Examine the radioactive decay of atomic nuclei (alpha, beta, gamma) and represent these decays as nuclear equations; use the decay law N = N_0 e^(-lambda t) and the concept of half-life T_1/2

What this dot point is asking

NESA wants you to describe the three principal types of radioactive decay (alpha, beta, gamma), balance nuclear equations using conservation of mass number and atomic number, use the decay law $N = N_0 e^{-\lambda t}$ along with the equivalent half-life form $N = N_0 (1/2)^{t/T_{1/2}}$, and connect $\lambda$ and $T_{1/2}$ via $\lambda T_{1/2} = \ln 2$.

The answer

What radioactive decay is

A radioactive nucleus is one that spontaneously transforms into another nuclear state, releasing energy as kinetic energy of the products and/or as electromagnetic radiation. Decay is a random process for any individual nucleus, but the statistics for large samples follow a predictable exponential law. The probability per unit time that a given nucleus decays is the decay constant $\lambda$, independent of how long the nucleus has existed.

Alpha decay

A heavy nucleus emits an alpha particle ($^4_2$He, two protons and two neutrons). The atomic number decreases by 2, mass number by 4.

Example:

Alpha decay typically occurs in heavy nuclei ($Z > 82$) where the Coulomb repulsion between protons becomes hard for the strong force to overcome. The alpha particle escapes by quantum tunnelling. Alphas have short range (a few cm in air, stopped by paper) but cause heavy ionisation per unit path.

Beta-minus decay

A neutron in the nucleus converts to a proton, emitting an electron (the beta particle) and an electron antineutrino. Atomic number increases by 1, mass number unchanged.

Example:

Beta-minus decay tends to occur in neutron-rich nuclei. The continuous energy spectrum of beta particles was the historical clue that a third particle (the antineutrino) carries away the missing energy.

Beta-plus decay

A proton in the nucleus converts to a neutron, emitting a positron and an electron neutrino. Atomic number decreases by 1, mass number unchanged.

Example: $^{22}_{11}\text{Na} \to ^{22}_{10}\text{Ne} + e^+ + \nu_e$. Beta-plus occurs in proton-rich nuclei. The competing process is electron capture, in which a proton absorbs an inner-shell electron and converts to a neutron plus a neutrino.

Gamma decay

The nucleus is left in an excited state after an alpha or beta decay (or after a nuclear reaction). It drops to a lower state by emitting a high-energy photon (gamma ray). No change in $Z$ or $A$.

Example: $^{60}_{27}\text{Co}^{\ast} \to ^{60}_{27}\text{Co} + \gamma$. Gamma rays are penetrating (centimetres of lead required to attenuate) but cause less local ionisation than alpha or beta.

Balancing nuclear equations

In any decay equation, two conservation laws must hold:

• mass number $A$ balances on both sides,
• charge $Z$ balances on both sides (counting an electron as $-1$ and a positron as $+1$).

Lepton number is also conserved, which is why an electron emitted in beta-minus decay is accompanied by an antineutrino, and a positron in beta-plus is accompanied by a neutrino.

The decay law

If $N(t)$ is the number of undecayed nuclei at time $t$, the rate of decay is proportional to $N$:

Solving with $N(0) = N_0$:

The activity (number of decays per unit time) is $A(t) = \lambda N(t)$, measured in becquerels (Bq, 1 decay per second).

Half-life

The half-life $T_{1/2}$ is the time for half the sample to decay. From $N(T_{1/2}) = N_0 / 2$:

So $\lambda = \ln 2 / T_{1/2}$.

The decay law in terms of half-life:

Half-lives of common isotopes:

Applications

• Radiometric dating. $^{14}$C for organic material up to 50000 y; $^{238}$U/$^{206}$Pb for rocks up to billions of years; $^{40}$K/$^{40}$Ar for igneous rocks.
• Nuclear medicine. $^{99\text{m}}$Tc for diagnostic imaging; $^{131}$I for thyroid therapy; positron emitters for PET.
• Smoke detectors. $^{241}$Am alpha source ionises air; smoke disrupts the ion current and triggers the alarm.
• Industrial gauging. Penetrating gammas measure thickness of steel sheets without contact.

Try it: Radioactive decay calculator to find remaining activity, time elapsed, or decay constant from half-life.

Worked example: dating a wooden artefact

A wooden bowl contains $^{14}$C at 25% of the modern atmospheric ratio. The half-life is 5730 y. Estimate the age.

$N / N_0 = 0.25 = (1/2)^2$, so the bowl is 2 half-lives old:

$t = 2 \times 5730 = 11460$ y.

Alternatively, using the decay law:

$0.25 = e^{-\lambda t}$, so $\ln 0.25 = -\lambda t$, giving $t = (\ln 4) / \lambda = 1.386 / (1.21 \times 10^{-4}) = 11460$ y.

Examples in context

Example 1. Lucas Heights $^{99m}$Tc supply chain for NSW hospitals. $^{99m}$Tc has a half-life of $T_{1/2} = 6.0 \text{ h}$. ANSTO's OPAL reactor extracts $^{99m}$Tc from $^{99}$Mo generators and ships it overnight to Sydney hospitals. Starting with $N_0 = 4.0 \times 10^{15}$ atoms at $6$ am, after $t = 12 \text{ h}$ (two half-lives), $N = N_0 (\tfrac{1}{2})^2 = 1.0 \times 10^{15}$, only $25\%$ remaining. The decay constant is $\lambda = \ln 2 / T_{1/2} = 0.693 / (6 \times 3600) = 3.21 \times 10^{-5} \text{ s}^{-1}$. Activity $A = \lambda N$. To deliver $1 \text{ GBq}$ at noon, the morning shipment must contain $A_0 / (\tfrac{1}{2})^1 = 2 \text{ GBq}$ at $6$ am.

Example 2. Carbon-14 dating of an Aboriginal artefact at the Australian Museum. A charcoal sample from a Sydney rock shelter contains $^{14}$C activity of $A = 5.0 \text{ Bq/g}$, compared to the present-day biosphere standard $A_0 = 13.6 \text{ Bq/g}$. The half-life is $T_{1/2} = 5730 \text{ yr}$, so $\lambda = \ln 2 / 5730 = 1.21 \times 10^{-4} \text{ yr}^{-1}$. The age is $t = (1/\lambda) \ln(A_0/A) = (5730/0.693) \ln(13.6/5.0) = 8270 \ln(2.72) = 8270 \text{ yr}$. The artefact dates to roughly 8300 years before present, placing it in the early Holocene period of Aboriginal occupation of the Sydney basin.

Try this

Q1. Distinguish between alpha, beta-minus and gamma decay in terms of the change to mass number $A$ and atomic number $Z$. [3 marks]

• Cue. $\alpha$: $A \to A-4$, $Z \to Z-2$. $\beta^-$: $A$ same, $Z \to Z+1$ (with $\bar\nu_e$). $\gamma$: $A$ and $Z$ unchanged (excited-state photon).

Q2. A sample contains $8.0 \times 10^{12}$ atoms of $^{32}$P ($T_{1/2} = 14.3 \text{ d}$). Calculate the activity in Bq. [3 marks]

• Cue. $\lambda = 0.693/(14.3 \times 86400) = 5.61 \times 10^{-7} \text{ s}^{-1}$; $A = \lambda N = 5.61 \times 10^{-7} \times 8.0 \times 10^{12} = 4.49 \times 10^6 \text{ Bq}$.

Q3. $^{226}$Ra decays by $\alpha$-emission with $T_{1/2} = 1600 \text{ yr}$. (a) Write the nuclear equation. (b) Calculate the fraction of the original $^{226}$Ra remaining after $4800 \text{ yr}$. (c) Find the activity of $1.0 \text{ g}$ of pure $^{226}$Ra in Bq. [2+2+3 marks]

• Cue. (a) $^{226}_{88}\text{Ra} \to {}^{222}_{86}\text{Rn} + \alpha$. (b) $4800/1600 = 3$ half-lives; $(1/2)^3 = 1/8 = 12.5\%$. (c) $N = 1.0/226 \times N_A = 2.66 \times 10^{21}$; $\lambda = \ln 2 / (1600 \times 3.15 \times 10^7) = 1.37 \times 10^{-11} \text{ s}^{-1}$; $A = 3.65 \times 10^{10} \text{ Bq}$ (= 1 Ci).