Average lifetime of a radioactive atom

Posted by David Zaslavsky on November 22, 2011 2:43 AM — Edited February 27, 2012 4:19 PM

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Some time ago I posted about the theoretical justification for exponential decay. In that post, I showed that you can quantify exponential decay with this equation:

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

where $N(t)$ is the number of undecayed atoms at time $t$ and $\lambda$ is a constant representing the decay rate. If you plug in $N(t) = \frac{1}{2}N(0)$, the condition for the half-life, you can find that

$$\lambda = \frac{\ln 2}{t_{\frac{1}{2}}}$$

But physicists usually write the formula like this,

$$N(t) = N(0)e^{-t/\tau}$$

where $\tau$ is called the time constant. We prefer this to using the decay rate because, as I wrote in Calculating Terminal Speed, it’s often best to write a physics question in terms of dimensionless ratios like $\frac{t}{T}$, where $T$ is some time characteristic of the physical system you’re studying. We could use the half-life for $T$, but the time constant $\tau$ is more appealing for a couple of reasons: it keeps that ugly factor of $\ln 2$ out of the formula, and more importantly, $\tau$ is physically meaningful because it’s the average lifetime of an individual atom.

If you wanted to calculate that, your first thought might be, as mine was, to use the usual formula for finding the expectation value of a variable: each possible value of the variable $t$ multiplied by its probability $P(t)\udc t$ (note that $P(t)$ itself is actually the probability density), integrated over the allowed domain.

$$\expect{t} = \int t P(t)\udc t$$

But that doesn’t quite work. Basically, the reason is that if the atom decays at some early time, it won’t be around to decay at a later time, so the earlier times “count” more than the later times. When you compute the expectation value, you have to incorporate a weighting function that weights the earlier times more heavily, so your integral winds up looking more like this:

$$\expect{t} = \int t w(t) P(t)\udc t$$

You might be able to guess what the weighting function should be, but let’s figure it out with a proper calculation.

Imagine a large collection of atoms, initially of size $N_0 \equiv N(0)$, and calculate the average decay time of these atoms. We can start with the expression

$$\expect{t} = \frac{1}{N_0}\sum_{n=1}^{N_0}t_n$$

where $t_n$ is the time of the decay of the $n$th atom, which sums the decay times of all atoms and divides it by the total number. Now imagine cutting the time domain $[0,\infty)$ into small ranges of width $\Delta t$, and grouping the terms in the sum according to which time range they fall in.

$$\expect{t} \approx \frac{1}{N_0}\sum_{i=0}^{\infty}(i\Delta t)[N(t) - N(t + \Delta t)]$$

$N(t) - N(t + \Delta t)$ is the number of atoms in each time range, and $i\Delta t$ is the approximate value of the decay time of each of those atoms. This approximation becomes exact in the limit $\Delta t \to 0$:

$$\expect{t} = \lim_{\Delta t\to 0}\frac{1}{N_0}\sum_{i=0}^{\infty}(i\Delta t)[N(t) - N(t + \Delta t)] = \frac{1}{N_0}\int_0^{\infty}(t)[-N'(t)\udc t]$$

That integral actually simplifies a bit, even without plugging in $N(t)$: you can integrate by parts to get

$$-\evalat{t N(t)}{0}{\infty} + \int_0^{\infty}N(t)\udc t$$

and if you assume $N(\infty) = 0$ (which should be the case for any decay process), the first term is zero and you’re just left with

$$\expect{t} = \frac{1}{N_0}\int_0^{\infty}N(t)\udc t$$

This is a nice general formula that gives you the average decay time of any one atom in terms of the function $N(t)$ that tells you how many atoms are left at any given time. Even if the decay is not exponential, this should still work (although I can’t think of any cases offhand in which things decay non-exponentially).

If you plug in the exponential function for $N(t)$, you get

$$\expect{t} = \int_0^{\infty}e^{-t/\tau}\udc t = \evalat{-\tau e^{-t/\tau}}{0}{\infty} = -\tau(0 - 1) = \tau$$

OK, good! This shows that the average lifetime of a radioactive atom actually is the time constant $\tau$.

Now, back to figuring out the weighting function. Let’s look at the probability density function, $P(t)$, which represents the probability per unit time that a given atom will decay. You can calculate it from the function $N(t)$: take a very small time interval $\Delta t$, and divide the number of atoms which decay in that time by the total number of atoms at the start of the interval. That should be equal to the probability per unit time, $P(t)$, times the length of the time interval:

$$P(t)\Delta t = \frac{N(t) - N(t + \Delta t)}{N(t)} = -\frac{N(t + \Delta t) - N(t)}{\Delta t}\frac{1}{N(t)}\Delta t$$

Then take the limit as $\Delta t$ becomes infinitesimally small, and you get

$$P(t)\udc t = -\frac{N'(t)}{N(t)}\udc t$$

Solve this for $N'(t)$ and you get

$$N'(t) = -P(t)N(t)$$

That can be substituted into the result $-\frac{1}{N_0}\int tN'(t)\udc t$ from earlier, and it gives you

$$\expect{t} = \frac{1}{N_0}\int_0^\infty t P(t)N(t)\udc t$$

Surprise surprise, the weight function is just the number of atoms, $N(t)$. This is the mathematical way to express that the earlier times, when there are more atoms, count more when you’re computing the average.