# Tag Info

4

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 ...

2

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 ...

2

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 ...

1

There is no process $\gamma \to n$ at all, nor $\gamma \to n \bar{n}$ with an on-shell photon. The first violated multiple quantum number conservation rules and the second conservation of four-momentum. The two-photon process $$\gamma + \gamma \to n + \bar{n} \,,$$ has allowed quantum numbers but will be exceedingly rare. It is worth noting that ...

1

On the Wikipedia page for the semi-empirical mass formula (based on the Gamow liquid drop model of nuclei) it basically says that the mass defect $\Delta M$ is given by the difference in the masses of the unbound protons and neutrons $Zm_p+Nm_n$ minus the actual mass of the nucleus, $^A_ZM$ (the rest of the semi-empirical formula is not needed here). The ...

1

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 ...

1

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 ...

1

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 ...

1

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" ...

1

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 ...

1

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 ...

1

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 ...

1

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 ...

1

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 ...

1

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, ...

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