Radioactive Decay Equations Revision Notes | A Level AQA Physics

1. Why Decay Equations Are the Language of Nuclear Physics

Every radioactive decay is a nuclear transformation — a nucleus changes its identity by emitting a particle. The question is: how do you track exactly what changes and what stays the same? That is precisely what decay equations do.

At A-level, you need to write these equations with precision, read nuclear energy level diagrams, and justify which decay mode a nucleus undergoes. This lesson builds all three skills from the ground up.

2. The Notation: Reading and Writing Nuclear Symbols

Before writing any decay equation, you need to be fluent in nuclear notation. Every nuclide is written as $^{A}_{Z}X$, where $X$ is the chemical symbol, $A$ is the nucleon number (protons + neutrons), and $Z$ is the proton number.

Consider uranium-238: written $^{238}_{92}\text{U}$. Here $A = 238$, $Z = 92$, so $N = 238 - 92 = 146$ neutrons. Every decay equation you write must conserve two quantities: the total nucleon number $A$ and the total proton number $Z$. Think of these as the two bookkeeping rules that make nuclear equations work.

• Top numbers (A values) must sum to the same total on both sides of the equation
• Bottom numbers (Z values) must sum to the same total on both sides of the equation
• N is derived from A and Z; it changes in a predictable way for each decay type, and its value is always determined by N = A - Z

3. Alpha Decay: Changes in N and Z

An alpha particle is a helium-4 nucleus: $^{4}_{2}\text{He}$. When a nucleus emits an alpha particle, it loses 2 protons and 2 neutrons in one go. The changes are: $Z$ decreases by 2, $A$ decreases by 4, and therefore $N$ decreases by 2.

Alpha decay is most common in heavy nuclei where both $Z$ and $N$ are large. Emitting an alpha particle reduces the size of the nucleus significantly, moving it towards a more stable region of the N-Z chart. The daughter nucleus is always a different element because $Z$ changes.

Worked example: Writing an Alpha Decay Equation

Uranium-238 undergoes alpha decay. Write the full nuclear equation and state the changes in Z, A, and N.

1. $^{238}_{92}\text{U}$ — A = 238, Z = 92, so N = 238 - 92 = 146 for the parent.
2. $^{238}_{92}\text{U} \rightarrow ^{?}_{?}Y + ^{4}_{2}\text{He}$ — The alpha particle always has A = 4 and Z = 2. Place it on the product side.
3. $238 = A_Y + 4 \Rightarrow A_Y = 234$ — Top numbers must balance: 238 on the left, so 234 + 4 = 238 on the right. [M1]
4. $92 = Z_Y + 2 \Rightarrow Z_Y = 90$ — Bottom numbers must balance: 92 on the left, so 90 + 2 = 92 on the right. [M1]
5. $Z = 90 \Rightarrow \text{Thorium (Th)}$ — Use the periodic table: element 90 is thorium. [A1]
6. $^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^{4}_{2}\text{He}$ — Changes: Z decreases by 2 (92 to 90), A decreases by 4 (238 to 234), N decreases by 2 (146 to 144).

Answer: $^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^{4}_{2}\text{He}$. Z: 92 to 90 (decrease by 2); A: 238 to 234 (decrease by 4); N: 146 to 144 (decrease by 2).

4. Beta-Minus Decay: A Neutron Becomes a Proton

Beta-minus decay ($\beta^-$) occurs when a nucleus has too many neutrons relative to protons. A neutron inside the nucleus transforms into a proton, emitting a fast-moving electron (the beta-minus particle) and an antineutrino.

The beta-minus particle is written $^{0}_{-1}\text{e}$ (or $^{0}_{-1}\beta$). The antineutrino is written $\bar{ u}_e$ and carries away energy and momentum, but has negligible mass and no charge. In the decay equation, the antineutrino must be included for completeness at A-level.

Notice that $A$ does not change — no nucleons are lost, because the neutron converts into a proton. However, $Z$ increases by 1, so the daughter is a different element. And because $N = A - Z$, with $A$ fixed and $Z$ rising by 1, $N$ decreases by 1.

Worked example: Writing a Beta-Minus Decay Equation

Carbon-14 undergoes beta-minus decay. Write the full nuclear equation and state the changes in Z, A, and N.

1. $^{14}_{6}\text{C}$ — A = 14, Z = 6, so N = 14 - 6 = 8 neutrons in the parent.
2. $^{14}_{6}\text{C} \rightarrow ^{?}_{?}Y + ^{0}_{-1}\text{e} + \bar{ u}_e$ — The beta-minus particle has A = 0 and Z = -1. Include the antineutrino.
3. $14 = A_Y + 0 \Rightarrow A_Y = 14$ — A is unchanged in beta-minus decay. [M1]
4. $6 = Z_Y + (-1) \Rightarrow Z_Y = 7$ — 6 on the left must equal Z_Y + (-1) on the right, so Z_Y = 7. [M1]
5. $Z = 7 \Rightarrow \text{Nitrogen (N)}$ — Element 7 is nitrogen. [A1]
6. $^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}\text{e} + \bar{ u}_e$ — Changes: Z increases by 1 (6 to 7), A unchanged (14), N decreases by 1 (8 to 7).

Answer: $^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}\text{e} + \bar{ u}_e$. Z: 6 to 7 (increase by 1); A: 14 (unchanged); N: 8 to 7 (decrease by 1).

5. Beta-Plus Decay: A Proton Becomes a Neutron

Beta-plus decay ($\beta^+$) is the mirror image of beta-minus. Here, a nucleus has too many protons relative to neutrons. A proton converts into a neutron, emitting a positron (the antiparticle of the electron) and a neutrino.

The positron is written $^{0}_{+1}\text{e}$ (or $^{0}_{+1}\beta$). The neutrino is $ u_e$. In beta-plus decay: $Z$ decreases by 1, $A$ stays the same, and $N$ increases by 1 — the opposite pattern to beta-minus.

A useful way to keep beta-plus and beta-minus straight: in beta-minus, the nucleus gains a proton (Z goes up); in beta-plus, the nucleus loses a proton (Z goes down). The emitted particle always carries away the charge that accounts for the change in Z.

6. Gamma Emission and Nuclear Energy Level Diagrams

After alpha or beta decay, the daughter nucleus is often left in an excited energy state — think of it like an atom with electrons in higher energy levels, but here it is the nucleus itself that holds excess energy. The nucleus then releases this energy as a gamma ray ($\gamma$), a high-energy photon.

Gamma emission does not change $Z$, $A$, or $N$ at all. It is purely an energy transition within the same nucleus. Gamma emission is not a primary decay mode in the sense that it does not change nuclear composition; it typically follows alpha or beta decay when the daughter is left in an excited state.

In an energy level diagram, the parent nucleus sits at the top (highest energy). An arrow pointing downward and to the right or left represents a decay, labelled with the particle emitted. If the daughter is produced in an excited state, a second downward arrow shows the gamma transition to the ground state. The vertical distance between levels represents the energy released.

AQA exam questions may show you a nuclear energy level diagram and ask you to identify the decay type from the arrow, or to complete a missing label. Use these rules: if both $Z$ and $A$ change, it is alpha decay; if only $Z$ changes (with $A$ unchanged), it is beta decay; if neither $Z$ nor $A$ changes, it is gamma emission.

7. Comparing All Decay Types: N and Z Changes at a Glance

Now that all three primary decay modes are in place, it helps to compare them side by side. This table is the kind of summary that earns marks in 'describe' and 'compare' questions.

Notice that beta-minus and beta-plus are mirror images in terms of their effect on $Z$ and $N$. Alpha decay is the only mode that reduces $A$. Gamma emission changes nothing about the nuclear composition — it is purely an energy release.

8. Completing Unknown Decay Equations

A common exam task gives you a partial equation with one nucleus missing and asks you to identify it. The method is always the same: apply conservation of $A$ and conservation of $Z$, then use the periodic table to name the element.

Worked example: Identifying an Unknown Daughter Nucleus

Complete the following decay equation and identify the decay type: $^{210}_{84}\text{Po} \rightarrow ^{206}_{?}\text{Y} + ^{4}_{2}\text{He}$. State the changes in Z, A, and N.

1. Emitted particle is $^{4}_{2}\text{He}$ — this is an alpha particle. — Recognising the emitted particle immediately tells you the decay type and the expected changes.
2. $210 = 206 + 4$ ✓ — Top numbers already balance: 206 + 4 = 210. A of the daughter is confirmed as 206. [M1]
3. $84 = Z_Y + 2 \Rightarrow Z_Y = 82$ — 84 on the left must equal Z_Y + 2 on the right, so Z_Y = 82. [M1]
4. $Z = 82 \Rightarrow \text{Lead (Pb)}$ — Element 82 is lead. [A1]
5. $^{210}_{84}\text{Po} \rightarrow ^{206}_{82}\text{Pb} + ^{4}_{2}\text{He}$ — Both A and Z balance. This is a classic alpha decay in the uranium decay series.
6. Parent: $N = 210 - 84 = 126$. Daughter: $N = 206 - 82 = 124$. — Z: 84 to 82 (decrease by 2). A: 210 to 206 (decrease by 4). N: 126 to 124 (decrease by 2).

Answer: $^{210}_{84}\text{Po} \rightarrow ^{206}_{82}\text{Pb} + ^{4}_{2}\text{He}$. Alpha decay. Z decreases by 2, A decreases by 4, N decreases by 2.

The same method works when the emitted particle is the unknown. If you are given the parent and daughter, subtract their $Z$ and $A$ values to find the emitted particle's numbers, then identify what particle that corresponds to.

9. Exam Technique: Writing Decay Equations for Full Marks

Decay equation questions are reliable mark-earners — but only if you avoid the traps that cost students marks every year. Here is what the mark scheme actually rewards and what it penalises.

For nuclear energy level diagram questions, the examiner may ask you to 'explain' a transition. Use this structure: name the decay type, state the particle emitted, then give the changes in Z, A, and N. That three-part answer covers the marking points for an explain question on this topic.

Every decay equation question on AQA papers rewards the same two conservation laws applied carefully. Master the notation, apply conservation of A and Z, identify the element, and state the changes in N — that sequence covers every mark available on this topic.