# Tag Info

21

The half-life of Uranium 238 is about the age of the Earth, so only about half of the original supply should have decayed by now. Also, there are some radioactive nuclei that get created by interactions with cosmic rays in the upper atmosphere (carbon-14) or decay from more stable nuclei (all of the daughter nuclei between U-238 and lead, for example).

21

It's also worth noting that there is nothing special about atoms. If you have any system where in every period of time an event has a certain chance of happening which only depends on internal effects of the object and no memory or communications with others - you will get the same decay curve. It's purely a matter of the statistics. If you have a ...

16

No, one doesn't need to measure the material for years - or even millions or billions of years. It's enough to watch it for a few minutes (for time $t$) and count the number of atoms $\Delta N$ (convention: a positive number) that have decayed. The lifetime $T$ is calculated from $$\exp(-t/T) = \frac{N - \Delta N}{N}$$ where $N$ is the total number of atoms ...

15

It's because the half life time is also incredibly long. The half-life of Uranium-238 is $4.5*10^9$ (=4.5 billion) years. Thorium-232 has $1.4*10^{10}$. Potassium-40 has $1.2*10^9$. These are all examples of primordial nuclides. Such half lives are of the order of the age of the universe. There's also the effect of having a decay chain, since decay ...

14

Actually, all the atoms are identical. The time at which it is observed to decay is not an intrinsic property of a given atom, but rather an effect of quantum mechanics. For any given time bin, there is some finite amplitude for a transition to a decayed state, which in turn corresponds to a finite probability, since the particle(s) emitted will escape from ...

14

We can show this by thinking about what is happening. Suppose we have a set of $N$ nuclei that are all radioactive. Each of these nuclei has a chance of decaying, $\lambda$. In people lifetimes, some people live longer and some live shorter than others, but there is an average lifetime; this is what $\lambda^{-1}$ represents for nuclei. Now how many ...

10

The reason why alpha particles heavily dominate as the proton-neutron mix most likely to be emitted from most (not all!) radioactive components is the extreme stability of this particular combination. That same stability is also why helium dominates after hydrogen as the most common element in the universe, and why other higher elements had to be forged in ...

10

The chance for a fixed nucleus to decay doesn't depend on the number of nuclei. In a fixed amount of time all the nuclei have a certain chance to decay. Increasing the number of nuclei will increase the number of nuclei that decay, but that's really just what you'd expect. It's like rolling lots dice, the number of dice showing a certain digit will be ...

10

Carbon-14 makes up about 1 part per trillion of the carbon atoms around us, and this proportion remains roughly constant due to continual production of carbon-14 from cosmic rays. The half life of carbon-14 is about 5,700 years, so if we measure the proportion of C-14 in a sample and discover it's half a part per trillion, i.e. half the original level, we ...

9

why is plutonium considered more dangerous than radioactive iodine? Because the press have heard of Plutonium and Pu=atomic bombs=bad Plutonium's danger is over stated, it's insoluble so hard to get into the food chain and even if ingested is going to go straight through you. Pu is only a real concern if breathed into the lungs as a fine dust. Iodine ...

9

The short half-life elements ocuring in nature come from the decay of long-half life elements. You can see examples of decay chains on this wikipedia page. For example, ²²⁴Ra (3.6 days half life) is produced by the decay of ²³²Th (14 billion years decay).

9

Are you worried that the cables that go to the Fukushima reactors will carry radioactivity out? The answer is No. You should read up a bit on radioactivity and educate yourself, since it is one of the facts of life. In the article you will see that it is atoms that are responsible for radioactivity whereas the current in the cables is due to electrons. The ...

9

The short answer is no: halflives are constant. However, let's discuss a situation in which that comment might have some kind of truth behind it. If you have a parent nucleus that decays to a radioactive daughter so that there will be two (or more) decays before stability. In general there are two possibilities for this: The daughter has a shorter ...

8

The simple answer is no, though as usual in Physics things are a bit more complicated than that. There are several ways in which radionucleotides decay: alpha decay, beta decay, gamma decay, and fission. These are all mediated by the weak and strong nuclear forces, though the electromagnetic force plays some part in alpha decay and nuclear fission. There is ...

7

Re: How do we know that C14 decay's exponentially compared to linear? Experiment. Re: Have there been any studies to verify this? Yes. You can read a nice writeup of some of the early studies of C14 at the nobel prize website: http://static.nobelprize.org/nobel_prizes/chemistry/laureates/1960/libby-lecture.pdf But to expand a bit on the underlying ...

6

The thing you're trying to get at is the "quality factor" or "Q value" of the radiation. Short short version photons, betas, muons: 1 protons, pions, other light mesons: 2 neutrons: 1--10 depending on energy alphas and fission fragments: 20 Note that alpha have a very high value, but it doesn't count starting from the surface of the skin, as at the ...

6

Qualitatively, isotopes that carry a lot of unnecessary extra energy are unstable - and it makes them more unstable with respect to alpha or beta decay as well as fission (the latter is useful for nuclear fission energy). However, the question why fissiles are radioactive (unstable) has a trivial answer. Only elements up to $Z=82$, $N=126$ i.e. $A=208$ may ...

6

How do we know that C14 decay's exponentially compared to linear Here's an argument that might help: suppose, temporarily, that radioactive decay was linear. Let's say you started out with a sample, call it sample #1, of a billion atoms in a box, 5700 years ago (that's one half-life). By the current day, half of them would have decayed, so you'd have ...

6

When trees grow, they add successive layers to a central core. After only a year or two, this middle core becomes established and stops growing. The living part of a tree is mostly in the bark and layer immediately below that (as well as leaves and roots). This is why woodpeckers don't kill trees but girdling one will. As a consequence, it is possible to ...

6

As @dmckee says the problem is complicated. It is complicated because it is not a solution of a potential describing one force, but a balance between electromagnetic forces and the strong force that is keeping the quarks within the nucleons. (In the nucleus the strong force is like a type of Van der waals potential, a higher order interaction, overflowing ...

6

This is really a comment, since I don't think there is an answer to your question, but it got a bit long to put in as a comment. If you Google for "Why is technetium unstable" you'll find the question has been asked many times in different forums, but I've never seen a satisfactory answer. The problem is that nuclear structure is much more complex than ...

5

Half life is defined as the period of time it takes for a substance undergoing decay to decrease by half. The name was originally used to describe a characteristic of unstable atoms (radioactive decay), but it may apply to any quantity which follows a set-rate decay. The shorter the life time the faster the material returns to normal levels of ...

5

To deduce this, You have to specify the kind of decay and the nature of the "compund" is it a crystal, a small molecule in gas phase, a organic material? Beta decay shifts the nucleus one position upward in PSE, thus any "compound" will be transformed into a cation by loss of an electron, and whre say a iodide Ion had been, there will be an Xe atom. ...

5

Yes. Have a look: “The ‘Reifenschweiler effect’ is the observation that the beta-decay of tritium half-life 12.5 years is delayed reversibly by about 25-30% when the isotope is absorbed in 15 nm titanium-clusters in a temperature window in between 160-275 C. Remarkably at 360 C the original radioactivity reappears. The effect is absent in bulk metal. ...

5

There is nothing magical about lead for this purpose. The driving factor is the number of electrons per unit volume, which reduces (to a first approximation) to the mass density. You get very good (better than lead) shielding performance from gold, tungsten, mercury, etc; and quite reasonable performance from iron or copper. Question for the student: why ...

5

From: NobelPrize.org "Her continued systematic studies of the various chemical compounds gave the surprising result that the strength of the radiation did not depend on the compound that was being studied. It depended only on the amount of uranium or thorium. Chemical compounds of the same element generally have very different chemical and physical ...

5

The beta-decay may be "locally" reduced to a decay of a proton or a neutron inside a nucleus. The beta-minus decay contains the microscopic process $$n\to p + e^- + \bar \nu_e + O(1{\rm \,MeV})$$ where the last term indicates the rough increase of the kinetic energy of the decay products. On the other hand, the beta-plus decay contains the process $$p + ... 5 Have a look at the paragraph "radioactive decay" . The half life is characteristic of each radioactive nucleus and depends on the basic interactions holding the nucleus together. It depends on the quantum mechanical probabilities of transition from one energy level to another, sometimes changing element in the periodic table. Thus, to affect the half ... 5 We come to the first formula by considering the differential equation which we can experimentally measure:$$\frac{dN}{dt}=-\lambda N\tag{1}$$We can solve differential equation (1) by rewriting as follows:$$\frac{dN}{N}=-\lambda\cdot dt$$We then integrate:$$\int{\frac{dN}{N}}=\int{-\lambda\:dt}\implies\ln{N}=-\lambda t+c_{1} Exponentiating both ...

5

The first answer (form Shaktal)is in my opinion perfect from the mathematical point of view (solving the differential equation) as well as from the physical point of view (having all units in order). However your probably conceive your approach more intuitive because you know this from calculating things with Interest i.e. \$N_{Money}=N_{0,Money}\times ...

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