Exercise 13 - Radioactive decay

Exercise 13 - Radioactive decay

Andrew Valentine, Louis Moresi, louis.moresi@anu.edu.au

Unstable atomic nuclei may undergo radioactive decay, losing energy (and particles) via radiation. This is a random process, but one which can be quantified by the concept of ‘half-life’ or ‘decay rate’. The half-life is the amount of time required for 50% of the atoms in an initially pure sample to decay. It is often represented by the symbol \(t_{1/2}\), and is related to the decay rate, \(\lambda\), by the formula

\[ t_{1/2} = \frac{\ln 2}{\lambda} \]

We can interpret \(\lambda\) as being the probability that each nucleus will decay per unit time. We can also show that radioactive decay obeys the differential equation

\[ \frac{\mathrm{d}N}{\mathrm{d}t} = -N\lambda \]

This means that if there are \(N\) radioactive particles at time \(t\), then we expect to see \(N \lambda \Delta t\) of them decay in the (infinitesimal) time interval \(\Delta t\).

The isotope radon-219 decays to a stable isotope of lead, principally via the following decay chain (part of the actinium series):

\({}^{219}\mathrm{Rn}\left[\to{}^{215}\mathrm{Po}\right]\to{}^{211}\mathrm{Pb}\to{}^{211}\mathrm{Bi}\to {}^{207}\mathrm{Tl} \to{}^{207}\mathrm{Pb}\)

The half-lives of each unstable isotope in this chain are:



Energy release (MeV)


3.96 s



1.7 ms



36.1 mins



2.14 mins



4.77 mins


Suppose you have a sample that is known to be pure, and contain N atoms of \({}^{219}\)Rn at some time \(t\). Provided \(\Delta t\) is small, you expect that at the time \(t+\Delta t\) you will have \(N(1-\lambda_{{}^{219}\mathrm{Rn}}\Delta t)\) atoms of \({}^{219}\)Rn, and \(N (\lambda_{{}^{219}\mathrm{Rn}}\Delta t)\) atoms of \({}^{215}\mathrm{Po}\). The amount of each isotope within the sample is a balance between its production, and its decay.

➤ Write code to simulate the decay chain. Starting with 10,000 \({}^{219}\mathrm{Rn}\) particles, produce a graph of the amount of each isotope present at any point in time. (Since the halflife of \({}^{215}\)Po is so short, we can neglect this step and treat the system as though \({}^{219}\)Rn decays directly to \({}^{211}\)Pb.)

# Try it here!

Try changing your timestep, \(\Delta t\). What happens as you make this bigger?

# Try it here

The energies reported above are for the decay of a single atom. To obtain the energy release in a given timestep, we need to multiply these by the number of decay events occurring.

➤ Plot the energy released by the sample over time.

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Suppose your sample now contains three times as much \({}^{207}\)Pb as \({}^{211}\)Pb. Show that it is 1 hour, 15 minutes and 8 seconds old.

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