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The penetrability factor is a carefully defined factor that separates the coulomb-force and the interesting nuclear-forces in the relationship between the partial decay width and the reduced width (for R-Matrix analysis) of a particular resonance. $$\Gamma = 2\gamma^2P(l,\rho,\eta)$$ Here $\Gamma$ is the partial decay width, $\gamma$ is the reduced width, ...


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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 ...


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You have exactly the right idea. The decay products are only useful in as much as we can compare a previous isotopic ratio to the current (measured) ratio. How this is done differs from dating method to dating method. As an example, C14 dating originally came about with the idea that atmospheric and therefore living plants have a near-constant ratio that ...


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One needs to account for lead that was formed as lead, rather than by radioactive decay. Wikipedia has a discussion. Lead-204 is not formed by decay, so if you know the primordial distribution of the isotopes, you can compute the primordial amount of each of the others. The rest comes from decay.


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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.


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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 ...


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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: ...


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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 ...


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Actually fission processes are stochastic, so we can't predict exact products of it. The most probable fission products are Cs-137 and Sr-90. The total amount of neutrons emitted is also quite unpredictable, but as I recall it usually lies in the range between 2 and 4.


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Just wanted to summarize my comments: Are you sure that it is Germanium-72? Actually, it is a stable isotope and it should not have any decay mode. Can you please tell us which table you are using to search for radionuclides? There is also Mn-54 which emits 834 keV gamma radiation, but it is relatively unstable so I doubt that it's easy to find its presence ...


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I was thinking of the H-bomb, I assume they split hydrogen atoms for that. For the H-bomb, Deuterium is fused into Helium. Deuterium is easier to fuse than Hydrogen. But I think the energy output of Hydrogen into Helium, which takes more than 1 step, is .68% of the rest mass. If we apply E=MC^2 to a mole of hydrogen (a bit over 1 gram, so not ...


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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 ...


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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 ...


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Lubos's answer covers very well and in a general way the different possibilities for reactants and products in the reactions of subatomic particles and nucleons. The basic equation is Energy of photons = energy of reactants - energy of products. Due to the enormous energies involved it is actually unlikely that visible light will be released, most ...


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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 ...


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What is meant by mass defect of a single neutron or a single proton? The reduction in its mass. The mass defect of a nucleus represents the mass of the energy binding the nucleus, and is the difference between the mass of a nucleus and the sum of the masses of the nucleons of which it is composed. It isn't quite that, in that binding energy is ...


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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.


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When an isotope of an element $A$ is fissioned, it breaks up in a number of pairs of daughter products, say $B$ and $C$, where $B$ and $C$ are isotopes of comparable atomic mass. Often 2 or 3 neutrons are also released (opening up the possibility of chain reactions). However, the total number of nucleons, that is protons plus neutrons, is preserved during ...


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Very hard to tell without knowing how the spectrum was produced (type and size of the detector, resolution, anticompton, ...). Anyway 180 counts do not seem so many. The single escape peak is normally weaker and the double escape even more, especially if you are just above the pair production threshold. Sounds reasonable that you may not have significant ...


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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 ...


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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 ...


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According to wikipedia, Vanadium-48 decays via $\beta^+$ (positron emission) to Titanium-48, which is a stable isotope. The emission of neutrons for Vanadium-48 isn't allowed becuase it doesn't conserve energy: Vanadium-48 has a mass of 47.9522537 u, and Vanadium-45 plus 3 neutrons have a total mass of 44.965776 u +3·1.00866491600 u = 47.991770 u. For a ...


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Well your reasoning is completely justified and valid. I am going to provide you with some elements to answer more what you want to know, but the exact response would depend on the case under study, and you will see why. Internal conversion happens mainly in heavy nuclei because they have electrons deeply bound, and their ionization energies are higher, ...


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You do this mainly because we want to calculate the spin-orbit interaction precisely for the electron. And we do this because we observe that movement of the electrons (within the validity of this view) is decoupled from the movement of the nucleon. This makes sense because the nucleus mass is about N*2000 times larger than that of the electron, where N is ...


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In theory this could happen provided the correct conditions would hold. But in reality fission will yield fragments most probably having 2/5 and 3/5 of the mass of the original fragment, and some 2 or 3 neutrons will take part of the kinetic energy. So that already imposes that the medium needs to be in some heated state so that nuclei have the required ...


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John answered it well. Just to add, I beleive that there can always be better design to extract more out of the system and the decision is usually around the trade-offs of the return with the well estabilshed methods at the time of imlementation. In future, we may find a simplistic easy way to extract heat from the waste for useful purposes. Thermocouples ...


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Spent fuel is extremely radioactive, and it will induce radioactivity in anything that is in contact with it such as the cooling water. So you'd need separate primary and secondary cooling circuits to avoid heating your houses with radioactive water. You could do this, but it starts getting expensive.


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In a dilute gas, the photon density should be the same as inside an evacuated black box of the same temperature (independent of the gas density). Edit: In other words, the massive particles in the Sun have a temperature based on their kinetic energies. The photons must be distributed as black-body radiation at the same temperature. The density of photons in ...


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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 ...


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The number of atoms in the sun is on the order of $10^{57}$, see here. The number of photons emitted per second is on the order of $10^{44}$, see here. The difference is on the order of $10^{13}$. So if photons emitted for $10^{13}$ seconds, 315,00 years, the number of photons would begin to overtake the number of atoms in the sun. The sun is much older ...


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With regard to your first question, the transmitted plane wave doesn't undergo any scattering from the potential. This is made explicit by the representation of the scattered wavefunction as a sum of an incident planewave and an outgoing spherical wave. As such, the transmitted wave doesn't have anything to tell us about the scattering event. All of that ...


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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 ...


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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 ...


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If I understand your questions correctly: Yes, it can be somehow the other way around. But we do "know": There are two sort of particles in here, one of them has a certain charge and is light, the other has the opposite charge and is heavy. You can then claim that the heavy ones rather stay in place and the light ones sprint around and make up the current. ...


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Generally it is said that current is due to the flow of electrons; how can we make this claim? If the wire is in a magnetic field the moving charges will move in a circle based on the magnetic force. This happens until enough charge imbalance develops on the edges of the wire to produce an equal and opposite electric force. But measuring the voltage ...


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From classical models, the electron and a proton revolve around their mutual center of mass, which approximately lies on the proton itself, because the proton has a significantly higher mass than the electron. This is why electrons revolve "around" the proton, and hence form the outer layer of an atom. Quantum mechanically, electrons could never form a ...


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As I see it, and, correct me if I'm wrong, but there is a way to do it. The problem with using pure Uranium or any other readily fissionable element is that, as one element decays, that releases 2 neutrons which can speed up the decay of nearby elements. If you had a ball of U235 or U238, say, the size of a planet or even a small moon, you'd have a ...


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In principle yes, though it would be a highly contrived situation and not one likely to arise naturally. A star works because hydrogen to helium fusion is energetically favourable. But the process has a huge activation energy so you need an environment as hot and dense as a star's core to provide that activation energy. Likewise, for any sufficiently heavy ...


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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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(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 ...


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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 ...


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The only thing I have to add is to answer your question about where the -3 and 1 come from. First, the spin operator is related to the Pauli matrices by $$s_1 = \frac{1}{2}\sigma_1$$ and so we have the relation $$\sigma_1 \cdot \sigma_2 = 4 s_1\cdot s_2$$ Next, we can evaluate $s_1\cdot s_2$ using $$ S^2 = (s_1+s_2)^2 = s_1^2 + s_2^2 + 2 s_1\cdot s_2$$ $$ ...


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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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, ...


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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 ...


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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 ...


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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 ...


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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 ...


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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.


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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 ...



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