The big idea: When a nucleus decays it turns into a new nucleus (the daughter) and shoots out a small particle. We write this as a nuclear equation.
Two numbers must balance — be the same on both sides:
- the nucleon number A (the top number — protons + neutrons) - the proton number Z (the bottom number — protons)
Balance those two and you have found the daughter nuclide.
New words, plainly: Parent = the nucleus before it decays.
Daughter = the new nucleus made by the decay.
Nuclide = a specific nucleus, written (top = nucleon number, bottom = proton number).
Conserved = the same before and after — it does not change.
- nucleon number (top) = protons + neutrons
- proton number (bottom) = number of protons (sets the element)
- the parent nuclide (before decay)
- the daughter nuclide (the new element formed)
The one rule that does everything: Top row adds up the same. Bottom row adds up the same.
That single check finds the daughter every time — you do not need to memorise long lists.
There are two decays you must be able to balance. Each emits a different particle, so each changes A and Z in its own fixed way.
Alpha (α) decay throws out an alpha particle — a helium-4 nucleus, 2 protons and 2 neutrons, written . So the top falls by 4 and the bottom falls by 2:
- parent nuclide before decay
- daughter: nucleon number falls by 4, proton number by 2
- alpha particle = a helium-4 nucleus (2 protons + 2 neutrons)
Beta-minus (β⁻) decay happens when a neutron turns into a proton inside the nucleus, firing out an electron (written ) and an antineutrino. The nucleon number stays the same, but the proton number goes up by 1:
- parent nuclide before decay
- daughter: nucleon number unchanged, proton number rises by 1
- beta-minus particle = an electron (created when a neutron becomes a proton)
- antineutrino — emitted with the electron (no charge, ≈ no mass)
How A and Z change: Lock in the two patterns — then you never have to think:
Alpha: A drops by 4, Z drops by 2.
Beta-minus: A stays the same, Z goes up by 1.
| Decay | What leaves the nucleus | Nucleon number A | Proton number Z |
|---|---|---|---|
| Alpha (α) | a helium-4 nucleus (2 p + 2 n) | falls by 4 (A → A − 4) | falls by 2 (Z → Z − 2) |
| Beta-minus (β⁻) | an electron + an antineutrino | unchanged (A → A) | rises by 1 (Z → Z + 1) |
Worked example — find the daughter of an alpha decay
Polonium-210 (A = 210, Z = 84) decays by alpha emission. Find the nucleon number and proton number of the daughter nuclide.
Solution
- Write the decay and use the rule top balances, bottom balances:
- Top (nucleon number): the daughter's A is the parent minus the alpha's 4:
- Bottom (proton number): the daughter's Z is the parent minus the alpha's 2:
Final answer
The daughter has nucleon number 206 and proton number 82 (Z = 82 is lead, so it is lead-206).
Worked example — find the daughter of a beta-minus decay
Caesium-137 (A = 137, Z = 55) decays by beta-minus emission. Find the nucleon number and proton number of the daughter nuclide.
Solution
- Write the decay, including the emitted electron:
- Top (nucleon number): the electron has A = 0, so the daughter's A is unchanged:
- Bottom (proton number): the electron has Z = −1, so the daughter's Z must be 1 MORE to balance:
Final answer
The daughter has nucleon number 137 and proton number 56 (Z = 56 is barium, so it is barium-137).
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How this is tested: Balancing decay equations is a guaranteed, quick-mark skill.
- Paper 1A (MCQ): a one-step determine — given a parent and the decay type, pick the daughter's A and Z; or follow a short chain (e.g. an alpha then a beta-minus) and find the final nucleus. - Paper 2: complete a nuclear equation — fill in the missing A, Z or particle — or identify the daughter element from its proton number.
Classic trap: in beta-minus decay, dropping Z by 1 instead of raising it. The electron's −1 charge means Z must go up by 1 to balance.
Decay chains: do one step at a time: If a nucleus emits two particles in a row, apply the changes one after the other.
Keep a running total of A and Z, updating after each emission — never try to do both at once.
IB-style question — (a) the daughter of the alpha step
Thorium-230 (A = 230, Z = 90) first emits an alpha particle, then the nucleus it forms emits a beta-minus particle. Find the nucleon number and proton number of the nuclide produced AFTER the alpha emission.
Solution
- Alpha emission: nucleon number falls by 4, proton number falls by 2. Apply both:
- So after the alpha step the nucleus has A = 226 and Z = 88.
Final answer
After the alpha emission: nucleon number 226, proton number 88 (this is radium-226).
IB-style question — (b) the daughter after the beta-minus step
Continue from part (a): the nuclide with A = 226 and Z = 88 now emits a beta-minus particle. Find the nucleon number, proton number AND neutron number of the FINAL nuclide.
Solution
- Beta-minus emission: nucleon number unchanged, proton number rises by 1:
- Neutron number = nucleon number − proton number (N = A − Z):
Final answer
The final nuclide has A = 226, Z = 89 and N = 137 (this is actinium-226). Always update A and Z one decay at a time.