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One more assumption must be made regarding the $^{222}Rn$ concentration at time zero (week ago). If you assume radioactive equilibrium between $^{222}Rn$ and the parent $^{238}U$ you can proceed like you did. But instead of taking one wall, you should take into account all the walls made of the $^{238}U$ contaminated material. On the other hand, if you ...


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Here is what wiki has to say about Polonium radiological toxicity: By mass, polonium-210 is around 250,000 times more toxic than hydrogen cyanide (the LD50 for 210Po is less than 1 microgram for an average adult (see below) compared with about 250 milligrams for hydrogen cyanide[66]). The main hazard is its intense radioactivity (as an alpha ...


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Let's suppose you have swallowed one of the Po-210 sources from this student kit. Its activity is 3700 Bq (0.1 ╬╝Ci). Based on the Table 6 in the meta-study [1], it is probably safe to ingest up to 0.02 MBq/kg of the Po-210. This means, that for 80 kg person, it is probably safe to ingest 1.6 MBq of the Po-210, so you "need" to eat approx. 400 of these ...


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here's a view from a pure mathematician in an abstract ideal world, untroubled by real-world complexities. I hope it's helpful even if it is a gross simplification of reality. Atomic shells feel evenly balanced when they are of a size equal to twice a square number, i.e. 2, 8, 18, 32, 50, 72, 98, 128 etc. Every period in the periodic table has a number of ...


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As @dmckee pointed out in his comment, Radium is just at the "top of the decay chain". Specifically the decay chain looks like this source Now these decays all look like $\alpha$ and $\beta$ emissions - none of which could be detected from any distance (since both are massive charged particles, they lose a lot of energy over a short range). We need to ...


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Short answer: turned ionization detectors are easy enough, but they are not "Geiger counters". The core of a Geiger counter is a gas ionization detector that runs in a saturated cascade mode. They respond to ionization in the gas and are very nearly digital in nature. Notice that Geiger counters are often rigged to emit an audible click when it respond, ...


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About the physical implications, what this exercise shows you, is that the radiactive nucleus as a 50% chance to decay during its half-life period, but if it has not decayed, it remains unchanged: it has same number of protons and neutrons, same instability, and so during the next period it still have a 50% probability to decay, and so on.


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The gas inside the geiger tube will have a different response according to what kind of radiation is entering and what energy it has. For example, here's Helium's absorption for electrons , and here's its X-ray absorption . The tube only gives you a particle count, but if you have a fixed distribution of incoming particles then you can calculate/calibrate a ...


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True, Geiger counters can be tuned to be more sensitive for certain types of decay and for the amount of radiation emitted. They can detect ionizing radiation in the form of alpha particles, beta particles and gamma rays. Given the different nature and energy of these types of radiations (alphas are Helium nuclei, betas are electrons or positrons, gammas ...


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It means that after every half-life of time there is a 50% probability that any given nucleus will decay. So after one half life, there is a 50% probability that a particular nucleus will have decayed. But after that time, if your particular nucleus has not decayed, then there is a further 50% probability that it will decay after another half life. Thus the ...


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Probably the best way to think of this, and the way I suspect it works in the minds of people who write exactly this sort of equation in physics textbooks, is that it's an equation describing a nuclear process, rather than a chemical process. The electrons initially bound to the nucleus are irrelevant spectators. A less ambiguous way to describe the ...


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The energy of the decay has little to say about whether the covalent bond will remain after the decay. The reason is because the $\beta$-decay electron (or positron since the question doesn't specify) will be moving so fast (compared to the orbital electrons that the cross-section for scattering will be quite small. Since scattering off the orbital ...


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I think I found the solution. The decay energy of radiocarbon is the following $ 0.156476 MeV = 2.50702189 \cdot 10^{-17} kJ $ ${2.51 \cdot 10^{-17} kJ} \cdot {6 \cdot 10^{23} \cdot 1/mol} = 1.51 \cdot 10^7 kJ/mol$ If we compare this decay energy to the energy of the chemical bonds table (~ 200-400kJ/mol), we will see that it exceeds that 100000 fold, ...


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First of all the $\beta $ particle emitting from nucleus is too energetic to be captured in atomic orbit to form atom.So it is definitely not the case. Infact thta's why we are sure that it is coming from nucleus not from atomic orbital. More over while writing for radio active decay we are looking for change in the nucleus as it is basically a nuclear ...



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