Nuclear Equations

When an unstable nucleus decays, it changes into a different nucleus and emits radiation. We can record exactly what happens using a nuclear equation — a kind of balance sheet for the nucleus, written with the same nuclide notation you met earlier. The power of a nuclear equation is that it must balance: the total mass number and the total atomic number (charge) are the same on both sides, because nucleons and charge are conserved. This lesson, part of Topic P6 (Radioactivity) of OCR Gateway Science A, shows how to write nuclear equations for alpha decay, beta-minus decay and gamma emission, and how to balance them and find an unknown product.

By the end of this lesson you should be able to write and balance nuclear equations for alpha, beta-minus and gamma decay, state how the mass number and atomic number change in each, apply the conservation of mass number and charge, and find an unknown nuclide in a decay.

The Symbols Used in Nuclear Equations

A nuclear equation uses the standard nuclide notation $^{A}_{Z}\text{X}$ , where $A$ is the mass number (top) and $Z$ is the atomic number (bottom). The particles you will meet are:

Particle	Symbol	Mass number ( $A$ )	Atomic number ( $Z$ )
Alpha particle	$^{4}_{2}\text{He}$ (or $^{4}_{2}\alpha$ )	$4$	$2$
Beta-minus particle	$^{\ \ 0}_{-1}\text{e}$ (or $^{\ \ 0}_{-1}\beta$ )	$0$	$-1$
Gamma ray	$^{0}_{0}\gamma$	$0$	$0$
Neutron	$^{1}_{0}\text{n}$	$1$	$0$
Proton	$^{1}_{1}\text{p}$ (or $^{1}_{1}\text{H}$ )	$1$	$1$

Notice that the beta particle is written with an atomic number of $-1$ . This is not because it contains a negative proton, but because giving it a "charge number" of $-1$ makes the equation balance correctly — its single negative charge has to be accounted for. The two golden rules for any nuclear equation are:

the sum of the mass numbers (top numbers) is the same on both sides, and
the sum of the atomic numbers (bottom numbers) is the same on both sides.

Exam Tip: Write the beta particle as $^{\ \ 0}_{-1}\text{e}$ — mass number $0$ , charge number $-1$ . The two rules to apply every time are: top numbers balance (mass number conserved) and bottom numbers balance (charge conserved). Check both before you finish.

Alpha Decay

In alpha decay, the nucleus emits an alpha particle ( $^{4}_{2}\text{He}$ ). Since the alpha particle carries away 2 protons and 2 neutrons, the original nucleus loses 4 from its mass number and loses 2 from its atomic number:

mass number: $A \to A - 4$
atomic number: $Z \to Z - 2$

Losing 2 protons means the atom becomes a different element, two places back in the periodic table. A classic example is the alpha decay of uranium-238 into thorium-234:

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

Check it balances. Mass numbers: $238 = 234 + 4$ . ✓ Atomic numbers: $92 = 90 + 2$ . ✓ Both sides balance, so the equation is correct.

Worked Example — Alpha decay of radium-226

Radium-226, $^{226}_{\ 88}\text{Ra}$ , decays by emitting an alpha particle. Write the nuclear equation and identify the daughter nucleus. (Radon has atomic number 86.)

Step 1 — alpha decay reduces the mass number by 4: $226 - 4 = 222$ .

Step 2 — alpha decay reduces the atomic number by 2: $88 - 2 = 86$ .

Step 3 — atomic number 86 is radon (Rn), so the daughter is $^{222}_{\ 86}\text{Rn}$ .

Step 4 — write the equation:

$^{226}_{\ 88}\text{Ra} \rightarrow\ ^{222}_{\ 86}\text{Rn} + {}^{4}_{2}\text{He}$

Step 5 — check: mass numbers $226 = 222 + 4$ ✓; atomic numbers $88 = 86 + 2$ ✓.

Answer: $^{226}_{\ 88}\text{Ra} \rightarrow\ ^{222}_{\ 86}\text{Rn} + {}^{4}_{2}\text{He}$ — radium-226 decays to radon-222.

Exam Tip: For alpha decay, do $A - 4$ and $Z - 2$ , then use the new atomic number to name the daughter element from the periodic table. Always finish by checking both totals balance.

Beta-Minus Decay

In beta-minus decay, a neutron inside the nucleus turns into a proton, and a fast electron (the beta particle) is created and emitted. The neutron-to-proton change is the heart of beta decay:

$^{1}_{0}\text{n} \rightarrow\ ^{1}_{1}\text{p} + {}^{\ \ 0}_{-1}\text{e}$

Because a neutron becomes a proton, the nucleus gains a proton but its total number of nucleons is unchanged (a neutron is simply swapped for a proton). So in beta decay:

mass number: unchanged ( $A \to A$ )
atomic number: $Z \to Z + 1$

Gaining a proton means the atom becomes a different element, one place forward in the periodic table. A standard example is the decay of carbon-14 into nitrogen-14:

$^{14}_{\ 6}\text{C} \rightarrow\ ^{14}_{\ 7}\text{N} + {}^{\ \ 0}_{-1}\text{e}$

Check it balances. Mass numbers: $14 = 14 + 0$ . ✓ Atomic numbers: $6 = 7 + (-1)$ , and $7 + (-1) = 6$ . ✓ This is exactly why the beta particle is given a charge number of $-1$ : it makes the bottom row balance.

Worked Example — Beta decay of strontium-90

Strontium-90, $^{90}_{38}\text{Sr}$ , is a beta-minus emitter. Write the nuclear equation and identify the daughter nucleus. (Yttrium has atomic number 39.)

Step 1 — beta decay leaves the mass number unchanged: it stays $90$ .

Step 2 — beta decay increases the atomic number by 1: $38 + 1 = 39$ .

Step 3 — atomic number 39 is yttrium (Y), so the daughter is $^{90}_{39}\text{Y}$ .

Step 4 — write the equation:

$^{90}_{38}\text{Sr} \rightarrow\ ^{90}_{39}\text{Y} + {}^{\ \ 0}_{-1}\text{e}$

Step 5 — check: mass numbers $90 = 90 + 0$ ✓; atomic numbers $38 = 39 + (-1) = 38$ ✓.

Answer: $^{90}_{38}\text{Sr} \rightarrow\ ^{90}_{39}\text{Y} + {}^{\ \ 0}_{-1}\text{e}$ — strontium-90 decays to yttrium-90.

Exam Tip: For beta-minus decay, the mass number stays the same and the atomic number goes up by 1 (a neutron becomes a proton). The daughter element is one place forward in the periodic table. The $-1$ charge on the electron makes the atomic numbers balance.

Gamma Emission

Gamma emission is different: no particles leave the nucleus, so neither the mass number nor the atomic number changes. Instead, the nucleus simply loses surplus energy as a gamma ray. It often happens immediately after an alpha or beta decay, when the new nucleus is left in an excited (high-energy) state and needs to settle:

mass number: unchanged ( $A \to A$ )
atomic number: unchanged ( $Z \to Z$ )

Because nothing changes except the energy, gamma emission leaves the same nucleus, just in a lower-energy state. We can show it by writing the gamma ray as $^{0}_{0}\gamma$ , which has mass number $0$ and atomic number $0$ , so it does not affect the balance at all. For example, an excited nucleus of an element X loses energy as a gamma ray:

$^{A}_{Z}\text{X}^{*} \rightarrow\ ^{A}_{Z}\text{X} + {}^{0}_{0}\gamma$

The asterisk simply denotes the "excited" higher-energy nucleus. Gamma emission is the nucleus's way of getting rid of energy without changing what it is.

Exam Tip: In gamma emission, nothing changes about the nucleus's mass number or atomic number — it stays the same element and isotope, just loses energy. The gamma ray $^{0}_{0}\gamma$ has zero mass number and zero charge.

Summary of the Changes

It helps to see the three decays side by side. The diagram and table together capture how each one changes the nucleus.

Nuclear Equations

Nuclear Equations

The Symbols Used in Nuclear Equations

Alpha Decay

Worked Example — Alpha decay of radium-226

Beta-Minus Decay

Worked Example — Beta decay of strontium-90

Gamma Emission

Summary of the Changes

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