When an atom changes its identity, two numbers tell the story
An examination question may show uranium-238 becoming thorium-234, then ask you to complete the nuclear equation. It is tempting to search your memory for a familiar element name. A more dependable method starts with just two numbers: the nucleon number, A, and the proton number, Z. A nuclear equation is complete only when those numbers balance on both sides and the particles make physical sense. This chapter takes you from a single nuclear change to connected chains in which several emissions occur, so that an unfamiliar nuclide cannot frighten you into guessing.
For the 2027 Singapore–Cambridge Secondary Education Certificate (SEC), G3 Physics is syllabus K323. Topic 20 includes the composition of the atom, nuclide notation and changes during alpha, beta-minus and gamma emission. Those are examination requirements; this guide is teaching practice rather than an official examination paper. The basic science of radioactivity, background radiation and half-life already has its own home in Learner’s Guide 0279. Here we concentrate on balancing and checking nuclear changes, especially when a question gives the daughter and asks for the parent or combines several stages.
The previous EMS induction review, Volume 0503 concluded a sequence about magnetic fields, experimental readings and public claims. We now move into a different section of the Physics syllabus: the nucleus. In both sections the important habit is the same. An attractive explanation cannot outrun the evidence. In nuclear questions, that means checking each symbol, not assuming an element name is correct because the arithmetic looks familiar. Readers moving from primary school can revisit the PSLE science lesson on tracking matter rather than declaring it lost; the secondary-level step is to track specified nuclear particles and numbers.
Quick method: balance before you name
First copy A and Z for the parent nucleus. Second identify the emitted particle: alpha carries A = 4 and Z = 2; beta-minus carries A = 0 and Z = −1; gamma carries A = 0 and Z = 0. Third subtract or add the two numbers using conservation across the entire equation. Fourth use the resulting Z to identify the daughter element from the periodic table, if a name is required. Fifth read the full equation again from left to right, checking that A totals match and Z totals match. Never balance only the top number.
In a chain, make a small two-column record after every emission. The next stage begins with the daughter of the previous stage; it does not begin with the original parent again. If a question is reversed, work backwards with the opposite number change. A repeated alpha decay changes A by eight and Z by four; a beta-minus decay changes Z upward by one without changing A. Do not let the word ‘decay’ mislead you into automatically subtracting from every number. A beta-minus daughter has one more proton than its parent.
Write a sentence after the arithmetic: ‘Alpha emission produces a daughter with nucleon number four less and proton number two less’; or ‘In beta-minus emission, a neutron changes into a proton and an electron is emitted, so the daughter Z increases by one while A stays constant.’ This final explanation is what turns a balanced equation into an intelligible Physics answer.
What the two nuclear numbers mean
Proton number Z counts protons in the nucleus. It defines the element. Any neutral atom of carbon has Z = 6. Nucleon number A counts protons plus neutrons. The neutron number is therefore A − Z. A carbon nucleus with A = 14 has six protons and eight neutrons. The numbers are counts, not measurements in kilograms and not decimal masses from a periodic-table box. Students sometimes read the relative atomic mass on a periodic table as though it were the nucleon number of one particular nucleus. That confuses a weighted average for a whole-number nuclear count.
Nuclide notation places A at the upper left and Z at the lower left of the chemical symbol. In plain text we write carbon-14, C-14, or describe it as A = 14, Z = 6. A full examination answer should use the notation requested in its question. Isotopes are atoms of the same element with the same Z but different neutron numbers and therefore different A. If Z changes, the daughter is a different element. If only A changes while Z is unchanged in a nuclear expression, check carefully what process the question describes instead of inventing a new kind of radiation.
Electrons belong to the electron arrangement of an atom, but a beta-minus particle is emitted in the nuclear transformation; it is not simply an ordinary orbital electron being shaken loose. An alpha particle is a helium nucleus with two protons and two neutrons. Gamma is high-energy electromagnetic radiation, not a proton or a neutron. These definitions prevent common category errors before the arithmetic even begins.
Alpha emission: subtract four from A and two from Z
Consider uranium-238, with Z = 92. An alpha particle leaves with A = 4 and Z = 2. The daughter therefore has A = 234 and Z = 90, which is thorium-234. Check A: 238 = 234 + 4. Check Z: 92 = 90 + 2. Both lines balance. The short form ‘the element loses four’ is weak because four of what? You must distinguish four nucleons from two protons. The daughter has two fewer protons and two fewer neutrons.
Here is an effective reverse check. Suppose a daughter is lead-206, Z = 82, after one alpha emission. The parent must have A = 210 and Z = 84, which is polonium-210. Work backwards by adding four and two. This exercise is not solved by memorising a decay series; the calculation itself proves what the missing numbers must be. The element name is checked afterwards using the periodic table or data provided.
Although a nuclear equation balances counts of nucleons and charge, do not claim that the rest masses of every product add up exactly in kilograms. Nuclear transformations can exchange small amounts of mass and energy. At G3 level the immediate task is to conserve A and Z as prescribed by nuclide notation, not to run an advanced energy calculation.
Beta-minus emission: A stays, Z rises by one
Beta-minus decay is the counterintuitive one. A neutron in an unstable nucleus transforms into a proton while an electron is emitted. The nucleus therefore gains one proton and loses one neutron. The total number of nucleons remains unchanged. If carbon-14, Z = 6, undergoes beta-minus decay, the daughter is nitrogen-14, Z = 7. Balancing proton numbers gives 6 = 7 + (−1). Balancing nucleon numbers gives 14 = 14 + 0. The negative sign is essential: it is not a suggestion that the daughter has fewer protons.
A fuller physical account includes an antineutrino in beta-minus decay, but many G3 nuclide-balancing questions use the simplified beta particle notation named in the syllabus. Follow the form and particles given in the question; do not treat an omitted advanced particle as a licence to break the required A and Z checks. If the exam explicitly supplies additional information, use it rather than forcing a memorised example onto a different process.
Do not confuse beta-minus with beta-plus emission. The 2027 K323 syllabus specifies beta particles as beta-minus for this topic. If you learn a broader nuclear-physics course later, you will encounter other transformations. Here, within the stated G3 treatment, a beta-minus emission keeps A unchanged and increases daughter Z by one. That single invariant is the fastest way to catch a reversed sign.
Gamma emission: the nucleus loses energy, not a counted nucleon
Gamma emission carries energy from an excited nucleus, but it has neither nucleon number nor electric charge. A and Z therefore remain unchanged. In symbolic terms, an excited state of a particular nuclide can release a gamma photon and become a lower-energy state of the same nuclide. If a student writes a gamma daughter with A smaller by one, that would imply a change in the nucleon count and is not justified by gamma emission alone.
Do not say that ‘nothing changes’. The energy state changes; that is why radiation is emitted. The accurate statement is narrower: the identity of the element and its proton and nucleon counts remain the same in gamma emission. This is the difference between a cautious statement and a false one. In mixed questions, gamma is also a useful control: it should not erase an earlier alpha or beta-minus number change.
An examination may test alpha, beta and gamma through their nature, ionising ability and penetrating ability as well as through nuclear equations. Keep these skills related but distinct. A particle’s penetration is not determined by the numerical changes to A and Z alone. Volume 0279 owns the broader radioactivity explanation; this page owns the numerical and symbolic consistency of the nuclear equation.
A full chain: uranium-238 to uranium-234
Begin with uranium-238, Z = 92. After one alpha emission, the daughter is thorium-234, Z = 90. After one beta-minus emission, the daughter is protactinium-234, Z = 91. After another beta-minus emission, the daughter is uranium-234, Z = 92. The numerical path is therefore (238,92) → (234,90) → (234,91) → (234,92). This portion of a known decay chain makes a useful classroom example because every step has a different accounting purpose.
Now inspect the entire chain without retracing every intermediate symbol. One alpha emission contributes A = 4 and Z = 2 to the emitted products. Two beta-minus emissions contribute A = 0 and total Z = −2. Across the full sequence the daughter has A lower by four and Z unchanged overall. This does not mean nothing happened or that the first alpha decay was reversed. Intermediate elements changed and radiation was emitted; only the final proton-number count happens to equal the first.
A candidate who calculates only the net change may still miss a question asking for the intermediate nuclide. The safer method is to show the chain stage by stage, then use the net change as a second check. If the problem asks only for the final A and Z after known emissions, a net calculation can be efficient. If it asks for a specific daughter, write that daughter’s numbers at the relevant stage.
Counting an unknown mix of alpha and beta-minus emissions
Suppose a hypothetical nucleus starts at A = 232, Z = 90 and ends at A = 220, Z = 90 after some alpha and beta-minus emissions. The fall in A is twelve. Every alpha reduces A by four, so there were three alpha emissions. Those three emissions reduce Z by six. To return Z to its starting value of 90, six beta-minus emissions are required. The answer is not guessed from a printed isotope chain; it follows directly from the two conservation equations.
In general let a be the count of alpha emissions and b the count of beta-minus emissions. Then final A = initial A − 4a, and final Z = initial Z − 2a + b. Therefore a = (initial A − final A)/4 and b = final Z − initial Z + 2a. A non-whole-number result signals an inconsistency with the assumed emission types or with the supplied numbers. A negative emission count also signals a model mismatch unless the task explicitly asks you to work backwards.
This general rule is a tool for checking a proposed sequence, not a claim that every mathematically balanced chain occurs naturally. Real decay pathways depend on nuclear stability and allowed transformations. The examination question may give a hypothetical chain so that you can test arithmetic. Distinguish a balanced symbolic possibility from evidence of a real physical sequence.
How to use command words and marks wisely
‘State’ usually asks for a direct fact such as the value of Z or the emitted radiation. ‘Explain’ asks for a cause linked to a consequence, for example why the proton number increases after beta-minus decay. ‘Deduce’ asks you to use the supplied information to reach a conclusion, often without memorising the isotope. ‘Complete the equation’ calls for correct nuclide numbers, symbols and balancing. ‘Compare’ may need both likeness and difference, such as alpha and beta-minus changes in A and Z.
The wording ‘hence’ often means the result from one part is needed in the next. Carry your own earlier values forward carefully, and if you discover an arithmetic error later, correct it clearly. Avoid writing two contradictory final daughters. No mark is helped by filling every line with terminology unrelated to the requested conservation check. A well-organised solution can be shorter and more convincing than an elaborate explanation.
During revision, practise a blank table with columns labelled emission, change in A, change in Z, daughter A, daughter Z. After sufficient practice you should be able to do a single emission mentally and a longer sequence by two short equations. The table is scaffolding, not a ritual. Keep it when it prevents errors; remove it only when you can still verify each stage independently.
A worked examination-style challenge
Question: An imaginary nuclide X has A = 224 and Z = 88. It undergoes two alpha decays followed by three beta-minus decays. Determine the final A and Z, explain the direction of each change, and state one reason why arithmetic alone cannot establish whether the sequence occurs in nature. Work the stages: after the first alpha (220,86); after the second (216,84); after three beta-minus events (216,87). Thus the final nucleus has A = 216, Z = 87. Check directly: 224 − 2×4 = 216 and 88 − 2×2 + 3 = 87.
Explanation: each alpha emission removes a helium nucleus, reducing A by four and Z by two; each beta-minus emission converts a neutron into a proton while emitting an electron, changing Z upward by one and leaving A unchanged. The physical caveat is that a numerically balanced sequence is not proof that the described nuclei are unstable in exactly that way. A real decay pathway requires nuclear evidence beyond the supplied arithmetic.
Mark your own answer against four questions. Did you calculate both nuclear numbers? Did you account for all five emissions? Did you name the mechanism of the changes rather than merely state the numbers? Did you distinguish nuclear bookkeeping from a statement about nature? If any response is missing, the work may be arithmetically right but incomplete for a multi-part question.
Practice clinic: twenty-four short nuclear investigations
These original questions are grouped by the kind of reasoning, not by the name of a chapter in a textbook. Cover the explanation, attempt the prompt, then compare your answer. All unlabelled chains are bookkeeping exercises: unless the question names an actual observed decay, they do not assert that a particular isotope follows that path.
1. Identify the numbers in oxygen-18.
Oxygen has Z = 8. Oxygen-18 therefore contains eight protons and ten neutrons. The neutral atom has eight electrons, but the nuclear equation deals with the nucleus. Write 18 for the nucleon number and 8 for proton number. If you wrote ten under the chemical symbol, you have accidentally used the neutron count instead of Z.
2. What makes carbon-12 and carbon-14 isotopes?
Both are carbon because each has six protons. Their nucleon numbers differ: carbon-12 has six neutrons while carbon-14 has eight. Isotopes differ in neutron count, not in proton number. A nuclide with seven protons is nitrogen, not another carbon isotope, even if its total nucleon number is fourteen.
3. Complete polonium-210 alpha decay.
Polonium has Z = 84. Subtract four from A and two from Z to obtain A = 206 and Z = 82: lead-206. Check that the daughter plus alpha totals 210 nucleons and 84 protons. An answer naming lead but using A = 210 is not balanced.
4. Reverse an alpha daughter given as radon-222.
Radon has Z = 86. Add four to 222 and two to 86 to reconstruct the parent: A = 226, Z = 88, radium-226. The reverse operation is adding, not subtracting. Label which way the emission goes before writing a symbol or you may swap the parent and daughter.
5. Complete carbon-14 beta-minus decay.
A stays 14 and Z rises from 6 to 7, so the daughter is nitrogen-14. The emitted beta-minus particle has A = 0, Z = −1. Check Z as 6 = 7 − 1. The daughter is not carbon-13: no nucleon has left in the beta-minus bookkeeping.
6. Reverse a beta-minus daughter of sodium-24.
Sodium has Z = 11. In beta-minus decay, the daughter has one more proton than its parent. Hence the parent has Z = 10, A = 24: neon-24. This exercise uses the numbers specified; the method does not require memorising an isotope’s observed decay behaviour.
7. What numbers change during gamma emission?
Neither A nor Z changes because the gamma photon carries neither counted nucleons nor electric charge. The nucleus can move to a lower-energy state, so avoid claiming there was no physical effect. Say precisely that its nuclear identity remains the same.
8. Two alpha emissions from (A = 240, Z = 94).
Each alpha subtracts four and two. After two emissions, A = 232 and Z = 90. A quick net check subtracts eight and four. A stage-by-stage answer could record (236,92) between the starting and final nuclides if an intermediate is requested.
9. Three beta-minus emissions from (A = 90, Z = 36).
The nucleon number stays 90. Each beta-minus step raises Z by one, so the final Z is 39. If the question names the final element, identify it from Z = 39, yttrium. Do not subtract three merely because the process has ‘decay’ in its name.
10. Alpha then beta-minus from (A = 212, Z = 84).
Alpha gives (208,82). Beta-minus then gives (208,83). The final values are A = 208 and Z = 83. Check the order; here the net counts match beta-minus then alpha, but if the question asks for the intermediate daughter, the order matters.
11. Beta-minus then alpha from the same starting values.
Starting (212,84), beta-minus first gives (212,85), then alpha gives (208,83). The final pair is the same as in question 10, while the intermediate pair differs. This illustrates why a final-number check cannot replace an explicitly requested intermediate equation.
12. One alpha, two beta-minus emissions; what is the net change?
A falls by four. Z first falls by two and then rises by two, so its net change is zero. A student who writes ‘no element change’ ignores the intermediate identities; say only that the starting and ending proton numbers are equal.
13. Find alpha count from A = 228 to A = 216.
The decrease is twelve. Each alpha accounts for four, so three alpha emissions are required if the given emissions are alpha and beta-minus only. Beta-minus and gamma do not change A. State the assumption: an unspecified neutron-emission process would change the bookkeeping model.
14. A falls by eight while Z falls by two; find alpha and beta-minus counts.
A loss of eight means two alpha emissions. Those would reduce Z by four. The observed reduction is only two, so two beta-minus emissions offset two units. Answer: two alpha and two beta-minus steps.
15. A falls by four while Z rises by one.
One alpha step reduces Z by two. Three beta-minus steps are needed to move the net change from −2 to +1. Thus one alpha and three beta-minus emissions fit the stated changes. Check by substitution rather than relying on intuition.
16. A falls by six with alpha and beta-minus only.
No whole-number count of alpha emissions gives a decrease of six, because each alpha reduces A by four. Therefore the assumed model does not fit the supplied data. A strong answer flags the inconsistency rather than fabricating half an alpha particle.
17. What if a calculated beta-minus count is negative?
Recheck the direction of the question, signs and emission list. A negative count cannot describe a forward sequence consisting only of positive counts of beta-minus emissions. It may indicate wrong given data or another process. Do not silently replace it with its absolute value.
18. Can three gamma emissions turn uranium into thorium?
No. Gamma emission does not change proton number, whereas uranium and thorium have different proton numbers. A gamma photon can change an excited nucleus’s energy state, not its Z. A common wrong answer confuses energy release with element change.
19. Why does an alpha particle count as helium but not neutral helium?
An alpha particle contains two protons and two neutrons but has no surrounding electron pair. It is a positively charged helium nucleus. Its nuclear numbers are A = 4 and Z = 2. Do not call it a neutral helium atom inside the nuclear equation.
20. Is a beta-minus particle emitted from an orbital?
No. In the nuclear process used here, beta-minus emission accompanies a neutron changing into a proton. It is not a regular electron falling out of its atomic shell. The distinction matters because the daughter’s proton number increases.
21. Which equation is more important: A or Z?
Both are necessary. An equation that balances A but not Z is incomplete, and an equation that balances Z but not A is also incomplete. The independent checks identify different mistakes. Treat them as two short proofs rather than two optional decorations.
22. Why might a balanced equation still be physically wrong?
Conservation of counted nucleons and charge is necessary, but not sufficient to establish an actual decay. The parent must be able to undergo that transformation; energy and nuclear stability matter. When an exam supplies a hypothetical chain, solve it as stated while resisting claims beyond the data.
23. How does this differ from a half-life calculation?
Nuclear equations identify the type of transformation and the daughter nuclide. Half-life concerns the statistical time scale of a radioactive population. One correct nuclide equation does not supply the elapsed time or the decay rate. For half-life and background correction, return to Volume 0279.
24. Explain one beta-minus step in one full sentence.
A suitable sentence is: ‘A neutron in the nucleus changes into a proton and a beta-minus particle is emitted, so the nucleon number is unchanged and the proton number increases by one.’ Check that your wording explains both numbers, not just one.
How to practise without relying on answer recognition
Work with three kinds of prompt on separate days: forward decay, reverse decay and unknown emission counts. First do four uncomplicated examples in which the radiation type is stated. Then combine two or three stages. Finally cover the emitted-particle label and infer it from the change in A and Z. When an unfamiliar element symbol appears, leave its name blank for a moment and calculate the numbers; the periodic table can supply the final name afterwards.
Keep an error log with four categories: wrong sign, incorrect neutron count, failure to balance both totals, and unsupported physical claim. Give each correction a complete explanation. Writing only ‘careless mistake’ does not tell your future self what to change. An error log becomes useful when it contains a specific action: check the β− sign; label A and Z separately; show the intermediate daughter; or state the hypothetical-chain condition.
After several days, try a ten-minute mixed set without notes. Your target is not merely a correct final number. You should be able to explain why the numbers change as they do, detect a mathematically impossible answer, and tell the examiner when the arithmetic cannot establish a real decay path. Accuracy followed by clarity is a more durable examination habit than memorising a handful of nuclide symbols.
Examination context, scope and further reading
SEAB’s 2027 school-candidate list assigns G3 Physics the code K323 (the earlier reference code is 6091). The K323 syllabus includes nuclear composition, alpha, beta-minus and gamma radiation, nuclide notation, radioactive decay, background radiation, half-life, applications and hazards. Its Physics assessment comprises Paper 1 multiple choice (one hour, 40 marks, 30%), Paper 2 structured and free response (one hour 45 minutes, 80 marks, 50%), and Paper 3 practical (one hour 50 minutes, 40 marks, 20%). A useful learner never claims a particular practice item is guaranteed to appear.
These descriptions were checked against the official 2027 K323 G3 Physics syllabus, the SEAB subject-code table and the SEC overview. The examples and clinic questions above are original teaching questions, not SEAB past questions. For the broader topic of half-life and radioactive evidence, follow Volume 0279. For the next mathematical skill, continue to the companion volume on reverse nuclear-change arithmetic once published.
Finish with a compact test. For each decay, can you write the parent A and Z, the emitted particle numbers, the daughter A and Z, and an explanation in ordinary English? Can you perform the reverse calculation? Can you name the limit of the conclusion? If you can do all three calmly, you understand the mechanism well enough to solve new questions rather than recognise old ones.
