# Tag Info

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There is an official convention for a positive muon being a nucleus, that a positive muon with one electron is muonium (Mu) and a positive muon with two electrons is muonide (Mu-). See Names for Muonic and Hydrogen Atoms and Their Ions. For a negative muon replacing an electron in helium, I see both $He\mu$ and $^{4.1}H$ in the same paper: Kinetic Isotope ...

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In the 1940s and 50s, physicists were trying to understand the formation of carbon-12 in stars. It was (correctly) proposed that two Helium-4 nuclei first fuse to produce beryllium-8, which then fuses with another Helium-4 to produce Carbon-12. This is known as the triple-alpha process. An apparent problem with this explanation was that the ground state ...

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For the academic source - someone directly involved with physics would just not be interested in this kind of text. But look for introductions for academic fields that are only remotely/lightly involved with nuclear physics, maybe medicine or biology. And: Not the right place for this I think.

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In analyzing mass-energy calculations involving beta decay, it is important to avoid simple bookkeeping errors. You look at the balanced nuclear reaction, and use tables of isotope masses to calculate the loss of mass (and its conversion into energy) The problem comes from the two terms "mass of the final atom" and "tabulated mass of the final atom" ...

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I presume you are referring to process of nuclear fission of uranium-235, which has the equation: $$^1_0\text{n}+^{235}_{\ \ 92}\text{U}\longrightarrow ^{236}_{\ \ 92}\text{U}$$ However, a subsequent reaction is: $$^{236}_{\ \ 92}\text{U}\longrightarrow^{144}_{\ \ 56}\text{Ba}+^{89}_{36}\text{Kr}+3^{1}_{0}\text{n}$$ The production of neutrons is a feature of ...

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First part of your question: part of the mass is used as kinetic energy for the electron/positron and (anti-)neutrino to leave the core. Therefore mass can't be conserved. For the beta plus decay: I don't know your textbook, but assume the mass of the atom includes the surrounding electrons. Then the core emits an positron (1st half of the mass loss), and ...

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Deuterium-Deuterium fusion without Tritium is very possible. Historically, D-D fusion was the first form of nuclear fusion mankind successfully achieved. It may be worthwhile remembering that, at the dawn of the nuclear age in 1952, an inertial confinement fusion experiment that used nuclear fission to produce the conditions for nuclear fusion called Ivy ...

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I don't know about the first part of your question, I think gas centrifuge specifications and capabilities may be classified which would make a reasonably exact estimate quite hard. The Fat Man nuclear bomb (20kt-yield, implosion design) required 6.2kg of plutonium (or, if the country is unwilling to take the extra step of converting its uranium to plutonium ...

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The problem of whether and how to include electrons/positrons in mass-energy calculations is based on the difference between the entities involved in the reaction, and the entities for which masses can be determined. This becomes more complicated when there may be both orbital electrons and nuclear-sourced electrons/positrons involved. Consider the last ...

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The question says "nuclei of hydrogen" ie. protons. In the middle of a star there isn't likely to be much atomic hydrogen around!

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Firstly, fusion doesn't happen in the way depicted in the question. Four protons don't participate in a 4-body reaction. Instead there are many intermediate steps: Each step has its own reaction rate. The overall reaction rate is determined by the rate limiting step. The proton-proton reaction is the rate limiting step in this case. It is important ...

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What is force? It is the transfer of momentum with respect to time. Momentum is the product of mass of a body, and it relative velocity with respect to something. If we imagine, for once, our world made up of tiny uniform particles, then we can take the mass of one particle as one unit mass. This simplifies the situation, as the momentum, 'mass x velocity' ...

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One model is to say that the atom is in an impenetrable spherical box, and solve for the wavefunctions. See Y P Varshni Accurate wavefunctions for the confined hydrogen atom at high pressures J. Phys. B: At. Mol. Opt. Phys. 30 No 18 (28 September 1997) L589-L593. The Fermi Contact Term (electron density at the nucleus) greatly increases as the size of the ...

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the main reason is that the space is uniform, and that there is nor absolute reference point in the universe. basically, what appears to be moving at constant speed to you, will be moving at a different constant speed or even not moving at all to another observer who is moving at constant speed in reference to you. since your point of view is not any better ...

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A very n-rich nucleus is unstable to beta decay. The neutron is more massive than the proton, there are therefore lower energy proton states available for neutrons to decay into (emitting a beta decay electron at the same time). Filling these states with protons (i.e. reducing the N/Z ratio) blocks this beta decay channel because the Pauli exclusion ...

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In a neutron star there are mostly "free" neutrons and the question then is why they don't all beta decay into electrons and protons? Well, some of them do, but the point is that when the electron (or proton, there are equal numbers of each) numbers build up then they become degenerate (meaning no more than two electrons can occupy the same energy state and ...

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When they show a decay with a certian amount of energy, this energy is net of the masses of the particles. So you get Co = $\beta$ + Ni + 0.31 MeV, the energy is attached to the ejected beta particle.

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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 ...

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Fast rotation keeps the paper-symmetry protons equivalent. High concentration makes for fast proton exchange in the alcohols (sharp line, no coupling to methylene protons). Integration identifies populations. The difference in chemical shifts tells you the temperature of the sample, ...

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You are right when you say that one peak is due to $\mathrm{OH}$ groups and the other one to $\mathrm{CH_2}$ groups. The protons in each groups are chemically equivalent and contribute to the same peak. You should compute the area below each peak. Since there are 4 $\mathrm{CH_2}$ protons and only 2 $\mathrm{OH}$ protons, my guess is that one peak's area ...

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Gamma emission is emission of a photon upon a nucleus transitioning from an excited state to a lower or ground state of the same nucleus. The number of neutrons and protons in the nucleus is exactly the same before and after the gamma photon is emitted. Beta decay results from a nucleus having too few or too many neutrons relative to the number of protons ...

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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 ...

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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.

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