Calculate the decay constant of a radioactive isotope of half-life 138.5 s.

Answer Details

Radioactive decay is random, so the number of undecayed nuclei falls exponentially: \(N = N_0 e^{-\lambda t}\), where \(\lambda\) is the decay constant, the probability per second that a given nucleus decays. The half-life \(t_{1/2}\) is the time for \(N\) to fall to \(N_0/2\). Putting \(N = N_0/2\) and \(t = t_{1/2}\) into the exponential law gives

\[\tfrac{1}{2} = e^{-\lambda t_{1/2}} \quad\Rightarrow\quad \lambda t_{1/2} = \ln 2 \quad\Rightarrow\quad \lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{t_{1/2}}.\]

Substituting the given half-life,

\[\lambda = \frac{0.693}{138.5\ \text{s}} = 5.004\times 10^{-3}\ \text{s}^{-1},\]

which to two significant figures is \(5.0\times 10^{-3}\ \text{s}^{-1}\). Note that the decay constant has the unit \(\text{s}^{-1}\), the reciprocal of time, because it is a rate per nucleus rather than a time.

The neighbouring values here are all within a couple of per cent of one another, so they are testing whether the constant \(0.693\) is used rather than a rounded \(0.7\) (which would give \(5.05\times 10^{-3}\)) or an inverted formula such as \(t_{1/2}/\ln 2\). Keep \(\ln 2 = 0.693\) and remember that a short half-life means a large decay constant, since the two are inversely proportional.