Preview

Nuclear Decay & Radioactivity

Q: After 3 half-lives, what fraction of the original sample remains?

After each half-life, half remains. Apply (½)³.

Q: ²³⁸₉₂U undergoes alpha decay. What are the mass number and atomic number of the daughter nucleus?

Alpha decay: A decreases by 4, Z decreases by 2.

Q: Carbon-14 has a half-life of 5730 years. A sample contains only 25% of its original C-14. How old is it?

25% = (½)² means 2 half-lives have elapsed. Age = 2 × T½.

Visualize alpha, beta, and gamma decay with half-life calculations

Paris, 1896. Henri Becquerel wraps a photographic plate in an opaque envelope and lays a uranium-rich mineral on top — and the plate darkens anyway, fogged by invisible rays the uranium emits continuously, with no light, no heating, no energy input at all. Marie Curie gives the phenomenon its name, radioactivity, and Ernest Rutherford sorts the rays by how a magnetic field bends them: heavy positive alphas, light negative betas, unbending gammas. But the deepest surprise is the clock inside every unstable nucleus. A sample never decays all at once, and never at a steady linear rate: in one characteristic time, half the nuclei go; in the next equal time, half of what's left goes; then half of that. Each single nucleus is a coin flip with no memory — one that survived a century is no closer to decaying than one formed a second ago — yet countless coin flips stack into the smoothest curve in physics, N(t) = N₀e^(−λt). The Geiger counter hears that curve as activity — A = λN, with λ = ln2/T½: every second each surviving nucleus flips its coin, so the click rate runs proportional to how many nuclei remain and inversely proportional to the half-life. Nothing you do to the sample changes the rhythm: radiation does not vary with chemical state, temperature, or pressure. And the rhythm spans eternity: polonium-214 halves every 163 millionths of a second, carbon-14 every 5730 years. That stubborn regularity turns decay into a timepiece — the Shroud of Turin kept 92% of living tissue's carbon-14, and the exponential priced 690 years, matching the historical record. In this lab the clock is yours: one glowing nucleus, a live decay curve, and three real isotopes whose half-lives — microseconds to millennia — are compressed into seconds. Watch the rungs fall: 1000, 500, 250, 125. And when you double the sample, watch what the activity does.

What you'll be able to do

Run the half-life ladder on the panel's own numbers: at the default state (N₀ = 1000, T½ = 20 s) the N row steps 1000 → 500 → 250 → 125 exactly as the Elapsed T½ row ticks 1.00 → 2.00 → 3.00 and the Remaining % row reads 50.0 → 25.0 → 12.5 — 'the number of radioactive nuclei decreases from N to N/2 in one half-life, then to N/4 in the next, and to N/8 in the next' (OpenStax CP2e §31.5, verbatim), so after 3 half-lives the fraction is (½)³ = 1/8 (the HTML quiz Q1's keyed answer), and 'many half-lives (not just two) pass before all of the nuclei decay'. The same ladder reads dates backwards: 25% remaining means two half-lives, so a C-14 sample at 25% is 2 × 5730 = 11,460 years old, and the Shroud of Turin's 92% prices t = 0.0834/λ = 690 years (CP2e §31.5, Example 31.4)
Price activity with A = λN = (ln2/T½)·N: at the default state λ = 0.693147/20 = 0.0346574 s⁻¹ and the Activity row reads 34.7 Bq; doubling N₀ to 2000 doubles the row to 69.3 Bq — 'activity R should be proportional to the number of radioactive nuclei, N, and inversely proportional to their half-life' (CP2e §31.5, verbatim, Eq. 31.48) — while sweeping T½ 1 → 100 s at fixed N₀ = 1000 walks the row down 693.1 → 6.9 Bq, the same inverse proportionality. Unit literacy included: 1 Bq = 1 decay/s, 1 Ci = 3.70×10¹⁰ Bq (CP2e §31.5, Eqs. 31.46–31.47), and the λ ↔ T½ conversion λ = 0.693/t½ is the Chemistry 2e §21.3 worked example (Co-60: 0.132 y⁻¹)
Keep the nuclear bookkeeping straight across the three decay modes the Decay Type select offers: alpha ejects ⁴He so A → A−4 and Z → Z−2 (Ra-226 → Rn-222, the HTML quiz Q3's keyed answer; Chem 2e §21.3's Po-210 → Pb-206 + α example), beta-minus converts a neutron to a proton so Z → Z+1 with A unchanged (I-131 → Xe-131, Chem 2e §21.3 verbatim equation — the I-131 preset's real mode), and gamma changes neither A nor Z ('no change in mass number or atomic number', Chem 2e §21.3 on Co-60*) — with the historical frame that Rutherford separated the three by how a magnetic field bends them and proved 'α radiation is the emission of a helium nucleus' (CP2e §31.1)

Formulas

N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}

Amount remaining after time t (half-life form)

N(t) = N_0 e^{-\lambda t}

Radioactive decay law with decay constant λ

\lambda = \frac{\ln 2}{T_{1/2}}

Relationship between decay constant and half-life

{}^A_Z X \rightarrow {}^{A-4}_{Z-2}Y + {}^4_2\text{He}

Alpha decay: mass number −4, atomic number −2

Make a prediction

Default state: N₀ = 1000 nuclei of an isotope with half-life T½ = 20 s — the Activity row reads 34.7 Bq. You drag the N₀ slider to 2000 (same isotope, same half-life) and the run restarts. What does the Activity row read now?

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Replay the doubling gate live: at T½ = 20 s, drag N₀ from 1000 to 2000 — the run restarts and the Activity row jumps 34.7 → 69.3 Bq exactly, the decay curve sits twice as tall, and the Elapsed T½ row marches at the unchanged pace. A = λN, felt in one drag (quiz Q2, live).
Ride the default run to three half-lives on the decay graph: watch the N row step 1000 → 500 → 250 → 125 as the Elapsed T½ row ticks 1.00 → 2.00 → 3.00 and Remaining % reads 50.0 → 25.0 → 12.5 — the (½)ⁿ ladder, never a linear slide to zero (quiz Q1's 1/8, earned rung by rung).
Visit the three isotope presets by hand and read one law at three tempos: Carbon-14 (50 s stands in for 5730 yr, β⁻, A₀ = 27.7 Bq) → Iodine-131 (30 s for 8.02 d, β⁻, 69.3 Bq) → Polonium-214 (5 s for 163.47 μs, α, 693.1 Bq). Same exponential, same halving ladder — only the clock speed and the particle signature change.
Pull the half-life lever at fixed N₀ = 1000: sweep T½ 1 → 100 s and watch the Activity row fall 693.1 → 6.9 Bq (A ∝ 1/T½) while the graph curve stretches toward flat — then flip the Decay Type select mid-run and read the particle vocabulary: fat gold alpha spheres, small cyan beta sparks, purple gamma rings.

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Your prediction

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Answer: 69.3 Bq — exactly double: λ = ln2/T½ is unchanged, and A = λN tracks the nucleus count one-for-one

Activity is a headcount rate, not an isotope trait: A = λN with λ = ln2/T½ = 0.693147/20 = 0.0346574 s⁻¹ pinned by the half-life, so at N₀ = 1000 the row reads 34.657 → '34.7 Bq' and at N₀ = 2000 it reads 69.315 → '69.3 Bq' — an exact doubling. The textbook says it in plain words: 'you would expect the activity of a source to depend on two things: the amount of the radioactive substance present, and its half-life… activity R should be proportional to the number of radioactive nuclei, N, and inversely proportional to their half-life' (OpenStax CP2e §31.5, verbatim; R = 0.693N/t½). Option B is the misconception this gate targets — and the sim's own quiz Q2 ships it as trap option a ('activity only depends on the isotope type'); its keyed feedback gives the mechanism: 'each nucleus has the same decay probability λ per second', so twice as many independent nuclei produce twice as many decays per second. Option C invents a 'decay budget' that nuclei compete for — there is none: each nucleus flips its own memoryless coin ('The probability of decay is the same no matter when you start counting', §31.5), independent of how many neighbors it has. The half-life itself is untouched: the Elapsed T½ row marches at exactly the same pace in both runs — only the height of the curve, and the activity, doubled.

Quiz (0/3)

After 3 half-lives, what fraction of the original sample remains?

²³⁸₉₂U undergoes alpha decay. What are the mass number and atomic number of the daughter nucleus?

Carbon-14 has a half-life of 5730 years. A sample contains only 25% of its original C-14. How old is it?

You can now

Run the half-life ladder on the panel's own numbers: at the default state (N₀ = 1000, T½ = 20 s) the N row steps 1000 → 500 → 250 → 125 exactly as the Elapsed T½ row ticks 1.00 → 2.00 → 3.00 and the Remaining % row reads 50.0 → 25.0 → 12.5 — 'the number of radioactive nuclei decreases from N to N/2 in one half-life, then to N/4 in the next, and to N/8 in the next' (OpenStax CP2e §31.5, verbatim), so after 3 half-lives the fraction is (½)³ = 1/8 (the HTML quiz Q1's keyed answer), and 'many half-lives (not just two) pass before all of the nuclei decay'. The same ladder reads dates backwards: 25% remaining means two half-lives, so a C-14 sample at 25% is 2 × 5730 = 11,460 years old, and the Shroud of Turin's 92% prices t = 0.0834/λ = 690 years (CP2e §31.5, Example 31.4)
Price activity with A = λN = (ln2/T½)·N: at the default state λ = 0.693147/20 = 0.0346574 s⁻¹ and the Activity row reads 34.7 Bq; doubling N₀ to 2000 doubles the row to 69.3 Bq — 'activity R should be proportional to the number of radioactive nuclei, N, and inversely proportional to their half-life' (CP2e §31.5, verbatim, Eq. 31.48) — while sweeping T½ 1 → 100 s at fixed N₀ = 1000 walks the row down 693.1 → 6.9 Bq, the same inverse proportionality. Unit literacy included: 1 Bq = 1 decay/s, 1 Ci = 3.70×10¹⁰ Bq (CP2e §31.5, Eqs. 31.46–31.47), and the λ ↔ T½ conversion λ = 0.693/t½ is the Chemistry 2e §21.3 worked example (Co-60: 0.132 y⁻¹)
Keep the nuclear bookkeeping straight across the three decay modes the Decay Type select offers: alpha ejects ⁴He so A → A−4 and Z → Z−2 (Ra-226 → Rn-222, the HTML quiz Q3's keyed answer; Chem 2e §21.3's Po-210 → Pb-206 + α example), beta-minus converts a neutron to a proton so Z → Z+1 with A unchanged (I-131 → Xe-131, Chem 2e §21.3 verbatim equation — the I-131 preset's real mode), and gamma changes neither A nor Z ('no change in mass number or atomic number', Chem 2e §21.3 on Co-60*) — with the historical frame that Rutherford separated the three by how a magnetic field bends them and proved 'α radiation is the emission of a helium nucleus' (CP2e §31.1)