# Tag Info

3

I personally doubt that the Compact Fusion Reactor as presented by Lockheed Martin last week can work, but I haven't seen enough information to be certain. And to some extent, you never know until you try. (As I understand it, they only have a very early prototype, I mean try as in a full scale prototype.) What I think I can say with certainty, is that it ...

0

If we re-designate all positive electric charges as negative and vice versa, while keeping their absolute value, the resulting physics would be the same. So exact choice is merely a matter of convention. It only matters that electric charge of proton and electron are opposite ($Q_\text{proton}=-Q_\text{electron}$). The charge is additive, meaning that if ...

1

In quantum theory of elementary particles (in a sense of irreducible representation of Poincare group with mass $m$ and spin (helicity) $s$) if some operator $\hat{Q}$ of internal symmetry commutes (like electric charge charge) with Hamiltonian $\hat{H}$ of given field theory, there must be  [\hat{Q},\hat{\varphi}^{\dagger}_{A}(\mathbf p)] | \rangle = ...

0

It might be possible, but it's uneconomical and probably also uses more electric energy than you get from the power plant. Do you really mean fisson? Fission is splitting a nucleus in two approximately equal parts. This is done with certain isotopes of e.g. uranium 235 which are only barely stable and sometimes fission spontaneously. Other isotopes such as ...

0

The difference is that angular momentum flux $\psi$ depends on the direction and scalar momentum flux does not, because is integrated over all directions: $\phi = \int_{4\pi} \psi d\Omega$. For the rest, they exhibit the same dependence on Energy. So $\psi$ gives the amount of neutrons that at a certain position $\bf r$ and at energy $E$ are moving in the ...

2

Elements heavier than iron are produced mainly by neutron-capture inside stars, although there are other more minor contributors (cosmic ray spallation, radioactive decay or even the collision of neutron stars). Neutron capture can occur rapidly (the r-process) and occurs mostly inside supernova explosions. The free neutrons are created by electron capture ...

3

1mol of protons is $6\times10^{23}$ particles. If proton decay has a half life of $6\times10^{33}$ years, then there should be one proton decaying per year per $10^{10}$mol. Since one mol of protons weighs one gram, that's a mere $10^7$kg of protons. Even at the density of liquid hydrogen (70kg/m$^3$), that's only $1.4\times10^6$m$^5$ of liquid volume. Take ...

1

"Free" electrons and protons can combine to form neutrons (and electron neutrinos) through weak interactions. The process would usually be energetically unfavoured in low density situations because the rest mass of the neutron is about 1.29MeV higher than the combined rest mass energies of the proton and electron. This reaction, between free protons and ...

3

Yes, they can be. The production of neutrons by deep inelastic scattering of electrons on protons was studied in HERA experiments in the 90's. No neutrino needed, at least not as one of the colliding particles. Here are some links: Deep-Inelastic Electroproduction of Neutrons in the Proton Fragmentation Region Measurement of Leading Proton and Neutron ...

0

In electron capture an inner atomic electron and a proton will together form a neutron and an electron neutrino, I'm not sure if that's what you're looking for however. Source: Young and Freedman Physics textbook Electron Capture Wikipedia article`

-1

Since a neutron decays into a proton, electron and an electron anit-neutrino you will need to get all three together at the same time. Good luck on that one!

0

To illustrate the difficulty of answering your question, here are the components of a local fallout model, as given in Stanford's Fallout models and radiological countermeasure evaluations (PDF link): Weapon models Condensation models Particle cloud models Distribution models Contamination models Dose models Resource models Transport system models Every ...

4

Please note, that this test was conducted by exactly the same group that did the previous test, lead by Guiseppe Levi, who is closely connected to Andrea Rossi. Also, you can read from the report, that Rossi himself was present in the test pulling the strings. Hardly independent testing, is it? The above facts alone are enough to make the report somewhat ...

2

Their latest press release says it has been confirmed by 6 independent observers. However, so far I cannot find who these are nor the status of their scientific credentials. If Rossi really has something working I do not see why he does not simply get a whole slew of patents and throw the whole technology wide open for inspection by qualified teams of ...

2

A link would be useful to see the context of the quote One often hears the phrase "most of nuclear physics is in the low energy regime of QCD, where strong coupling constant is large ...", As it is it is wrong. This is correct The nuclear force is now understood as a residual effect of the even more powerful strong force, or strong interaction, ...

1

Here are some "order-of-magnitude" arguments: Quoting https://en.wikipedia.org/wiki/Decay_heat#Spent_fuel : After one year, typical spent nuclear fuel generates about 10 kW of decay heat per tonne, decreasing to about 1 kW/t after ten years Now since this is heat, you can't convert it to electricity with 100% efficiency, the maximum possible ...

0

There are many reasons why they are not used, the reasons or my explanations may not or may not be good/useful. in no particular order Alpha voltaic's are prob the best, with Pu238 or Am241 being likely candidates, though, Cm-243,244 are also options. They are superior to batteries in almost every way, but cost limits them to micro power devices. uses are ...

1

This is perhaps a comment to the @Floris answer. (I can move it there.) First about T1, It not only describes the decay of the Z-magnetization. But if you suddenly turn on a B field it also describes how long it takes the spins to become polarized in that direction. (Spins are not immediately polarized.) Concerning the decay of x-y magnetization. (I ...

3

I don't consider myself an expert in MRI, but let me try (since nobody else has stepped up in the last hour...) You are right with your first assertion: the spin precesses about the B vector (this is why you get resonance in the first place). However, on average there is a net component of the magnetic moment aligned with the B field. This is what gives ...

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You are not required to regard decay as quantum mechanical (QM) phenomenon. Regard it as a nuclear reaction, which occurs spontaneously. There are similar reactions in chemistry, like ozone decay. Modern QM is unable to calculate decay probabilities and half-lives, they are only measured experimentally. In principle it is possible to model nucleus in the ...

1

I don't know what "penetration power" is or why quantum tunneling needs to be invoked. Sr-90 decays entirely via beta emission with up to $0.546\ \mathrm{MeV}$ given to the electron, and its daughter isotope similarly decays with up to $2.28\ \mathrm{MeV}$ given to the electron. These energy ranges are right around the $1.71\ \mathrm{MeV}$ of P-32, whose ...

3

Yes, beta decay of Sr-90 produces a 546keV beta (note that Y-90, the daughter nucleus, is also a beta emitter, but at 2.284MeV). This energetic electron can then produce bremsstrahlung x-rays from interactions with electrons. For a given x-ray energy, lead will have some absorption coefficient - really this is no different than visible light interactions ...

2

I wonder if your number 48% comes from the typical Carnot efficiency of a heat engine - see for example a detailed description at http://www.visionofearth.org/industry/fusion/how-do-we-turn-nuclear-fusion-energy-into-electricity/ When you want to use heat to create electricity, you typically convert the heat into motion (for example by rotating a turbine, ...

0

Here is a translation of the original paper by Bethe: Bethe article

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