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While I was reading up about mean life I came across a common definition.

It is the average time taken by an arbitrary radioactive nucleus to undergo decay (since different particles may take different times). But radioactivity is an exponential function or phenomenon. Therefore I feel that there will always be that one nucleus that hasn't decomposed, making its life infinity. Therefore the mean life (being the average value of individual lives) must also always be infinity, right?

Can someone tell me what's wrong in this logic?

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  • $\begingroup$ infinity is not a number, it is a limit to be approached, and the limit has been given the name "infinity". $\endgroup$ – anna v Feb 19 '18 at 9:51
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Perhaps looking at the probability distribution function which characterises a radioactive decay process might help?

The basic premise when dealing with radioactive decay is that there is a constant parameter $\lambda$ locked away in a nucleus which dictates the decay of the nucleus.

You perhaps first meet it when it is stated that the rate at which nuclei decay $\dfrac {dN}{dt}$ is proportional to the number of undecayed nuclei $N$ which leads to the expression $\dfrac{dN}{dt} = - \lambda \,N$.

Another way of expressing this is to use a probability distribution function for the decay.

$F(t) = \lambda \, e^{-\lambda t}$

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You observe one nucleus and start a clock ($t=0$).

The probability of the nucleus decaying within a time $t$ after starting the clock is $\displaystyle \int^t_0 \lambda \, e^{-\lambda t} \, dt$ ie the area under the probability distribution curve between time $0$ to time $t$.

Another way of stating this that after a time $t$ the probability of the nucleus decaying in the next interval of time $dt$ is $ \lambda \, e^{-\lambda t} \, dt$.

As an example if I want to know how long one has to wait $\tau$ for the probability of a decay to be $\frac 12$.

$\displaystyle \int^\tau_0 \lambda \, e^{-\lambda t} \, dt = \dfrac 12$

which produces the relationship between $\lambda$ and $\tau$

$\lambda\,\tau = \log_{\rm e} 2$ and $\tau$ is called the half life of the decay process.

Now note that $\displaystyle \int^\infty_0 \lambda \, e^{-\lambda t} \, dt=1$ which seems to say that if you wait for an infinite length of time the nucleus is certain to decay and the mean time for a decay is $\langle t \rangle = \displaystyle \int^\infty_0 t\, \lambda \, e^{-\lambda t} \, dt= \dfrac 1 \lambda$.

Put another way this is what you would get if you started with an infinite number of nuclei and watched them decay over an infinite length of time.

Now look at the evaluation if the time is $\frac {20}{\lambda}$ (about 14 half lives) then you get the probability of a decay in that time is $1-2\times 10^{-9}$ and the mean life is $\dfrac {1-4 \times 10^{-8}} {\lambda}$.

What this shows is that you are working out a weighted average and the weighting for large values of time are very small so much so as for them to be insignificant on any realistic time scale.

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The usual expression for the number of nuclei remaining after a time $t$ is:

$$ N(t) = N_0 e^{-t/\tau} $$

where $\tau$ is related to the half life by:

$$ t_{1/2} = \tau\ln 2 $$

And as you point out this implies that $N(t)$ never reaches zero so there must be at least one nucleus remaining.

But this equation is a statistical one and only applies when there are a large number of nuclei. When we get down to one nucleus this obviously isn't a large number of nuclei so the equation above doesn't apply. So you cannot argue that the equation means there must always be one nucleus remaining.

If we have just one nucleus then all we can say is that there is a $50$% chance that the nucleus will decay within a time $t_{1/2}$. The probability that the nucleus remains undecayed goes to zero as $t$ goes to infinity.

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  • $\begingroup$ So assuming i start the decay with an infinite number of particles.... The mean life will be infinite?! $\endgroup$ – Adithya Eshwarla Nov 1 '17 at 7:32
  • $\begingroup$ Also... Doesn't this contradict the property of the decay constant, which is not supposed to change with initial amount of substance taken.. Because by derivation...mean life=1/(decay constant).and as we have discusse here....the cases for a finite and infinite initial amount give different values of mean life $\endgroup$ – Adithya Eshwarla Nov 1 '17 at 7:35
  • $\begingroup$ @AdithyaEshwarla: The decay constant is related to the probability of a given nucleus decaying and it is a constant. But the equation for the number of nuclei using the decay constant $N(t) = N_0 e^{-\lambda t}$ is an approximation that only holds for large numbers of nuclei. $\endgroup$ – John Rennie Nov 1 '17 at 8:18
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If initially you have a finite number of nuclei, there will not "ALWAYS be that ONE nucleus that hasn't decomposed". If initially you have an infinite number of nuclei, there will "ALWAYS be that ONE nucleus that hasn't decomposed", but that would not mean that the halflife is infinite, as you need to average over an infinite number of nuclei.

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  • $\begingroup$ Ok i get your point. But assuming i start with an infinite number of nuclei..... Averaging even an infinite number of nuclei with 1 of the lives as infinity will not give me a finite value right?! $\endgroup$ – Adithya Eshwarla Nov 1 '17 at 7:29
  • $\begingroup$ @AdithyaEshwarla: It can give you a finite number, as the probability of infinite life is zero. $\endgroup$ – akhmeteli Nov 1 '17 at 8:02
  • $\begingroup$ You can't easily apply logic that applies to finite sets to an infinite set. For a finite set, there will be no nucleus that does not decay. For an infinite set ... well there's no such thing. One can ask what would happen in the limit as the set becomes infinite. One would probably be lead to the use of continuous functions and probability densities. The average of the continuous distribution would not be infinite. $\endgroup$ – garyp Nov 1 '17 at 15:32

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