(mc^2)^2 + (pc)^2}$, where $p$ is momentum. This equation describes any physical system. Take a box and put in whatever you want - compressed springs, balls ...

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There is a lot of confusion over the critical mass of uranium 235 as there are two aspects to be considered. One is the theoretical calculation and the other is the numerical value. The numerical value is dependent on the various values of the physical parameters known at the time and vary wildly. There is also no actual critical mass unless you specify the ...

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