# Tag Info

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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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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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Actually, in a large solar flare particle energies can get up to 1 GeV, but the top energy of some particles is not really the issue. The issue is the flux of these high energy particles. A 10 MeV proton or electron pretty much rips through most spacecraft bodies, thus, their electronics are effectively exposed to particles at these energies. The often ...

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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: ...

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 ...

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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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Is it the threat of having to eat five servings of fruits and vegetables a day? When the magnetic field deflects charged particles the energy is deflected and this means less radiation and less ionizing radiation from he particles that get deflected. A solar electron, proton, or alpha particle with 10keV would be capable of ionizing molecules and atoms. ...

0

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 ...

1

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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When a nucleus decays, there is usually excess energy (binding energy of the nucleons) liberated, and this energy is carried away by any of the "fragments" - daughter nuclei, emitted particles (alpha, beta, gamma, neutrino, ...). The amount of energy involved can be quite large (think atom bomb), so when the energy liberated is released in the form of ...

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Note: I'm aware this answer needs some work: Whilst I'm sure there will be a plethora of much more detailed answers I'll give a simple answer to question one. This is more about internally understanding how the process works than an extremely accurate description. Gamma ray is an electromagnetic wave. Basically we have a set of forces called electromagnetic ...

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The Sievert is a derived measure of stochastic health risk. It's used only in cases of low dosage ionizing radiation. High dosages that produce deterministic health effects are measured in the Gray (Gy), a purely physical term which represents the actual deposit of one joule of energy in one kilogram of matter. Unlike the Gray, the Sievert does not ...

1

First, what is an "EM field"? At any point in space we can define an electric field (E field). It points in the direction that a positive charge would accelerate if we released it from that point. At any point we can also define a magnetic field (B field). It is a little more complicated, but for now let's just say it points in the direction that a ...

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Any electrically conductive material will block IR. The greater the conductivity, the greater the blocking. Aluminum foil will kill all IR, bot high range and low. Most plastics allow IR to pass through. Glass will bock low frequency IR (red hot), but allow the passage of high frequency (white hot) IR. Hence, the heat of the sun will easily pass into a ...

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I'm guessing this is related to your earlier question, Will neutral particles be affected by EM waves?, and you're puzzled that Rayleigh scattering by air and reflection/refraction by solids and liquids seem so very different. The answer is that in Rayleigh scattering each scattering object behaves as an independant scattering centre, so the scattering is ...

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The solution for oscillating charges or plasmons has an evanescent field that exponential decays in the z direction (out into the "air"). This is currently being postulated as a way to subvert the diffraction limit. For more info regarding the lens : http://www.sciencemag.org/content/308/5721/534.

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To expand on this, you would point your telescope at an empty patch of sky (as mentioned). You expect the CMB to be thermal to a very high degree ($\sim 1$ part in $10^5$). Also, you can point your telescope to any direction on the sky and get (statistically) the same observation (if you have the resolution to pick up the primordial fluctuations that are ...

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The instrument should be pointing at an empty ( no stars no galaxies) region of the sky and be able to record very low frequencies Other radiation comes in much higher frequencies from stars , and would not overlap with the low frequency part. >Discovered accidentally in 1964 by Penzias and Wilson (Nobel Prize, 1978), the CMB is a remnant of the hot, ...

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