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How do we write balanced nuclear equations for alpha, beta and gamma decay?

The nature and properties of alpha, beta and gamma radiation, their ionising and penetrating power, and writing balanced nuclear equations for alpha decay, beta-minus decay and gamma emission, conserving mass number and atomic number.

A focused answer to OCR Gateway GCSE Physics A topic P6 on radioactive emissions and nuclear equations, covering the nature and properties of alpha, beta and gamma radiation, their ionising and penetrating power, and writing balanced nuclear equations that conserve mass number and atomic number.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. Alpha, beta and gamma radiation
  3. Writing balanced nuclear equations
  4. Worked example of alpha decay
  5. Try this

What this topic is asking

OCR wants you to describe the nature and properties of alpha, beta and gamma radiation, compare their ionising and penetrating power, and write balanced nuclear equations for alpha decay, beta-minus decay and gamma emission. This is part of topic P6.1 of the OCR Gateway Physics A (J249) specification.

Alpha, beta and gamma radiation

Ionising power and penetrating power are inversely related: alpha ionises strongly but is easily stopped, while gamma ionises weakly but passes through far more material. This explains both their hazards and their uses.

Writing balanced nuclear equations

The three decays change the numbers in fixed ways:

  • Alpha decay: the nucleus loses an alpha particle (24He^{4}_{2}\text{He}), so the mass number falls by 4 and the atomic number falls by 2, forming a different element.
  • Beta-minus decay: a neutron changes into a proton and a fast electron is emitted (−10e^{0}_{-1}\text{e}), so the mass number is unchanged and the atomic number rises by 1.
  • Gamma emission: only energy is carried away, so the mass number and atomic number are both unchanged; gamma usually follows alpha or beta decay as the nucleus loses surplus energy.

Worked example of alpha decay

A typical question gives the parent nucleus and asks for the daughter. The general form for alpha decay is ZAX→ Z−2A−4Y+ 24He^{A}_{Z}\text{X} \rightarrow\ ^{A-4}_{Z-2}\text{Y} +\ ^{4}_{2}\text{He}, and for beta-minus decay ZAX→ Z+1AY+ −10e^{A}_{Z}\text{X} \rightarrow\ ^{A}_{Z+1}\text{Y} +\ ^{0}_{-1}\text{e}. Always check that the top numbers and the bottom numbers each add up to the same total on both sides.

Try this

Q1. State what a beta-minus particle is and how it is produced in the nucleus. [2 marks]

  • Cue. A fast-moving electron, produced when a neutron in the nucleus changes into a proton.

Q2. Uranium 92238U^{238}_{92}\text{U} emits an alpha particle to form thorium (Th). State the mass number and atomic number of the thorium nucleus. [2 marks]

  • Cue. Mass number 238−4=234238 - 4 = 234; atomic number 92−2=9092 - 2 = 90, so 90234Th^{234}_{90}\text{Th}.

Exam-style practice questions

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

OCR 20184 marksCompare alpha, beta and gamma radiation in terms of what each one is, its ionising power, and what material stops it.
Show worked answer →

A P6 question worth four marks comparing the three radiations. Alpha (α\alpha) is a helium nucleus (2 protons and 2 neutrons), the most ionising and least penetrating, stopped by a sheet of paper or a few centimetres of air (1-2 marks). Beta (β\beta) is a fast-moving electron from the nucleus, moderately ionising and penetrating, stopped by a few millimetres of aluminium (1 mark). Gamma (γ\gamma) is a high-energy electromagnetic wave, the least ionising but most penetrating, only reduced by thick lead or concrete (1 mark). Markers reward the identity, ionising power and stopping material for each, especially the order of penetration (alpha least, gamma most). A common error is to say alpha is the most penetrating.

OCR 20213 marksA nucleus of radium, 88226Ra^{226}_{88}\text{Ra}, decays by emitting an alpha particle to form radon (Rn). Write the balanced nuclear equation for this decay.
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A P6 Calculate-style question on balancing a nuclear equation. An alpha particle is 24He^{4}_{2}\text{He} (or 24α^{4}_{2}\alpha), with mass number 44 and atomic number 22 (1 mark for the alpha particle). The new nucleus has mass number 226−4=222226 - 4 = 222 and atomic number 88−2=8688 - 2 = 86, which is radon: 86222Rn^{222}_{86}\text{Rn} (1 mark for the daughter nucleus). The full equation is 88226Ra→ 86222Rn+ 24He^{226}_{88}\text{Ra} \rightarrow\ ^{222}_{86}\text{Rn} +\ ^{4}_{2}\text{He} (1 mark for the balanced equation). Markers reward the correct alpha particle, the correct daughter nucleus, and the totals balancing (226=222+4226 = 222 + 4 and 88=86+288 = 86 + 2). A common error is to change the mass or atomic number the wrong way.

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