# Tag Info

2

You should read the article in wikipedia on nuclear force. Various models exist that describe the behavior of nuclear forces, which are the result of a spill over of the strong force, the force that exists within the proton and the neutron. From the link Force (in units of 10,000 N) between two nucleons that experience the nuclear force, as a ...

1

This is a nice idea that has been studied thoroughly, and continues to being studied. You are talking about a type of sub-critical reactors referred to as Accelerator Driven Systems (ADS). Here is a nice review Review:Basics of accelerator driven subcritical reactors, where references to previous studies are given. For modern advances you might like to check ...

14

A nuclear winter would be a result of large amounts of smoke blocking light from the Sun. The smoke would be from the fires started by nuclear bombs on cities, not directly from the bombs. Most bomb tests have been underground, and the above ground tests were mainly done where there wasn't much to burn, for example in the Nevada desert so they didn't ...

0

I'm not that firm with the concept of blazars, but as with all super-massive astrophysical objects like neutron stars, black holes and all their subclasses, their heaviness forces the particles to highly-relativistic states. There you just can't use the hypothesis of "charged billiard balls" for atoms any more - a neutron star for example is only stable ...

0

Explaining the nuclear structure (...and therefore any other properties like half-life or even mass) is still an open topic of research. The models at disposal (see for example Two-Center Shell Model) which are used for super heavy nuclei searches are generally non ab-initio, therefore no fundamental conclusions can be drawn regarding the underlying physics. ...

4

I want to cite the original paper of Walecka where he awnsers your question: From Annals of Physics 83/2 (1974) p. 429 "A theory of highly condensed matter "The reader might object to the fact that there is no one-pion exchange tail in this interaction; however, the strong spin and isospin dependence of the potential arising from the exchange of an ...

1

For a fission chain reaction to spontaneously happen in uranium there are some requirements to be fulfilled. A) there should be at least 4% U-235 in the mix of uranium isotopes. B) there should be a moderator to decrease the energy of the neutrons which result from the fission of a uranium core. 2 billion years ago the percentage of U-235 was large ...

4

There isn't really a difference between "natural" or "artificial" reactions. All reactions are just "things that can happen". Some things only happen in certain circumstances, and those circumstances may be very unlikely to occur without being specifically engineered, but there is no reason in principle why they could not happen naturally. There is evidence ...

3

They don't. Here's a figure from Wikipedia: Typically there's daughter with mass around 95, a daughter with mass around 140, and two or three extra free neutrons. In discussion of environmental contamination after nuclear accidents, you hear a lot about iodine-133 and strontium-90, because they are relatively long-lived and biologically active. Iodine-133 ...

2

No, there are many nuclei produced by fission. You should show the mass numbers in any case-you have only specified the division of the protons, not the neutrons. This Wikipedia article discusses the range of fission products and states that each decay is not predictable, but the probability distribution is measurable. For the second question, no, ...

4

First, note that we are quite sure what the overall nuclear spin is; we are not sure how to obtain it mathematically from available models. Due to the phenomenon of color confinement, there are no gluons at low energies in QCD (the theory underlying nuclear physics). Importantly, you can't say there are this or that many gluons in any proton or neutron. ...

0

Only half of the total number of nuclear reactions release energy. The other half are the reversed processes that absorb energy. However almost the totality of the spontaneous processes are the ones that release energy. Processes that requires some external energy to be inserted into the system are much more rare as can take place only when this energy is ...

3

Very interesting question! In chemistry you spend lots of time discussing exothermic and endothermic reactions: when you put your reagents together, sometimes the reaction heats things up, and sometimes the reaction cools things down. Nuclear reactions are very different, in that essentially all spontaneous reactions studied in laboratories are exothermic. ...

10

If they didn't release energy, they wouldn't happen. The alternative, nuclear reactions that require energy, clearly need said amount of energy, which has to come from somewhere, e.g. kinetic energy involved in the collision of two nuclei (even ones that release energy usually have a "barrier" and some amount of initial kinetic energy is needed to overcome ...

0

Yes jim is right wikipedia article will answer your most of the questions. Still ill list few facts about nuclear reactions here hope those will help you Nuclear reaction basically has two major types fision reaction and fusion reaction (There are other types also i am just listing only two to answer your quenstion in simple words) In fusion reaction two ...

3

(The following extends Georg's remarks earlier, where K-capture refers to K-electron capture.) Beta-plus decay competes with electron capture, but there are few positrons around for beta-minus decay to compete with, so even when beta-plus decay is possible, its branching ratio may be small or overwhelmed by EC. Moreover, in EC (versus beta-plus decay) the ...

5

At first, consider two particles decay: $A\rightarrow B + e^-$ Where A is initially rest. So $\vec{p_B}+\vec{p_{e^-}}=0$ now \begin{align} \frac{p_B^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released} \\ \frac{p_{e^-}^2}{2m_B}+\sqrt{p_{e^-}^2+m_{e^-}^2}&=E_{released} \tag{1} \end{align} see here you have uncoupled equation (equ.1) for $p_{e^-}$ ...

1

You're correct: the unique thing about beta decay is that there's a three-body final state. In the reference frame where the decay takes place at rest, the daughter nucleus, beta particle, and neutrino share the momentum roughly equally, and because of the mass scales the beta and the neutrino take the bulk of the energy. It's pretty straightforward to show ...

1

Try exploring the National Nuclear Data Center. If I search the Evaluated Nuclear Structure Data File for "incident particle: g" and "outgoing particle: g" I get datasets for most, but not all, nuclei. There are other databases hosted by the NNDC as well, and all are well-referenced to the experimental literature.

1

This is a hard question to answer, in the end. However, be assured that long, long, ago we started looking for nuclear expositions by looking for X and gamma radiation using the Vela satellites http://en.wikipedia.org/wiki/Vela_(satellite) . These did not find much in the way of violations of the nuclear test ban treaties but did discover astronomical ...

2

This would ultimately be more a problem of signal processing than physics. The situation is detecting a signal at a very low signal to noise ratio. At the broadband level, the noise (starlight) is several orders of magnitude more intense than the signal (the explosion). The only hope would be some sort of spectral technique , taking advantage of spectral ...

3

To calibrate our expectations, consider the largest nuclear weapon ever detonated, the Tsar Bomba. It's yield was at most about $58$ megatons TNT equivalent, or about $2.43\times10^{24}$ erg. Now, let's consider a smallish star, something like Gliese 581, which is reasonably nearby, small and faint, and has a planetary system (of some sort: the number of ...

0

If you're in the pool the pressure wave could crush you depending on strength of blast. Water can't compress. So there's a two fold issue to entertain your idea, heat and pressure. Radiation will be your next concern if you survive the initial blast.

1

If you only have two particles, they only have a mutual angular momentum — there's only one $L$. If you have many particles orbiting a central potential, like electrons in an atom, or if you can get away with pretending like you do, as in the nuclear shell model, then the eigenvalue of the system under parity inversion is the product of the eigenvalues of ...

1

The final state would be hot, high pressure gases inside, if it is a sealed container. Examples of gases would be nitrogen and carbon dioxide, but depend upon the explosive material.

1

Explosions such as you describe are exothermic reactions, releasing their stored chemical energy in the form of light, heat and the acceleration of the mass making up the outside of the device. Because energy is conserved, and because your question stipulates that the explosion is surrounded by an adiabatic barrier, the inside of the container would ...

2

To answer your question, let's look at how this equation can be derived. Say, at some time $t$, there are $N(t)$ nuclei. Let $p_t(Δt)$ be the probability that any one nucleus has not decayed (this probability is assumed to be the same for all nuclei) after an additional time $Δt$. If we also assume that there are a lot of nuclei (this is important), we can ...

2

A more general way to state the physics here is something like The probability that any given nucleus will have already decayed after a time $t$ has elapsed is $P_{decayed} = (1 - e^{-t/\tau})$ and is independent of the fate of all other nuclei. Which remains valid for any number of nuclei and from which you can easily extract the large numbers law ...

2

Let us take Newton's law for a particle in special relativity $$F^\mu = \frac{d p^\mu}{d \tau} = \frac{d m_0}{d\tau} u^\mu + m_0\frac{du^\mu}{d\tau}$$ where $F^\mu = - \partial^\mu V$, $p^\mu = m_0 u^\mu = m_0 \gamma (c,\vec{v})$ and in our lab frame $V(x,t) = V(x)$. Taking $m=m_0 \gamma$ and a transformation from the proper time of the particle to our lab ...

1

It's the total product. The famous example is the spin of the deuteron. We have evidence that the two-nucleon isospin triplet with $I=1$ is unbound because we do not observe diprotons or dineutrons in nature, so we expect the deuteron to have isospin $I=0$. We know that the deuteron has positive parity, so we require $L$ even; by antisymmetry the deuteron ...

2

An age estimate of the Earth is not used to determine half-life of an isotope. Instead, the rate of decay (decays per time unit) relative to the amount of sample (number of nuclei), and first order kinetics are used to determine half-life. For example, in 1932 Kovarik and Adams calculated the Uranium-238 half life based upon 24,770 alpha particle decays ...

5

It isn't possible to measure potential energy because it has a (global) gauge symmetry. It's like trying to measure the height of a mountain - this could be the height above sea level, the height relative to the deepest sea trench, the height relative to the centre of the earth and so on. Any measurement can only measure the change in potential energy, and ...

0

Chemical forces are the result of the higher order moments, mainly of the electric field, of the neutral atoms. Look at these orbitals: The five d orbitals in ψ(x, y, z)2 form, with a combination diagram showing how they fit together to fill space around an atomic nucleus. These anisotropic shapes in space allow for spill over attractive and ...

1

In both cases the potential energy for two interacting unit charges is $$U = -\alpha\frac{\hbar c}{r}$$ The strong force is between color charges and has $\alpha \approx 1$, while the electric interaction is between electric charges and has $\alpha \approx 1/137$. However the gluons, which carry the color force, are themselves charged. This means that ...

3

Tabulated values of nuclear mass defects do in fact include an arbitrary offset. Usually the standard is that one mole of carbon-12 weighs exactly twelve grams, so that a bare proton or neutron has a positive mass defect, while a tightly-bound nucleus like iron or nickel has a negative mass defect. In computing the $Q$-values of decays it is only the ...

3

If you use relativity (which the use of $E = mc^2$ implies), we cannot choose the potential arbitrarily, because the relation between energy and mass makes absolute values of the energy measurable through the gravitational forces exerted by stored energy. EDIT: Since you asked, I will explain it in somewhat more detail: Let \$V : \mathbb{R}^4 \rightarrow ...

-1

You are confusing the strength of the bond with the amount of energy released when the bond is broken. The strong force is about 100 times stronger than electromagnetic attraction, although, when nuclear bonds are broken matter is converted into energy, but when a chemical bond is broken energy is only given off from a chemical change. The strength of a ...

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