Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: allan marsh on 10/07/2014 10:49:38
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imagine i have in my hand one atom of say Tritium. As a group of atoms of tritium they have a joint half life of some 11 years
BUT how about this one tritium atom I have in my hand. When will THIS ATOM metaphorically go pop and decay.
It could be now or perhaps any time from now to thousands of years.
SO
without using the word probability,!! please would someone explain the mechanism that determines , WHEN the atom goes pop?? ie is the pop caused by some internal nuclear mechanism or by a universal space time field that we don't understand?
if the mechanism is internal, you obviously know the reason, so I await the answer with interest?
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SO
without using the word probability,!! please would someone explain the mechanism that determines , WHEN the atom goes pop??
No one knows an answer to that. I will point out that the way you phrase the question presumes that such a mechanism exists and that there is a definite time when a specific nucleus (not atom, to be pedantic) will "go pop." Modern physics always talks about probability because there is currently no evidence to suggest any mechanism behind quantum phenomena like radioactive decay that makes them actually deterministic.
That doesn't mean that such a mechanism doesn't exist. It does mean that physicists are going to want any proposed mechanism to be testable. It's also a pretty safe bet that if a mechanism exists, it's going to be a pretty strange one.
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Thanks, after working for 7 year in the old UKAEA I was wondering if my work was now understood and I,m really happy the someone really says " we don't know "
I,m a Yorkshireman and knows where there is muck there is money. I happy to think in my dotage that the pile of high level radioactive waste will eventually be valuable... Perhaps.!
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some nuclear "waste" contains valuable elements such as Polonium our uranium 294 but the cost of extraction is very high and would not justify extracting elements such as Radium or trying to use the heat generated by the waste en mass to generate electricity.
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When will THIS ATOM metaphorically go pop and decay.
I’m sure that JP will do much better than I can; it’s quite literally impossible to make that determination.
It could be now or perhaps any time from now to thousands of years.
That is correct.
SO without using the word probability,!! please would someone explain the mechanism that determines ,…
I’ll do my best Allan,
… and WHEN the atom goes pop??
You’re asking for the impossible because one has to use quantum mechanics to address questions such as this. Radioactive decay is a completely non-deterministic phenomenon. Think of the nucleons held together by the strong force and the alpha particles bouncing off the wall. The more they bounce off the wall the more likely to tunnel through the barrier.
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I,m very interested in all your replies but so far no one had suggested that the action is produced by non internal to the atom activity and as likely universal action not within the atom.
Mixture of superposition and entanglement across the space time field!
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I,m very interested in all your replies but so far no one had suggested that the action is produced by non internal to the atom activity and as likely universal action not within the atom.
Mixture of superposition and entanglement across the space time field!
I’m sorry, Allan, but we haven’t suggested it because observations demonstrate that such is simply not true. I explained the nature of nuclear decay in my post above. First off the reason for the various types of decay is different. Therefore the reason for the decay of tritium is different. The one I gave was that of alpha decay, which is something I know about. It was proposed by George Gamow. You wanted know about the decay of tritium. But in no case is it for what you think it has to do with, of, i.e. spacetime, superposition and entanglement.
Let:
(3,1)T = A nucleon having 3 protons and 1 neutron and therefore is an isotope of tritium.
(3,2)He+ = positively charged ion of helium having 3 protons and 2 neutrons and therefore is a an isotope of helium
e(-) = electron
anti-v(e) = electron antineutrino
Then https://en.wikipedia.org/wiki/Tritium#Decay
(3,1)T -> (3,2)He+ + e(-) + anti-v(e)
However this requires understanding high-energy particle physics, which I’m not skilled at. That’s why I used something I was familiar with, i.e. alpha decay, which is what I gave
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Let:
(3,1)T = A nucleon
Ionized Tritium (triton) is nucleus, not nucleon.
http://en.wikipedia.org/wiki/Nucleon
having 3 protons and 1 neutron and therefore is an isotope of tritium.
Tritium has 1 proton and 2 neutrons...
Atom with 3 protons is Lithium.
However this requires understanding high-energy particle physics, which I’m not skilled at. That’s why I used something I was familiar with, i.e. alpha decay, which is what I gave
Decay energy calculation is very simple.
It has been described in this thread on real examples that you can repeat in OpenOffice SpreadSheet:
http://www.scienceforums.net/topic/83451-radioactive-decay-and-information-split-from-what-is-real-in-physics/?p=808149
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First, find parent isotope mass. They're in f.e. wikipedia articles such as http://en.wikipedia.org/wiki/Isotopes_of_hydrogen (last portion of link is name of element).
For instance Carbon-14 has 14.0032419894 u
Then, you need to multiply it by 1u = 931.494 MeV to receive total energy of nucleus + electrons.
Then you need to get rid of mass-energy of electrons. Multiplying 0.510999 MeV by quantity of protons/electrons. Subtract total energy of electrons from energy of isotope.
Repeat it for daughter isotope. In Carbon-14 case it's Nitrogen-14.
Subtract N-14 nucleus energy from C-14 nucleus energy.
Subtract electron energy (it's emitted during decay together with antineutrino)
Result will be Decay Energy of isotope (kinetic energy of particles + energy of neutrino in our case).
Decay Energy must be positive value.
If you will calculate Decay Energy from stable isotope such as Deuterium, Helium-3, Helium-4 for all possible decay modes you will see that D.E. would be negative value. Thus these particles are stable.
For Carbon-14 it's 0.156 MeV.
As you can see this value does match article:
http://en.wikipedia.org/wiki/Carbon-14
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If you will calculate decay energy from Tritium you will receive 18.6 keV energy carried by electron (kinetic energy) and/or antineutrino.
(3,2)He+ = positively charged ion of helium having 3 protons and 2 neutrons and therefore is a an isotope of helium
Yet another wrong. Helium-3 has 2 protons, and 1 neutron.
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Ionized Tritium (triton) is nucleus, not nucleon.
http://en.wikipedia.org/wiki/Nucleon
FYI, just in case this happens again: Yes, yes. I know that all too well.
Tritium has 1 proton and 2 neutrons...
Atom with 3 protons is Lithium.
Again, something I know all too well.
The mistakes above were not mistakes I made because I don’t understand these very simple things. They’re mistakes I made because I have insomnia and haven’t slept in several days. For that I thank you for pointing them out.
Now that my bout with insomnia is over, this bout with error making is over too. Now, on to correct your bout with mistakes. :)
However this requires understanding high-energy particle physics, which I’m not skilled at. That’s why I used something I was familiar with, i.e. alpha decay, which is what I gave
Decay energy calculation is very simple.
Well, UT, the problem with that is that nobody ever asked about that. Alan Marsh, i.e. the person who created this thread, didn’t show any interest in the decay energy and I myself am certainly not interested in it for the purpose of this thread. You confused my comment above with something having to do with decay energy, which it doesn’t. I was explaining to Allan that in order to determine the when a nucleus will decay you need to know high-energy particle physics and don’t know that branch of physics to that level to answer that question of physics, i.e. to understand what the probability of it happening is. That’s why I used alpha decay above, i.e. because it gives one an idea of the nature of how such a probability of happening.
Yet another wrong.
So is the phrase “Yet another wrong.” :)
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Calculation of decay energy is key method to check whether some isotope is stable, or not stable.
It fits perfectly to thread with title "radioactive decay", regardless whether somebody asked for it or not.
Original poster might even not know about it, so I mentioned it just in case.
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Calculation of decay energy is key method to check whether some isotope is stable, or not stable.
Yes, yes, of course. But then again if you feel obliged to talk about “key” method of checking whether an isotope is stable, why not go yet another step yet again and mention what the criteria is for determining why any decay process occurs? A decay process that does not violate conservation of the energy or momentum laws and perhaps other particle conservation laws is permitted to happen (although not all have been detected).
I didn’t imply that you shouldn’t mention it. I stated that you’re wrong if you implied that I should have. That’s all.
I know that decay energy determines whether an atom is stable or not. But that doesn't pertain to the question. The question is that, given than atom isn't stable, when will it decay. However given that this is something that can't be known the question then becomes, what is the half life?
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Sorry folks I just selected tritium as part of my life has been spent firing deuterons electrically at tritium with a predictable result ! For a specific result!
No, the question relates to any radio isotope, decaying alpha, beta, or gamma, neutron , neutrino or any other bit.
The question was what causes the single atom to decay... ( POP!). Ie. WHEN!
The don't know answer is fine !
What appears clearly to me, it that TIME is a factor, and time is my obsession !
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Sorry folks I just selected tritium as part of my life has been spent firing deuterons electrically at tritium with a predictable result ! For a specific result!
But (usual) collision between D+ + T+ is fusion.
D+ + T+ -> He-4 + n0 + 17.6 MeV
If instead of bombarding D-T, you would bombard proton -proton
p+ + p+ -> D+ + e+ + Ve + 0.42 MeV
decay energy 0.42 MeV is randomly spread as kinetic energy of positron and neutrino.
One event positron takes more energy with itself, and neutrino less,
other time positron takes less energy, and neutrino more.
Probability.
No, the question relates to any radio isotope, decaying alpha, beta, or gamma, neutron , neutrino or any other bit.
The question was what causes the single atom to decay... ( POP!). Ie. WHEN!
The don't know answer is fine !
What appears clearly to me, it that TIME is a factor, and time is my obsession !
The all answers that are not related to probability is treated on the most of physics forums as non mainstream physics..
You can be banned straight away for such on many forums.
But I will tell you about electron capture - if we will ionize unstable isotope that has exclusive decay mode through electron capture, it won't be able to decay.
It's even mentioned in article
http://en.wikipedia.org/wiki/Electron_capture
Have you read this?
http://www.economist.com/blogs/babbage/2012/08/neutrinos-and-solar-storms
Jere Jenkins and Ephraim Fischbach detected that decay rate for a couple isotopes are varying depending on Solar activity.
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The question was what causes the single atom to decay... ( POP!). Ie. WHEN!
The don't know answer is fine !
Allan - The answer has already been given to you. Please don't confuse "don't know" with "it is impossible, in principle, to determined it" because they are very different things. The former is rooted in ignorance while the latter is rooted in the probablistic nature of quantum mechanics.
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Have you read this?
http://www.economist.com/blogs/babbage/2012/08/neutrinos-and-solar-storms
Jere Jenkins and Ephraim Fischbach detected that decay rate for a couple isotopes are varying depending on Solar activity.
Very interesting! I see this article is from 2012, has there been additional confirmation or follow-up since?
I guess I can believe that neutrino flux could have an effect on certain types of radioactive decay. They interact so weakly with other particles, that they can get very, very close to the nucleons. If they do happen to interact with the nucleus, it is through the weak force, and is very unlikely to occur, so an increase of flux a few orders of magnitude should have some small effect on the rate of decay. But I don't know enough about this to do the back-of-the-envelope calculations to know what the rate increase says about the interaction cross-section (probability), or the extent of activation (what fraction of nuclei that interact with a neutrino undergo decay?). Are non-linear effects are possible with high neutrino fluxes from solar storms (or supernovae???)
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dear PmbPhy may I refer you to Richard Feynman's famous saying
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sorry i know the forum is serious!!!! but I had to add this, which may come out a bit blurred.... Sorry!
I had to call up help from Mr Faraday.
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dear PmbPhy may I refer you to Richard Feynman's famous saying
I know it all too well and its one of the most misinterpreted quotes in all of physics. It gives the false impression that we don't know what can and what can't be determined and also gives the false impression that we don't know all the laws of quantum mechanics that can be known or that we don't know how to use them to their fullest extent. What it does mean is that physicists find nature at the quantum level to be very strange when compared to physics at the macroscopic level.
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Faraday? But add Donald Rumsfeld . It's the unknown unknowns that make it interesting.Great fun!
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To Ultimate Theory.
Ref tritium and deutrerons result fast neutrons
Ref Elliot automation Borehamwood 1960s. Called. "L tube" otherwise you need to go to Christmas Island 1960's
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To Ultimate Theory.
Ref tritium and deutrerons result fast neutrons
Didn't I write
"D+ + T+ -> He-4 + n0 + 17.6 MeV"
That summarize it all?
Ref Elliot automation Borehamwood 1960s. Called. "L tube" otherwise you need to go to Christmas Island 1960's
Sorry, but I don't know what you wanted to say..
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dear PmbPhy may I refer you to Richard Feynman's famous saying
This guy must be my hero.