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41

If protons decay, then what you say is true: all atomic nuclei are indeed unstable, and a so-called "stable" nucleus simply has too long a half-life for its decay to be observed. The most tightly bound nucleus is $^{62}$Ni, with a binding energy per nucleon of 8.79 MeV [source], which is less than 1% of the mass of a nucleon. On the other hand, the decay of ...


35

While they work on the same principles, the detonation of an atomic bomb and the meltdown of a nuclear plant are two very different processes. An atomic bomb is based on the idea of releasing as much energy from a runaway nuclear fission reaction as possible in the shortest amount of time. The idea being to create as much devastating damage as possible ...


23

We are never 100% certain of anything. The scientific method falsifies wrong theories, but it does not verify those we colloquially call "correct" or "true" If we tomorrow detect a normal oxygen atom decaying, we'll have to devise new theories to explain it. But we don't expect the things we call stable to ever decay (that's why they're called stable). We ...


19

Heavy water is easy to separate from regular water because the difference in mass is quite large. The molar mass of heavy water is 11% heavier that regular water. However if we take uranium separation, then the percentage weight difference between $^{235}$UF$_6$ and $^{238}$UF$_6$ is only 0.9%, so the relative difference is far smaller. So it's a lot ...


15

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 ...


10

this just pushes back the mystery back one step for me: why are nuclei with even atomic number more stable? There is no big mystery about this. There is a pairing interaction in nuclei. It's loosely analogous to the Cooper pairs in a superconductor. It's easy to construct an argument as to why, if you're looking for an element with no stable isotopes, ...


10

Short answer: A nuclear power plant contains a lot more nuclear material than an atomic bomb. The "Little Boy" bomb was detonated at 1968 feet (600m) over Hiroshima with the nuclear material dispersed quickly in the air; the Chernobyl meltdown contaminated its environment for decades. Long answer: http://en.wikipedia.org/wiki/Background_radiation Total ...


10

A quick calculation brings some of the points in the other answers into clear focus. Consider a big power station, like Fukishima before its demise. Its output was at a whopping rate of $5GW$. From here I get the conversion factor that 1 kiloton of TNT equivalent is taken to be $4.184\times 10^{12}$ joules. Assuming the Nagasaki bomb let slip 20 kiloton ...


8

In regards to a single element, this actually happened. Natural Bismut is about 100% 209Bi, which shows no obvious indication of being radioactive. But it turned out that all existing bismuth is radioactive. The isotope 209 had been suspected to be unstable before, but that was experimentally verified as recently as 2003, finding a half life of 1.9×1019 ...


7

After 30 minutes most of the tea will still be in your stomach. Drink a full cup of very salty warm water then barf it all back up again (note to commenters - yes, there are better emetics. Salt is readily available). Repeat until the ambulance arrives and have someone tell the HAZMAT cleanup crew that it's radioactive. Lay on your left side to discourage ...


7

From the binding energy given experimentally, using precise QM calculations or using a given formula, one should first check for "stability in particles", if the binding is negative, you will of course not have stability. Then the next thing, if you have a formula, is to check for each type of stability. For example, to check for stability against a given ...


6

The funny thing is that in the case described in the paper, pretty safe isotope (I-129) was converted into deadly dangerous I-128, I would prefer to leave it as is :-) Transmutating random mix of isotopes would give you more random mix of isotopes (garbage in->garbage out), and it's a big question if this would lead to lower integral activity. So 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 ...


5

First order reaction kinetics only states that the system obeys the decay equation $$\frac{d[X]}{dt}=-k[X],\quad k>0.$$ The solution is usually written $$[X(t)]=[X_0]e^{-kt},$$ where $[X_0]$ is the starting concentration (or simply number of particles), but in terms of the half-life, $k=(\ln 2)/\tau_{1/2}$, so $$[X(t)]=[X_0]2^{-t/\tau_{1/2}}.$$ So using ...


4

The requirement for a material to be fissile (to be able to sustain a chain fission reaction) is not simply that it have neutrons, but that each fission releases enough neutrons of the right energy to trigger further reactions. As it so happens (and I know of no simple reason for this, it's just the way the equations work), U-238 isn't fissile. You can ...


4

Thanks to @dmckee, and the link he suggested: interactive table of the isotopes. Looking at that table, it seems to me that there is not a reliably direct relationship between number of neutrons and radioactivity. Using Calcium (Ca) as an example (assuming I'm reading the chart correctly): Ca-40: stable Ca-41: radioactive (with a relatively long ...


4

Chart of Nuclides? Or like this


3

Neither of the answers addresses what fissile really means, why to use fissile materials, and why not to non-fissile materials. Fissioning any fissionable isotope inherently releases a probabilistic number of neutrons, the average of which ranges from 2-4 per fission (bottom table (v bar)), and this number is not a major criterion for choosing an isotope. ...


3

Effectively there is a maximum number, or rather producing more and more neutron-rich isotopes requires energy. You can think of it this way. Identical particles are affected by the Pauli exclusion principle; this applies to neutrons in the nucleus. Therefore a stack of neutrons will fill up the lowest energy states, but will then have to occupy higher ...


3

Your confusion comes from the oft-made, but not strictly true, assumption that global potential energy minima must be deep local minima. Stability is a function of the local depth of a potential energy well in a potential curve, and excited states in atoms/molecules have different potential curves, above those of their ground states. It is however possible ...


2

It depends on the context. For a given number of nucleons, there are typically one or two charge states that have much lower energy than any of their neighbors. For example, in the $A=12$ isobar, the weak decays $^{12}\text{B}\to{}^{12}\text{C}$ and $^{12}\text N\to{}^{12}\text C$ both release around 15 MeV excess binding energy. Carbon-12 is at the ...


2

First part: From the formula for the radius, and the fact that magnetic field is the same in both cases, you get: $$ B = \frac{m_1 v_1}{q_1 r_1} = \frac{m_2 v_2}{q_2 r_2} $$ Because you don't know the velocities, you want to get them from the potential difference. You also have $$ v = \sqrt\frac{2q V}{m} $$ You put that back into the first equation, and ...


2

I may illustrate the already mentioned problem with mixture of many isotopes. Why only I-129 is chosen? In the paper it is compared with only two isotopes with life times actual for period more than 100 000 years after incident. If we talk about periods comparable with our time scale, the list of isotopes is rather big, e.g. for melting of a nuclear reactor ...


2

You're assuming that nuclei with exactly a magic number of neutrons are more stable than all their non-magic neighbors in the chart of nuclides, but there's no reason to think that. If a nucleus has a magic number of neutrons, that means one shell is completely full, and the next shell is empty. Therefore the next neutron you add (going to magic+1) will ...


2

If you mix D$_2$O and H$_2$O you quickly get DHO due to the Grotthuss mechanism. I assume this is what you mean by heavy water being contaminated on exposure to humid air. Obviously it wouldn't be contaminated by exposure to dry air because there's no hydrogen present in dry air. The hydrogen atoms in polyethylene aren't mobile and won't react with D$_2$O. ...


2

Sure you can do chemistry with unstable isotopes. But in the case of I-131, the halflife of only 8 days implies that if you have stored the salts for length of time the remaining activity will be very small. Nor do you find any radioactive iodine in environmental iodine except in the neighborhood of recent radioactive events of various kinds.


2

The key difference is the complexity scale. In a typical every day reaction involving water, the process is thermodynamically driven; the difference in free energy between the reactants and products is much greater than any effect the extra neutron may have. In short, things happen mostly because there is a loss of energy or gain of entropy; and all the ...


1

I think you want to have something that is short-lived but not too short-lived. You do not want your patients to go around radioactive for several weeks after their examination. Regarding the precise choice of of radioactive material, that is likely a medical question. Different compounds will move around in the body in different ways and affect your ...


1

The problem with using a chemical approach is that isotopes have nearly identical chemical properties. Anything you can do with water can be done identically with heavy water. The only real differences are mass and any radioactivity the isotope provides. Thus the only known effective way to separate isotopes is to rely on mass differences. Unfortunately, ...


1

It's difficult to get hard figures without exposing people and seeing how many die! Then the problem is that for low levels of exposure you have to workout how many extra people have died The Nation Cancer Institute has some calculators and papers based on US nuclear testign exposure



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