Radioactivity: Synthesis and Exam Skills

You have now met every idea in Topic P6: the nuclear model of the atom, isotopes and radioactive decay, the three types of radiation, nuclear equations, half-life, the uses and hazards of radiation, and nuclear fission and fusion. This final lesson pulls the topic together — gathering the key rules in one place, working through a multi-step problem that draws on several ideas at once, and warning you about the mistakes that most often cost marks in this topic. The aim is to help you move confidently between balancing a nuclear equation, doing a half-life calculation, and comparing fission with fusion. This lesson is part of Topic P6 (Radioactivity) of OCR Gateway Science A.

By the end of this lesson you should be able to recall and apply the decay rules for nuclear equations, carry out half-life calculations, compare fission and fusion confidently, recognise and avoid common P6 exam mistakes, and work through a multi-step radioactivity problem.

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The Key Rules of P6 in One Place

Most of Topic P6 rests on a small set of rules and definitions. Knowing exactly what each one says is half the battle.

Idea	The rule to remember
Nuclide notation	$^{A}_{Z}\text{X}$ZA​X: $A$A = mass number (protons + neutrons), $Z$Z = atomic number (protons); neutrons $= A - Z$=A−Z
Alpha decay	$A \to A-4$A→A−4, $Z \to Z-2$Z→Z−2 (emit $^{4}_{2}\text{He}$24​He)
Beta-minus decay	$A$A unchanged, $Z \to Z+1$Z→Z+1 (emit $^{\ \ 0}_{-1}\text{e}$−1  0​e; a neutron → a proton)
Gamma emission	$A$A and $Z$Z unchanged (emit $^{0}_{0}\gamma$00​γ; energy only)
Half-life	Time for half the undecayed nuclei (or the count-rate) to decay; fraction left after $n$n half-lives $= \left(\tfrac{1}{2}\right)^{n}$=(21​)n
Activity	Decays per second, in becquerel (Bq); falls over time
Fission	A large nucleus splits (after absorbing a neutron) into two smaller nuclei + 2–3 neutrons + energy
Fusion	Two light nuclei join into a heavier nucleus + energy; powers the Sun and stars

Exam Tip: The two rules you will use most are the decay changes (alpha $A-4, Z-2$A−4,Z−2; beta $A$A same, $Z+1$Z+1) and the half-life halving (fraction left $= (\tfrac{1}{2})^n$=(21​)n). Have both at your fingertips — they appear on almost every P6 paper.

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Balancing Nuclear Equations: A Quick Recap

Every nuclear equation must have the mass numbers balanced (top) and the atomic numbers balanced (bottom). To find a missing particle, subtract the totals.

Worked Example — Identifying an unknown emission

A thorium nucleus decays as $^{234}_{\ 90}\text{Th} \rightarrow\ ^{234}_{\ 91}\text{Pa} + {}^{A}_{Z}\text{?}$ 90234​Th→  91234​Pa+ZA​?. Find the missing particle and name the type of decay.

Step 1 — balance the mass numbers: $234 = 234 + A$234=234+A, so $A = 0$A=0.

Step 2 — balance the atomic numbers: $90 = 91 + Z$90=91+Z, so $Z = 90 - 91 = -1$Z=90−91=−1.

Step 3 — a particle with mass number $0$0 and charge $-1$−1 is a beta-minus particle $^{\ \ 0}_{-1}\text{e}$−1  0​e.

Step 4 — the mass number stayed the same and the atomic number rose by 1: this is beta-minus decay.

Answer: the missing particle is a beta-minus particle and the decay is beta-minus decay.

Exam Tip: To name an unknown particle: $(4, 2)$(4,2) → alpha, $(0, -1)$(0,−1) → beta-minus, $(0, 0)$(0,0) → gamma. Then read off the type of decay from how $A$A and $Z$Z change.

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Half-Life Calculations: A Quick Recap

The method never changes: find the number of half-lives, then halve the starting value that many times (or use $\left(\tfrac{1}{2}\right)^{n}$(21​)n).

Worked Example — Amount remaining

A source has an activity of $3200 \text{ Bq}$3200 Bq and a half-life of $5$5 hours. What is its activity after $20$20 hours?

Step 1 — number of half-lives: $20 \div 5 = 4$20÷5=4.

Step 2 — halve four times: $3200 \to 1600 \to 800 \to 400 \to 200 \text{ Bq}$3200→1600→800→400→200 Bq.

Answer: the activity after $20$20 hours is $200 \text{ Bq}$200 Bq. (Check: $3200 \times \left(\tfrac{1}{2}\right)^{4} = 3200 \div 16 = 200 \text{ Bq}$3200×(21​)4=3200÷16=200 Bq.)

Worked Example — Finding the half-life

A radioactive sample falls from $960 \text{ Bq}$960 Bq to $60 \text{ Bq}$60 Bq in $36$36 days. Find its half-life.

Step 1 — halve from $960$960 until you reach $60$60: $960 \to 480 \to 240 \to 120 \to 60$960→480→240→120→60 — that is 4 halvings.

Step 2 — the $4$4 half-lives took $36$36 days, so one half-life $= 36 \div 4 = 9$=36÷4=9 days.

Answer: the half-life is $9$9 days.

Exam Tip: If a question gives before and after activities and a total time, first count the halvings to find the number of half-lives, then divide the total time by that number to get the half-life.

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A Multi-Step Worked Problem

Exam questions often combine several P6 ideas in one. Working through one carefully shows how the parts of the topic connect.

Worked Example — From decay to half-life to safety

A medical physics lab receives a sample of technetium-99m, a gamma-emitting isotope used as a tracer. It has a half-life of $6$6 hours and an initial activity of $800 \text{ MBq}$800 MBq.

(a) The sample is produced by the beta-minus decay of molybdenum-99, $^{99}_{42}\text{Mo}$4299​Mo. Write the nuclear equation for this decay. (Technetium has atomic number 43.)

Step 1 — beta-minus decay: mass number unchanged ($99$99), atomic number $+1$+1 ($42 \to 43$42→43).

Step 2 — atomic number 43 is technetium (Tc):

$^{99}_{42}\text{Mo} \rightarrow\ ^{99}_{43}\text{Tc} + {}^{\ \ 0}_{-1}\text{e}$4299​Mo→ 4399​Tc+−1  0​e

Step 3 — check: mass numbers $99 = 99 + 0$99=99+0 ✓; atomic numbers $42 = 43 + (-1) = 42$42=43+(−1)=42 ✓.

(b) A patient is given the tracer. After $24$24 hours, what fraction of the activity remains, and what is the activity?

Step 4 — number of half-lives: $24 \div 6 = 4$24÷6=4.

Step 5 — fraction remaining: $\left(\tfrac{1}{2}\right)^{4} = \tfrac{1}{16}$(21​)4=161​.

Step 6 — activity: $800 \div 16 = 50 \text{ MBq}$800÷16=50 MBq.

(c) Explain why a gamma emitter with this short half-life is suitable for use as a tracer inside the body.

Step 7 — gamma is penetrating, so it escapes the body to be detected, and is only weakly ionising inside the body. The short half-life ($6$6 hours) means the activity falls quickly (to $\tfrac{1}{16}$161​ in a day), so the patient's exposure is limited.

Answer: (a) $^{99}_{42}\text{Mo} \rightarrow\ ^{99}_{43}\text{Tc} + {}^{\ \ 0}_{-1}\text{e}$4299​Mo→ 4399​Tc+−1  0​e; (b) $\tfrac{1}{16}$161​ remains, an activity of $50 \text{ MBq}$50 MBq; (c) gamma escapes the body for detection and the short half-life limits the patient's exposure. This problem links beta decay, half-life calculation and choosing an isotope — exactly the synoptic thinking P6 rewards.

Worked Example — Fission versus fusion in context

State whether each is fission or fusion, and where it occurs: (a) uranium-235 absorbs a neutron and splits into two daughter nuclei in a power station; (b) hydrogen nuclei join to form helium in the Sun.