Tag Info

New answers tagged

2

Most atoms have an ionization energy of a few tens of electron volts. Beta decay electrons typically have a range of energies, with the mean and maximum energies typically a few million electron volts; the probability that the electron energy is small enough to be captured is pretty small. In addition, the daughter atom cannot capture the decay electron ...


2

No, Chemical Processes and Mechanical Processes can also release more energy than is input. See Exothermic Process and Catalysis. Think of an explosion of, say, dynamite: for the low energy input of lighting a wick which can be done with a lighter, you can output a large explosion. Also, Palladium is not Fissile, though it is a Fission product (aka ...


1

Yes, the atom (along with the surrounding material) may have a lasting electron deficit after the beta particle emission because harvesting this surplus charge is how betavoltaics work. Beta particles are more penetrating than charged nuclei like alpha particles, and because of that I believe they are better at carrying the charge away. The exact reasons ...


3

The semi-empirical mass formula is derived from a fit to the measured masses. If you know the numbers of protons and neutrons then the idea is that the SEMF should give you a good estimate of that mass (there are of course small residuals of the order 0.2 MeV to the fit). $$M(A,Z)c^2 = (A-Z)m_n c^2 + Zm_pc^2 - AE_b,$$ where $A,Z$ are the mass number and ...


2

The semi-empirical mass formula (SEMF) is much more involved than this, and tries to take into account various other factors that will affect the mass/energy of the nucleus. For any nucleus, of course if you KNOW its mass beforehand then simply subtracting the individual masses of protons and neutrons will give you ($\pm$)the binding energy, but the SEMF at ...


2

Good question! I can maybe guide you in the right direction, although I only found this post because I wanted clarity myself. The reduced width idea comes from the R-matrix formalism (a good paper is by Descouvemont and Baye here). The most basic understanding of it is that the most general cross-section for an interaction of two nuclei (in which a ...


1

I'll give this a shot. I think I follow what you're asking. I'm thinking of the to-be-fissioned-away material as a mass traveling at the speed of light in some sense, In a sense that's true, but it's probobly good to keep in mind that time isn't a dimension quite like the other 3 and traveling through time isn't exactly moving, so in a sense it's ...


0

Let us take a uranium nucleus being hit by a neutron .At the rest mass system of the two bodies there is an invariant mass m described by E=m*c^2. The CM system, seen as an excited U236 in the diagram below, is not moving in three dimensions nor in any other dimensions, velocity needs a dx/dt. An induced fission reaction. A neutron is absorbed by a ...


0

E=mc2 applies equally to ordinary chemical bombs, except there is less "m" turned into "E", by around a factor of a million.


1

What you're missing is the difficulty of actually getting the nuclei that you are working with to actually hit each other. Nuclei are tiny, so if you try to aim them at each other, you will probably miss. This page suggests that at the energies in the core of the sun, only 1 in every $10^{26}$ collision events actually fuses. Now this isn't pure D-D, and ...


3

The amount of energy liberated per gram of material per second in the fusion reactions depends on the density, the mass fraction (hydrogen, $X$, helium, $Y$, and all others $Z$) and temperature: $$ \epsilon = \epsilon(\rho,X, Y, Z, T) $$ Typically we express the energy generation rate as a power law, $$ \epsilon\propto\rho^\alpha T^\delta. $$ though the ...


0

No, they don't. Only some of negative particles (electrons) on the top shell could do that. Atomic nuclei together with inner electronic shells do not move (almost). In metals they form kind of crystalline cubic grid.


2

Short answer: We can measure their energy and momentum distribution functions in the nucleus. We do this by interacting with them individually, either knocking them out of a nucleus left otherwise undisturbed (quasi-elastic scattering) or by exciting them to higher energy states inside the nucleus (many inelastic scattering reaction backed up by data from ...


5

Evidence that there are distinct protons and neutrons in nuclei starts with the Pauli term (pairing term) in the semiempirical mass formula of the liquid drop model. Furthermore, all nuclei with even numbers of protons and neutrons have nuclear spin of zero. This is consisent with shells being filled with spin up and spin down pairs of nucleons, each pair ...


0

Splitting a helium atom requires energy, whereas fusing two deuterium atoms into helium liberates energy. As it can be seen from this graph: the energies you were talking about will be the same (since they involve the same number of nucleons), but the sign will be different. Note that for small nuclei, energy is released by fusing them, while for large ...


0

The energy generated during fusion or fission can be seen with this graph: When a light atom is made into a heavier one by adding nucleons, it will lead to a greater output in energy; but when you reach Iron you can no longer gain energy through fusion. For heavier elements, you begin to lose energy when you fuse them and the way to gain energy is to split ...


1

As an analogy consider an atom. Atoms are objects made up of particles (electrons and nuclei) and they have a variety of excited states, which we can with some effort calculate using quantum mechanics. Transitions between these excited states release energy (as light). The excited states of a hydrogen atom are not the same as the excited states of a helium ...


2

The energy produced in a nuclear reaction is proportional to the difference in rest masses between daughter nuclei (final products) and the initial nuclei (in your example one nucleus and a neutron). Because different nuclei decay into different nuclei, and because the binding energies that determine the rest masses of nuclei are of complicated nature, it ...


3

For those curious I was able to find an answer. IT stands for Isomeric Transition. A metastable state emits a photon to decay to a lower energy List of decay modes: http://ie.lbl.gov/education/decmode.html


3

The Kola borehole seems impressively deep, but compared to the thickness of the crust it is but a scratch. There's nothing special about the rock at a depth of 12km (except that it's hot - 180ºC!). Setting off a nuclear blast at the bottom of the borehole would be little different to any underground nuclear test. The seismic waves might propagate a bit ...


1

The superscript is not mass, it is the mass number $A$ (i.e. the number of nucleons). It is is sometime called the "atomic mass", but that should be understood as a shorthand for "atomic mass number". A positraon, being a fundamental particle in its own right, has zero nucleons and as such has a mass number of zero. The notation used there is problematic, ...


0

Note that this is not a nuclear reactor, but a Fusor. Fusors work by using a voltage drop to accelerate ions to high velocities and a few may fuse. Due to inefficiencies, they cannot be used for power generation. They are relatively simple to build (they are basically beefed-up plasma globes).


-1

I think there's a few questions in there. First, Atoms can (kind of) be seen by electron microscope, so it's not entirely accurate that they can't see Uranium. But scientists had figured out there were atoms long before they actually saw one, through experimentation. This is long, but the first few parts move pretty quick and it covers the discovery of ...


1

In nuclear fission, splitting atoms is a exact calculation or probability(like we 1 gram of uranium it will contain millions of atoms, some of them will split)? In nuclear decays , when a nucleus splits into fragments, yes, there is a probability distribution characterized by the lifetime of the state. Eventually an unstable mass will decay into its ...


0

Nuclear fission uses a chain reaction. With a large enough mass, as one atom decays it will set off other atoms to decay which will lead to the release of a large amount of heat. This heat boils water which turns a turbine to generate electricity.


0

Let's assume that there is a "shadow shield" protecting the ship from the nuclear reactor, this means that the shielding is only in the direction of the rest of the ship (protecting the crew while saving weight). When the unshielded, uncontrollable, reactor hits the enemy ship, the radiation will kill everything on board. The reactor will not explode, ...


0

Building a nuclear reactor is very large-scale investment, and as thorium reactors are unproven, and there is already a large infrastructure in place for the uranium fuel cycle (mining, purification, enrichment, rod fabrication, etc...) a uranium reactor is considered a safer investment.


1

I believe you have the basic ideas correct. The binding energy is the energy required to create Z separate protons and N=A-Z separate neutrons from a (A,Z) nucleus in its ground state. Another way to think about it is binding energy is the mass energy which is missing from a nucleus compared to the mass energy of the individual nucleons. When talking ...


0

It's made up of many particles. I would assume some of them "orbit" the center. I know many electron orbitals have angular momentum. The orbitals of nuclei are less well-understood, but I would expect the same is true.


0

The angular momentum of the nucleus is the combined contribution of the spin-orbit angular momenta of the constituent particles. In order for an entity to have orbital angular momentum of its own it must some conceptual orbit: electrons in the atom, protons and neutrons in the nucleus, atoms in a molecule. That's why the angular momentum of a nucleus is ...


1

The energy Eigenstates of the final nucleus (after the neutron has been captured) form a complete set. That means that any wave function can be written as a superposition of these states; in particular we can express the incoming neutrons wave function in terms of these states, $$ \psi_{in}(x,t) = \sum_{n=1}^\infty a_n(t) \psi_n(x) e^{i E_n t}.$$ Lets say ...


4

This is because U-235 is fissile, that is you only have to deliver the neutron to the nucleus for the magic to happen. Unlike U-238 where just delivering it doesn't do the job, there you also have to impart the nucleus with the neutrons kinetic energy. Once we know this, it becomes clear that for low energy neutrons, their de Broglie wavelength is very big. ...


1

You might be interested in the extensive Wikipedia article on U/Pb dating. Your equations are a little too simplistic; what's really happened is: at time $t=0$, we had some amounts $N_{\text{Pb}}(0)$ and $N_{\text{U}}(0)$; but due to the $\text{U}\rightarrow\text{Pb}$ decay mode this has changed to: $$ N_{\text U}(t) = N_{\text U}(0) \exp(-t/\tau_{\text ...


0

It initially contains no lead, so $N_{0(Pb)} = 0$. Lead is an eventual byproduct of uranium. The statement "The half lives of all the decays leading to Lead are many orders of magnitude smaller than the half life of the initial alpha decay of U." is there to mean that you can assume for practical purposes that Uranium decays into lead, even though it ...


3

$^{56}Ni$ is produced in silicon-fusion stars. The fusion process doesn't "stop" at $Fe$. Several A=56 nuclides show up. See the Wiki-pedia article on :Silicon burning. Also, Introductory Nuclear Physics by Krane, Chapter 19, Section 4.


-1

From Wikipedia: Because the nuclear force is stronger than the Coulomb force for atomic nuclei smaller than iron and nickel, building up these nuclei from lighter nuclei by fusion releases the extra energy from the net attraction of these particles. For larger nuclei, however, no energy is released, since the nuclear force is short-range and ...


0

When someone refers to the quadrupole moment of a nucleus, they usually assume cylindrical symmetry and are referring to the $zz$ component of the tensor $Q_{ij}$: $$<Q_{zz}> = 3z^{2}-r^2$$ For a spherical nucleus, $<Q_{zz}>=0$, but this doesn't mean it has no size. So no, in general you can't get an estimate of the size of the nucleus from the ...



Top 50 recent answers are included