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The radioactive decay formula is

$$ N(t) = N_{0}e^{- \lambda t} $$

My understanding is that $\lambda$ indicates the probability of decay per unit time. Let's say $\lambda = 0.5 s^{-1}$. So we would intuitively expect half of the atoms in a population to have decayed after 1 second. Let's say we started with 1000 atoms, so there would be 500 undecayed ones left. However, the formula above gives:

$$ N(t) = 1000 e^{- 0.5} = 606.5 $$

It is consistently overestimating how many are undecayed. Please could anyone explain where the problem is with this reasoning?

Thank you

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  • $\begingroup$ because $e > 2$. $\endgroup$
    – JEB
    Commented Mar 19, 2022 at 20:08

4 Answers 4

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Let's say $\lambda = 0.5 s^{-1}$. So we would intuitively expect half of the atoms in a population to have decayed after 1 second.

Those don’t follow. To model a system where half the atoms decay after one second, use the half-life $\tau_{1/2} = 1\,\rm s$, and the relation

$$ N(t) = N_0 \cdot 2^{-t/\tau_{1/2}} $$

This is the same as your relation if

\begin{align} e^\lambda &= 2^{1/\tau_{1/2}} \\ \lambda &= \frac{1}{\tau_{1/2}} \ln 2 \end{align}

For a half-life $\tau_{1/2}=1\rm\,s$, this relation gives $\lambda = 0.69\rm\,s^{-1}$. You are underestimating the decays because your decay constant is too small.

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  • $\begingroup$ Thank you @rob, this is a useful and interesting angle to approach it from. However, I'm finding this has now flipped the problem around. If the decay constant is now 0.69 per second, and I go through each of the thousand atoms, and there is a 0.69 chance of each one decaying per second, doesn't this result in 690 atoms decaying after one second, and not 500 as the half-life suggests? $\endgroup$
    – Academo
    Commented Mar 19, 2022 at 15:58
  • $\begingroup$ You can't interpret the decay constant that way. Suppose the decay constant were $\lambda = \rm 6.9\,s^{-1}$. Would that mean a given atom has a 690% chance of decaying in one second? It would not. It would mean instead that the half-life is one-tenth of a second. Grab your calculator and do the arithmetic: you should get $e^{-0.69} = \frac12$. $\endgroup$
    – rob
    Commented Mar 19, 2022 at 16:14
  • $\begingroup$ Many thanks for the clarification, @rob. I would like to pursue the idea of interpreting the decay constant as a probability, so can I just check that if the decay constant was indeed 6.9 $s^{-1}$ then the probability an atom would be undecayed after 1 second would be $e^{-6.9} = 0.001$? $\endgroup$
    – Academo
    Commented Mar 19, 2022 at 16:40
  • $\begingroup$ Right. Ten half-lives gives a survival probability of $2^{-10}\approx 10^{-3}$. As your other answers say, you can think of $1/\lambda$ as the “mean life” of a particle from your ensemble. $\endgroup$
    – rob
    Commented Mar 19, 2022 at 16:57
  • $\begingroup$ Great, thank you. $\endgroup$
    – Academo
    Commented Mar 19, 2022 at 16:59
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Because your interpretation of "probability per unit time" as meaning "half the atoms have decayed in one second" is not quite correct, even if that probability "per unit time" is $0.5\ \mathrm{s}^{-1}$. The reason for this is probabilities are not additive with time. If they were, you'd eventually get probabilities larger than $1$, which makes no sense. Your interpretation would result not only in less than zero atoms remaining after two seconds, but in other cases of probability, probabilities quickly exceeding 100%.

Instead, you should think about probability per unit time "differentially". In particular, if there is a PPUT $v_p$ ("velocity of probability") that some event happens, the chance to happen over a time $\Delta t$ is not $v_p\ \Delta t$, but if time $\Delta t$ is very very very short, then it is approximately $v_p\ \Delta t$, with this becoming exact in the limit $t \rightarrow 0$. Try your formula with a time of one microsecond, i.e. $10^{-6}\ \mathrm{s}$. Intuitively, you should expect $0.0005\%$ of the particles to have decayed. The formula actually gives $1-e^{-0.5 \times 10^{-6}} \approx 0.00049999875\%$ of them to have decayed. As you can see, that's a far smaller discrepancy.

Hence, the problem is your interval of one second is much too large for this rate in order for the non-exponential formula to work.

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The expectation value of a particles lifetime is given by $$\begin{aligned} \langle t\rangle &= \frac{\int_0^\infty t N(t) dt }{\int^\infty_0 N(t)dt}\\ \langle t \rangle &=\frac{1/\lambda^2}{1/\lambda }=\frac{1}{\lambda }\\ \lambda &=\frac{1}{\langle t\rangle } \end{aligned}$$ The expectation value of the lifetime tells us what the average value of lifetime will converge to if we increase our samples of lifetime measurements. The rate $\lambda$ tells us what the average decay rate will converge to. So $\lambda$ does not directly tell you how many particles decay per second, it tells you what the average decay value will be if you measure the decays per second many times and the average over them. You would have to measure the decay rate for varying $t$ values and then average over them to get $\lambda$.

It also doesn't tell you directly how many particles will be left after one second unless you are able to evaluate the exponential function in your head. You do know that your average population will be decayed to a percentage of $N(1s)/N_0 =\exp(-\lambda)$ particles (where $\lambda$ stands for the numerical value of $\lambda$ in units of seconds).

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For $t$ much smaller than $1/\lambda$, a proportion $1-e^{-\lambda t}\approx\lambda t$ of atoms decay within a time $t$. This approximation is actually an overestimate; the Taylor series continues $-\frac12\lambda^2t^2$. You can't use small-$t$ approximations for large reductions in the particle count.

Related: $1/\lambda$ is the mean lifetime; the half-life is the median lifetime, $\ln 2/\lambda$.

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