# Tag Info

91

Since gold is much more abundant in the universe than is uranium (by a factor of about 20:1)1, why is the situation reversed in the Earth's crust (by a factor of about 1:600)2? The answer lies in chemistry. Uranium is chemically active. It readily oxidizes (pitchblende) and it readily combines with silicates. Uranium is a lithophile (literally, rock-loving) ...

20

I'll be using the tabulated values from a Wikipedia article of abundance of elements in the Earth's crust. Gold has a tabulated value of 0.0031 ppm in mass for crustal abundance. Uranium has a tabulated value for 1.8 ppm in mass for crustal abundance. These figures are interesting in that uranium has an abundance of almost 500 times more ppm than that of ...

15

Short summary In fact, in typical reactor, neutron needs to travel quite a lot before it initiates next fission, if during these travels it encounters control rod it is "lost" and chain reaction slows down. Neutron needs to travel because it needs to lose energy (or in other words slow down), this is because modern reactors are designed in such way that ...

12

Assume that you have a fission of an atom of $U^{235}$, and that we look at one of the neutrons produced. Although the neutron itself is sub-atomic, the "size" of the space needed for the fission neutron to slow down through collisions with the moderator atoms, avoid capture by control rods or reactor structure, find another atom of $U^{235}$, collide ...

9

It's not true that the atom is electrically neutral afterward. If you have a single atom, isolated in a vacuum, and all that is emitted is an alpha particle, then as you say, it has a net charge of -2e. In reality, alpha decay is a violent process that is likely to knock out some electrons as well. Furthermore, if the atom is in a solid, then electrons are ...

6

WP says: Of all the common nuclear fuels, Pu-239 has the smallest critical mass. A spherical untampered critical mass is about 11 kg (24.2 lbs),1 10.2 cm (4") in diameter. Using appropriate triggers, neutron reflectors, implosion geometry and tampers, this critical mass can be reduced by more than twofold. This optimization usually requires a large ...

4

The constant is a function of the stability of the nucleus, and is experimentally determined for every isotope. In other words - every kind of nucleus has its own value of $\lambda$ and there is no way (that I know) to get an accurate value for it, other than measurement. But there are some nuclear physicists roaming who will put me out of my misery, I'm ...

4

Generally the particles that enter these ionization chambers have such high kinetic energy that they can pass through walls. The gas inside the ionization chamber is not travelling nearly so fast. Kinetic energies of the gas would be at least a million times lower and often much lower than that. The gas cannot escape through he walls. High energy ...

3

Every particle collider does that. You shoot two very fast particles at each other, and the (sum of the) mass of the many resulting particles after collision is greater than the rest mass of the initial two particles.

3

The transition probability per unit time of a nucleus from an initial state i to a final state f, representing the decayed system, is modeled by Fermi's Golden Rule: $$\lambda=T_{i\rightarrow f} = \frac{2\pi}{\hbar}\left|\left\langle i\left|H'\right|f\right\rangle\right|^2\rho$$ Where $T_{i\rightarrow f}$ is the transition probability from state $i$ to state ...

3

Here is a table of isotopes versus lifetimes the color code of the lifetimes on the right hand column: Isotope half-lives. Note that the darker more stable isotope region departs from the line of protons (Z) = neutrons (N), as the element number Z becomes larger Modeling a nucleus is a many body problem and also a many forces problem. There exists ...

3

A decay destroys the electromagnetic wave function of the atom, the one that generates the energy levels which keep the electrons bound to it. The new nucleus, after the deacy, will have a new potential whose solutions will have binding levels for n-2 electrons. The two extra will be left in the lattice ( or in the gas) free to join up in the energy ...

3

This is a very interesting question: By the time the neutron gets out of the fuel to the point where it can strike the control rod and be absorbed by it, isn't it out of the fuel by that point and not going to cause any more chain reactions anyway? It reflects a natural intuition about the role of the fuel and the coolant which just isn't how it works. ...

3

At a first glance, W bosons are not needed at all in those scenarios: you could model the interaction by supposing a direct coupling exists between the 4 particles (neutron, proton, electron, neutrino) -- which is what was initially proposed by Fermi (see Fermi's 4-point Interaction). Thus, both the $n+\nu \rightarrow p + e$ interaction and the decay ...

3

The nucleon-nucleon interaction has a short range, roughly 1 fm. Therefore if there were to be a bound dineutron, the neutrons would have to be confined within a space roughly this big. The Heisenberg uncertainty principle then dictates a minimum uncertainty in their momentum. This amount of momentum is at the edge of what theoretical calculations suggest ...

2

I think the plot you show is the estimated abundance of the interstellar medium from which the Sun has formed. The chemical abundances of the interstellar medium change with time, so you have to define some point in time at which to estimate them. As the initial chemical abundance in the Universe is basically H, He, with traces of D, Li and Be, then it ...

2

You're right that in the context of radioactivity, antineutrinos are pretty much only released when a neutron turns into a proton, ${}_0^1n\to {}_1^1p+{}_{-1}^{\;0}e+\bar{\nu}$. They can also be consumed when a proton turns into a neutron and a positron, ${}_1^1p + \bar{\nu}\to {}_0^1n + {}_{1}^{0}e$. There are some other processes that involve ...

2

Why can't nuclear fusion be done with just helium 4? Helium-4 is the product of fusion, where protons are the fuel. The helium-3 is produced as an intermediate step. As an analogy, think of wood in a fire: the wood turns to charcoal, then to ash. You can douse the fire and pull the charcoal out to burn later, but you can't burn the ashes any more. ...

2

Here are the neutron decay Feynman diagram : A free neutron will decay by emitting a W-, which produces an electron and an antineutrino. and the diagram for neutrino neutron scattering : This interaction is the same as the one at top since a W+ going right to left is equivalent to a W- going left to right. In the quantum mechanical ...

2

Are we able to touch the atomic orbital of an element ? If we define "touch" as the exchange of virtual photons between our hand and the object, yes. The Pauli exclusion principle, to start with, does not allow the electron orbitals in our hand to overlap with the electron orbitals of the surface unless very specific quantum mechanical conditions are ...

2

An ultra relativistic electron has a very small wavelength. The quarks and gluons in the proton have very small energies with respect to this ultra relativistic energy. Another way of looking at "frozen", is to think of them on a lattice. It is also true that the electron will only hit one parton each time, and so this seems identical to a multi-slit ...

2

You're right, directly after the emission it will have charge -2e if it was neutral before, i.e. be an ion. But within a gas or liquid electrons are very easily exchanged and ejected. This is of course at a much lower energy energy scale than the nuclear emission and therefore less noticeable.

2

It might not actually answer your question, but to throw it into the bowl: There are some advances in MRI using permanent magnets and even conventional electromagnets with static magnetic fields of about 0.5 Tesla. As far as I know one can do imaging with a reasonable resolution with these devices without the need for extensive cooling. They are used for ...

2

The notation $\lvert \phi \rangle \lvert \psi \rangle$ is shorthand for $\lvert \phi \rangle \otimes \lvert \psi \rangle$. What you are doing is flipping a tensor product around. Though, in general, $A \otimes B = B \otimes A$, it is not the case that $a \otimes b = b \otimes a$ for $a \in A, b \in B$ (since the left and right hand side don't even live in ...

2

A chain reaction happens when an isotope, hit by a neutron, undergoes fission with more than one neutron being produced in the process - and where the neutrons produced have a sufficiently high probability of themselves creating further fission. Now there is nothing in the above that requires the isotope to be unstable to begin with - as long as you hit it ...

2

Not possible in practice, even though neutrinos emitted by the Plutonium might be used in principle if we ever found a way of intercepting them with almost 100% efficiency. However, there is/was a scheme to use neutrino analysis to determine whether a reactor is being used to create Plutonium

1

If we ignore neutrinos, which are weakly interacting, radioactivity is still classified as alpha beta and gamma. The energies are of order MeV. Of these three, only gamma is neutral and has a chance to cross the atmosphere . Then one has to take into account the 1/r**2 diminution of the flux for the large distances . To localize a source another ...

1

Sufficiently high energy photon can spontaneously convert into a pair of leptons as long as there is a heavy charged spectator (i.e. an atomic nucleus) nearby to absorb some momentum. This is the largest energy-loss mechanism for very energetic photons in high $A$ materials. You can also get two-photon interaction to ends with massive particles in the final ...

1

Yes it has been defined in a coherent way and used many times. You are following a correct reasoning: once you treat a nucleus as an ensemble of nucleons is natural to associate the chemical potential as the energy involved in adding/removing a nucleon from the nucleus. This is generally the case for excitation energies higher than 1 MeV/A when the shell ...

1

The best place to look is the Evaluated Nuclear Structure Data File, hosted by the National Nuclear Data Center at Brookhaven National Lab. For example, the data file for helium-4, shows that the first two states with isospin $T\neq 0$ are the negative-parity states centered at 23.3 and 23.6 MeV. Be warned that at modest proton number $Z$ the assumption ...

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