Does Temperature Influence Radioactive Decay Rates?

In summary, according to statistically observed decay rates, a unit of the particle's former affiliation falls apart post-escape somewhat, and the atom is changed into an isotope or some other element. This process is called barrier tunnelling, and it is indeed how nuclei decay.
  • #1
zforgetaboutit
46
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Question #1: Can I speak of an atom decaying because one or more of its particles exceeded "escape velocity" according to statistically observed decay rates and a unit of the particle's former affiliation fell apart post-escape somewhat, and the atom was changed into an isotope or some other element? I was wondering, in the case of a single atom, what causes the breakdown.

Does this lead to Einstein's [problem with quantum physics] quote (paraphrased) "God does not play dice with the universe".

Question #2: Does temperature affect half-life? I.E. theoretically, at zero K, would decay still occur at the same rate as STP?
 
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  • #2
As far as your second question goes

If you are talking about nuclear fission, I would think it is pretty much insensitive to temperature over a wide range up to some really high temperature. The half life for fission of U235, for instance, probably doesn't change from absolute zero to room temperature. Induced fission (as opposed to spontaneous fission) ought to occur when temperature is high enough that nucleons or small nuclei can get close enough to the nucleus in question to induce it to split.
 
  • #3


Originally posted by Janitor
The half life for fission of U235, for instance, probably doesn't change from absolute zero to room temperature.

With all due respect, "probably" is not a robust reply. :-)
 
  • #4
Hey, no respect is due to little old me!

Since I was speaking off the top of my head (read: I was too lazy to try to look it up), I figured I had better take the path of caution. Someone more knowledgeable than I will surely weigh in here.
 
  • #5
Originally posted by Janitor
Hey, no respect is due to little old me!

... I figured I had better take the path of caution.

My preferred "path of caution" is "I don't know". :smile:
 
  • #6
Janitor's not a bad guy. He's smarter than he looks.
 
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  • #7
Originally posted by zforgetaboutit
Question #1: Can I speak of an atom decaying because one or more of its particles exceeded "escape velocity" according to statistically observed decay rates and a unit of the particle's former affiliation fell apart post-escape somewhat, and the atom was changed into an isotope or some other element? I was wondering, in the case of a single atom, what causes the breakdown.

It's not a good idea to look at it in terms of "escape velocity". The escape velocity of a planet is derived classically, using conservation of energy. The "classical" part implies that the object (say, a rocket) absolutely, positively must be moving at that velocity in order to both conserve energy and escape.

But nuclei are not subject to the rules of classical mechanics, they are subject to the Rules of Quantum Mechanics.

Quantum mechanics predicts that energy can be conserved, even when a particle is found in a classically forbidden region. The process is called barrier tunnelling, and this is indeed how nuclei decay.

Does this lead to Einstein's [problem with quantum physics] quote (paraphrased) "God does not play dice with the universe".

It is a manifestation of it, because in a sample of radioactive nuclei, we have no way of predicting which particular nucleus will decay. We can only predict the decay rate.

Question #2: Does temperature affect half-life? I.E. theoretically, at zero K, would decay still occur at the same rate as STP? [/B]

Calculate the thermal energy of those two temperatures (E=kT, k=Bolzmann's constant, T=absolute temp). Compare it to the energy needed to induce fission (you'll have to look it up). Is the thermal energy high enough to cause a reaction?
 
  • #8
Hey, thanks for the compliment, Michael... I think. :wink:

One item I glossed over is that fission of U235 and certain other nuclei can be induced by "thermal neutrons," i.e. free neutrons at room temperature. Since the neutron is electrically neutral, it can approach a heavy nucleus without electrostatic repulsion coming into play. Thus a neutron does not have to be in an ultra-hot environment to do its thing on a nucleus. Italian physicists such as Enrico Fermi were heavily into investigating this phenomenon in the years prior to the Second World War.
 
  • #9
Originally posted by Janitor
One item I glossed over is that fission of U235 and certain other nuclei can be induced by "thermal neutrons," ...

(snip)

I want to correct an error that was made by both Janitor and myself earlier. I saw the word "fission" somewhere, and as I wrote my answer I had fission on the brain. But upon reading the original post again, I see that fission was not mentioned anywhere in it!

zforgetaboutit was asking about simple radioactive decay, not fission. We do not need to concern ourselves with thermal neutrons.

So, with regards to my earlier answer to Question #2:

Calculate the thermal energy of those two temperatures (E=kT, k=Bolzmann's constant, T=absolute temp). Compare it to the energy needed to induce fission (you'll have to look it up). Is the thermal energy high enough to cause a reaction?

We want to correct that to comparing the ambient thermal energy to the difference in energy levels of the nucleus (which again, you'll have to look up).
 
  • #10
Good point Tom.

When he asked about "an atom decaying" in his first post, I thought of several possible things he could mean by that. One such would be for an excited electron state to decay with release of photons. That is why my first response to him started with "If..."

Of course he also spoke of isotopes, implying fission, fusion, alpha decay, beta decay or what have you.
 
  • #11


Originally posted by Tom
It's not a good idea to look at it in terms of "escape velocity".
Keep in mind the reason I quoted the phrase was to use an analog of escape velocity. To be more verbose, I could also say the particle was too far removed in energy/position to remain a part of the larger organization of particles.

Quantum mechanics predicts that energy can be conserved, even when a particle is found in a classically forbidden region. The process is called barrier tunnelling, and this is indeed how nuclei decay.
Sounds like we don't have a mechanism for deterministic tunneling yet. We say the particle played on the wrong side of the tracks, but don't know why it crossed over. Maybe we should ask John Edwards.

Calculate the thermal energy of those two temperatures (E=kT, k=Bolzmann's constant, T=absolute temp). Compare it to the energy needed to induce fission (you'll have to look it up). Is the thermal energy high enough to cause a reaction?
If I were a physicist or physics student, I certainly would but I'm not. I'm an interested lay person :-) Besides, at absolute zero T, in your formula, would be zero so E would be zero. There would not be any thermal energy. Now what? The answer would seem to be "no" decay would/could occur. The atom could last forever in this isolated state.

I doubt if Boltzman made any observations anywhere near absolute zero (AZ), when determining his constant because from what I read,
the liquefaction of helium at 4.2 Kelvin (minus 268.8 °C) was done in 1908 by Heike Kamerlingh Onnes
but Boltzmann unfortunately died in 1906.

Boltzmann's classical theory may break down at AZ because maybe there is a thermal analog to quantum effects that becomes prominent, and the macro scale rules of statistical mechanics no longer apply.

I'm asking here if may is still an appropriate term to use, or has the idea already been settled empirically?

I'm skeptical of the fast answer at nature's boundaries, without myself having a good education in physics, so please bear with me.
 
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  • #12


Originally posted by zforgetaboutit
Sounds like we don't have a mechanism for deterministic tunneling yet. We say the particle played on the wrong side of the tracks, but don't know why it crossed over. Maybe we should ask John Edwards.

The "mechanism" for tunneling is the wave-like nature of paticles like electrons. Do not confuse this idea with the notion that we do not have a clear description of tunneling. It is SO clear that we USE this effect so many times. There certainly is no need nor justification to invoke pseudoscience into this.

If I were a physicist or physics student, I certainly would but I'm not. I'm an interested lay person :-) Besides, at absolute zero T, in your formula, would be zero so E would be zero. There would not be any thermal energy. Now what? The answer would seem to be "no" decay would/could occur. The atom could last forever in this isolated state.

Can we, once and for all, be VERY clear on the fact that energy ISN'T zero at T~0K? To say that atoms and molecules simply stop moving (making their positions and momenta to be well defined simultaneous) is simply erroneous not just on principle grounds, but also experimentally. Please look up the significance of the de Boer parameter for the solidified nobel gasses. These are experimentally verified parameter that comes into the thermodynamics internal energy of these solids as T -->0. It clearly shows the presence of the zero-point energy AS predicted by QM. Furthermore, if one has had even minimal lessons in QM, one would have easily noticed that for the quantum harmonic oscillator (which can be applied to many solids), the LOWEST energy state is NOT zero (a popular question in physics GRE).

I'm skeptical of the fast answer at nature's boundaries, without myself having a good education in physics, so please bear with me.

You also need to keep in mind that when you ASK a physics question, you tend to get a response that involves phrases and terms that are common and WELL-DEFINED in physics. You should be aware that there is a distinct possibility, due to your lack of physics education, that these things may be interpreted the wrong way at your end, while they appear obvious to most who have had a physics education. This can be made worse if the question you asked isn't as trivial as it appears, and this is especially true if you tend to want to dig deeper into it, as it seems to be occurring here. You were offered quantum tunneling as the phenomena resposible for radioactivity, and you were only given a scant description of this since I think there was a sense that maybe anything more would make it incomprehensible at this level. This is where someone with a physics background would have understood why.

However, since you want to dig deeper, somehow "tunneling" is now something that John Edwards might be appropriate to handle. Having spent 3 out of the 5 years of my graduate work working on tunneling spectroscopy in high-Tc superconductors, you will understand if I find such comment to be a bit annoying.

I'm not saying you shouldn't ask questions. I'm saying that you should at least be aware that at some point, you have to do your own leg work in trying to decipher the answer you were given.

Zz.
 
  • #13


Originally posted by ZapperZ
There certainly is no need nor justification to invoke pseudoscience into this. ... these things may be interpreted the wrong way at your end ... Having spent 3 out of the 5 years of my graduate work working on tunneling spectroscopy in high-Tc superconductors, you will understand if I find such comment to be a bit annoying.

I don't understand your annoyance. My intention in cleverly (if I may say so) mentioning JohnE was to be topically humorous. You know, as in "A sign above a bed at a travellers inn proudly boasting - 'Heisenberg may have slept here'".

But that's OK - sometimes jokes don't tunnel well in forum writing, although you may have had a more pleasant reaction to it if you read it in a book by a science-for-the-masses author. Sigh.

I hope you soon stop assigning emotion and attitude to my posts and get back to viewing them as neutral questions, not personal attacks on your particular research sacred cow.

You also need to keep in mind that when you ASK a physics question, you tend to get a response that involves phrases and terms that are common and WELL-DEFINED in physics.
Yeah, I remember reading about the last time things were so well-defined in physics at the turn of the century. And the time before that ... And ...

I wasn't previously complaining was I?

I think there's some wiggle room in most physics ideas.

In which specialties will today's obvious facts be corrected by future researchers? One might be yours. Stay tuned.

You also need to keep in mind that when you ASK a physics question, you tend to get a response that involves phrases and terms that are common and WELL-DEFINED in physics.
Not this time. With all respect I think the equation provided was inappropriate for absolute zero.

I know the readers/posters here are a mixed lot from varied backgrounds. I'm mainly interested in the underlying physical principles, not the math models. I'll state what I think I understand about the reply and then ask to be brought to the next level. What's wrong with that?

If somebody wants to explain a wacky physical phenomenon with a math model that flies in the face of everyday experience, well yes, I have a problem with that.

If the universe is reasonable then let's hear some logical reasons, right or wrong, and we'll see which ones fit our observations. If you want to have a nice model, do that also. The pleasure in trying to understand, is equal to what is gained from shrugging off the stifling yokes of religion.

If the universe is not reasonable it will be boring and there will not be any good models anyways.


Just for the record, I can accept the idea of Hawking Radiation as being a viable mechanism for black holes "evaporating". Maybe it's inaccurate, but at least it's plausible to me. It's a well-known fact that nothing can escape a black hole. It's a well-known fact that some things can escape black holes. It's a well-known fact that gravity escapes black holes. Theories Du Jour.

Please look up the significance of the de Boer parameter...
I did 3 lookups. They all mentioned a Monte Carlo simulation, not observations. What's your point? If a particle/wave/wavicle had zero energy, it would have no momentum, right? It wouldn't be a particle any more either, would it? It would cross over to non-particleness and wouldn't have any business appearing on any graph relating to particles.

I thoroughly enjoyed the audio recording "Particle Physics for Non-Physicists: A Tour of the Microcosmos" by http://www.teach12.com/store/professor.asp?ID=251&d=Steven+Pollock . A thrilling story of the evolution and revolution of QM narrated with much excitement, awe, and enthusiasm. I listened to it non-stop and found his account to be fair and balanced inasmuch as he liberally uses the phrases "maybe" and "we believe" and "the model predicted and the entity was found". Unless the topic is simple mathematics, this is the best way to declare things.

That QM models can generate observed results with the accuracy they've done so far strikes me as very wonderful and very odd at the same time. If these models are congruent to the notion of predictions due to the "The Conservation of Property fubar", ummm OK. No problem.

I'm currently working on the predicted discovery of the J.Lo particle.
 
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  • #14


Originally posted by zforgetaboutit
Yeah, I remember reading about the last time things were so well-defined in physics at the turn of the century. And the time before that ... And ...

You're confusing "well-defined" with "correct". The FACT that they are well-defined is the very same thing that allows us to notice when there are experiments that indicate that their properties go beyond these definitions. When I use the term "tunneling", I can't simply make things up as I go along, because that word has a set of physical and mathematical description. I simply cannot use it to explain things that it wasn't to be used with.

Some time in the future somebody will say "they had it somewhat/mostly/all wrong back then". It happened before and will continue to happen.

In which specialties will today's classical theory be corrected by future researchers? It might be yours. Stay tuned.

And you are also forgetting that to be able to know when something is not quite right, or if it is new, you have to first and foremost study things that we already know and already can explain. If one is ignorant of the state of current knowledge, one will never know if something new pops up even if it comes up and bites one's rear end. A lot of deviations from present-day understanding are extremely subtle involving higher-order corrections.

I know the readers/posters here are a mixed lot from varied backgrounds. I'm mainly interested in the underlying physical principles, not the math models. I'll state what I think I understand about the reply and then ask to be brought to the next level. What's wrong with that?

These "physical principles" ARE the mathematical "models". If I describe to you the photoelectric effect, I am putting into words via the english language, what the mathematical description is trying to say. The FULL, accurate, and unambiguous description of this phenomena can only be given via mathematics. Everything else are simply analogies, examples, specific cases, etc. This means that without understanding the mathematical description, you are only getting 2nd hand interpretation of the actual description. Are you telling me that you are willing to settle for getting only bits and pieces of something without having to look at the whole picture? How confident are you that the conclusions and deductions you make out of these bits and pieces are actually accurate, much less, valid?

I did 3 lookups. They all mentioned a Monte Carlo simulation, not observations. What's your point?

Ashcroft and Mermin, "Solid State Physics", p.412 (Saunders, 1976).

I thoroughly enjoyed the audio recording "Particle Physics for Non-Physicists: A Tour of the Microcosmos" by http://www.teach12.com/store/professor.asp?ID=251&d=Steven+Pollock . A thrilling story of the evolution and revolution of QM narrated with much excitement, awe, and enthusiasm. I listened to it non-stop and found his account to be fair and balanced inasmuch as he liberally uses the phrases "maybe" and "we believe" and "the model predicted and the entity was found". Unless the topic is simple mathematics, this is the best way to declare things.

As Integral has pointed out, studying physics isn't the same as studying about physics. Pop-science books do not teach you physics. If they do, we would be using them in our undergraduate physics classes. All they do is try to impart some flavor of the basic principles via analogies and examples. Analogies and examples of a principle is not the principle it is full glory.

I'm currently working on the predicted discovery of the J.Lo particle.

Somehow, I expected nothing nothing less...

Zz.
 
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  • #15


Originally posted by ZapperZ You're confusing "well-defined" with "correct".


I don't think so. Last century, the physicists were already using words like FACT and CORRECT and WELL-DEFINED. Some of them had it wrong. Now you're using the same terms as if only you are privy to the Grand Unified Theory.

A lot of deviations from present-day understanding are extremely subtle involving higher-order corrections.
I know some are.

Are you telling me that you are willing to settle for getting only bits and pieces of something without having to look at the whole picture?
A lot of electronic devices work for me in spite of my bits and pieces understanding of them. When I studied math I had to know every last detail down to the initial postulates before I felt comfortable. With physics topics I don't have those same time resources available.

studying physics isn't the same as studying about physics. Pop-science books do not teach you physics.
I didn't say they did.

If only we could return to the original topic, without the constant noise of the self-trumpeting how you're a High Priest of Physics and I'm not ... but I expect too much from you.

Janitor, Tom, and Michael D. Sewell were much more friendly and understanding than you were. Or can be.

I'll restate my original question in the next posting. You needn't participate at the monstrous expense of your self-dignity, but I hope you reconsider.
 
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  • #16
Why atoms decay? - continued

Let us consider the following delicious thought experiment:

Zapperz is in a normal closed room, when suddenly, all the air molecules migrate, in a statistically random way, to the other side of the room for an extended period of time; just long enough to silence his disgust at having to read my postings, and perhaps, as a bonus, long enough for him to fall unconscious for an extended time.

I can understand that there is a (very low) but non-zero probablility of the molecules motion happening in this way.

Is atomic decay sort of the same way - sub-atomic particles sometimes "migrate" in a probalistic way to the "outside of the atom" in such a way that they can't return because the atom doesn't have any walls per se to increase the chance of the particle returning to its original locale?

And furthermore, because of the actual particle recipe, decay rates can be more or less accurately predicted via a model?

I.E. if we ever encounter some atom/isotope containing this particular combination of particles we expect it to have a half-life of x.

I've read of elements (some stable, some unstable) being predicted and observed but don't remember any comments about their expected decay rates.
 
  • #17


Originally posted by zforgetaboutit
I don't think so. Last century, the physicists were already using words like FACT and CORRECT and WELL-DEFINED. Some of them had it wrong. Now you're using the same terms as if only you are privy to the Grand Unified Theory.

But it is BECAUSE they defined it very well that we KNOW when they go wrong. That was all I'm saying. So some reason, you keep missing this.

A lot of electronic devices work for me in spite of my bits and pieces understanding of them. When I studied math I had to know every last detail down to the initial postulates before I felt comfortable. With physics topics I don't have those same time resources available.

Then you shouldn't be making flippant remarks about tunneling phenomena as an explanation for radioactivity. I would have never gotten into this mess if it weren't for that. Maybe it's just me, but if I don't know something, I wouldn't be so quick to make snide comments about it.

And oh, I apologize for intruding into this string...

Zz.
 
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  • #18


Originally posted by zforgetaboutit
Keep in mind the reason I quoted the phrase was to use an analog of escape velocity. To be more verbose, I could also say the particle was too far removed in energy/position to remain a part of the larger organization of particles.

But that would still not be accurate. For instance, in no nuclear model does an alpha particle escape its parent nucleus simply because it "wandered too far from home". It escapes because of barrier tunneling.

Sounds like we don't have a mechanism for deterministic tunneling yet. We say the particle played on the wrong side of the tracks, but don't know why it crossed over. Maybe we should ask John Edwards.

Barrier tunneling is the mechanism. What you seem to be looking for here is a "hidden variable" theory, which does not exist.

If I were a physicist or physics student, I certainly would but I'm not. I'm an interested lay person :-)

Google, buddy! Google!

Besides, at absolute zero T, in your formula, would be zero so E would be zero. There would not be any thermal energy. Now what? The answer would seem to be "no" decay would/could occur. The atom could last forever in this isolated state.

Ah, no. That's not what I meant.

The decay would still occur at T=0 K. Remember, quantum motion does not cease, even at absolute zero! What I meant for you to think about was if the higher temperature would supply the needed energy to help the decay rate along. But if you calculate the thermal energy at room temperature, you will see that it is nowhere near the energy needed to kick a daughter nucleus "out of the house".

I doubt if Boltzman made any observations anywhere near absolute zero (AZ), when determining his constant because from what I read, but Boltzmann unfortunately died in 1906.

Doesn't matter--k is temperature-independent.

Boltzmann's classical theory may break down at AZ because maybe there is a thermal analog to quantum effects that becomes prominent, and the macro scale rules of statistical mechanics no longer apply.

No, there's nothing like that going on. The classical partition function is OK for all temperatures. The point at which it breaks down is the point at which we can't ignore the difference between bosons and fermions.
 
  • #19


Originally posted by Tom What I meant for you to think about was if the higher temperature would supply the needed energy to help the decay rate along. But if you calculate the thermal energy at room temperature, you will see that it is nowhere near the energy needed to kick a daughter nucleus "out of the house".
When you say room temperature, do you also include the interior of the atom (if such a thing has meaning) or just the exterior - the room. I want to clear up the semantics here.

When I had a wrist watch with a tritium cavity which illuminated the display, it worked at "room temperature". There was decay via electrons occurring. I worry this contradicts what you just wrote about STP being too low in energy to support decay.

About kicking a daughter nucleus, it may be an electron first got kicked, then the daughter nucleus - missing her electron pal - left in disgust. In my queries I've never mentioned the nature of the decay - alpha or beta or other particles, and I trust the comments I'm reading apply to all known cases.
========================================

When you say "hidden variable" do you mean as in "Why did the thrown die land with a 6 on top? There must be a reason. A hidden variable is lurking somewhere". I was looking for a lower-order influence to explain decay.

The die landed the way it did because of an infinite number of influences, in the QM sense. If we repeated the same toss exactly (not that such a thing is possible) don't we expect the same result? I do, in an ideal sense.

Without a hidden variable QM indicates an unpredictable precise result - luck of the draw. It may not end up as a 6. I think if the result is different, it's because of a 1st lower-order influence not mere probability. And a 2nd lower-order influence to that affects the 1st lower-order influence. And so forth. I don't think there's an end, but I wanted to reach the level lower than barrier tunnelling which sounds like "some how it crossed the barrier - we're not sure why".

At first it seems reasonable (a tricky term): why did the material decay? Particles left the locale of the nucleus. How come? Barrier tunneling. How come? Bad luck!(QM)

==================
About Googling I found this http://www.swc.cc.ca.us/~jveal/PHYSICS/Phys274/matter_wave.htm
An interesting aspect of uncertainty in quantum mechanics is barrier tunneling. Classically, if a particle is moving in a potential well and has less total energy than the height of the well, it cannot leave the well. In quantum mechanics, we consider uncertainty as an aggregate property. Therefore, even when the classical barrier penetration is forbidden, due to uncertainty there is often some probability of finding a particle beyond the barrier. One can consider how the particle crossed the barrier by studying the particle position. However, this adds to the uncertainty in momentum, and therefore of energy. Thus, in quantum mechanics, the question as to how the particle crossed the barrier is not a meaningful one. The probability for crossing the barrier is called the transmission coefficient. Some interesting technology, such as the scanning electron microscope discussed, is associated with this barrier penetration phenomenon.

In the end, quantum mechanics is elusive. The beautiful statistical patterns that it predicts have been well-verified. But the important question as to how the "subjective" has produced these patterns is not a meaningful question."
I think this is a fair and balanced statement. The author is saying in QM there is no lower order influence. There may be another chapter to our understanding of these things. Maybe not. QM may be the terminating case. Time will tell.
 
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  • #20


Originally posted by zforgetaboutit
When you say room temperature, do you also include the interior of the atom (if such a thing has meaning) or just the exterior - the room. I want to clear up the semantics here.

When I had a wrist watch with a tritium cavity which illuminated the display, it worked at "room temperature". There was decay via electrons occurring. I worry this contradicts what you just wrote about STP being too low in energy to support decay.

I am referring to the ambient (macroscopic) temperature that gives rise to (microscopic) particle motion. I had meant for you estimate the kinetic energy of particles or photons that could possibly collide with radioactive nuclei, and to compare it with the difference in energy levels of the nuclei themselves. Upon doing that, you would have found that the energy of the moving particles is not nearly enough to affect the nucleus.

About kicking a daughter nucleus, it may be an electron first got kicked, then the daughter nucleus - missing her electron pal - left in disgust. In my queries I've never mentioned the nature of the decay - alpha or beta or other particles, and I trust the comments I'm reading apply to all known cases.

Actually, they don't. The "barrier tunneling" comments refer only to strong decays, whereas beta decay is a weak decay.

When you say "hidden variable" do you mean as in "Why did the thrown die land with a 6 on top? There must be a reason. A hidden variable is lurking somewhere". I was looking for a lower-order influence to explain decay.

That's what I mean by "hidden variable" theory. It would be a theory that could predict not only the decay rate, but also which nucleus in the sample was going to decay. No such theory exists.

About Googling I found this

My "Google" comment was in reference to the difference in energy levels of a typical nucleus. Since you didn't find it, I'll just tell you: the difference is on the order of a few MeV, which is way more than the KE of particles moving around at room temperature.

edit: fixed a typo
 
  • #21
external dependence of decays

Alpha decay has been experimentally shown to be robust against temperature and chemical changes; I supposse that some more try against anti-zeno effect must be in progress, or will be in the near future.

The same can be said of fission processes.

Beta decay has some chemical/temperature dependence in a few experiments, if I remember well the biggest variation of half-live is about 15%-20%. (I am quoting from memory, I hope I have no exchanged in my mind the results from alpha and beta)

The only claims for changes beyond one order of magnitude come from alchemists. They believe that some solid state configuration either alters the half-life or induces a chain reaction.
 

1. What is radioactive decay?

Radioactive decay is the process through which an unstable atom releases energy in the form of radiation in order to become more stable. This process can occur naturally or be induced through artificial means.

2. What causes radioactive decay?

Radioactive decay is caused by the instability of an atom's nucleus. This instability can be due to an excess of neutrons or protons, or a combination of both. The nucleus then releases particles or energy in order to reach a more stable state.

3. How is the rate of radioactive decay calculated?

The rate of radioactive decay is measured using a unit called half-life, which is the amount of time it takes for half of the unstable atoms in a sample to decay. This rate can be calculated using mathematical formulas that take into account the type of radioactive element and its half-life.

4. What are the types of radioactive decay?

There are three main types of radioactive decay: alpha decay, beta decay, and gamma decay. Alpha decay involves the release of an alpha particle (two protons and two neutrons), beta decay involves the release of a beta particle (an electron or positron), and gamma decay involves the release of high-energy photons.

5. What are the practical applications of radioactive decay?

Radioactive decay has many practical applications in various fields such as medicine, energy production, and dating geological materials. It is used in medical imaging techniques, like PET scans, to track the movement of radioactive tracers in the body. In energy production, nuclear reactors use the heat produced by radioactive decay to generate electricity. Additionally, radioactive dating methods use the rate of decay to determine the age of rocks and other materials.

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