A-level Physics

8.3 Radioactive decay and half-life — Learn

Prerequisites: exponential decay from Section 5.4; the graph of an exponential from Chapter 3; reading a gradient; natural logarithms; SI prefixes.

By the end of this section you can: define activity and half-life; use A=λNA = \lambda N; find a decay constant from a half-life; use N=N0e−λtN = N_0 e^{-\lambda t} and A=A0e−λtA = A_0 e^{-\lambda t}; read a decay curve; and write a balanced nuclear equation for alpha, beta and gamma decay.

1. The nucleus and its notation

An atom has a small, dense nucleus. The nucleus contains protons and neutrons, which are together called nucleons. The number of protons is the proton number ZZ, and it decides which element the atom is. The number of nucleons is the nucleon number AA. Atoms of the same element with different numbers of neutrons are isotopes.

A nuclide is written with these numbers:

ZAX^{A}_{Z}\text{X}

For example, 918F^{18}_{9}\text{F} has 9 protons and 18 nucleons, so it has 18−9=918 - 9 = 9 neutrons.

Some nuclei are unstable. They change into a more stable nucleus and release radiation. This is radioactive decay. The decay is random: you cannot predict which nucleus will decay next, or when. It is also spontaneous: nothing outside the nucleus has to set it off.

2. Activity

The activity of a source is the number of nuclei that decay each second.

A=ΔNΔtA = \frac{\Delta N}{\Delta t}

The unit of activity is the becquerel, Bq\text{Bq}. One becquerel is one decay per second.

A detector measures a count rate, not the activity. Not every decay sends a particle into the detector, so the count rate is smaller than the activity. The detector also records a background count rate from nearby rocks, building materials and cosmic rays. Subtract the background before you use the count rate:

corrected count rate=measured count rate−background count rate\text{corrected count rate} = \text{measured count rate} - \text{background count rate}

3. Half-life

The half-life T1/2T_{1/2} of a source is the average time for half of the unstable nuclei in a sample to decay. It is also the time for the activity to fall to half of its starting value.

After nn half-lives, the fraction that remains is (12)n\left(\tfrac{1}{2}\right)^n:

A=A02nA = \frac{A_0}{2^n}

Worked example 1. The hospital tracer starts the morning with an activity of 8.0×108 Bq8.0 \times 10^{8}\,\text{Bq}. Its half-life is 110 minutes. Find the activity after 330 minutes.

The time is three half-lives, because 330110=3\dfrac{330}{110} = 3.

A=A023=8.0×1088=1.0×108 BqA = \frac{A_0}{2^3} = \frac{8.0 \times 10^{8}}{8} = 1.0 \times 10^{8}\,\text{Bq}

After 330 minutes the activity is 1.0×108 Bq1.0 \times 10^{8}\,\text{Bq}.

4. The decay constant and the exponential law

Half-lives describe whole steps. For any time, the decay follows an exponential curve. The decay constant λ\lambda is the probability that a given nucleus decays in one second. It has the unit s−1\text{s}^{-1}.

N=N0e−λtA=A0e−λtN = N_0 e^{-\lambda t} \qquad A = A_0 e^{-\lambda t}

The activity is proportional to the number of undecayed nuclei, which gives

A=λNA = \lambda N

The decay constant and the half-life are linked:

T1/2=ln⁡2λλ=ln⁡2T1/2T_{1/2} = \frac{\ln 2}{\lambda} \qquad \lambda = \frac{\ln 2}{T_{1/2}}

Here ln⁡2=0.693\ln 2 = 0.693. Because the decay rate depends only on the nucleus, the half-life does not change with temperature or pressure. It is a property of the nuclide.

Worked example 2. Find the decay constant of fluorine-18, whose half-life is 110 minutes.

First change the half-life to seconds.

T1/2=110×60=6600 sT_{1/2} = 110 \times 60 = 6600\,\text{s} λ=ln⁡2T1/2=0.6936600=1.05×10−4 s−1\begin{aligned} \lambda &= \frac{\ln 2}{T_{1/2}} = \frac{0.693}{6600} \\ &= 1.05 \times 10^{-4}\,\text{s}^{-1} \end{aligned}

Worked example 3. The tracer has an activity of 8.0×108 Bq8.0 \times 10^{8}\,\text{Bq} at the start. Find the number of fluorine-18 nuclei that are still undecayed at that moment.

Rearrange A=λNA = \lambda N.

N=Aλ=8.0×1081.05×10−4=7.6×1012\begin{aligned} N &= \frac{A}{\lambda} = \frac{8.0 \times 10^{8}}{1.05 \times 10^{-4}} \\ &= 7.6 \times 10^{12} \end{aligned}

About 7.6×10127.6 \times 10^{12} nuclei are undecayed.

5. Nuclear equations

The nucleon number and the proton number are conserved in every decay. That is how a decay equation is balanced.

Worked example 4. Write the balanced equation for the beta-plus decay of fluorine-18, 918F^{18}_{9}\text{F}.

Beta-plus decay emits a positron. To balance the nucleon numbers, the product nucleus also has 18 nucleons. To balance the proton numbers, it has 9−1=89 - 1 = 8 protons, which is oxygen.

918F→818O++10e+ν^{18}_{9}\text{F} \rightarrow {}^{18}_{8}\text{O} + {}^{0}_{+1}\text{e} + \nu

The neutrino ν\nu carries away some energy and momentum. It is why the emitted positron has a range of kinetic energies rather than one fixed value.

6. The decay curve

A graph of activity against time for a radioactive source is a falling exponential curve. It is steep at the start, when many nuclei remain, and it flattens as the sample runs out.

You can read the half-life from the curve. Find the starting activity, halve it, and read the time at which the activity reaches that value. Repeat for the next halving to check.

A straight-line graph is easier to use. Taking logarithms of A=A0e−λtA = A_0 e^{-\lambda t} gives

ln⁡A=ln⁡A0−λt\ln A = \ln A_0 - \lambda t

A graph of ln⁡A\ln A against tt is a straight line with gradient −λ-\lambda. The half-life follows from T1/2=ln⁡2λT_{1/2} = \dfrac{\ln 2}{\lambda}.

Watch out. Half-life does not mean "the time for all the atoms to decay". After one half-life, half remain. After two, a quarter remain. After ten, about one thousandth remain, and the sample is still slightly radioactive. A second trap is to use different units for the time and the half-life. If the half-life is in minutes, convert the time to minutes, or convert both to seconds. A third trap is to think that a temperature change alters the half-life. It does not: the decay is a property of the nucleus.

7. Try it: the decay curve

The figure draws the activity of a source against time. Two sliders set the half-life and the starting activity. The readout gives the activity after the chosen time and the number of half-lives that have passed.

Predict first. Do not move the sliders yet. Write down your answers.

  1. After three half-lives, what fraction of the starting activity remains?
  2. If the half-life is halved, does the curve fall faster or more slowly?
  3. Does doubling the starting activity change the half-life?

Test. Set a half-life and a starting activity, then read the activity at one, two and three half-lives. Change one slider at a time and check each prediction.

Explain. Why is the curve steepest at the start? Write two or three sentences. Use the idea that the activity is proportional to the number of undecayed nuclei, and that many nuclei remain at the start.

Progress check

Return to the hospital tracer. It starts the morning with an activity of 8.0×108 Bq8.0 \times 10^{8}\,\text{Bq} and a half-life of 110 minutes.

  1. Find the activity after 330 minutes, and after 550 minutes.
  2. Find the decay constant in s−1\text{s}^{-1}.
  3. Find the time at which the activity has fallen to 2.0×108 Bq2.0 \times 10^{8}\,\text{Bq}.
  4. Explain why the team cannot predict which nucleus will decay next, even though they can predict the activity of the whole sample well.

If question 1 goes wrong, count the half-lives and divide by 2n2^n. If question 2 goes wrong, change 110 minutes to seconds and use λ=ln⁡2T1/2\lambda = \dfrac{\ln 2}{T_{1/2}}. If question 3 goes wrong, find how many halvings take the activity from 8.0×1088.0 \times 10^{8} to 2.0×1082.0 \times 10^{8}. If question 4 goes wrong, think about the size of the sample and the meaning of a probability.

You should now be able to

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