Activity, decay series and dating

How many decays each second?

A Geiger counter clicks each time a particle from a decay goes through it. How fast it clicks depends on how many nuclei decay each second. That is what we call the activity.

The activity $R$ of a sample is the number of decays in each second:

NCERT writes the activity as $R$. A-Level and IB write $A$ for activity, and then $A$ must not be mixed up with the mass number. The SI unit is the becquerel, $1\ \text{Bq} = 1$ decay per second. An older unit is the curie, $1\ \text{Ci} = 3.7 \times 10^{10}\ \text{Bq}$. Substituting $N = N_0e^{-\lambda t}$:

So the activity falls with the same half-life as the number of nuclei. A counter placed near the source reads the count rate, which is smaller than the activity, since the counter catches only some of the particles.

Worked example: a gram of radium

Radium-226 has a half-life of $1600$ years. Find the activity of $1.0$ g of it. The curie was first defined as this activity.

1. The number of atoms. The molar mass is $226$ g/mol, so $N = \dfrac{1.0\ \text{g}}{226\ \text{g/mol}} \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 2.66 \times 10^{21}$.
2. The decay constant. $T = 1600 \times 3.156 \times 10^{7}\ \text{s} = 5.05 \times 10^{10}\ \text{s}$. So $\lambda = \dfrac{0.693}{5.05 \times 10^{10}\ \text{s}} = 1.37 \times 10^{-11}\ \text{s}^{-1}$.
3. The activity. $R = \lambda N = 1.37 \times 10^{-11}\ \text{s}^{-1} \times 2.66 \times 10^{21} = 3.66 \times 10^{10}\ \text{Bq}$.

That is $0.99\ \text{Ci}$. A huge number of decays from one gram, and it is why radium was so dangerous. It fits, since the curie was defined from radium.

Safe use

Radiation from decay damages living cells, and the damage depends on the dose. The safe way to work with a source is simple. Spend less time near it, stay far from it, and put a shield in between. The intensity of the rays from a point source goes down as $\dfrac{1}{r^2}$. Doubling the distance cuts it to a quarter.

Decay series

The daughter of a decay may itself be radioactive. A chain goes on through many steps until a stable nucleus is made. We call it a decay series. In an alpha decay $A$ falls by $4$, and in a beta decay $A$ does not change. So along a series $A$ changes only by multiples of $4$, and the remainder of $\dfrac{A}{4}$ is the same all along. This gives four series.

• Thorium series: $A = 4n$. It starts at thorium-232 and ends at lead-208.
• Neptunium series: $A = 4n + 1$. It starts at neptunium-237 and ends at bismuth-209.
• Uranium series: $A = 4n + 2$. It starts at uranium-238 and ends at lead-206.
• Actinium series: $A = 4n + 3$. It starts at uranium-235 and ends at lead-207.

The picture shows the main path of the uranium series. Count the steps: $8$ alpha decays take away $32$ nucleons and $16$ protons, and $6$ beta decays give back $6$ protons.

Take thorium-232 to lead-208. The mass number falls by $24$, so there are $\dfrac{24}{4} = 6$ alpha particles. These take away $12$ protons, from $90$ to $78$. Lead has $82$, so there must be $4$ beta-minus particles.

Secular equilibrium. Suppose a long-lived parent decays into a short-lived daughter. After a long time the daughter is made as fast as it decays, so $\lambda_1N_1 = \lambda_2N_2$ and the two activities are equal. Then $\dfrac{N_1}{N_2} = \dfrac{T_1}{T_2}$. In a rock that has had uranium-238 ($4.47 \times 10^{9}$ y) for a very long time, the ratio of the numbers of uranium-238 to radium-226 atoms is $\dfrac{4.47 \times 10^{9}}{1600} = 2.8 \times 10^{6}$.

Dating old things

Radiocarbon dating. Cosmic rays make carbon-14 from nitrogen high in the air. Carbon-14 has a half-life of $5730$ years, and it is mixed through the air as carbon dioxide. A living plant takes in carbon from the air, so the fraction of its carbon that is carbon-14 stays the same. When the plant dies, it takes in no more, and its carbon-14 decays. So the activity of one gram of carbon from a dead plant falls with $T = 5730$ years. The age comes from the activity ratio:

Suppose the activity of a sample is $60\ \%$ of that of a living plant. Then $t = 5730\ \text{y} \times \log_2\dfrac{1}{0.60} = 5730\ \text{y} \times 0.737 = 4223$ y. After about ten half-lives, about $50\,000$ years, too little carbon-14 is left to measure.

Uranium-lead dating. For rocks we use the very long half-life of uranium-238, $4.47 \times 10^{9}$ years. Every uranium-238 atom ends as a lead-206 atom. If a rock began with no lead, and none left or entered, the numbers of atoms now are $N_U$ and $N_{Pb}$. The uranium that was there at the start was $N_U + N_{Pb}$, so

For $N_{Pb} : N_U = 1 : 3$, $t = \dfrac{4.47 \times 10^{9}\ \text{y}}{0.693} \times \ln\dfrac{4}{3} = 1.86 \times 10^{9}$ y. Take care: the ratio must be of atoms and not of mass.

Keep the half-life at $5730$ years. At what age is a quarter of the activity left? Now choose a half-life of $1600$ years. How does the curve change?

Practise with new numbers

Try it in Python

The program finds the age of four samples from the fraction of the carbon-14 activity that is left.

Common mistakes

• Mixing up activity and the number of nuclei. $R = \lambda N$. The two have the same half-life, but they are not the same quantity.
• Mixing up the units. $\lambda$ must be in per second to give becquerel. A half-life in years needs a conversion.
• Using the mass ratio in the uranium-lead age. The law needs the ratio of atoms, so divide each mass by its molar mass.
• Dividing by $\ln 2$ twice, or not at all. For an age, $t = \dfrac{T}{\ln 2}\ln\dfrac{R_0}{R}$ has one $\ln 2$.
• Counting beta decays as changes of $A$. Along a series only the alpha decays change $A$.
• Saying the count rate is the activity. The counter catches only part of the particles.

Remember these

• $R = \lambda N = R_0e^{-\lambda t} = R_0\,2^{-t/T}$. $1\ \text{Bq} = 1\ \text{s}^{-1}$ and $1\ \text{Ci} = 3.7 \times 10^{10}\ \text{Bq}$.
• Along a decay series $A$ falls by $4$ in each alpha decay and stays the same in each beta decay.
• $\dfrac{N_1}{N_2} = \dfrac{T_1}{T_2}$ in secular equilibrium.
• Age by carbon-14: $t = T\log_2\dfrac{R_0}{R}$, with $T = 5730$ y. Age by uranium-lead: $t = \dfrac{T}{\ln 2}\ln\left(1 + \dfrac{N_{Pb}}{N_U}\right)$.
• Fewer decays happen with a longer time, a longer distance and a shield between the source and you.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A source has an activity of $800$ Bq. Its half-life is $2.0$ min. What is its activity after $6.0$ min?

2. A wooden object has one eighth of the carbon-14 activity of living wood. How old is it?

3. Why must the daughter of a very long-lived parent decay as fast as it is made, after a long time?

4. A series starts at ${}^{235}_{92}\text{U}$ and ends at ${}^{207}_{82}\text{Pb}$. How many alpha and how many beta-minus particles are emitted?

References

• NCERT Physics Part II (Class 12, rationalised edition), Chapter 13 Nuclei, on radioactivity and the law of decay.
• OpenStax University Physics Volume 3, Chapter 10 Nuclear Physics, on radioactive decay and dating.
• OpenStax Chemistry 2e, Chapter 21 Nuclear Chemistry, on radiocarbon dating and decay series.