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Electron capture Electron capture (K-electron capture, also K-capture, or L-electron capture, L-capture) is a process in which the proton-rich nucleus of an electrically neutral atom absorbs an inner atomic electron, usually from the K or L electron shell. This process thereby changes a nuclear proton to a neutron and simultaneously causes the emission ...


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The decay of potassium-40 to argon-40 is either a $\beta^+$ decay in which what is emitted is not an electron but a positron $$ {}^{40}{\rm K} \to {}^{40}{\rm Ar} + e^+ + \nu_e $$ or, more frequently (if we have whole atoms), an electron capture that you mentioned in which no charged leptons are emitted at the end! About 11% of the potassium-10 decays ...


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RAD, Radiation Absorbed Dose 1 RAD = 100 ergs absorbed per gram. In my opinion there is no simple formula to calculate the dose in a very straight forward manner. To calculate the dose you have to calculate the energy deposition in a mass. For this you need range vs energy deposition curves. In the case of heavy particles they show a so called Bragg peak i....


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Wikipedia's article on radioactive decay gives the equation that describes a two-step decay chain, together with the corresponding solutions. In your case, you'll have $$\begin{align} N_{\text{Kr}}(t) &= N_{\text{Kr}}(0)e^{-\lambda_{\text{Kr}}t} & N_{\text{Rb}}(t) &= N_{\text{Kr}}(0)\frac{\lambda_{\text{Kr}}}{\lambda_{\text{Rb}} - \lambda_{\text{...


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I figure that given the rather extensive question that it does deserve some answer. I don't know if there is stack exchange on the history of science or physics, but that might in fact be a more appropriate place for this. The grandfather of quantum mechanics was Max Planck. His assumption that distributions of energy occurred in discrete units is what lead ...


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During proton therapy, most of the damage is actually done in the last few mm before the beam stops - at the point called the Bragg Peak Yes, the penetration distance is largely determined by the energy above a few MeV; as the particle slows down, it starts to dump more energy per unit length. Quoting from "The physics of protons for patient treatment" (...


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Since no nuclear bomb has been used since since Nagasaki we can and indeed do only infer "applications" if you are talking about dropping a nuclear weapon on a human population. These estimates based upon the criteria you have given appear very effective as no theronuclear weapon has been detonated on a population center since Nagasaki, Japan in 1945. I ...


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The so-called TNT equivalent of a nuclear weapon is an unambiguous way of quantifying how much energy is released by the nuclear weapon. There's nothing 'wrong' about it. The only caveat is that the damage caused by, say, Little Boy versus 15 kilotons of TNT would not be identical despite having an equivalent yield (for various practical reasons). ...


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The assumption of straight field lines in the deceleration shell has to do with the geometry of the derivation's setup, as shown in the figure below (from Purcell simplified; please keep in mind that not all assumptions are mentioned in the caption): 1) The width of the shell is much less than the radius of the shell. This is because the field is observed ...


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The very flexibility that you mention in your post is a bit of a problem in an experimental context. In order to understand the signal that a Cf-Be calibration source would generate in your detector and tease useful information out of it, you're going to have to model all three channels that generate neutrons and the gammas that escape the source. This means ...


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there are two types of vacuum in physics. one with and one without radiation (radiation being any kind of electromagnetic interaction, or photons). Sloppy scientist call a matterless volume vacuum, although it still has radiation (and therefor contains energy). You can imagine that it is very hard to achieve a true vacuum in which only the vacuumfluctuation ...


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Anything can emit radiation if it has mass just from acceleration alone. The question is can there really be a mass that is incapable of adsorption. Also blackbody radiation does not depend on absorption.


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The lower level of laser transition is above the ground state,this means that the fast non-radiative transfer from the lower laser level to ground state keeps a reasonable population in the upper laser level(3rd) which is fundamental for lasing.Hence its more effective than 3-level system


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Yes, there is a direct quantum mechanical connection between absorption and emission through something called the Einstein coefficients. See https://en.m.wikipedia.org/wiki/Einstein_coefficients If a material has zero probability of making an absorptive radiative transition between quantum state 1 and quantum state 2 (where state 2 has the higher energy), ...


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Lead is a chemical element in the carbon group with symbol Pb (from Latin: plumbum) and atomic number 82. It is a soft, malleable and heavy post-transition metal. Freshly cut, solid lead has a bluish-white color that soon tarnishes to a dull grayish color when exposed to air; the liquid metal has shiny chrome-silver luster. Lead has the highest atomic ...


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Temperature is a measure of how much energy photons have $E = h\nu$ $= kT$, and the temperature reflects the energy of the photon incident on the detector. The total irradiance or EM power incident on an area $A$ is $P = \epsilon\sigma AT^4$ from Rayleigh-Jeans law. Here $\sigma$ is the Stefan-Boltzmann constant and $\epsilon$ the emissivity of the body. ...



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