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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 ...


12

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

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 ...


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 ...


6

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, ...


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

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

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

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.


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


1

Have you considered the problem in light of the semi-empirical mass formula? $$ m = Zm_p + Nm_n - \frac{E_B}{c^2} $$ where the binging energy is $$ E_B = a_VA - a_SA^{2/3} - a_C\frac{Z(Z-1)}{A^{1/3}} - A_A\frac{(A - 2Z)^2}{A} +\delta(A,Z)$$ and all the coefficients are experimentally tuned values. What you are looking for is a pair where the mass of ...


1

As @Keenan rightly pointed out, the main decay branch for all of my examples is alpha-decay. Alpha particles consist of two protons and two neutrons. And it's easy to remove two neutrons from a completely full ("magic") neutron shell than from a "magic-1" neutron shell. Because a "magic-1" neutron shell is more neutron-deficit than a "magic" one. Maybe that ...


1

Exploring the Table of Isotopes is my go-to site. There are a few others out there, but when I moved from my last computer those were some of the bookmarks I didn't bring, 'cause I never used them... You haven't specified what you want in your CVS file. I think this site only offers transition energies and branching fractions by radiation type. For ...


1

Why can you not use this plot? It is given relative to the radiation level.Assume a radiation level and multiply by the percentage, and you will get the absolutes for shortly after the explosion as the table is labeled. Your question then becomes : What is the radiation level versus time. This paper has activity concentrations for Chernobyl versus time ...


1

These "metastable states" are excited states of the nucleus that have a non-trivial lifetime (most nuclear excited states decay very quickly).


1

Both isotopes are isovalent so electronically they are identical (in essence your not going to get that much difference in bond angle). However in the asymmetric well approximation since deuterium is heavier the $D-O$ bond is lowered down the well i.e. it has a lower ZPE than the $H-O$ bond and is therefore a stronger bond. This means it has a smaller ...


1

Wow its Dec2014 now. Hello Physics StackExchange again! I have long wondered about why certain elememts nuclei are more unstable the the others. Technetium is one example. It is odd atomic numbered but has no reason to be less stable then other fermionic-like atoms, some heavier but stable like gold. I shall explain further (as I am also unsure) but I will ...


1

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 ...



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