52

The white flashes seen in the video are the result of radioactive contamination of the original film. Airborne radioactive particles were deposited onto the film and caused a local overexposure of the negative, quite similar to the historic experiment of Antoine Henri Becquerel. That's why the flashes only appear in individual spots in individual frames of ...


47

Important naturally-occurring radioactive lead nuclides are: $\ce{^{214}Pb}$ $\left(t_{1/2}=26.8\ \mathrm{min}\right)$ from $\ce{^{238}U}$ $\ce{^{210}Pb}$ $\left(t_{1/2}=22.3\ \mathrm{a}\right)$ from $\ce{^{238}U}$ $\ce{^{211}Pb}$ $\left(t_{1/2}=36.1\ \mathrm{min}\right)$ from $\ce{^{235}U}$ $\ce{^{212}Pb}$ $\left(t_{1/2}=10.64\ \mathrm{h}\right)$ ...


41

Theoretically, a radioactive material will still be radioactive at absolute zero, and its rate of decay will be $100.00\%$ of that at room temperature. Practically, at the lowest achievable temperatures we observe the same thing: radioactivity is still there, not affected the slightest bit. Nuclear motion does not slow down as we approach absolute zero, ...


38

An article by Snell and Pleasanton, 'The Atomic and Molecular Consequenses of Radioactive Decay', (J. Phys. Chem., 62 (11), pp 1377–1382, $1958$) supports Ben Norris's comment. It is clear ... that $\ce{C^{14}O2}$ remains predominantly bound as $\ce{NO2+}$, a result that is perhaps not surprising. [This occurs in] $81$% of the decays. In $\ce{C^{14}O2 -&...


26

There are four main decay chains for actinides and superheavy elements. This is a simple consequence of the fact that one of the main processes to increase a heavy nucleus' stability is the emission of alpha particles, which have a mass number of 4 ($\ce{^4_2\alpha}$); notice that if you take the isotope's mass number and divide it by 4, the remainder of ...


25

It is possible to modify nuclear decay rates using chemistry, though it is rare and the effect is usually very small. Here I summarize the information available in this link. You may want to see the references within. There is a type of nuclear decay called electron capture, where a nuclide directly captures an electron from the innermost electron shells ...


24

The antineutrino is not completely irrelevant. The decay energy is shared between the beta particle and the neutrino. Therefore, the beta particles appear with an energy distribution that ranges from zero to the maximum beta energy. The maximum beta energy for the decay of $\ce{^14C}$ is $E_\text{max}=0.1565\ \mathrm{MeV}$; the average beta energy is $E_\...


20

Interesting idea, but it has already been done, and not cheaply - read on. How could we get a great quantity of $\ce{_{82}^{197}Pb}$ ? There would be two problems with getting a large amount of $\ce{_{82}^{197}Pb}$. First, the parent nuclide of $\ce{_{82}^{197}Pb}$ is $\ce{_{83}^{197}Bi}$ which is unstable and has a half-life of only 9.33 minutes - so ...


18

Water, when hit by radioactive particles, will most likely suffer ionization (or radiolysis). The ejected electron very quickly leaves the vicinity of the affected molecules (in the case of exposure to gamma or beta radiation) or a very chemically stable neutral helium atom is formed (in the case of exposure to alpha radiation), so you obtain short-lived ...


18

The usage in ionizing smoke detectors requires a radioactive isotope to work. In addition to a sufficient half-life to make a smoke detector with a suitable service life, Am-241 emits mostly alpha radiation (which is used for the ionizing mechanism) with relatively little of the more hazardous and useless gamma radiation. See the Applications section in ...


17

Short Answer: They don't. As the comments above indicate, the decay chains will terminate at a stable nuclei. For the series you mentioned in your question, lead has the first stable isotope. A really useful tool for looking at different decay chains is available from Oak Ridge National Laboratory's Risk Assessment Information System. You can pick from ...


16

There are two factors here: Radon is a gas and can thus enter the body very easily, through inhalation. $\ce{^222Rn}$ and its first 4 decay products are extremely radioactive, with halflives of miliseconds to a few days, releasing alpha and beta radiation. When the two are combined, we have a problem. Even if the quantity is low, inhaling radon places an ...


15

There are plenty of good sources online explaining the principle behind radiocarbon dating. For instance, the wikipedia explains: During its life, a plant or animal is in equilibrium with its surroundings by exchanging carbon either with the atmosphere, or through its diet. It will therefore have the same proportion of $\ce{^14C}$ as the atmosphere, or in ...


14

One radioactive isotope of lead is $\ce{^{210}_{82}Pb}$, which has a half-life of 22.20 years. The only reaction that $\ce{^{210}_{82}Pb}$ undergoes is: $$\ce{^{210}_{82}Pb ->^{210}_{83}Bi +^{0}_{-1}e}$$ The $\ce{^{0}_{-1}e}$ released in the reaction is a high speed electron, otherwise known as a $\beta^-$ particle. The fact a $\beta^-$ particle is ...


13

There seem to be three processes to consider here: As the high-energy electron exits, it creates a rapidly varying electrical current, which produces intense electromagnetic fields for a short time. This may act on the other charged particles that are present, possibly creating further ionization and/or breaking the chemical bond. The nitrogen nucleus ...


13

Of course it would break, just like you said; also, a high-energy $\beta$ particle would kill quite a lot of bystander molecules. Also, if not for other reason, the resulting molecule would no longer be DNA , since the decayed atom would no longer be $\rm P$. Also, the product would no longer be radioactive, so we wouldn't be able to detect it anyway. The ...


12

No, deuterium is completely stable. I found the answer at Hyperphysics, and it has to do with the mass energies of the products and reactants of this hypothetical reaction. The decay of deuterium would be $$\ce{D -> P + N + e + \bar{\nu}}_e$$ where $\ce{D}$ is deuterium, $\ce{P}$ is a proton, $\ce{N}$ is a neutron, $\ce{e}$ is an electron and $\ce{\bar{\...


12

For all radioactive decay (or other nuclear reaction) of a nuclide into other nuclides, the atomic number $Z$ and mass number $A$ need to be conserved. $$\ce{_{Z_1}^{A_1}X -> _{Z_2}^{A_2}Q + _{Z_3}^{A_3}R }$$ $$Z_1 = Z_2 + Z_3$$ $$A_1 = A_2 + A_3$$ Additionally, the charges must be conserved. If $\ce{_{Z_3}^{A_3}R }$ is an alpha particle $\ce{_2^2\...


11

Technetium does occur in nature. From the Los Alamos page on technetium: Technetium was the first element to be produced artificially. Since its discovery, searches for the element in terrestrial material have been made. Finally in 1962, technetium-99 was isolated and identified in African pitchblende (a uranium rich ore) in extremely minute quantities as ...


10

There are several reasons that neptunium may not be used. The first is the abundance of neptunium. From Wikipedia, it states the best source is from spent fuel rods. Basically, someone would need to handle very radioactive material to obtain a small amount of neptunium. Next would be the likelihood of fission when struck with a neutron. If it is the same ...


10

For isobars (i.e. nuclides with a constant mass number $A$), the binding energy $E$ as a function of the atomic number $Z$ describes a parabola, the so-called valley of β-stability (see also Weizsäcker’s formula or semi-empirical mass formula). $$E = a \cdot Z^2 + b \cdot Z + c \pm d/A^{3/4}$$ The most stable nuclides lie at the bottom of the valley (but ...


10

As noted in the comments, the premise is wrong. Actually: Lead is the heaviest stable element. Why are heavier elements unstable? The atomic nucleus consists of neutrons and protons. Without going into detail, these are held together by short range nuclear forces. The protons also repel each other due to their positive charge. As we add more protons, ...


10

Large elements of my answer is drawn from my own notes that I use for teaching General Chemistry II. What is −0.693? Radioactive decay processes follow first-order kinetics. A first-order reaction is one where the rate depends only on the concentration of one of the reactants raised to the first power. So consider the following reaction: $$\mathrm{A} \...


10

No. Radioactivity means that the nucleus of a particular atom is unstable, and the nucleus will decay to a more stable state either by decomposing or emitting energy (as a photon). This instability will not make other nuclei unstable unless the decay of one of the radioactive atoms causes a fission process in one of the atoms in the desk. This is a secondary ...


9

I'd like to add to the other answers. Although not directly an answer to your question, I do feel it's relevant. Radioactive decay can cause damage to crystalline solids. Take the example of apatite $\left(\ce{Ca5[PO4]3[F,Cl,OH]}\right)$, which can host a bit of uranium in the crystal structure. When this uranium decays, the resulting particle damages the ...


9

Defintions Occur on Earth differentiates between all isotopes in the universe and isotopes which exist or have existed on Earth. The passage identifies that there are a total of 308 isotopes occurring on Earth, but it is worth noting that there are more than 308 total isotopes in the universe. This discrepancy is because some isotopes have extremely short ...


8

Radioactivity is not at all dependent on the atomic or molecular structure: it a purely nuclear phenomenon. The only thing that is relevant is the structure of the nucleus. The nucleus contains only protons and neutrons and the distinguishing feature that makes nuclei radioactive is, crudely, the ratio of the two. Some arrangements are unstable for a ...


8

Radioactive isotopes have similar chemical properties to that of nonradioactive element of same atom. They actually do from chemical bond with compound. You even heard the name of radio-pharmaceutical that involved the complex of radioactive metal with ligand or the radioactive isotope is attached to the pharmaceutical through covalent bond. The simplest ...


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