Worked example 1
iodine-131 has a half-life of 8.0 days. How much of a 40.0 g sample remains after 32 days?
Try it first: How many 8-day intervals fit into 32 days?
0 of 3 steps revealed.
What you'll be able to do: Use half-life to find how much of a sample remains after a given time, and work backwards from the fraction remaining to the elapsed time or the half-life.
Best after: Types of Nuclear Decay and Balancing Nuclear Equations
Radioactive decay is the classic first-order process: a fixed fraction decays in every equal interval, never a fixed amount. Once you see that, most half-life questions collapse into repeated halving.
These are recommended, not required. You can start this lesson at any time.
The half-life, t₁/₂, is the time it takes for half of the radioactive nuclei present to decay. It is a constant for a given nuclide. Start with 80 g or 8 g and the half-life is the same; only the amount left differs. Ordinary temperature, pressure and bonding changes have negligible effects on most decay half-lives; electron-capture rates can show small electronic-environment effects, so independence is a very good introductory approximation rather than an absolute universal law.
A half-life is a statistical population property, not a scheduled lifetime for each nucleus. A nucleus that has survived one half-life has the same decay probability over the next interval under the constant-λ model. Large samples closely follow the exponential expectation; a tiny sample shows random fluctuations. The continuum curve approaches zero asymptotically, but a real finite sample consists of discrete nuclei and can eventually have none left.
If the elapsed time is a whole number of half-lives, do not use logarithms at all. Divide the total time by the half-life to get n, then multiply the starting amount by (1/2)ⁿ. This handles most exam questions in one line.
N = N₀ × (1/2)ⁿ, where n = t / t₁/₂
Radioactive decay is first order, so the integrated rate law applies. First convert the half-life into a rate constant with k = 0.693 / t₁/₂, then use ln(N/N₀) = −kt. The same two equations solve for N, for t or for t₁/₂ depending on which quantity is missing.
Write f = N/N₀ = 2−t/t₁/₂ = e−λt. Then t = −ln f/λ, λ = ln 2/t₁/₂ and t₁/₂ = t ln 2/[−ln f]. If 30.0% remains after 50.0 years, λ = −ln(0.300)/50.0 ≈ 0.0241 yr⁻¹ and t₁/₂ ≈ 28.8 years. The logarithm uses the fraction remaining, not percent decayed. Time and λ must use reciprocal units: a decay constant in s⁻¹ needs seconds.
k = 0.693 / t₁/₂; ln(N/N₀) = −kt
Living things exchange carbon with the atmosphere and hold a steady ¹⁴C level. Once the organism dies, uptake stops and the ¹⁴C decays with a half-life of 5730 years. Measuring the fraction of ¹⁴C remaining and solving for t gives the age. Because roughly ten half-lives leaves under 0.1% of the original, the method runs out at about 50,000 years.
The simple carbon-14 calculation assumes a known initial isotope ratio and no later carbon exchange or contamination. Atmospheric ratios vary, so real radiocarbon ages require calibration rather than interpreting every idealized 5730-year calculation as a precise calendar date. Newer carbon contamination makes an old sample appear too young in the simple model; an initially low carbon-14 fraction can make it appear older. A daughter isotope or decay chain can also complicate a measured activity; identify what isotope the detector is actually counting.
Each nucleus has the same probability of decaying in the next second, so the number of decays per second is proportional to the number of nuclei present.
As the sample shrinks the decay rate shrinks with it, which is exactly what makes the decay exponential rather than linear.
A nuclide's half-life is fixed by its decay constant, independent of how large the sample is. Activity is the number of decays per unit time, A_activity=λN, so a larger sample of the same nuclide has proportionally higher activity. Both samples lose the same fraction over one half-life, but they lose different absolute numbers of nuclei.
Activity Aactivity = λN is measured in becquerels, 1 Bq = 1 decay/s. Convert a half-life in days to seconds before calculating activity in Bq. For a pure single isotope with constant detector efficiency and negligible background, activity ratios equal remaining-nucleus ratios. For a measured count rate, subtract background and account for detector efficiency, sample geometry and other emitting isotopes before making that inference. Equal masses of different radioisotopes need not have equal activities because their λ values and nuclei per gram differ.
The time required for half of the radioactive nuclei in a sample to decay.
Half-life is independent of sample size and ordinarily insensitive to temperature, pressure and chemical form; electron-capture and extreme-environment exceptions require qualification.
After n half-lives, the fraction remaining is (1/2)ⁿ.
k = 0.693 / t₁/₂ connects the half-life to the rate constant.
The measured activity in decays per second is proportional to the number of undecayed nuclei, so activity halves on the same schedule.
N = N₀ × (1/2)ⁿ
k = 0.693 / t₁/₂
ln(N/N₀) = −kt
iodine-131 has a half-life of 8.0 days. How much of a 40.0 g sample remains after 32 days?
Try it first: How many 8-day intervals fit into 32 days?
0 of 3 steps revealed.
A wooden artifact retains 25% of the ¹⁴₆C found in living wood. carbon-14 has a half-life of 5730 years. How old is it?
0 of 2 steps revealed.
strontium-90 has a half-life of 28.8 years. What percentage of a sample remains after 50.0 years?
0 of 4 steps revealed.
Two pure samples of the same radioisotope contain 8.0 g and 2.0 g. Compare their initial activities and half-lives.
Try it first: Their decay constants are the same.
0 of 3 steps revealed.
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Assuming a fixed mass decays in each half-life, so two half-lives means the sample is gone. | Each interval halves the expected remaining parent population. The continuum model approaches zero; a real finite sample is discrete and can eventually contain no parent nuclei. | Write out the halving sequence explicitly: 100%, 50%, 25%, 12.5%. |
| Using the half-life directly in place of k in ln(N/N₀) = -kt. | The equation requires a rate constant with units of inverse time; the half-life is a time. | Always compute k = 0.693 / t₁/₂ as a separate first step. |
| Expecting heating a sample or forming a compound to speed up decay. | Most decay rates are negligibly affected by ordinary conditions; electron capture involves atomic electrons and can have small environmental effects. | Treat half-life as a fixed property of the nuclide. |
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After the lesson · Question bank
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17 topic questions across 2 difficulty levels.
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For a constant decay constant λ, the expected parent population obeys N/N₀ = e−λt = 2−t/t₁/₂, with λ = ln 2/t₁/₂. Activity is λN, so sample size changes the activity but not the half-life. Most decay rates are negligibly affected by ordinary chemical conditions, with electron-capture qualifications. Dating requires an initial isotope estimate and a defensible closed-system history; real finite samples and detector counts also show statistical fluctuations.
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