Tag Info

Hot answers tagged

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 ...


4

Such information can be inferred from the differential cross section. Different spins and parities lead to different angular distributions of decay products / scattering partners. You find these correlations by using one decay product to define an axis and measuring the distribution of the other decay products with respect to that. This requires a lot more ...


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. ...


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.


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 ...


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 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 ...


2

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 ...


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 ...


2

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 ...


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 ...


2

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

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

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 ...


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 ...


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 ...


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, ...


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 ...


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 ...



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