Radioactivity

Lessons

Senior 6 · Chemistry

Radioactivity

5 lessons

Lesson 3 of 5

Rate of Decay and Half-Life

25 min

A clock that cannot be reset

Dig up a piece of charcoal from an ancient cooking fire and you can find out, within a few decades, when the tree it came from died. No written record is needed: the charcoal carries its own clock in the form of carbon-14 nuclei, which decay at a rate that no heat, pressure or chemical change can alter. In Unit 13 you met first-order reactions, whose half-life does not depend on how much reactant there is. Radioactive decay is the most perfect first-order process in chemistry.

Definition

Activity, $A$ : the number of nuclei in a sample that decay per second. Unit: the becquerel, $1\,\mathrm{Bq} = 1$ decay per second. Example: living plant material has a carbon-14 activity of about $0.23\,\mathrm{Bq}$ per gram of carbon.

Decay constant, $\lambda$ : the first-order rate constant for decay, the probability per unit time that a nucleus decays; unit $\mathrm{s^{-1}}$ (or $\mathrm{y^{-1}}$ ). Activity and number of nuclei are linked by $A = \lambda N$ .

Half-life, $t_{\frac{1}{2}}$ : the time for half of the radioactive nuclei in a sample to decay, and so for the activity to halve. $t_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$ . Example: iodine-131, $t_{\frac{1}{2}} = 8.02$ days.

Decay is first order

Each nucleus decays at random, independently of the others, so the number decaying per second is proportional to the number present:

-\frac{dN}{dt} = \lambda N \quad\Rightarrow\quad N = N_0\,e^{-\lambda t} \quad\text{and}\quad A = A_0\,e^{-\lambda t}

This is exactly the first-order law of Unit 13 with $\lambda$ in place of $k$ . After $n$ half-lives ( $n = \frac{t}{t_{\frac{1}{2}}}$ ) the fraction left is $\left(\frac{1}{2}\right)^n$ , which also applies to the mass of the radioisotope and to its activity. Because the activity measures $N$ , you can follow decay by counting, and $\ln A$ against $t$ is a straight line of gradient $-\lambda$ .

Radioisotope	Decay	Half-life	Main use
technetium-99m	γ	6.01 h	medical imaging
radon-222	α	3.82 d	a natural hazard in buildings
iodine-131	β⁻, γ	8.02 d	thyroid diagnosis and treatment
phosphorus-32	β⁻	14.3 d	tracer in plant and fertiliser studies
cobalt-60	β⁻, γ	5.27 y	radiotherapy, sterilising, food irradiation
carbon-14	β⁻	5730 y	radiocarbon dating, tracer
uranium-238	α	$4.47 \times 10^{9}$ y	dating rocks; nuclear fuel

Half-lives from the NNDC NuDat database, rounded; 5730 y is the carbon-14 half-life used in dating calculations, as in Senior 6 Physics.

Worked example 1: mass remaining

Problem. A research station receives $8.0\,\mu\mathrm{g}$ of phosphorus-32 ( $t_{\frac{1}{2}} = 14.3$ d). How much is left (a) after 42.9 days, (b) after 30 days?

Solution. (a) $\frac{42.9}{14.3} = 3$ half-lives: $8.0 \times \left(\frac{1}{2}\right)^3 = 1.0\,\mu\mathrm{g}$ .

(b) Not a whole number of half-lives, so use the exponential: $\lambda = \frac{0.693}{14.3\,\mathrm{d}} = 0.0485\,\mathrm{d^{-1}}$ .

m = 8.0\,e^{-0.0485 \times 30} = 8.0\,e^{-1.454} = 8.0 \times 0.234 = 1.9\,\mu\mathrm{g}

Worked example 2: activity from mass

Problem. Calculate the activity of $1.0\,\mu\mathrm{g}$ of phosphorus-32 ( $M = 32\,\mathrm{g\,mol^{-1}}$ , $t_{\frac{1}{2}} = 14.3$ d).

Solution.

N = \frac{1.0 \times 10^{-6}\,\mathrm{g}}{32\,\mathrm{g\,mol^{-1}}} \times 6.022 \times 10^{23}\,\mathrm{mol^{-1}} = 1.88 \times 10^{16}

\lambda = \frac{0.693}{14.3 \times 86\,400\,\mathrm{s}} = 5.61 \times 10^{-7}\,\mathrm{s^{-1}}

A = \lambda N = 5.61 \times 10^{-7} \times 1.88 \times 10^{16} = 1.1 \times 10^{10}\,\mathrm{Bq}

One microgram, far too little to see, gives ten billion decays a second. This is why radioisotopes work as tracers in tiny amounts, and why they must be handled with care.

Radiocarbon dating

Carbon-14 forms continuously in the upper atmosphere (Lesson 2), is oxidised to $\mathrm{CO_2}$ , and enters plants by photosynthesis and animals through food. While an organism lives, it keeps exchanging carbon with its surroundings, so its carbon-14 activity stays at the same level as in the air, about $0.23\,\mathrm{Bq}$ per gram of carbon. When it dies, the exchange stops and the activity falls with a half-life of 5730 years (Figure 3.1). Measuring the activity per gram of carbon in a sample gives its age:

t = \frac{t_{\frac{1}{2}}}{\ln 2}\,\ln\frac{A_0}{A}

Figure 3.1: The carbon-14 activity of dead organic material falls with age.

The method works for organic material up to about 50 000 years old (about nine half-lives); older samples have too little carbon-14 left to measure reliably. Its results are corrected (calibrated) using tree rings of known age, because the carbon-14 level in the air has varied slightly over time. Rocks are dated with much longer-lived nuclides, such as uranium-238 decaying to lead-206.

Worked example 3: where does 0.23 Bq per gram come from?

Problem. In living material about $1.2 \times 10^{-12}$ of all carbon atoms are carbon-14. Show that the activity is about $0.23\,\mathrm{Bq}$ per gram of carbon ( $M(\mathrm{C}) = 12.0\,\mathrm{g\,mol^{-1}}$ ).

Solution. Atoms of C in 1 g: $\frac{1}{12.0} \times 6.022 \times 10^{23} = 5.02 \times 10^{22}$ . Carbon-14 atoms: $5.02 \times 10^{22} \times 1.2 \times 10^{-12} = 6.0 \times 10^{10}$ .

\lambda = \frac{0.693}{5730 \times 3.156 \times 10^{7}\,\mathrm{s}} = 3.83 \times 10^{-12}\,\mathrm{s^{-1}}, \qquad A = \lambda N = 3.83 \times 10^{-12} \times 6.0 \times 10^{10} = 0.23\,\mathrm{Bq}

Worked example 4: dating charcoal and a rock

Problem. (a) Charcoal from an old hearth has a carbon-14 activity of $0.17\,\mathrm{Bq}$ per gram of carbon. Estimate its age. (b) A mineral contains 0.50 atoms of lead-206 for every atom of uranium-238. Assuming all the lead came from uranium decay, estimate the age of the rock ( $t_{\frac{1}{2}} = 4.47 \times 10^{9}$ y).

Solution.

(a) $t = \frac{5730}{0.693}\ln\frac{0.23}{0.17} = 8270 \times 0.302 = 2.5 \times 10^{3}$ years.

(b) For every 1.00 U atom left there were originally $1.00 + 0.50 = 1.50$ U atoms, so $\frac{N_0}{N} = 1.50$ :

t = \frac{4.47 \times 10^{9}}{0.693}\ln 1.50 = 6.45 \times 10^{9} \times 0.405 = 2.6 \times 10^{9}\,\text{years}

Warning

Common mistakes

Thinking that after two half-lives nothing is left. Half goes in the first half-life, a quarter of the original in the second: after two half-lives a quarter remains.
Mixing time units: if $\lambda$ is in $\mathrm{s^{-1}}$ , $t$ must be in seconds. Convert days or years first ( $1\,\mathrm{y} = 3.156 \times 10^{7}\,\mathrm{s}$ ).
Using radiocarbon dating on a rock or a metal. It works only for material that once lived and took in carbon from the air.
Saying the decay products "disappear". The mass of the radioisotope falls, but the daughter nuclides remain in the sample.

Tip

If the time is a whole number of half-lives, halve repeatedly; it is quicker and less error-prone. Otherwise use $N = N_0\,e^{-\lambda t}$ , or $\ln\frac{N_0}{N} = \lambda t$ when the time is the unknown.

Professor Dave Explains: Radiometric Dating, Carbon-14 and Uranium-238. Watch how the fraction of the parent nuclide left is turned into an age, and why carbon-14 suits once-living material while uranium-238 suits rocks.

✦

Practice

State the relationship between half-life and decay constant, and show how it follows from $N = N_0\,e^{-\lambda t}$ .
What fraction of a sample of iodine-131 remains after 24 days ( $t_{\frac{1}{2}} = 8.02$ d)?
5.0 g of cobalt-60 ( $t_{\frac{1}{2}} = 5.27$ y) is sealed in a radiotherapy source. What mass of cobalt-60 remains after 15.81 years, and what has the rest become?
How long does it take for the activity of a cobalt-60 source to fall to 80 % of its original value?
A technetium-99m dose ( $t_{\frac{1}{2}} = 6.01$ h) is injected for a scan. What percentage of its activity remains after 24 hours?
A wooden tool has a carbon-14 activity of $0.12\,\mathrm{Bq}$ per gram of carbon. Estimate its age.
Explain why carbon-14 cannot be used to date a 2-million-year-old fossil bone, and name a better method.

Show answersHide answers

$t_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$ : put $N = \frac{1}{2}N_0$ in $N = N_0\,e^{-\lambda t}$ to get $\frac{1}{2} = e^{-\lambda t_{\frac{1}{2}}}$ , so $\lambda t_{\frac{1}{2}} = \ln 2$ .
$\left(\frac{1}{2}\right)^{\frac{24}{8.02}} = 0.126$ , about one eighth (24 d is almost exactly 3 half-lives).
$\frac{15.81}{5.27} = 3$ half-lives: $5.0 \times \frac{1}{8} = 0.63\,\mathrm{g}$ . The rest (about 4.4 g) has become nickel-60 by β⁻ decay.
$t = \frac{5.27}{0.693}\ln\frac{1}{0.80} = 7.60 \times 0.223 = 1.7$ years.
$100 \times \left(\frac{1}{2}\right)^{\frac{24}{6.01}} = 6.3\,\%$ .
$t = \frac{5730}{0.693}\ln\frac{0.23}{0.12} = 8270 \times 0.651 = 5.4 \times 10^{3}$ years.
2 million years is about 350 half-lives of carbon-14, so no measurable carbon-14 remains (the limit is about 50 000 years). Use a long-lived nuclide in the surrounding rock, such as potassium-40 decaying to argon-40, or uranium–lead dating.

Sources and credits

Half-lives: NuDat 3, National Nuclear Data Center, Brookhaven National Laboratory (public data, values rounded by Ganzaa).
Video: Radiometric Dating: Carbon-14 and Uranium-238, Professor Dave Explains, embedded from YouTube.
The text, table, graph, worked examples and Practice items were written and drawn by Ganzaa.

1A clock that cannot be reset

2Decay is first order

3Radiocarbon dating

A clock that cannot be reset

Decay is first order

Radiocarbon dating