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The formula for radioactive decay, $$N = N_0 \ e^{- \lambda t}$$ is derived from the following assumption: $$\frac{dN}{dt} = - \lambda N.$$

I understand an assumption has been made that no external factor can affect the probability of decaying, so the probability becomes identical for all atoms.

Here is the question: Do all radioactive decays follow that relationship in the real world?

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    $\begingroup$ So your question is essentially just "Can we effect the rate of radioactive decay?" $\endgroup$ – BioPhysicist Sep 24 '19 at 4:49
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    $\begingroup$ Related: "Changing decay rates", "Radioactive decay", Wikipedia. $\endgroup$ – Nat Sep 24 '19 at 10:30
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The equation $\frac{dN}{dt} = -\lambda N$ holds if

  1. the rate of the decay of any atom at any point of the time is constant and independent of any external factors, like the number of atoms left
  2. there are no other effects influencing the number of atoms other than the radioactive decay

The second one can be easily broken, for example by having the atoms to be produced in another process (as is the case with $\,^{14}C$). The first one is usully true, as the atoms aree usually far enough from each other to not affect their decays. However, in some cases, when they are more concentrated, it is possible for them to affect one another, and something like a chain reaction may happen (even in natural situation, like in the uranium deposits in Oklo.

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On a trivial level it might be that the rate of change of $N$ may not follow such a relationship because (a) there are multiple decay routes, (b) another process/decay is producing $N$ at the same time.

A less trivial point is that the relationship may only be strictly true in a vacuum. For instance, if we take the example of beta decay, the electron that is produced must occupy a quantum energy/momentum state according to the total energy it has after the decay. If this state is already occupied then the Pauli Exclusion Principle would forbid that particular channel.

Of course the decay probability is actually an integral over all possible states, but nevertheless, if a nucleus is surrounded by a degenerate electron gas, then it may not be possible to beta decay if the electron Fermi energy is high enough.

This situation pertains in the crusts and interiors of neutron stars, where the degenerate free electrons have Fermi energies of tens to hundreds of MeV. This results in the presence of very massive, neutron-rich nuclei (in the crust) and a fluid composed mostly of neutrons (in the interior), because the beta decay process is highly suppressed.

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The statement holds for particles decaying in their center of mass system. For a sample of atoms either in a solid or in a liquid or gas at rest, it holds true.

Once high velocities are considered one has to use special relativity, and then the apparent lifetime as measured in the lab will change, according to the equations of special relativity because of time dilation. So a relativistically moving radioactive atom will seem to live longer, and in that way would have a different λ .

Note the definitions of lifetime and λ, as @AgniusVasiliauskas says:

λ is decay rate defined as λ=1/τ, where τ is mean lifetime, defined as time at which the population of the assembly is reduced to 1/e

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  • $\begingroup$ Can you please elaborate on what you mean by decaying in their center of mass system? $\endgroup$ – curious Sep 24 '19 at 6:19
  • $\begingroup$ Also I have one more question. Is $\lambda$ the probability of disintegration per atom? $\endgroup$ – curious Sep 24 '19 at 6:21
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    $\begingroup$ @curious $\lambda$ is decay rate defined as $ \lambda = 1/ \tau$, where $\tau$ is mean lifetime, defined as time at which the population of the assembly is reduced to $1/e$ $\endgroup$ – Agnius Vasiliauskas Sep 24 '19 at 6:36
  • $\begingroup$ center of mass means that the atom is at rest, average momentum zero. This is true within errors for any bulk matter where classical equations hold. , i.e. non relativistic velocities $\endgroup$ – anna v Sep 24 '19 at 7:36
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When comparing dates of artifact extracted from $^{14}\text{C}$ isotope decaying method and from dendrochronology method - one sees that radiocarbon dating method gives very inaccurate dates of samples which are older than $3000$ years. Standard explanation to this is that at ancient times atmospheric ratio of $ {^{14}C}/{^{12}C}$ was different than that of now. However, there are other approaches to this problem too. This paper in 2012 year proposes alternative explanation,- that on long time scales, slow decay process exhibits a non-linear behavior, thus so attributing to incorrect radiocarbon dating conclusions when decay law is analyzed only exponentially. More concretely - they proposes such decay law : $$ \frac{dN}{dt} = -\alpha N + g(N,\theta,t) $$ where term $g()$ attributes to non-linear decay part, which affects long-term, slow-mode decay. However :

  • I haven't seen in their paper exact form of their $g$ function, seems that this was on-going reasearch at moment
  • It's very hard to validate such hypothesis, because you can't employ atmospheric radiocarbon tests in a lab, nor try to test slow-decay of atoms, because simply we don't have so much time for a test. The only thing which we can - try to "rollback" the time and construct some hypothesis about past and validate it against some observation evidence. But this is not an easy task.

EDIT:

BTW, just for curiosity - How atomic bomb tests were affecting $^{14}C$ isotope ratio in atmosphere: enter image description here

Thus for creatures who have lived and died in atomic bomb test places roughly in period $ 1955-1995 $,- radiocarbon creature age dating method will not work too, at least without including a-bomb tests effects.

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  • $\begingroup$ On the contrary, radiocarbon dating is used for materials with origins in the years 1960+ because the graph above is well defined - and is known as "bomd pulse dating". en.wikipedia.org/wiki/Bomb_pulse $\endgroup$ – Rob Jeffries Sep 24 '19 at 14:13
  • $\begingroup$ Have you noticed my sentence part "at least without including a-bomb tests effects" ? Seems, not. If you would perform classical carbon dating, then it will not work, because it assumes $^{14}C$ ground level of 100%. So, you must take into account increased levels of radiocarbon at that time for classical carbon dating method. BTW, also notice phrase "Bomb pulse dating should be considered a special form of carbon dating" in your mentioned link in section "Difference_with_classical_radiocarbon_dating". So, it's not classical carbon dating - that was the thing I was trying to say $\endgroup$ – Agnius Vasiliauskas Sep 24 '19 at 15:58

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