Atomic nuclei isotopes and radioactivity

An atom consists of an electron cloud and a nucleus; the nucleus contains protons with charge $+1e$ and neutrons with no charge, and the mass number $A$ equals the number of nucleons. A nucleus is written as $^{A}_{Z}X$, where $Z$ is the number of protons (atomic number), so the neutron number is $N=A-Z$. Isotopes are nuclei of one element with the same $Z$ but different $N$; therefore their electron shells and chemical behavior are nearly identical, while stability and nuclear properties can differ greatly. Radioactivity is the spontaneous decay of unstable nuclei: in $\alpha$ decay a helium nucleus $^{4}_{2}\mathrm{He}$ is emitted, in $\beta^-$ decay a neutron turns into a proton with emission of an electron and an antineutrino, and $\gamma$ radiation accompanies a transition of an excited nucleus to the ground state without changing $A$ and $Z$. In every decay, charge and nucleon number are conserved, which allows nuclear equations to be balanced. If the half-life $T_{1/2}$ is known, then after time $t=nT_{1/2}$ the remaining fraction of original nuclei is $(1/2)^n$, that is, $N=N_0(1/2)^n$.

The uranium-238 nucleus is written as $^{238}_{92}\mathrm{U}$. How many protons, neutrons, and nucleons does it contain? What is the nuclear charge in units of $e$?

Radium-226 ($^{226}_{88}\mathrm{Ra}$) undergoes $\alpha$ decay. Write the decay equation and identify the daughter nucleus.

A radon-222 sample has an initial activity of $2000\,\mathrm{Bq}$ and a half-life $T_{1/2}=3.8$ days. What will the activity be after $11.4$ days? What fraction of the nuclei will have decayed by that time?

Safe self-study with a simulation. Open PhET “Radioactive Dating Game” (or a similar radioactive decay simulation) and choose one isotope. Do not use real radioactive sources — make all observations on the screen.

1. Start the decay and record the fraction of remaining nuclei at equal time intervals.
2. Find the half-life $T_{1/2}$ — the time after which half of the nuclei remain.
3. Check the prediction $N=N_0(1/2)^n$ for $n=1,2,3$ and explain why the graph becomes smoother when the number of nuclei is large.

Show the connection between nuclear notation, proton and neutron numbers, and the changes in $\alpha$ and $\beta$ decay, making the conservation of $A$ and $Z$ visible.

Diagram: on the left, a nucleus $^{A}_{Z}X$ with $Z$ protons and $N=A-Z$ neutrons indicated; on the right, a decay chain with arrows, where the emitted object ($\alpha$, $\beta^-$, $\gamma$) is above each arrow and the new $A$ and $Z$ are below it.

A banana of about $120\ \mathrm{g}$ contains roughly $0.4\ \mathrm{g}$ of potassium. In natural potassium, $0.0117\ \%$ of atoms are the radioactive isotope $^{40}\mathrm{K}$ with a half-life of $1.25 \times 10^9$ years. So the banana holds about $7 \times 10^{17}$ atoms of $^{40}\mathrm{K}$, and its activity is roughly $13\ \mathrm{Bq}$: that many nuclei decay every second.

Two ideas are at work: isotopes — atoms of the same element with different neutron numbers ($^{39}\mathrm{K}$ and $^{40}\mathrm{K}$), and radioactivity — the spontaneous $\beta^-$ decay of $^{40}\mathrm{K}$ into $^{40}\mathrm{Ca}$ with emission of an electron and an antineutrino. Activity in becquerels directly counts decays per second.

The body of a $70\ \mathrm{kg}$ person contains about $140\ \mathrm{g}$ of potassium and about $18\ \mathrm{kg}$ of carbon. Owing to $^{40}\mathrm{K}$ ($0.0117\ \%$ of natural potassium) and $^{14}\mathrm{C}$ (about $0.23\ \mathrm{Bq/g}$ of carbon), the body's activity is roughly $8\ \mathrm{kBq}$ — that is, about $8000$ nuclei decay every second while a footballer runs across the pitch.

The example shows that radioactivity is not an external 'contagion' but a property of specific isotopes. $^{40}\mathrm{K}$ and $^{14}\mathrm{C}$ are radioactive isotopes of ordinary elements; their nuclei are unstable and decay emitting $\beta$ particles. The total activity is the sum of the contributions of all such isotopes in the body.

A household smoke detector contains about $0.3\ \mu\mathrm{g}$ of americium-241. This isotope undergoes $\alpha$ decay with a half-life of $432$ years and an $\alpha$-particle energy of $5.5\ \mathrm{MeV}$. The source activity is about $37\ \mathrm{kBq}$; the $\alpha$ particles ionize the air in the chamber, creating a current of about $10^{-10}\ \mathrm{A}$, and smoke reduces this current and triggers the alarm.

Three aspects of the topic matter here: the isotope $^{241}\mathrm{Am}$ is chosen because it emits $\alpha$ particles that do not escape the casing; $\alpha$ decay turns the $^{241}\mathrm{Am}$ nucleus into $^{237}\mathrm{Np}$ (the nuclear charge decreases by $2$ and the mass number by $4$); and activity in becquerels determines how many ions are created per second.