# Tag Info

92

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) ...

29

Your understanding is pretty much correct and your question quite a natural one. The core did react: the release of energy heated the core and shells quickly, thus changing the neutron capture cross section for the plutonium in the core. A plutonium (or any fissionable) atom's ability to capture a neutron and undergo fission is weakly dependent on ...

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 ...

13

No. The decay products of a certain particle are not equivalent to its constituents. This is evident especially in the context of fundamental particles: quarks can decay into other particles, but that does not mean that a quark is not elementary (see my answer to this question). Nuclei are made of neutrons and protons, which in turn consist of quarks and ...

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 ...

9

No. The atoms are protons, electrons and neutrons. The fact that neutrons beta decay into a proton + electron + electron antineutrino does not mean that neutrons are made of a proton and electron and a neutrino.

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

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 ...

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 ...

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

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 ...

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

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

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

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

I can only answer qualitatively: The experiment where the death occurred was on a subcritical mass of plutonium, and reflectors were being used to bring the number of neutrons to the ones required for criticallity. The mean generation time, Λ, is the average time from a neutron emission to a capture that results in fission and l is the the prompt ...

2

When a charged particle comes into the vicinity of another, it's path is deflected. It decelerates in one direction, and accelerates in another. All charged particles that are accelerated/decelerated by another charged particle, or a magnetic field, emit radiation. See: Bremsstrahlung. Synchrotron radiation. Cyclotron radiation.

2

Both single beta-decay and double beta-decay may occur with $e^+$ as well as $e^-$. However, in both cases, the emission of $e^-$ is predicted to appear (and in the single beta case, is also observer to appear) in a larger number of processes essentially because it's energetically easier for neutrons to decay to protons plus electrons; than it is for ...

2

I think there can be confusion around what 'binding energy' and 'mass excess' mean. The Wikipedia entry on Deuterium has links to explanation them, which may clear it up if the following doesn't. IF you could start with isolated protons and neutrons and assemble your own nucleus, the mass (energy) balance of the result would be the sum of the isolated ...

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

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

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

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

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

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 ...

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