# Tag Info

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I'm sure this has been answered elsewhere on the site, so I won't provide too much detail. The nucleus is held together by the strong nuclear force between nucleons (protons and neutrons). The force is short range and effectively only acts between adjacent nucleons. At the same time, the Coulomb repulsion between protons acts over a long range - so that all ...

8

I'll approach this slightly differently. The abundance of Li in the solar system and in the Earth's crust is low compared with elements like carbon, oxygen, silicon and iron. The solar system lithium is created partly (only 10%) by primordial nucleosynthesis, a bit by spallation reactions of cosmic rays on nuclei in the interstellar medium, but mainly in ...

6

This is a special example of "what will happen" under given circumstances. Almost all of physics – and natural science – is about answering such questions. But they're really very many very different questions and one must be a little bit more specific about what the question is. Your general question "what forms of energy will result" is so general that it ...

4

The peaks at $E = 1099\:\mathrm{keV}$ ($P = 56.5\:\%$) and $E = 1292\:\mathrm{keV}$ ($P = 43.2\:\%$) are the most important gamma lines for Fe-59. If a significant Fe-59 activity is present in your sample and your detector is sufficiently sensitive in this energy range, you should be able to see both peaks. The intensity of both peaks might be reduced due to ...

4

Considering the neutron/proton as a single atom, the mass defect is by definition zero, as there are no binding energies, which tie your particle to something else.

4

The word thing you are looking for is "Black Dwarf" stars. Which are White Dwarf stars which have cooled to match the temperature of the cosmic background. Since this is likely to take more than the current age of the universe, there aren't any. These will exist forever, unless hypothetical proton decay finishes them off or a hypothetical Big Rip due to Dark ...

4

(By stars I'm assuming you're implying stars like the Sun, which are a majority of the stars we see. @Dirk Bruere's answer about Black Dwarves is correct. ) No, I don't think they can. The primary process that 'fuels' stars is nuclear fusion. In the process of nuclear fusion, lighter elements fuse together, releasing a tremendous amount of energy (because ...

3

The answer is already on page 2 of your link above: "Among the large number of radionuclides of medical interest, Sc-44 is promising for PET imaging. Either the ground-state Sc-44g or the metastable-state Sc-44m can be used for such applications, depending on the moleculeused as vector." So the metastable state Sc-44m decays to the ground state Sc-44g.

3

Fission reaction produce a distribution of several different nuclei. Wikipedia (User:JWB) has a nice graph that shows the relative probabilities. However, in the case of a single reaction, where it is given that a Cs-137 nucleus is produced, you can probably be more specific because there are correlations between the fission products, but it will require ...

3

It's a bit of a puzzling question. I'll try to work it out, but one of the tricky parts is that atoms absorb and emit photons all the time, higher temperatures emit higher wavelengths. Photons created in the sun (per second) can be estimated, but those are fusion gamma rays. The sun burns about $564$ million tons of hydrogen per second. (Source), and 1 ...

2

The "bullet" in Thin Man was a conical shell, which would go *over" the target, reducing contact time incredibly. It was molded on an old cone clutch design from an automobile, with two conical rings of plutonium meeting at high speed. The problem was contamination. In effect, they would have to cast the core pieces, install, and drop within 28 days, else ...

2

A typical implosion assembly of a Pu pit is time limited by the maximum detonation velocity of the high explosive used. As a ball park figure, you are not going to get a shock a compression rate of more than 10 km/s because of that limit. So in theory a gun assembly might work if you could accelerate the projectile to twice that velocity (the implosion is ...

2

The only split you can do is to ionize the atom, separating the proton and electron. That requires 13.6 eV, the amount of energy one electron acquires on falling through a potential of 13.6 Volts. In ordinary terms, this is a minuscule amount of energy. It is absorbed, not produced. $\phantom{This is here to add characters to make the edit long enough to ... 2 For light elements, roughly up to iron (Fe), splitting atoms actually costs energy, rather than energy being released. Only the really heavy atoms from an atomic mass number of about 150 release energy during fission (atom splitting). Hydrogen is in any case a bit unique in that the nucleus only contains one nucleon (a proton) so that splitting that nucleus ... 2 As indicated in the answer to this physics.stackexchange question the total amount of nucleons is preserved during fission. As a result the atomic number of a daughter product can be predicted if another was already known. In the case of$U_{92}^{235}$fissioning to$Cs_{55}^{137}$, we know the atomic number of the second fission product is given by: ... 2 The occurrence of pair production within the detector depends on gamma energy and detector material. The occurrence of escape peaks in addition to a given full-energy peak depends on detector geometry and sample geometry. If the gamma energy is large enough to make pair production relevant, the photon may disappear and be replaced by an electron and a ... 2 Fission reaction tend to produce a range of potential daughter nuclei, so the decay path is not just one single pair. There is a broad literature on fission yields, much of it from the 1950's and 60's (not surprisingly). For example, the 1965 IAEA symposium on Physics and Chemistry of Fission is available on-line at the IAEA. A search on Cs137 finds papers ... 2 Be-7 is common atmospheric radionuclide produced by cosmic ray spallation of nitrogen and oxygen. Ground level concentration of Be-7 is in order of ~mBq per cubic meter of air. Main deposition process of Be-7 is a wet scavenging which yields to ~Bq per litre of rainwater. It is therefore possible to find Be-7 in background (depends on location of ... 2 There are, of course, limitations to the independent particle model (IPM) version of the shell model. First of all, the configuration you get from the shell model depends on the assumed ordering of the single-particle orbits. If one uses single-particle orbits calculated for 208Pb, then the ordering for neutron orbits above N=50 is 1g7/2, 2d5/2, 2d3/2, ... 2 You have surely seen slit experiments, where waves which pass through the slit and scatter to produce a pattern on a screen placed behind the slits. In the far-field approximation (also known as Fraunhofer diffraction) this pattern is precisely the Fourier transform of the slits. Perhaps you even remember that waves passing by lines produce the same pattern ... 2 A turbine is machine in which the kinetic energy of a moving fluid is converted to mechanical power by the impulse or reaction of the fluid with a series of buckets, paddles, or blades arrayed about the circumference of a wheel or cylinder. The mechanical power typically has the form of a torque on a rotating axis. A motor is generic term for a machine ... 2 The protons themselves are not elementary particles, they are composed of smaller particles, called quarks, and it is these that are involved in a proton proton collision. What exactly emerges from a collision depends on the energies involved when the quarks hit each other. The higher the level of energy the quarks have, the greater the probability that ... 1 I think the answer Acid Jazz gave is very good, but since you don't like it, I can give a few examples. Because protons are positively charged they repel each other, so it takes very high energy to get 2 protons to actually collide at all. This happens in the center of the sun and in very specialized circumstances on earth, like particle colliders. The ... 1 Isospin is useful approximation when dealing with groups or families of particles with nearly the same mass. For example, the proton and neutron are practically degenerate (have nearly the same mass, why we call them nucleons) but differ by charge. If you ignore the mass difference you can say that each nucleon has isospin 1/2, and the difference between ... 1 Isospin is useful because the strong interaction which is important in the creation of nuclei is charge blind. The strong interaction sees a nucleon and does not see its charge. The effect of the charges enters as higher order corrections to how a nucleus is formed. As a first order approximation treating the proton and the neutron as two faces of a nucleon ... 1 I would argue that this is like saying that the main difference between roses and rosemaries is the smell, in the sense that they differ in many things while both being flowers, and you are arbitrarily selecting a difference to be the outstanding one. The differences between direct reaction (DR) and compound nucleus reaction (CNR) are: the time duration: ... 1 The metal plate is typically attached to a circuit which collects the ejected electrons making the system net neutral overall just with a current flowing. Also they are usually more easily ejected because the plate is at a potential attached to a battery. However, if the plate were suspended by an insulator in a vacuum and it continuously lost electrons ... 1 From a practical point of view, when performing nuclear structure calculations one often stores matrix elements of the nuclear interaction in the form$\left\langle a b | V | c d \right\rangle$. Where$a,b,c,d$label different orbitals. If we distinguish protons and neutrons, then we need to store separate matrix elements for$\left\langle pp | V | pp ...

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The range of attraction between two protons is short. So if you have a large number of protons only the long range repulsive Coulomb force will dominate and the nucleus will not be stable. So you need neutrons which are free from this repulsive force.

1

Answer from astrophysicist. Point 1. The problem with eternal shining is that star loses energy with photons (and some material) and it'll need income of energy from somewhere. There are brown dwarfs, black dwarfs, black holes etc., which are just remaining of stars and don't 'shine' (or, in case of white dwarfs, fade out to the point we cannot detect ...

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