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My friend told me about this idea.

Let there be $N_0$ atoms of a particular radioactive material. Then the number of atoms at time $t$, $N_t$, we already know, is given by:

$$N_t = N_0 e^{-t/\tau},\; where\; \tau = \frac{t_{1/2}}{\ln(2)}$$

where Half-life of the material: $t_{1/2}$.

Is the following quantity useful in any physics or nuclear engineering or something:

$$T_{decay} = \tau(1 + \ln{N_0}) = \dfrac{t_{1/2}}{\ln{2}} (1 + \ln{N_0})$$

This is basically the sum of the time it takes for the number of atoms/nuclei to reduce to $1$ and the mean lifetime of a single atom/nucleus.

Isn't $T_{decay}$ the time it takes for all nuclei to have decayed/changed?

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$T_{decay}$ as you've defined it is a reasonable rough estimate of how long it will take before the last nucleus has decayed. However, you can't say that that's how long it will take for sure, because $N_t = N_0 e^{-t/\tau}$ is only accurate in a probabilistic sense. Also, that expression for $N_t$ is only accurate when $N_t$ is large enough that it's reasonable to treat it as a real number instead of an integer.

You can get a more accurate value for $T_{decay}$ by treating the problem probabilistically explicitly:

Given that the probability that a given initially undecayed nucleus $i$ will remain undecayed after a period $t$ is given by $$\bar{p}_i(t)=e^{-t/\tau}\ \ ,$$ the probability that all $N_0$ initially undecayed nuclei will have decayed after a period $t$ is given by $$p(t)=\prod_{i=1}^{N_0}\left[1-\bar{p}_i(t)\right]=\left ( 1-e^{-t/\tau} \right )^{N_0}\ \ .$$ Solving that expression for $t$ gives that the time required for there to be a probability $p$ that all of the nuclei have decayed is $$T_{decay}(p)=-\tau \ln \left ( 1-p^{1/N_0}\right )\ \ .$$

Note that that expression diverges in the limit $p \to 1$, reflecting the fact that you can never quite be 100% certain that all of the nuclei have decayed, no matter how long you wait.

The above can be related to the value of $T_{decay}$ given in the question by noting that

$$p\left(\tau(1+\ln N_0)\right)=\left(1-\frac{1}{e N_0}\right)^{N_0}\ \ .$$

For large $N_0$ this approaches

$$\lim_{N_0\to\infty}p\left(\tau(1+\ln N_0)\right)=e^{-1/e}=0.692201\ \ .$$

I.e., after a period of $T_{decay}$ as given in the question, there's about a 69.22% chance that all of the nuclei have decayed.

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  • $\begingroup$ What is $p(\tau(1 + \ln{N_0}))$? $\endgroup$ – PhyEnthusiast Feb 23 '18 at 7:07
  • $\begingroup$ @PhyEnthusiast I have augmented my answer to address your follow-up question. $\endgroup$ – Red Act Feb 23 '18 at 14:47
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According to the cited decay law for the decay of a very large number of radioactive nuclei, there is no finite time until all nuclei will have decayed. You can chose any large time $t$ and you will always still have a finite number of undecayed nuclei. You haven't given an explanation how you arrive at this quantity $T_{decay}$. $T_{decay}$ doesn't seem to have any significance.

Note: The exponential decay law is a probabilistic law. You can take $N/N_0=e^{-t/\tau}$ to represent the probability at time $t$ that a single nucleus has not yet decayed. Therefore, even if you have only one nucleus, you cannot give a time until it will have decayed!

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  • $\begingroup$ The OP is using the discrete nature of the number of nuclei and the fact that the continuous exponential approximation must break down once there are only a few nuclei left. They are saying that $\tau \ln N_0$ is the amount of time it takes for the expected number of nuclei to reach 1, and then the single additional $\tau$ is the time for that final nucleus to decay. $\endgroup$ – tparker Feb 22 '18 at 18:21
  • $\begingroup$ @tparker - The exponential decay law is a probabilistic law. You can take $N/N_0$ for the probability at time $t$ that a single nucleus has not yet decayed. Therefore, even if you have only one nucleus, you cannot give a time until it will have decayed. $\endgroup$ – freecharly Feb 22 '18 at 18:40
  • $\begingroup$ I know, I was just clarifying their (incorrect) reasoning for you. $\endgroup$ – tparker Feb 22 '18 at 18:43
  • $\begingroup$ @tparker - Thank you for the explanation of the OP's reasoning. You are right, I will extend my answer. $\endgroup$ – freecharly Feb 22 '18 at 18:47
  • $\begingroup$ @freecharly The formula only has $N_0$ because $N_t$ is taken to be $1$. As tparker put it, $\tau \ln N_0$ is the amount of time it takes for the expected number of nuclei to reach $1$, and then the single additional $\tau$ is the time for that final nucleus to decay. I have explained this within the question. $\endgroup$ – PhyEnthusiast Feb 23 '18 at 6:56

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