radioactive decay law and half-life

The decay of a single nucleus is a random event, but for a macroscopic sample the number of nuclei that have not yet decayed decreases in a regular, exponential way:

$$N(t)=N_0 e^{-\lambda t},$$

where $N_0$ is the initial number of nuclei and $\lambda$ is the decay constant, that is, the probability that one nucleus decays per unit time.

The half-life $T_{1/2}$ is the time during which half of the available nuclei decay; it is related to the decay constant by

$$T_{1/2}=\frac{\ln 2}{\lambda}\approx\frac{0{,}693}{\lambda}.$$

It follows from the decay law that after every $T_{1/2}$ the number of nuclei is halved: $N_0 \to N_0/2 \to N_0/4 \to N_0/8$, and so on, so instead of $e^{-\lambda t}$ it is often convenient to write $N=N_0\left(\tfrac12\right)^{t/T_{1/2}}$. The exponential never reaches zero in a finite time, so we speak not of complete decay but of the decayed fraction. The decay rate is described by the activity $A=\lambda N=A_0 e^{-\lambda t}$, measured in becquerels ($1\ \text{Bq}=1\ \text{decay/s}$), and the quantity $\tau=1/\lambda$ is called the mean lifetime of a nucleus.

A sample contains $N_0=8{,}0\cdot10^{20}$ nuclei of an isotope with a half-life $T_{1/2}=5{,}27$ years. How many nuclei remain after 15.81 years?

At $t=0$ the activity of a sample is $4000$ decays/s, and after $6{,}0$ h it is $500$ decays/s. Find the half-life of this isotope.

The carbon-14 activity of a sample of ancient wood is 0.30 of the activity of the same amount of carbon in a living tree. The half-life of $^{14}$C is 5730 years. Estimate the age of the sample and the mean lifetime of a $^{14}$C nucleus.

Carry out a safe simulation of decay without using real radioactive sources. Take 100 identical dice, pour them from a cup onto a table and count as "decayed" those that show, for example, a six (probability $1/6$ per throw).

1. Record the throw number and the number of remaining dice, then put them back into the cup.
2. Repeat the throw 10–12 times.
3. Plot the number of remaining dice $N$ against the throw number.
4. Check that in equal intervals the remaining number decreases by roughly the same factor, and estimate the "half-life" in throws.
5. To improve accuracy, pool data with 3–4 partners or compare the result with an online radioactive decay simulation.

To show that the decrease in the number of nuclei follows an exponential law rather than a linear one, and that exactly half of the nuclei remain after each half-life.

A graph of $N/N_0$ against time $t$ measured in half-lives $T_{1/2}$: a smooth decreasing exponential with marked points $(0;\,1)$, $(T_{1/2};\,0{,}5)$, $(2T_{1/2};\,0{,}25)$, $(3T_{1/2};\,0{,}125)$ and dashed lines to the axes; for comparison, a straight line corresponding to linear decrease is shown alongside. In addition, a sketch of 16 squares-«nuclei» in which half of the remaining squares are shaded at each step: $16\to8\to4\to2$.

A wooden beam is found in an old house. A carbon sample of mass $1.00\ \text{g}$ taken from it gives an activity of $0.150\ \text{Bq}$; the same amount of carbon in living wood gives $0.230\ \text{Bq}$. The half-life of $^{14}\text{C}$ is $5730\ \text{yr}$. The age of the beam is to be estimated from these two numbers.

Activity is proportional to the number of undecayed nuclei, so the activity ratio $0.150/0.230=0.652$ equals $N/N_0$. The decay law $N=N_0 e^{-\lambda t}$ together with $\lambda=\ln 2/T_{1/2}$ gives $t=\frac{T_{1/2}}{\ln 2}\ln\frac{N_0}{N}=\frac{5730}{0.693}\ln 1.53\approx 3500\ \text{yr}.$ What does the work here is the exponential law and the link between the decay constant and the half-life, not a linear fall-off.

At $08{:}00$ a hospital syringe holds $800\ \text{MBq}$ of technetium-$99\text{m}$ with a half-life of $6.01\ \text{h}$. The patient must be scanned before the activity drops below $100\ \text{MBq}$, otherwise too little of the tracer remains in the dose.

Every half-life cuts the activity exactly in half: $800\to 400\ \text{MBq}$ by $14{:}00$ and $400\to 200\ \text{MBq}$ by $20{:}00$. The relation $A=\lambda N$ with $\lambda=\ln 2/T_{1/2}$ gives $\lambda=\frac{0.693}{6.01\ \text{h}}=0.115\ \text{h}^{-1}=3.2\times 10^{-5}\ \text{s}^{-1}$, and $A=A_0 e^{-\lambda t}$ then predicts directly that after $12\ \text{h}$ the activity has fallen to $200\ \text{MBq}$.

A smoke detector contains an $^{241}\text{Am}$ source with an activity of $33\ \text{kBq}$ and a half-life of $432\ \text{yr}$. The task is to estimate how many nuclei, and what mass of americium, sit inside such a tiny capsule.

The decay constant follows from the half-life: $\lambda=\frac{\ln 2}{432\ \text{yr}}=\frac{0.693}{1.36\times 10^{10}\ \text{s}}=5.1\times 10^{-11}\ \text{s}^{-1}$. Then $N=A/\lambda=\frac{3.3\times 10^{4}}{5.1\times 10^{-11}}\approx 6.5\times 10^{14}$ nuclei, and the mass is $m=\frac{N M}{N_A}=\frac{6.5\times 10^{14}\cdot 241}{6.02\times 10^{23}}\approx 2.6\times 10^{-7}\ \text{g}=0.26\ \mu\text{g}$. Both $A=\lambda N$ and $\lambda=\ln 2/T_{1/2}$ are used: a huge number of nuclei giving a modest activity is exactly what a very long half-life means.