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.

Common traps

Forgetting the antineutrino in beta decay. Charge balances without it, but lepton number does not. Always include $\bar{\nu}_e$ for beta-minus and $\nu_e$ for beta-plus.

Confusing $\lambda$ and $T_{1/2}$. They are related by $\lambda T_{1/2} = \ln 2$, not $\lambda T_{1/2} = 1$. The numerical factor matters.

Using arithmetic decay (subtracting a fixed amount) instead of exponential. Each half-life removes half the remaining sample, not half the initial sample. After two half-lives the sample is 25% of the original, not 0%.

Confusing $A$ and $Z$ in alpha decay. Alpha decay reduces $A$ by 4 and $Z$ by 2. Both must change together.

Saying gamma decay changes the element. Gamma decay changes the nuclear excitation level only. The element is unchanged.

In one sentence

Radioactive decay proceeds via alpha (loss of $^4_2$He), beta (n to p with $e^-$ and $\bar{\nu}_e$, or the reverse with $e^+$ and $\nu_e$), and gamma (photon from de-excitation) processes; the number of undecayed nuclei follows $N = N_0 e^{-\lambda t}$, with half-life $T_{1/2} = (\ln 2)/\lambda$.