Radioactive decay and emissions

Deep inside a lump of pitchblende, or inside the smoke detector on your ceiling, some nuclei are simply too heavy or too lopsided in their proton-to-neutron ratio to stay together forever. Without any external trigger — no heating, no hammering, no chemical reaction — they spontaneously break apart, hurling out radiation and turning into a different element in the process. This is radioactive decay, and it comes in exactly three flavours: alpha ( $\alpha$ ), beta ( $\beta$ ), and gamma ( $\gamma$ ). Each behaves differently in a magnetic field, penetrates matter to a different depth, and changes the parent nucleus according to its own fixed bookkeeping rule. Master those three rules and you can predict, on paper, exactly which element any unstable nucleus turns into.

The core insight

An unstable nucleus sheds $\alpha$ , $\beta$ , or $\gamma$ radiation to become more stable, and each emission type changes the mass number $A$ and atomic number $Z$ by its own fixed, conserved amount.

The core idea

A nucleus is unstable when its proton-to-neutron ratio is unfavourable or when it is simply too massive to hold together. To move toward stability it spontaneously emits one of three kinds of radiation, transforming into a new nucleus (the daughter). Recall the labels you already know: isotopes share $Z$ but differ in $A$ , isobars share $A$ but differ in $Z$ , and isotones share the number of neutrons ( $A - Z$ ).

Alpha ( $\alpha$ ) particle — a bare helium nucleus, $^{4}_{2}\text{He}$ , with charge $+2e$ and relative mass 4. It is the heaviest and most charged of the three, so it interacts strongly with anything it passes near: highest ionising power, but because it loses energy so fast it also has the lowest penetrating power, stopped by a sheet of paper or a few centimetres of air. Being heavy and positive, it deflects only slightly in electric or magnetic fields, always toward the negative side.

Beta ( $\beta$ ) particle — a fast electron, charge $-e$ , with negligible mass. It is lighter and less charged than $\alpha$ , so it ionises moderately but penetrates moderately too, needing a few millimetres of aluminium to stop it. Because it is so light, it deflects more sharply than $\alpha$ in a field, and always toward the positive side (opposite direction to $\alpha$ , since the charges are opposite).

Gamma ( $\gamma$ ) ray — electromagnetic radiation of very short wavelength, with no charge and no rest mass. It barely interacts with matter, giving it the lowest ionising power but the highest penetrating power — several centimetres of lead or thick concrete are needed to stop it. Having no charge, it is not deflected by electric or magnetic fields at all. Each emission changes the identity of the nucleus by a fixed rule:

\alpha\text{-decay:}\quad {}^{A}_{Z}X \rightarrow {}^{A-4}_{Z-2}Y + {}^{4}_{2}\text{He}

where: $X$ is the parent nucleus, $Y$ is the daughter, $A$ is the mass number, and $Z$ is the atomic number. The nucleus loses 2 protons and 2 neutrons, so $A$ falls by 4 and $Z$ falls by 2.

\beta\text{-decay:}\quad {}^{A}_{Z}X \rightarrow {}^{A}_{Z+1}Y + {}^{0}_{-1}e

where: the ejected electron carries away charge $-1$ but essentially no mass number, so $A$ is unchanged while $Z$ rises by 1 — a neutron inside the nucleus has converted into a proton plus an emitted electron.

\gamma\text{-emission:}\quad {}^{A}_{Z}X^{*} \rightarrow {}^{A}_{Z}X + \gamma

where: the asterisk marks an excited (higher-energy) nucleus. Neither $A$ nor $Z$ changes; the nucleus simply drops to a lower energy state and releases the difference as a gamma photon. Gamma emission almost always rides along with an $\alpha$ or $\beta$ decay rather than occurring by itself, because that is usually what leaves the daughter nucleus in an excited state to begin with.

Figure

Three rows comparing alpha beta and gamma radiation for charge mass penetrating power and field deflection

Derivation ABalancing a Nuclear Decay Equation

The goal is to show, using only conservation of mass number and atomic number across a nuclear reaction, that once you know the emission type ( $\alpha$ , $\beta$ , or $\gamma$ ) the identity of the daughter nucleus is completely fixed — no guessing, just bookkeeping on two totals that must stay equal on both sides of the arrow.

Step 1. State the two conservation rules. In any nuclear decay, the total mass number $A$ (protons plus neutrons) and the total atomic number $Z$ (protons, i.e. charge) must be the same on the left and right of the equation, because nucleons and charge cannot simply vanish. Write the parent as ${}^{A}_{Z}X$ and the unknown daughter as ${}^{A'}_{Z'}Y$ .

Step 2. Fix what the emitted particle carries away. An $\alpha$ particle is ${}^{4}_{2}\text{He}$ , so it removes 4 from the mass-number total and 2 from the atomic-number total. A $\beta$ particle is ${}^{0}_{-1}e$ , so it removes 0 from the mass number but adds $-1$ to the charge total on the right, which forces $Z$ to rise by 1 to keep the left-hand total unchanged. A $\gamma$ photon carries no mass number and no charge, so it removes nothing from either total.

Step 3. Solve for the daughter's $A'$ and $Z'$ by matching totals. For $\alpha$ -decay: $A = A' + 4$ and $Z = Z' + 2$ , giving $A' = A-4$ , $Z' = Z-2$ . For $\beta$ -decay: $A = A' + 0$ and $Z = Z' + (-1)$ , giving $A' = A$ , $Z' = Z+1$ . For $\gamma$ -emission: $A' = A$ , $Z' = Z$ , since nothing is removed at all.

Step 4. Read off the element. Once $Z'$ is known, the periodic table gives the daughter element directly (atomic number is the element's identity card), and $A'$ tells you which isotope of that element it is. This is exactly the shortcut used in both worked examples below — no equation needs to be solved from scratch each time; the three rules from Step 3 are applied directly.

Worked example

A sample of uranium, $^{238}_{92}\text{U}$ , decays in two steps: it first emits an alpha particle, and the resulting nucleus then emits a beta particle. Find the final nucleus and name both elements formed.

Step 1 — apply the alpha-decay rule to uranium. Using $A' = A - 4$ and $Z' = Z - 2$ : $A' = 238 - 4 = 234$ and $Z' = 92 - 2 = 90$ . Atomic number 90 is thorium (Th), so

{}^{238}_{92}\text{U} \rightarrow {}^{234}_{90}\text{Th} + {}^{4}_{2}\text{He}

Check: mass numbers balance, $238 = 234 + 4$ ; atomic numbers balance, $92 = 90 + 2$ .

Step 2 — apply the beta-decay rule to the thorium daughter. Using $A' = A$ and $Z' = Z + 1$ : $A' = 234$ and $Z' = 90 + 1 = 91$ . Atomic number 91 is protactinium (Pa), so

{}^{234}_{90}\text{Th} \rightarrow {}^{234}_{91}\text{Pa} + {}^{0}_{-1}e

Check: mass numbers balance, $234 = 234 + 0$ ; atomic numbers balance, $90 = 91 - 1$ .

Step 3 — state the overall result. Starting from $^{238}_{92}\text{U}$ , the two decays produce, in order, thorium-234 and then protactinium-234. Final nucleus: $^{234}_{91}\text{Pa}$ , reached via the intermediate $^{234}_{90}\text{Th}$ — the first two steps of uranium's natural decay series.

Competitive stretch

If a third step emitted a gamma ray from an excited $^{234}_{91}\text{Pa}$ nucleus, nothing about $A$ or $Z$ would change at all — the equation would simply read ${}^{234}_{91}\text{Pa}^{*} \rightarrow {}^{234}_{91}\text{Pa} + \gamma$ , the nucleus staying protactinium-234 throughout, only shedding excess energy.

Common traps

Forgetting that in $\beta$ -decay the mass number stays the same — only $Z$ changes, because the ejected electron carries away charge but essentially no mass.
Believing $\gamma$ emission changes the element — it changes neither $A$ nor $Z$ ; it only releases excess energy from an already-formed excited nucleus.
Swapping penetrating power and ionising power — $\alpha$ is the most ionising but least penetrating, and $\gamma$ is the exact opposite; the two properties always run in reverse order.
Assuming $\alpha$ and $\beta$ deflect the same way in a field — they bend toward opposite plates because their charges are opposite, while $\gamma$ does not deflect at all.

Ask the tutor: if you want to practise chaining three or four decay steps together to trace a full natural decay series, ask the AI tutor to generate a fresh starting nucleus and walk through it with you.

Why does emitting a light electron change the atomic number so much in beta decay?

It is not the mass of the electron that matters here, it is its charge. Inside the nucleus, a neutron converts into a proton plus an electron, and the electron is then ejected as the beta particle. That conversion adds one extra proton to the nucleus, so $Z$ rises by 1, even though the ejected particle itself has almost no mass and therefore does not change $A$ . Think of it as a charge transaction, not a mass transaction.

If gamma rays do not change the element, why are they still dangerous?

Because "dangerous" here is about energy delivered to tissue, not about changing which element is present. A gamma photon carries high energy and, thanks to its very high penetrating power, can travel deep into the body and disrupt cells along its path, even though the nucleus that emitted it does not transmute into a different element. Its low ionising power per unit path length is offset by how far it can travel before losing all its energy.

How do I know whether a decay problem wants alpha or beta rules?

The question always tells you the emission type directly, or it gives you enough of the daughter nucleus that you can work backwards. If mass number drops by exactly 4 and atomic number drops by exactly 2, it was an alpha emission. If the mass number is unchanged but the atomic number has risen by 1, it was a beta emission. If both $A$ and $Z$ are identical before and after, it was a gamma emission releasing pure energy.

Radioactive decay and emissions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads