Atomic Nuclei and Radioactive Decay

Lessons

Senior 6 · Physics

Atomic Nuclei and Radioactive Decay

5 lessons

Lesson 4 of 5

The Decay Law and Half-Life

25 min

Unpredictable one by one, predictable in crowds

Hold a GM tube near a weak source and listen. The clicks come irregularly: two close together, then a pause, then three. No experiment can say WHEN a particular nucleus will decay; heating, pressure or chemical change make no difference at all. Yet a large sample behaves with perfect regularity: a sample of iodine-131 always loses half its nuclei in 8.02 days, whatever its size and however old it is. This lesson turns that randomness into a law, and uses the law to measure time, from minutes in a school laboratory to thousands of years in archaeology.

The decay constant and activity

Radioactive decay is random (we cannot predict which nucleus decays next) and spontaneous (it is not triggered by outside conditions). Every nucleus of a given nuclide has the same probability $\lambda\,\Delta t$ of decaying in a short time $\Delta t$ . The constant $\lambda$ is the decay constant.

If $N$ undecayed nuclei are present, the expected number decaying in $\Delta t$ is $\lambda N\,\Delta t$ . The rate of decay, called the activity, is therefore proportional to $N$ :

A = -\frac{dN}{dt} = \lambda N

Activity is measured in becquerels: $1\,\mathrm{Bq} = 1$ decay per second. A GM counter records only a fraction of the decays, so its count rate is proportional to the activity, not equal to it.

Deriving the decay law

Separate the variables in $\frac{dN}{dt} = -\lambda N$ and integrate from $N_{0}$ nuclei at $t = 0$ to $N$ at time $t$ :

\int_{N_0}^{N}\frac{dN}{N} = -\lambda\int_0^t dt \quad\Rightarrow\quad \ln\frac{N}{N_0} = -\lambda t \quad\Rightarrow\quad N = N_0e^{-\lambda t}

Since $A = \lambda N$ , the activity and the count rate fall the same way: $A = A_0e^{-\lambda t}$ . So does the mass of the undecayed nuclide.

Figure 4.1: Exponential decay. In each half-life, half of the remaining nuclei decay.

Half-life

The half-life $T_{\frac{1}{2}}$ is the time for $N$ (or $A$ ) to fall to half its value. Put $N = \frac{N_0}{2}$ in the decay law: $\frac{1}{2} = e^{-\lambda T_{\frac{1}{2}}}$ , so $\lambda T_{\frac{1}{2}} = \ln 2$ and

T_{\frac{1}{2}} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}

After $n$ half-lives the fraction left is $\left(\frac{1}{2}\right)^n$ , which is often the quickest way to calculate: $N = N_0\left(\frac{1}{2}\right)^{n}$ , where $n = \frac{t}{T_{\frac{1}{2}}}$ is the number of half-lives. The curve never reaches zero, but after 10 half-lives less than one nucleus in a thousand remains ( $2^{10} = 1024$ ).

Nuclide	Decay	Half-life	Use or occurrence
technetium-99m	γ	6.01 h	medical imaging
iodine-131	β⁻, γ	8.02 d	thyroid treatment
cobalt-60	β⁻, γ	5.27 y	sterilising, radiotherapy
carbon-14	β⁻	5730 y	radiocarbon dating
radium-226	α	1600 y	member of the U-238 series
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 usually used in dating calculations.

Definition

The decay constant $\lambda$ is the probability per unit time that a given nucleus decays (unit $\mathrm{s^{-1}}$ ). The activity $A = \lambda N$ is the number of decays per second (unit becquerel, Bq). The half-life $T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$ is the time for half the undecayed nuclei in a sample to decay. Example: iodine-131 has $T_{\frac{1}{2}} = 8.02\,\mathrm{d}$ , so $\lambda = 0.0864\,\mathrm{d^{-1}} = 1.00 \times 10^{-6}\,\mathrm{s^{-1}}$ .

Worked example 1: iodine-131

A hospital receives a sample of iodine-131 ( $T_{\frac{1}{2}} = 8.02$ days) with an activity of $5.0 \times 10^{8}\,\mathrm{Bq}$ . Find (a) its decay constant in $\mathrm{s^{-1}}$ , (b) its activity 20 days later.

Solution

(a) $\lambda = \frac{0.693}{8.02 \times 86\,400\,\mathrm{s}} = 1.00 \times 10^{-6}\,\mathrm{s^{-1}}$ .

(b) 20 days is $\frac{20}{8.02} = 2.49$ half-lives, so $A = 5.0 \times 10^{8} \times \left(\frac{1}{2}\right)^{2.49} = 5.0 \times 10^{8} \times 0.178 = 8.9 \times 10^{7}\,\mathrm{Bq}$ .

(Or $A = A_0e^{-\lambda t}$ with $\lambda = 0.0864\,\mathrm{d^{-1}}$ and $t = 20\,\mathrm{d}$ : $e^{-1.73} = 0.178$ , the same.) Keep $\lambda$ and $t$ in matching units.

Worked example 2: the activity of one gram of radium

Find the activity of 1.00 g of radium-226 ( $T_{\frac{1}{2}} = 1600$ years; $N_A = 6.02 \times 10^{23}\,\mathrm{mol^{-1}}$ ; 1 year $= 3.16 \times 10^{7}$ s).

Solution

Number of nuclei: $N = \frac{1.00}{226} \times 6.02 \times 10^{23} = 2.66 \times 10^{21}$ .

$\lambda = \frac{0.693}{1600 \times 3.16 \times 10^{7}} = 1.37 \times 10^{-11}\,\mathrm{s^{-1}}$ .

$A = \lambda N = 1.37 \times 10^{-11} \times 2.66 \times 10^{21} = 3.7 \times 10^{10}\,\mathrm{Bq}$ .

This is the old unit the curie (Ci), originally defined as the activity of one gram of radium.

Worked example 3: radiocarbon dating

Living wood has a carbon-14 activity of about 0.23 Bq per gram of carbon, kept steady because the plant keeps exchanging carbon with the air. When it dies the exchange stops and the carbon-14 decays. Charcoal from an ancient hearth has an activity of 0.080 Bq per gram of carbon. Estimate its age ( $T_{\frac{1}{2}} = 5730$ y).

Solution

From $A = A_0e^{-\lambda t}$ : $t = \frac{1}{\lambda}\ln\frac{A_0}{A} = \frac{T_{\frac{1}{2}}}{\ln 2}\ln\frac{A_0}{A}$ .

t = \frac{5730}{0.693}\ln\frac{0.23}{0.080} = 8270 \times 1.06 = 8.7 \times 10^{3}\,\mathrm{years}

Conditions: the method assumes the carbon-14 fraction in the air was the same when the wood died as it is now. It was not exactly constant, so laboratories correct ages with calibration curves. It works for material up to about 50 000 years old; after about 9 half-lives too little carbon-14 is left to measure against the background.

Experiment 4.1: a model of decay with dice

Apparatus: 100 six-sided dice (or cubes marked on one face) and a tray.

Method: Throw all the dice. Remove every die showing a six: these nuclei have "decayed". Record the number left. Throw the remaining dice again, and repeat for 15 throws. Plot the number left against the throw number, then plot ln(number left) against throw number.

What it shows: each die has a probability $\frac{1}{6}$ of "decaying" per throw, whatever its history, just as each nucleus has a probability $\lambda\,\Delta t$ . The first graph is roughly exponential and the second roughly a straight line. The "half-life" should be close to $\frac{\ln 2}{\ln\frac{6}{5}} \approx 3.8$ throws; the points scatter because 100 is a small number, just as count rates scatter for a weak source.

With a real source, your teacher may use a protactinium generator, a sealed bottle in which protactinium-234m (half-life about 70 s) forms and can be separated into a layer next to a GM tube. The next example analyses such readings.

Worked example 4: half-life from a graph

Background-corrected count rates from protactinium-234m were (illustrative data):

t / s	0	30	60	90	120	150
A / min⁻¹	600	446	331	246	183	136
ln A	6.397	6.100	5.802	5.505	5.209	4.913

Figure 4.2: Taking logarithms turns the exponential into a straight line.

Solution

Taking natural logs of $A = A_0e^{-\lambda t}$ gives $\ln A = \ln A_0 - \lambda t$ : a straight line of gradient $-\lambda$ . From the triangle, gradient $= \frac{-1.19}{120\,\mathrm{s}} = -9.9 \times 10^{-3}\,\mathrm{s^{-1}}$ , so $\lambda = 9.9 \times 10^{-3}\,\mathrm{s^{-1}}$ and

T_{\frac{1}{2}} = \frac{0.693}{9.9 \times 10^{-3}} = 70\,\mathrm{s}

A straight line confirms that the decay is exponential; the intercept gives $\ln A_0$ . A straight-line graph uses all the readings, so it is more reliable than reading one half-life off a curve.

Warning

Common mistakes. (1) Thinking that after two half-lives everything has decayed: a quarter is left. (2) Mixing units: if $T_{\frac{1}{2}}$ is in days, $\lambda$ is in $\mathrm{d^{-1}}$ , and $A = \lambda N$ is then decays per DAY, not becquerels. (3) Using the count rate as the activity: a detector catches only a fraction of the decays. (4) Forgetting to subtract the background before taking logarithms. (5) Believing a nucleus "ages": its chance of decaying in the next second is the same however long it has existed.

Tip

When the time is a whole number of half-lives, just halve repeatedly. Otherwise use $\left(\frac{1}{2}\right)^{n}$ with $n = \frac{t}{T_{\frac{1}{2}}}$ directly: no need to find $\lambda$ first. To find a TIME, take logarithms: $t = \frac{T_{\frac{1}{2}}}{\ln 2}\ln\frac{N_0}{N}$ .

Radioactivity: Expect the unexpected, by Steve Weatherall (TED-Ed). Watch how the random decay of single nuclei still gives a sample a fixed half-life.

✦

Practice

A sample contains $8.0 \times 10^{12}$ nuclei of a nuclide with a half-life of 5.0 hours. How many remain undecayed after 20 hours?
Find the decay constant of cobalt-60 ( $T_{\frac{1}{2}} = 5.27$ y) in $\mathrm{s^{-1}}$ , and the fraction of a cobalt-60 source remaining after 10 years.
A source's activity falls from 800 Bq to 50 Bq in 12 days. Find its half-life.
Derive $T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$ from $N = N_0e^{-\lambda t}$ .
Find the mass of iodine-131 in a source of activity $4.0 \times 10^{8}\,\mathrm{Bq}$ ( $\lambda = 1.00 \times 10^{-6}\,\mathrm{s^{-1}}$ ; molar mass 131 g mol⁻¹).
A bone has a carbon-14 activity per gram of carbon that is 30% of that in living tissue. Estimate its age.
Background-corrected count rates of a source are 480, 340, 240, 170 and 120 per minute at 1-minute intervals. Show that the decay is exponential and find the half-life.
Byiringiro says: "Carbon dating can find the age of a rock that formed 10 million years ago." Evaluate his statement.

Show answersHide answers

20 h is 4 half-lives: $8.0 \times 10^{12} \times \frac{1}{16} = 5.0 \times 10^{11}$ .
$\lambda = \frac{0.693}{5.27 \times 3.16 \times 10^{7}} = 4.16 \times 10^{-9}\,\mathrm{s^{-1}}$ ; 10 years is $\frac{10}{5.27} = 1.90$ half-lives, so the fraction $= \left(\frac{1}{2}\right)^{1.90} = 0.27$ .
$\frac{800}{50} = 16 = 2^4$ : four half-lives in 12 days, so $T_{\frac{1}{2}} = 3.0$ days.
$\frac{N_0}{2} = N_0e^{-\lambda T_{\frac{1}{2}}} \Rightarrow e^{\lambda T_{\frac{1}{2}}} = 2 \Rightarrow \lambda T_{\frac{1}{2}} = \ln 2$ .
$N = \frac{A}{\lambda} = 4.0 \times 10^{14}$ ; mass $= \frac{4.0 \times 10^{14}}{6.02 \times 10^{23}} \times 131 = 8.7 \times 10^{-8}\,\mathrm{g}$ , less than a millionth of a gram.
$t = \frac{5730}{0.693}\ln\frac{1}{0.30} = 8270 \times 1.20 = 1.0 \times 10^{4}$ years.
Ratios of successive readings: $\frac{340}{480} = 0.71$ , $\frac{240}{340} = 0.71$ , $\frac{170}{240} = 0.71$ , $\frac{120}{170} = 0.71$ : a constant ratio in equal times means exponential decay. It halves (480 to 240) in 2.0 min, so $T_{\frac{1}{2}} = 2.0$ min.
Wrong. A rock contains no carbon from living things, and after 10 million years (about 1700 half-lives of carbon-14) no measurable carbon-14 would remain; the method works only up to about 50 000 years for once-living material. Rocks are dated with long-lived nuclides such as uranium-238 or potassium-40.

Key points

Decay is random and spontaneous; $A = -\frac{dN}{dt} = \lambda N$ , in becquerels.
$N = N_0e^{-\lambda t}$ , $A = A_0e^{-\lambda t}$ ; $T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$ ; after $n$ half-lives $\left(\frac{1}{2}\right)^n$ remains.
A graph of $\ln A$ against $t$ is a straight line of gradient $-\lambda$ .
Radiocarbon dating: $t = \frac{T_{\frac{1}{2}}}{\ln 2}\ln\frac{A_0}{A}$ , valid for once-living material up to about 50 000 years.

Sources and credits

10.4: Radioactive Decay, from University Physics Volume 3 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Half-lives: NuDat 3, National Nuclear Data Center, Brookhaven National Laboratory.
Video: Radioactivity: Expect the unexpected, Steve Weatherall, TED-Ed, embedded from YouTube.

The two graphs, the dice experiment, the worked examples (the protactinium data are illustrative) and the practice exercises are written by Ganzaa.

1Unpredictable one by one, predictable in crowds

2The decay constant and activity

3Deriving the decay law

4Half-life

5Experiment 4.1: a model of decay with dice

Unpredictable one by one, predictable in crowds

The decay constant and activity

Deriving the decay law

Half-life

Experiment 4.1: a model of decay with dice