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How do you write balanced nuclear equations for alpha and beta decay?

Nuclear decay equations: how the mass and atomic numbers change in alpha, beta-minus and beta-plus decay and gamma emission, and balancing nuclear equations.

A focused answer to Edexcel GCSE Physics on nuclear decay equations, covering how the mass (nucleon) number and atomic (proton) number change in alpha, beta-minus and beta-plus decay and gamma emission, and how to balance a nuclear equation, with worked examples.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Alpha decay
  3. Beta-minus decay
  4. Beta-plus decay and gamma emission
  5. How Edexcel examines this
  6. Try this

What this dot point is asking

Edexcel wants you to write balanced nuclear equations for alpha decay, beta-minus decay, beta-plus decay and gamma emission, showing how the mass (nucleon) number and atomic (proton) number change, and to balance the top and bottom numbers across the equation.

Alpha decay

Because the alpha particle carries away 2 protons and 2 neutrons, the parent nucleus becomes a different element with a smaller mass. The equation must balance: the mass numbers (top) and the atomic numbers (bottom) each add up to the same total on both sides. For example, 92238U 90234Th+ 24α^{238}_{92}\text{U} \rightarrow\ ^{234}_{90}\text{Th} +\ ^{4}_{2}\alpha.

Beta-minus decay

The mass number stays the same because the total number of nucleons (protons plus neutrons) is unchanged; a neutron has simply turned into a proton. The new proton raises the atomic number by one. The emitted electron is written 10e^{0}_{-1}\text{e} or 10β^{0}_{-1}\beta, with mass number 00 and "atomic number" 1-1 so the equation balances.

Beta-plus decay and gamma emission

Beta-plus is the mirror image of beta-minus: a proton becomes a neutron, the positron carries away the positive charge, and the atomic number drops by one. Gamma emission usually happens just after another decay, as the nucleus sheds surplus energy; because gamma has no mass and no charge, the nucleus stays the same element with the same mass number.

How Edexcel examines this

This dot point is examined on both tiers as a "write the balanced nuclear equation" question, usually worth two or three marks, for alpha or beta decay. The mark scheme rewards the correct changes to the mass and atomic numbers and a correctly balanced equation, with marks often allocated separately to the new mass number, the new atomic number and the emitted particle. For beta-minus, examiners also reward explaining that a neutron changes into a proton; for beta-plus, that a proton changes into a neutron. The most penalised errors are altering the mass number during beta decay and reversing the direction of the atomic-number change, so fix firmly that beta-minus moves the element one place forward and beta-plus one place back. Always verify that the top numbers balance and the bottom numbers balance across the arrow; this self-check catches most mistakes. Gamma emission may appear as a statement that neither number changes because only energy is lost.

Try this

Q1. State how the mass and atomic numbers change in alpha decay. [2 marks]

  • Cue. Mass number decreases by 4; atomic number decreases by 2.

Q2. In beta-minus decay, what happens to the atomic number? [1 mark]

  • Cue. It increases by 1 (a neutron becomes a proton).

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20203 marksA nucleus of radium-226 (88226Ra^{226}_{88}\text{Ra}) decays by emitting an alpha particle to form radon (Rn). Write the balanced nuclear equation, giving the mass number and atomic number of the radon nucleus.
Show worked answer →

In alpha decay the nucleus loses 2 protons and 2 neutrons, so the mass number falls by 4 (2264=222226 - 4 = 222) and the atomic number falls by 2 (882=8688 - 2 = 86) (1 mark each for the new mass number and atomic number). The equation is 88226Ra 86222Rn+ 24α^{226}_{88}\text{Ra} \rightarrow\ ^{222}_{86}\text{Rn} +\ ^{4}_{2}\alpha (or 24He^{4}_{2}\text{He}) (1 mark for the balanced equation). Markers reward subtracting 4 from the mass number and 2 from the atomic number, and a correctly balanced equation where the top and bottom numbers each balance across the arrow.

Edexcel 20223 marksA nucleus of carbon-14 (614C^{14}_{6}\text{C}) decays by beta-minus emission to form nitrogen (N). Write the balanced nuclear equation and explain what happens inside the nucleus.
Show worked answer →

In beta-minus decay a neutron changes into a proton (and an electron, the beta particle, is emitted) (1 mark). The mass number stays the same (14), but the atomic number increases by 1 (676 \rightarrow 7) because there is now one more proton (1 mark). The equation is 614C 714N+ 10β^{14}_{6}\text{C} \rightarrow\ ^{14}_{7}\text{N} +\ ^{0}_{-1}\beta (1 mark). Markers reward the neutron-to-proton change, the atomic number rising by 1 with the mass number unchanged, and the correctly balanced equation including the beta particle written as 10e^{0}_{-1}\text{e} or 10β^{0}_{-1}\beta.

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