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Remember that plasma is a phenomenon of both high temperature and low pressure. Plasmas are ionized fluids, which occur either when the temperature is so high compared to atomic ionization energies, or when the fluid is so diffuse that charged particles are unlikely ever to find each other. A common tool in ordinary vacuum systems is the cold cathode ...


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This from the ITER page In the heart of the Korean tokamak KSTAR, in operation since 2008, a plasma pulse burns brightly. But don't be fooled—the brightest areas of the photo are in fact the coolest. At 150 million °C (the temperature in the centre), the plasma doesn't emit in the spectrum of visible light. © National Fusion Research Institute, Korea ...


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I guess the plasma in the ionosphere is typically invisible: the plasma frequency is much lower than the frequency of visible light, so almost all light passes through the ionosphere.


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Hydrogen plasma is probably invisible because all atomic lines of hydrogen are in the ultraviolet region. Somewhat related - hydrogen flame burning in air is completely invisible.


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Dark lighting is (evidently), the result of Bremsstrahlung radiaiton from fast moving free electrons in the air. It's not the usual re-attachment emission that I'm used to seeing in lab plasmas. But briefly, yes you could categorize this as a cold plasma (it's got a degree of net-neutral ionization). There are also plasmas that might just not be visible to ...


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You seem to have hit upon the challenge. Converting between the two is situation dependent. It depends on the source of the radiation and the amount of protection. One source might deliver energy as 50% alpha and 50% beta another might be 100% gamma. Put a sheet of paper in the way and the first source will shift and deliver a higher percentage of energy ...


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The Stefan-Boltzmann law for net power radiated pertains to the object. That is, we're simply asking, how much radiation leaves this object (this depends on the object's emissivity), and how much radiation is absorbed by this object (this depends on the objects absorptivity). The emissivity and absorptivity in the equation you present thus pertain to the ...


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The two existing answers have noted that EM radiation (X-rays, gamma) is effectively stopped by electrons. There are at least 4 other types common types of radiation: Alpha particles (2 protons, 2 neutrons - essentially He4 2+) Beta particles (single electron) Neutrons Ions (other than alpha particles) The first thee are commonly generated by nuclear ...


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Radiation can be several things, but since you specifically mentioned lead shielding, let's look at X-rays - a lot of what you learn applies to other radiation as well. To stop radiation it needs to interact with "something" that makes it give up its energy and momentum. This is how you get the radiation to stop going in the direction it was going. Now ...


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There are several different things labeled "radiation". Gamma rays are electromagnetic radiation, similar to visible light but at a higher frequency. X- rays are also electromagnetic radiation. For electromagnetic radiation, elements with heavy nuclei are good shielding. See this Wikipedia article on protection against electromagnetic radiation. Also called ...


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For the first question, yes. Because the surface of the sun is close to a blackbody emitter, it radiates at all wavelengths below the peak. So radio waves are included. However, the longer the wavelength, the less the power that is put into that portion of the spectrum. Radio is so far from visible light on the EM spectrum that the solar radiation in ...


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The diffusion approximation is one solution to the radiative transfer equation. In general, the choice of applying this particular solution depends on the optical limit, as you say. For an optically thin medium, radiation will travel and may interact along the way. This is not characterized as a diffusive process, because the beam can interact with the ...


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An optically thick medium is one for which the mean free path of a photon is low. This means that a photon won't be able to travel very far before it interacts with the matter than makes up the medium. The measure of optical thickness, optical depth, does depend on the volume of material in the medium. For example, for a material with a fixed density, ...


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Well, I would say that you are confused about the quantization idea. For quantization to be taken into account, physics of particles should be examined. For a simple example, particle in a box problem given in the beginning of quantum mechanics courses. So, quantization does not mean only atom, it can be examined for many situations in the real life. ...


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This link gives a clear account of the difference between the classical and the need for the quantum mechanical formulation . Blackbody radiation" or "cavity radiation" refers to an object or system which absorbs all radiation incident upon it and re-radiates energy which is characteristic of this radiating system only, not dependent upon the type of ...


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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 $p_{e^-}$ ...


2

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 ...


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There, in fact, is such a way to convert heat produced from radioactive decay into electrical energy. Many systems doing just this have already been designed and used. The most straightforward device is a Radioisotope Thermoelectric Generator, which does exactly what you are asking. However, in principle, almost any system could be implemented that uses ...


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The gray model results from starting with the assumption that we have a plane-parallel slab: The light ray from the source (i.e., the star's atmosphere) travels at some angle, $\theta$, from normal, $z=0$. Since the light is coming from an angle, we need to account for that by modifying the radiative transfer equation to have a vertical optical depth, ...


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Radiation is only damaging to DNA when it is ionising (contains enough energy to remove electrons from atoms). This occurs in frequencies higher than visible light (UV, x-rays, gamma rays). Anything below that, such as radio waves that RC toys use, won't cause any harm with that little power.


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The remote car toy radiation are radio waves. So it's as dangerous as using a radio... The power of thoses emitter are not enought to be dangerous. So it's safe for your kids.


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The cross section $\sigma$ is related to the absorption coefficient $\alpha$ by: $$ \sigma = \frac{\alpha}{N} $$ where $N$ is the number density of the scattering medium i.e. the number of particles per unit volume. This is described in more detail in the Wikipedia article on the absorption cross section. If you want $\alpha$ in units of cm$^{-1}$ you ...


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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.



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