# Tag Info

## New answers tagged nuclear-physics

1

We really can't tell very much at all about the presolar epoch by looking at Uranium in the Earth. However much is known for looking at more "pristine" materials from the early solar system. The Sun formed in an interstellar medium containing contributions from lots of different sources and lots of different stars. We know this from the study of presolar ...

0

I think that whatever answer is provided, is hypothetical and untestable. So I answer: No, a supernova is too large a phenomenon to affect. Yes, if we can find the appropriate "seed". For example, huge storm clouds can be triggered to rain, by seeding the cloud with silver nitrate crystals, dumping their content to the ground. Do I think humankind can ...

0

Any method that turns off the core will work (again, if the star is massive enough), the collapse and bouncing back of the shell will do the rest. For instance, suck up the core to the outside (of course we are no near to have that technology yet). Or alternatively, you can pump in antimatter to the core instead (protected in magnetic vessel/pipe so it ...

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 ...

28

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 ...

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 ...

1

There are many types of nuclear decay, and many techniques for estimating half-lives. For beta decay of states in spherical nuclei, calculation of decay rates is a classic application of the (spherical) nuclear shell model. For gamma decay, there are generic estimates that are based on the energy and multipolarity of the transition. (The term to google on ...

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 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 ...

-1

As far as I'm aware, this is just a phenomenological constant. That is, you get a few people looking (well, not exactly "looking") at a few mols of X radioactive element and. After they have enough measurements, they will get said constant through boring data fitting. Worked in it myself! It does not have a clear dependance on anything (and we have ...

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 ...

0

Yes. After beta decay the atom has charge +1. Electron has energy about 1MeV but due to scattering on the another atoms lose energy and stop.

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 ...

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

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 ...

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 ...

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

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 ...

-1

Under certain conditions, such as having a high charge/mass ratio (not enough neutron constituents) the nucleus will cast off some charge, but doesn't have enough energy to eject a whole proton. Effectively, in the field of a large mass, a positron and neutrino are ejected, the atomic number (Z) of the daughter nucleus is reduced by 1 compared to the ...

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 ...

0

For nuclear reactions we commonly talk about the Q of the reaction. for the reaction A(a,b)B where "A" and "a" are reacts (A is generally the target, "a" is the projectile) and "b" and "B" are the products, $Q = \left(m(\mathrm{A})+m(\mathrm{a})-m(\mathrm{B})-m(\mathrm{b})\right)c^2.$ Here, $m(A)$ is the ${nuclear}$ mass of A. If $Q>0$, the reaction ...

0

Just because $^{25}$Mg isn't in your table doesn't mean it doesn't exist or hasn't been measured. Do a search for "mass of Mg-25" or "chart of nuclides". The mass (in u) will be there. 1u = 931.5 Mev/c$^2$. What table are you using that doesn't have $^{25}$Mg? Rather than dealing with excess masses I have found it easier to use $Q = \epsilon_{reactants} - ... 0 An alpha particle really GETS OUT from the unstable nucleus, and what remains is the so-called daughter nucleus. See for instance the reaction 108Te -> 104Sn + alpha, where the numbers are the atomic masses of the respective elements. The daughter nucleus, here Sn, remains in some cases in its ground state, and in other cases, also in an excited state and ... 1 The best place to look is the Evaluated Nuclear Structure Data File, hosted by the National Nuclear Data Center at Brookhaven National Lab. For example, the data file for helium-4, shows that the first two states with isospin$T\neq 0$are the negative-parity states centered at 23.3 and 23.6 MeV. Be warned that at modest proton number$Z$the assumption ... 1 Yes it has been defined in a coherent way and used many times. You are following a correct reasoning: once you treat a nucleus as an ensemble of nucleons is natural to associate the chemical potential as the energy involved in adding/removing a nucleon from the nucleus. This is generally the case for excitation energies higher than 1 MeV/A when the shell ... 1 Sufficiently high energy photon can spontaneously convert into a pair of leptons as long as there is a heavy charged spectator (i.e. an atomic nucleus) nearby to absorb some momentum. This is the largest energy-loss mechanism for very energetic photons in high$A$materials. You can also get two-photon interaction to ends with massive particles in the final ... 0 Photons, which have zero rest mass, can be "collided" to form particles with nonzero rest mass if the original photons had sufficiently high energy (in the gamma range), see two-photon physics. See this article for discussion of some experimental confirmation. 3 Every particle collider does that. You shoot two very fast particles at each other, and the (sum of the) mass of the many resulting particles after collision is greater than the rest mass of the initial two particles. 2 It might not actually answer your question, but to throw it into the bowl: There are some advances in MRI using permanent magnets and even conventional electromagnets with static magnetic fields of about 0.5 Tesla. As far as I know one can do imaging with a reasonable resolution with these devices without the need for extensive cooling. They are used for ... 0 For those wondering, to find the total number detected over some time period: Let the number incident/unit time =$N_I$, and the number detected/unit time =$N_D$We need to consider the number of gold nucleons per unit area in the target. To find this we do:$\rho \over m_{Au}$The flux of the incident particles,$J$is simply$J$=$N_I$Let the ... 0 So the atom is left with two more electrons than protons and therefore has a negative net charge. This repels the two electrons, so they leave, and fast. And now the atom is neutral. 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 ... 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. 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 ... 1 I think I caught that show on the science channel too and I think they represented that particular point poorly. First, I'm quite sure the sun already has a healthy amount of iron in it already, because all the inner planets do. There's no logical reason why the sun wouldn't. Iron is reasonably plentiful in the universe. What happens when a star's ... 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 ... 0 Hirosima was 15kt and had a volume of$1.5m^3$, it produced a mushroom cloud of about$17000m$if things scaled linearly the coffee mug would produce a cloud of$0.0005m$in height However I really doubt it would be linear, even on the site that you linked there doesn't seem to be much of a correlation between kt and mushroom cloud height, not to mention ... 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 ... 0 Neutrons provide nothing to nuclear spin due to even number, but uncouple proton in 1d5/2.To maximize the Iz (uncouple proton) it should be in 3/2 state because couple protons in 5/2.So I=5/2 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) ... 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 ... 0 When we touch things we are actually feeling the electromagnetic force of repulsion. There are atoms in our hands and in every object. This force (one of the four fundamental) is responsible for what we feel. This force means that we come nowhere near to the orbits of the atoms in the object, at least in comparison. What we feel is the repulsion force. 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 ... 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. ... 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 ... 1 Actually, you forget the important duty of the water between rods. The neutron produced by a fission reaction will have a very high energy. However, if you want it been able to produce an other fission reaction efficiently in the thermal neutron reactor, you need to "slow it down". And this is why you have water. So the neutron will first go through water, ... 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 ... 1 The next atom that splits in a chain reaction doesn't care where the neutron that split it came from, or what path it took to get there. It could have been emitted from an atom right next door and take the shortest possible path, or it might have been emitted from an atom way over on the other side of the reactor and had really good luck with it's flight ... 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 ...

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 ...

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