Half-Life and Mean Life

The time to halve

The decay constant is exact but hard to picture. A more natural measure is the time a sample takes to fall to half. This is the half-life \(T_{1/2}\). Because an exponential falls by the same fraction in every equal interval, the half-life does not depend on how much there was to start with. In each half-life the sample halves again: to \(\tfrac12\), \(\tfrac14\), \(\tfrac18\) of the start, in equal steps of time.

Fig 4.1 shows the steps for a technetium-99m dose, whose half-life is \(6\,\text{h}\). The first \(400\,\text{MBq}\) go in six hours, the next \(200\,\text{MBq}\) in the next six, then \(100\,\text{MBq}\). The amount lost in each step shrinks, but the time for each halving never changes.

To link the half-life to \(\lambda\), put \(N = N_0/2\) at \(t = T_{1/2}\) into the decay law. Fill in the missing step.

How long does a nucleus live, on average?

Some nuclei of a sample decay in the first second, and a few survive for many half-lives. The average lifetime of all of them is the mean life \(\tau\). To find it, count the nuclei that decay between \(t\) and \(t + dt\). There are \(\lambda N_0 e^{-\lambda t}\,dt\) of them, and each lived for a time \(t\). Average \(t\) over all \(N_0\) nuclei.

The mean life is longer than the half-life, \(\tau = T_{1/2}/0.693 = 1.44\,T_{1/2}\). The few nuclei that survive for many half-lives pull the average up. At \(t = \tau\) the decay law gives \(N = N_0e^{-1} = 0.37\,N_0\). So after one mean life about 37 % of the nuclei are left, a little less than the half left after one half-life.

Fig 4.2 shows a second meaning of \(\tau\). The tangent at the start has slope \(-\lambda N_0\), so it reaches zero at \(t = N_0/(\lambda N_0) = 1/\lambda = \tau\). The mean life is the time the sample would take to decay completely if it kept its first rate. The rate falls instead, so the sample outlasts this estimate. The same holds at every point, as the tangent figure of the last lesson showed.

Counting in half-lives

Since \(\lambda T_{1/2} = \ln 2\), the decay law can be written with the half-life alone. Put \(\lambda = \ln 2/T_{1/2}\) into \(N = N_0e^{-\lambda t}\). Then \(e^{-\lambda t} = e^{-(t/T_{1/2})\ln 2} = 2^{-t/T_{1/2}}\), so \[ N = N_0\left(\tfrac12\right)^{n}, \qquad n = \dfrac{t}{T_{1/2}} \] Here \(n\) is the number of half-lives that have passed. It need not be a whole number.

After 1 half-life \(\tfrac12\) is left and \(\tfrac12\) has decayed. After 3 half-lives \(\tfrac18\) is left and \(\tfrac78\) has decayed. After 5 half-lives only \(\tfrac1{32}\) is left, and \(\tfrac{31}{32}\) has decayed. For a fraction of a half-life, use the power directly: after \(9\,\text{h}\), 1.5 half-lives of technetium-99m, the dose is \(800 \times 2^{-1.5} = 283\,\text{MBq}\).

A fraction that is not a power of \(\tfrac12\) needs a logarithm. To fall to one tenth, \(e^{\lambda t} = 10\), so \(t = \ln 10/\lambda\). With \(\ln 10 = 2.303\) and \(\lambda = 0.693/T_{1/2}\), this is \(t = (2.303/0.693)\,T_{1/2} = 3.32\,T_{1/2}\). A nuclide with a half-life of 3 days falls to a tenth of its activity in about 10 days.

Twenty-four hours after the morning measurement, the technetium dose of the last lesson is down to \(800/16 = 50\,\text{MBq}\), four half-lives later. The waste of a day's work is therefore harmless within a few days. That is why hospitals prefer a short-lived tracer: it does its work in an afternoon and is gone by the end of the week.